Ramin S Esfandiari, Bei Lu - Modeling and Analysis of Dynamic Systems, Third Edition (Solutions, Instructor Solution Manual) (2018, CRC Press) - libgen li

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1 
 
Review Problems 
1 Evaluate the function /22
3( , ) cos( 1)xf x y e y−= − for 0.23, 2.7x y= − = 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
Solution 
(a) 
>> f = sym('2*exp(-x/2)*cos(y-1)/3'); 
>> x = -0.23; y = 2.7; 
>> double(subs(f)) 
 
ans = 
 
 -0.0964 
 
(b) 
>> F = matlabFunction(f); 
>> F(-0.23,2.7) 
 
ans = 
 
 -0.0964 
 
 
2 Evaluate the function 1
2( , ) sin(2 1) tan( )g x y x y= + − for 0.45, 1.17x y= = − 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
Solution 
(a) 
>> g = sym('sin(2*x+1)*tan(y-1/2)'); 
>> x = 0.45; y = -1.17; 
>> double(subs(g)) 
 
ans = 
 
 9.5076 
 
(b) 
>> G = matlabFunction(g); 
>> G(0.45,-1.17) 
 
ans = 
 
 9.5076 
 
3 Evaluate the vector function 
1
( , )
2
xy
v x y
x y
+ 
=  − 
 for 1.54, 2.28x y= = 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
 
2 
 
 
Solution 
(a) 
>> v = sym('[x*y+1;x-2*y]'); 
>> x = 1.54; y = 2.28; 
>> double(subs(v)) 
 
ans = 
 
 4.5112 
 -3.0200 
 
(b) 
>> V = matlabFunction(v); 
>> V(1.54,2.28) 
 
ans = 
 
 4.5112 
 -3.0200 
 
 
4 Evaluate the matrix function 
3
( , )
1 sin 2
y x y
m x y
x y
− 
=  + 
 for 2, 3.35x y= − = 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
Solution 
(a) 
>> m = sym('[y 3*x-y;x+1 sin(2*y)]'); 
>> x = -2; y = 3.35; 
>> double(subs(m)) 
 
ans = 
 
 3.3500 -9.3500 
 -1.0000 0.4048 
 
(b) 
>> M = matlabFunction(m); 
>> M(-2,3.35) 
 
ans = 
 
 3.3500 -9.3500 
 -1.0000 0.4048 
 
5 If 3 /5( ) ln( 1)tf t e t t−= + + , evaluate /df dt when 4.4t = 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
 
 
3 
 
Solution 
(a) 
>> f = sym('exp(-3*t/5)+t*log(t+1)'); 
>> df = diff(f); t = 4.4; 
>> double(subs(df)) 
 
ans = 
 
 2.4584 
 
(b) 
>> dF = matlabFunction(df); 
>> dF(4.4) 
 
ans = 
 
 2.4584 
 
6 If 3 1( ) 2 cosx xg x e x− −= + , evaluate /dg dx when 1.37x = 
(a) Using the subs command. 
(b) By conversion into a MATLAB function. 
 
Solution 
(a) 
>> g = sym('2^(3*x-1)+exp(-x)*cos(x)'); 
>> dg = diff(g); x = 1.37; 
>> double(subs(dg)) 
 
ans = 
 
 17.6539 
 
(b) 
>> dG = matlabFunction(dg); 
>> dG(1.37) 
 
ans = 
 
 17.6539 
 
7 Solve the following initial-value problem, and evaluate the solution at 3.5x = . 
 
 ( 1) 2 , (2) 1x y xy x y′− + = = 
Solution 
>> y = dsolve('(x-1)*Dy+x*y=2*x','y(2)=1','x'); 
>> y = matlabFunction(y); y(3.5) 
 
ans = 
 
 1.9107 
 
 
 
4 
 
8 Solve the following initial-value problem, and evaluate the solution at 2.8t = . 
 
 , (1) 1ttx x e x+ = = − 
Solution 
>> x = dsolve('t*Dx+x=exp(t)','x(1)=-1'); 
>> x = matlabFunction(x); x(2.8) 
 
ans = 
 4.5451 
 
9 Plot /3 1
1 2( ) cos( )ty t e t−= and 2 ( ) ( 1) ty t t e−= + versus 0 10t≤ ≤ in the same graph. Adjust the 
limits on the vertical axis to 0.3− and 1.1. Add grid and label. 
 
Solution 
>> syms t 
>> y1 = exp(-t/3)*cos(t/2); y2 = (t+1)*exp(-t); 
>> ezplot(y1,[0,10]) 
>> hold on 
>> ezplot(y2,[0,10]) 
 
 Figure Review1No9 
 
10 Plot /2 1
1,2,3 3( ) sin( )atx t e t−= , corresponding to 1, 2,3a = , versus 0 5t≤ ≤ in the same graph. 
Adjust the limits on the vertical axis to 0.05− and 0.3 . Add grid and label. 
 
Solution 
>> syms t 
>> x1 = exp(-t/2)*sin(t/3); 
>> x2 = exp(-t)*sin(t/3); 
>> x3 = exp(-3*t/2)*sin(t/3); 
>> ezplot(x1,[0,5]) 
>> hold on 
>> ezplot(x2,[0,5]) 
>> hold on 
>> ezplot(x3,[0,5]) 
0 1 2 3 4 5 6 7 8 9 10
-0.2
0
0.2
0.4
0.6
0.8
1
t
y2
y1
 
 
5 
 
 
 
 Figure Review1No10 
 
11 Plot 
1
sin
t
x te xdx−∫ versus 1 3t− ≤ ≤ , add grid and label. 
Solution 
>> syms t x 
>> integ = int(exp(x-t)*sin(x),x,1,t); 
>> ezplot(integ,[-1,3]) 
 
 
 Figure Review1No11 
12 Plot 2 (2 )
0
( )
t
t xx t e dx− −+∫ versus 1 3t≤ ≤ , add grid and label. 
Solution 
>> syms t x 
>> integ = int((x+t)^2*exp(-(2*t-x)),x,0,t); ezplot(integ,[1,3]) 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-0.05
0
0.05
0.1
0.15
0.2
0.25
t
x2
x3
x1
-1 -0.5 0 0.5 1 1.5 2 2.5 3
-2
-1.5
-1
-0.5
0
0.5
t
 
 
6 
 
 
 Figure Review1No12 
 
13 Plot ( 1)( , ) (1 cos ) tz x t x e− −= + versus 0 10x≤ ≤ for two values of 1.5, 2.5t = in a 1 2× tile. 
Add grid and title. 
 
Solution 
x = linspace(0,10); t = [1.5,2.5]; 
for i = 1:2, 
 for j = 1:100, 
 z(j,i) = (1+cos(x(j)))*exp(-(t(i)-1)); % 100 values of z for each t 
 end 
end 
 
% Initiate figure 
subplot(1,2,1), plot(x,z(:,1)), title('t = 1.5') 
subplot(1,2,2), plot(x,z(:,2)), title('t = 2.5') 
 
 
 Figure Review1No13 
1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
t
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t = 1.5
x
z
0 2 4 6 8 10
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
t = 2.5
x
z
 
 
7 
 
14 Plot ( , ) sin(2.7 )cos(3.2 )z x t x t= versus 0 5x≤ ≤ for four values of 1, 1.5, 2, 2.5t = in a 2 2× 
tile. Add grid and title. 
 
Solution 
x = linspace(0,5); t = 1:0.5:2.5; 
for i = 1:4, 
 for j = 1:100, 
 z(j,i) = sin(2.7*x(j))*cos(3.2*t(i)); % 100 values of z for each t 
 end 
end 
% Initiate figure 
subplot(2,2,1), plot(x,z(:,1)), title('t = 1') 
subplot(2,2,2), plot(x,z(:,2)), title('t = 1.5') 
subplot(2,2,3), plot(x,z(:,3)), title('t = 2') 
subplot(2,2,4), plot(x,z(:,4)), title('t = 2.5') 
 
 Figure Review1No14 
 
15 Given 3 /2( ) cos( 1)xf x e x−= + − , plot ( )f x′ versus 0 8x≤ ≤ . Add grid and title. 
 
Solution 
>> f = sym('exp(-3*x/2)+cos(x-1)'); df = matlabFunction(diff(f)); 
>> ezplot(df,[0,8]) 
 
 Figure Review1No15 
0 1 2 3 4 5
-1
-0.5
0
0.5
1
t = 1
x
z
0 1 2 3 4 5
-0.1
-0.05
0
0.05
0.1
t = 1.5
x
z
0 1 2 3 4 5
-1
-0.5
0
0.5
1
t = 2
x
z
0 1 2 3 4 5
-0.2
-0.1
0
0.1
0.2
t = 2.5
x
z
0 1 2 3 4 5 6 7 8
-1
-0.5
0
0.5
1
x
f'(
x)
 
 
8 
 
 
16 Write a user-defined function with function call F=temp_conv(C) that converts the 
temperature from Celsius C to Fahrenheit F. Execute the function for the case of C = 27. 
 
Solution 
function F = temp_conv(C) 
F = 9*C/5+32; 
end 
 
>> F = temp_conv(27) 
 
F = 
 80.6000 
 
17 Write a user-defined function with function call s=speed_calc(x,t0) that calculates the 
speed of a moving object, whose position is described by a symbolic function x, at a 
specified time t0. Execute the function to find the speed (at 2.5t = ) of an object whose 
position is described by 3 2( ) 0.75 2 1.4x t t t= − + . 
 
Solution 
function s = speed_calc(x,t0) 
v = matlabFunction(diff(x)); 
s = abs(v(t0)); 
end 
 
>> x = sym('0.75*t^3-2*t^2+1.4'); 
>> s = speed_calc(x,2.5) 
 
s = 
 4.0625 
 
18 Write a user-defined function with function call fval = f_eval(f,a,b) where f is an 
anonymous function, and a and b are constants such that a b> f = @(x)(x*exp(-x)); 
>> fval = f_eval(f,-3,4) 
 
fval = 
 
 -29.7884 
 
 
9 
 
 
19 Repeat the example considered in this chapter involving a signal and its integral, but this 
time the sine wave has amplitude 0.8 and frequency 1.5 rad/sec , and the integrated signal 
gets amplified by a factor of 2 ; see Figure 1.22. Build the model, run the simulation and 
generate a plot that can be imported into a document. 
 
 Figure 1.22 Problem 19. 
Solution 
 
 
 Figure Review1No19 
 
 
20 Create the Simulink model shown in Figure 1.23. Select the signal generator as a square 
wave with amplitude 0.5 and frequency 0.5 rad/sec . To flip the gain block, right click on it, 
then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum 
(Commonly Used Blocks) allows control over the list of desired signs. Run simulation   
A I v 0 v v v 
(b) 
>> A = [1 1 0;3 3 0;0 0 -2]; [V,D] = eig(A) 
 
V = 
 
 -0.7071 -0.3162 0 
 0.7071 -0.9487 0 
 0 0 1.0000 
 
 
D = 
 
 -0.0000 0 0 
 0 4.0000 0 
 0 0 -2.0000 
 
 
5. 
1 0 0
1 4 0
2 1 1
− 
 =  
 − 
A 
Solution 
(a) Lower-triangular matrix; ( ) 1, 4,1λ = −A . For 1 1λ = − we solve 
 
 
 
70 
 
 [ ]
EROs
1 1 1 1
0 0 0 0 0 0 0 0 10
 1 5 0 0 1 5 0 0 2
2 1 2 0 0 11 2 0 11
−         
        + = ⇒ = ⇒ = ⇒ =        
        − −         
A I v 0 v v v 
For 2 4λ = we solve 
 [ ]
EROs
2 2 2 2
5 0 0 0 1 0 0 0 0
4 1 0 0 0 0 0 0 0 3
2 1 3 0 0 1 3 0 1
−         
        − = ⇒ = ⇒ = ⇒ = −        
        − −         
A I v 0 v v v 
For 3 1λ = we solve 
 [ ]
EROs
3 3 3 3
2 0 0 0 1 0 0 0 0
 1 3 0 0 0 1 0 0 0
2 1 0 0 0 0 0 0 1
−         
        − = ⇒ = ⇒ = ⇒ =        
        −         
A I v 0 v v v 
(b) 
>> A = [-1 0 0;1 4 0;2 -1 1]; [V,D] = eig(A) 
 
V = 
 
 0 0 0.6667 
 0 0.9487 -0.1333 
 1.0000 -0.3162 -0.7333 
 
 
D = 
 
 1 0 0 
 0 4 0 
 0 0 -1 
 
 
6. 
3 4 2
1 4 1
2 6 1
− − 
 = − 
 − − 
A 
Solution 
(a) ( ) 3, 2,1λ =A . For 1 3λ = we solve 
 
 [ ]
EROs
1 1 1 1
0 4 2 0 0 2 1 0 1
3 1 1 1 0 1 1 0 0 1
2 6 4 0 0 0 0 0 2
− − −         
        − = ⇒ − = ⇒ = ⇒ =        
        − − −         
A I v 0 v v v 
 
For 2 2λ = we solve 
 
 
 
71 
 
[ ]
EROs
2 2 2 2
1 4 2 0 1 0 0 0 0
2 1 2 1 0 0 2 1 0 1
2 6 3 0 0 0 0 0 2
− −         
        − = ⇒ − = ⇒ = ⇒ =        
        − − −         
A I v 0 v v v 
 
For 3 1λ = we solve 
 [ ]
EROs
3 3 3 3
2 4 2 0 1 0 1 0 1
 1 3 1 0 0 1 0 0 0
2 6 2 0 0 0 0 0 1
− − −         
        − = ⇒ − = ⇒ = ⇒ =        
        − −         
A I v 0 v v v 
(b) 
>> A = [3 -4 -2;-1 4 1;2 -6 -1]; [V,D] = eig(A) 
 
V = 
 
 -0.4082 -0.0000 0.7071 
 0.4082 0.4472 0.0000 
 -0.8165 -0.8944 0.7071 
 
 
D = 
 
 3.0000 0 0 
 0 2.0000 0 
 0 0 1.0000 
 
 
 
7. 
1 2 4
1 2 2
5 10 2
− 
 = − 
 − − 
A 
Solution 
(a) ( ) 3,0, 2λ = −A . For 1 3λ = we solve 
 
 [ ]
EROs
1 1 1 1
2 2 4 0 1 1 2 0 1
3 1 1 2 0 0 0 0 0 1
5 10 5 0 0 1 1 0 1
− − −         
        − = ⇒ − − = ⇒ = ⇒ =        
        − − −         
A I v 0 v v v 
 
For 2 0λ = we solve 
 
 [ ]
EROs
2 2 2 2
1 2 4 0 1 2 0 0 2
 1 2 2 0 0 0 1 0 1
5 10 2 0 0 0 0 0 0
− −         
        = ⇒ − = ⇒ = ⇒ =        
        − −         
A v 0 v v v 
 
 
 
72 
 
For 3 2λ = − we solve 
 
 [ ]
EROs
3 3 3 3
3 2 4 0 1 0 2 0 2
2 1 4 2 0 0 1 1 0 1
5 10 0 0 0 0 0 0 1
−         
        + = ⇒ − = ⇒ = ⇒ =        
        − −         
A I v 0 v v v 
(b) 
>> A = [1 -2 4;-1 2 2;-5 10 -2]; [V,D] = eig(A) 
 
V = 
 
 -0.5774 -0.8944 -0.8165 
 -0.5774 -0.4472 -0.4082 
 -0.5774 0.0000 0.4082 
 
 
D = 
 
 3.0000 0 0 
 0 -0.0000 0 
 0 0 -2.0000 
 
 
 
8. 
0 2 2 0
1 5 2 2
0 3 0 3
1 2 2 1
− 
 − − − =
 − −
 − − 
A 
Solution 
(a) ( ) 2, 1,0, 3λ = − − −A . For 1 2λ = − we solve 
 
[ ]
EROs
1 1 1 1
2 2 2 0 0 1 1 0 0 0 1
1 3 2 2 0 0 0 1 0 0 1
2 
0 3 2 3 0 0 1 0 1 0 0
1 2 2 1 0 0 0 0 0 0 1
−         
         − − − −        + = ⇒ = ⇒ = ⇒ =        − −              −         
A I v 0 v v v 
 
For 2 1λ = − we solve 
 [ ] 2 2 2
1 2 2 0 0 2
1 4 2 2 0 1
 
0 3 1 3 0 0
1 2 2 0 0 1
− −     
     − − −     + = ⇒ = ⇒ =    − −         − −     
A I v 0 v v 
 
For 3 0λ = we solve 
 
 
73 
 
 [ ] 3 3 3
0 2 2 0 0 1
1 5 2 2 0 1
 
0 3 0 3 0 1
1 2 2 1 0 1
−     
     − − − −    = ⇒ = ⇒ =    − − −        − −     
A v 0 v v 
For 4 3λ = − we solve 
 [ ] 4 4 4
3 2 2 0 0 0
1 2 2 2 0 1
3 
0 3 3 3 0 1
1 2 2 2 0 0
−     
     − − −     + = ⇒ = ⇒ =    − −         −     
A I v 0 v v 
(b) 
>> A = [0 2 -2 0;-1 -5 2 -2;0 -3 0 -3;1 2 -2 -1]; [V,D] = eig(A) 
 
V = 
 
 0.5774 -0.8165 -0.5000 0.0000 
 -0.5774 0.4082 0.5000 0.7071 
 0.0000 -0.0000 0.5000 0.7071 
 0.5774 -0.4082 -0.5000 0.0000 
 
 
D = 
 
 -2.0000 0 0 0 
 0 -1.0000 0 0 
 0 0 0.0000 0 
 0 0 0 -3.0000 
 
 
 
 
 In Problems 9 through 16, find the eigenvalues, eigenvectors, algebraic and geometric 
multiplicity of each eigenvalue, and decide whether the matrix is defective or not. Then 
transform the matrix into either a diagonal or a Jordan matrix, whichever applicable. 
9. 
2 1
6 7
− 
=  
 
A 
Solution 
>> A = [2 -1;6 7]; [V,D] = eig(A) 
 
V = 
 
 -0.4472 0.3162 
 0.8944 -0.9487 % A is not defective 
 
 
D = 
 
 4.0000 0 
 0 5.0000 % Eigenvalues are distinct, hence each has AM=1=GM 
 
 
74 
 
 
 
>> V\A*V 
 
ans = 
 
 4.0000 0 
 0.0000 5.0000 % A is transformed into a diagonal matrix D 
 
10. 
0 1
1 2
− 
=  
 
A 
Solution 
>> A = [0 -1;1 2]; [V,D] = eig(A) 
 
V = 
 
 -0.7071 -0.7071 
 0.7071 0.7071 % Indicates that A is defective 
 
 
D = 
 
 1.0000 0 
 0 1.0000 % AM(1)=2 but GM(1)=1 
 
 
 
Switching to "jordan", we find 
 
>> [V,J] = jordan(A) 
 
V = 
 
 -1 1 
 1 0 
 
 
J = 
 
 1 1 
 0 1 
 
>> V\A*V 
 
ans = 
 
 1 1 
 0 1 % A is transformed into Jordan matrix J 
 
 
 
 
75 
 
11. 
3 0 0
0 3 1
5 0 0
 
 =  
  
A 
Solution 
>> A = [3 0 0;0 3 1;5 0 0]; [V,D] = eig(A) 
 
V = 
 
 0 0 0.0000 
 1.0000 -0.3162 -1.0000 % 1st and 3rd columns are the same 
 0 0.9487 0.0000 % thus, there is a generalized eigenvector 
 
 
D = 
 
 3 0 0 
 0 0 0 % Eigenvalues are 3, 3, 0 
 0 0 3 % AM(3)=2, AM(0)=1, GM(3)=1, GM(0)=1 
 
 
Matrix A is defective. Therefore, we switch to jordan: 
 
>> [V,J] = jordan(A) 
 
V = 
 
 0 0 1.0000 
 0.5556 1.6667 -0.5556 
 -1.6667 0 1.6667 
 
 
J = 
 
 0 0 0 
 0 3 1 
 0 0 3 % V\A*V = J (Jordan matrix) 
 
 
 
 
12. 
4 1 0
0 4 1
1 1 3
 
 =  
 − − 
A 
Solution 
>> A = [4 1 0;0 4 1;-1 -1 3]; [V,D] = eig(A) 
 
V = 
 
 0.5774 + 0.0000i 0.5774 + 0.0000i 0.5774 + 0.0000i 
 -0.5774 + 0.0000i -0.0000 + 0.5774i -0.0000 - 0.5774i 
 0.5774 + 0.0000i -0.5774 + 0.0000i -0.5774 - 0.0000i 
 
 
76 
 
 
 
D = 
 
 3.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 
 0.0000 + 0.0000i 4.0000 + 1.0000i 0.0000 + 0.0000i 
 0.0000 + 0.0000i 0.0000 + 0.0000i 4.0000 - 1.0000i 
 
% A is not defective because it has distinct eigenvalues 3, 4 ± i 
 
>> V\A*V 
 
ans = 
 
 3.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 + 0.0000i 
 0.0000 - 0.0000i 4.0000 + 1.0000i -0.0000 + 0.0000i 
 -0.0000 + 0.0000i -0.0000 - 0.0000i 4.0000 - 1.0000i 
 
This, as expected, is the diagonal matrix D. 
 
 
 
13. 
2 3 1
1 2 1
1 1 0
− 
 = − 
 − 
A 
Solution 
Use the eig command to findeigenvalues and eigenvectors of matrix A . 
 
>> A = [-2 3 1;-1 2 1;-1 1 0]; [V,D] = eig(A) 
 
V = 
 
 -0.7071 0.7071 0.5774 
 -0.7071 -0.0000 0.5774 
 0.0000 0.7071 -0.5774 
 
 
D = 
 
 1.0000 0 0 % Eigenvalues are 1, -1, 0 (distinct) 
 0 -1.0000 0 % AM=1 for all and GM=1 automatically 
 0 0 0.0000 % V\A*V = D (diagonal matrix) 
 
 
 
14. 
2 2 1
0 1 0
0 0 1
 
 =  
  
A 
Solution 
>> A = [2 2 1;0 1 0;0 0 1]; [V,D] = eig(A) 
 
 
 
77 
 
V = 
 
 1.0000 -0.8944 -0.7071 
 0 0.4472 0 
 0 0 0.7071 
 
 
D = 
 
 2 0 0 
 0 1 0 
 0 0 1 
 
AM(2)=1, AM(1)=2. But since the columns of V are linearly independent, we 
conclude that GM(1)=2. We know GM(2)=1 automatically. 
 
 
>> D = V\A*V 
 
D = 
 
 2 0 0 
 0 1 0 
 0 0 1 % As expected, A is transformed into a diagonal D 
 
 
15. 
1 0 0
0 2 1
0 1 2
 
 = − − 
 − 
A 
Solution 
Use the eig command to find eigenvalues and eigenvectors of matrix A . 
 
>> A = [1 0 0;0 -2 -1;0 1 -2]; [V,D] = eig(A) 
 
V = 
 
 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i 
 0.7071 + 0.0000i 0.7071 + 0.0000i 0.0000 + 0.0000i 
 0.0000 - 0.7071i 0.0000 + 0.7071i 0.0000 + 0.0000i 
 
 
D = 
 
 -2.0000 + 1.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 
 0.0000 + 0.0000i -2.0000 - 1.0000i 0.0000 + 0.0000i 
 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i 
 
 
 
% λ(A)= -2 ± i, 1 
% A is not defective 
% V\A*V = D diagonal 
 
 
 
78 
 
16. 
5 0 0
2 5 1
0 0 6
 
 =  
  
A 
Solution 
>> A = [5 0 0;2 5 1;0 0 6];[V,D] = eig(A) 
 
V = 
 
 0 0.0000 0 
 1.0000 -1.0000 0.7071 
 0 0 0.7071 
 
 
D = 
 
 5 0 0 
 0 5 0 
 0 0 6 
 
The first two columns of V represent the same eigenvector, hence A is 
defective. Swith to "jordan": 
 
 
>> [V,J] = jordan(A) 
 
V = 
 
 0 0 -0.5000 
 1.0000 -1.0000 -1.0000 
 1.0000 0 0 
 
 
J = 
 
 6 0 0 
 0 5 1 
 0 0 5 
 
 
>> V\A*V 
 
ans = 
 
 6 0 0 
 0 5 1 
 0 0 5 % A is transformed into a Jordan matrix J 
 
 
17. Prove that if 2 2×A has a repeated eigenvalue λ , then A must be defective. 
Solution 
If the eigenvalue is repeated, we have AM( ) 2λ = . To determine the GM we must solve the 
eigenvalue problem ( )λ− =A I v 0 . But λ−A I is 2 2× and singular. Therefore, it must have 
 
 
79 
 
exactly one zero row in its REF. This means GM( ) 1λ = , which implies that A has a 
generalized eigenvector and is therefore defective. 
 
 
18. Prove that a singular matrix must have at least one zero eigenvalue. 
Solution 
The determinant of a matrix is the product of its eigenvalues. If the determinant is zero, then at 
least one of the eigenvalues of the matrix must be zero. 
 
 
19. Show that if the eigenvalues of A are 1 2, , ... , nλ λ λ , then the eigenvalues of a−A I are 
1 2, , ... , nλ a λ a λ a− − − . 
Solution 
( )a λ− = −A I v Av v . But λ=Av v . Insert into the last equation to obtain 
 ( ) ( )a a λ a λ a− = − = − = −A I v Av v v v v 
This describes the eigenvalue problem associated with a−A I , and proof is complete. 
 
 
20. Show that the eigenvalues of A and TA are the same. 
Solution 
It suffices to show that any eigenvalue of A is also an eigenvalue of TA . If λ is an eigenvalue 
of A , it must satisfy 0λ− =A I . But a matrix λ−A I and its transpose [ ]T Tλ λ− = −A I A I 
have the same determinant, which implies 0T λ− =A I . This is the eigenvalue problem 
associated with TA . Therefore, λ must be an eigenvalue of TA . 
 
 
Review Problems 
1. Find the value of a such that the following two matrices have the same determinant. 
 
 
1
2
1 3 0 4 0 0
2 0 , 0 3 2
0 0 4 0 1
a a
  
  − −  
   −   
 
Solution 
Equating the two determinants, 4( 6) 4(3 )a a+ = − , yields 3
2a −= . 
 
2. Determine whether the following vectors are linearly independent. 
 
 
3 5 2
1 , 4 , 5
4 2 2
     
     −     
     −     
 
 
 
80 
 
Solution 
 Assemble the vectors so that they form the rows of a matrix: 
 
3 1 4
5 4 2
2 5 2
− 
 
 
 − 
 
The determinant of this matrix is zero, which implies the rows are linearly dependent. 
 
 
3. Prove that the product of two symmetric matrices is not necessarily symmetric. 
Solution 
Let A and B be symmetric so that T =A A and T =B B . But ( )T T T= = ≠AB B A BA AB . 
 
4. If A is m m× and symmetric, and B is a general m n× matrix, show that TB AB is n n× 
and symmetric. 
Solution 
TB AB is clearly n n× . Also ( )TT T T T= =B AB B A B B AB , and the proof is complete. 
 
5. Given 
 1
3
0 1 0
0 0 1 , 2 0
2 1 2
 
   = =   
 − − − 
A C 
 find the rank of 
 2( )T T T T T 
 C A C A C 
Solution 
 
2 4
3 3
2 5
3
1 2
3 3
2
( ) 0 0 rank 3
3
T T T T T
−
−
 
   = ⇒ =  
  
C A C A C 
 
6. Determine a such that rank( ) 3=A , where 
 
 
3 7 2 1
1 4 3 2
1 4 5 3
7 12 1 5
a
− 
 − − =
 −
 
 
A 
Solution 
Using EROs, we find 
 
 
81 
 
 
EROs EROs
38 95
80 200
1 4 3 23 7 2 1 1 4 3 2
0 2 51 4 3 2 0 2 5
 
0 0 11 51 4 5 3 0 19 11 7
0 0 20 197 12 1 5 0 40 20 19
a a
a a
aa
a − −
− −
− − − − −   
    − −     = → →
 − +   − −
     − +−       
A 
 
For the rank to be 3, the ratios of the last two entries in the third column and those in the fourth 
column must be the same: 
 
 238 11 95 7 69 138 0 0, 2
80 20 200 19
a a a a a
a a
− − − +
= ⇒ − = ⇒ =
− − − +
 
 
Inserting 0a = reveals that rank( ) 4=A , hence rejected. It is easily verified that 2a = is the 
only acceptable value. 
 
7. Find the inverse of the rotation matrix 
 
 
cos sin 0
sin cos 0
0 0 1
θ θ
θ θ
 
 = − 
  
R 
Solution 
Noting 1=R , we find 
 1
cos sin 0
1 adj( ) sin cos 0
0 0 1
θ θ
θ θ−
− 
 = =  
  
R R
R
 
 
8. Find the value(s) of a for which the following system only has a trivial solution. 
 
 
2
13
2
3
1 2 0
2 1 0
1 1 5 0
x
a x
x
 −    
     − =    
    −    
 
Solution 
The homogeneous system has a trivial solution if and only if the coefficient matrix is non-
singular. Since 3 3a= +A , we must have 1a ≠ − . 
 
9. Solve the following linear system using Cramer’s rule; a is a parameter. 
 
1
2
3
4
3 0 1 2 2 4
2 2 0 2 2
 , , , 
1 3 1 1 2 2
2 1 2 3 2
x a
x
x a
x a
− +    
    −     = = = =    − − −        −    
Ax b A x b 
 
 
82 
 
Solution 
The determinant of the coefficient matrix is 
 
3 0 1 2
2 2 0 2
84
1 3 1 1
2 1 2 3
D
−
−
= = −
− −
−
 
Since 0D ≠ , we proceed by evaluating 
 1 2
2 4 0 1 2 3 2 4 1 2
2 2 0 2 2 2 0 2
84 , 84
2 2 3 1 1 1 2 2 1 1
2 1 2 3 2 2 2 3
a a
D D a
a a
a a
+ − + −
− −
= = − = = −
− − − − −
−
 
 
 3 4
3 0 2 4 2 3 0 1 2 4
2 2 2 2 2 2 0 2
84 , 84
1 3 2 2 1 1 3 1 2 2
2 1 2 3 2 1 2 2
a a
D D a
a a
a a
+ − +
−
= = = = −
− − − − −
− −
 
Finally, 
 1 2 3 41 , , 1 , x x a x x a= = = − = 
 
 
10. Solve the linear system in Problem 9 using the inverse of the coefficient matrix. 
Solution 
We find 1−A as 
 1
4 22 12 8
18 6 30 61 1adj( )
28 14 0 2884
22 26 18 2
−
− − − 
 − − = =
 − −−
 − − − 
A A
A
 
 
The solution is then obtained as 
 1
4 22 12 8 2 4 1
18 6 30 6 21
28 14 0 28 2 2 184
22 26 18 2 2
a
a
a
a a
−
− − − +     
     − −     = = =    − − − −−         − − −     
x A b 
 
 
11. Show that any matrix with distinct eigenvalues is non-defective. 
Solution 
Since the eigenvaluesare assumed distinct, each has AM 1= . But GM AM≤ so that each 
eigenvalue also has GM 1= . Therefore, there is no generalized eigenvector and the matrix is 
non-defective. 
 
 
83 
 
 
12. (a) Find all eigenvalues and eigenvectors of 
3 2 0
0 1 0
1 1 2
− − 
 = − 
 − − 
A . 
 (b) Confirm the results of (a) in MATLAB. 
Solution 
(a) A is block lower triangular, hence its eigenvalues are those of the upper-left corner block 
and the single block of 2− . Therefore, 1, 2, 3λ = − − − and the corresponding eigenvectors 
are found as 
 1 2 3
1 0 1
1 , 0 , 0
2 1 1
     
     = − = =     
     −     
v v v 
(b) 
 
>> A = [-3 -2 0;0 -1 0;1 -1 -2]; [V,D] = eig(A) 
 
V = 
 0 0.7071 -0.4082 
 0 0 0.4082 
 1.0000 -0.7071 -0.8165 
 
 
D = 
 -2 0 0 
 0 -3 0 
 0 0 -1 
 
13. Find the eigenvalues and the algebraic and geometric multiplicity of each, and decide 
whether the matrix is defective. 
 
1 0 0 0
2 3 0 0
1 0 1 1
0 2 0 1
 
 − =
 −
 − 
A 
Solution 
A is block lower triangular, comprised of two 2 2× blocks, one block is lower triangular, the 
other upper triangular. Therefore 1, 1,1,3λ = − − . For the two eigenvalues that occur only once, 
the AM and GM are both 1. Let us inspect 1λ = − as follows. 
 
 
Elementary
row operations
1 0 0 0 0
0 0 0 0 0
( ) 
0 0 0 1 0
0 1 0 0 0
a
b
c
d
     
     
    + = ⇒ =                  
A I v 0 
 
 
 
84 
 
 It is clear that there is only one free variable, which means only one independent eigenvector can 
be found for this eigenvalue. Therefore, AM = 2 while GM = 1. This implies there is a 
generalized eigenvector, hence the matrix is defective. 
 
14. Prove that the eigenvalues of a matrix are preserved under a similarity transformation, that is, 
if 1− =S AS B , then the eigenvalues of A and B are the same. 
Solution 
Let λ be an eigenvalue of B so that 0λ =I - B . Substitute for B : 
 
 1 1 10 ( ) 0λ λ λ λ− − −− = ⇒ − = − = − =I B I S AS S I A S S I A S 
 
But 1 1− =S S . Therefore 0λ − =I A , which means λ is an eigenvalue of A . 
 
 
15. Find the modal matrix and use it to transform A into a diagonal or a Jordan matrix: 
 
 
1 4 1 2
0 1 1 4
0 0 1 0
0 0 2 5
− − 
 
 =
 −
 − 
A 
Solution 
>> A = [-1 4 -1 2;0 1 1 4;0 0 -1 0;0 0 2 -5]; [V,D] = eig(A) 
 
V = 
 
 1.0000 0.8944 0.1374 1.0000 
 0 0.4472 -0.5494 -0.0000 
 0 0 0 0.0000 
 0 0 0.8242 0.0000 
 
 
D = 
 
 -1 0 0 0 
 0 1 0 0 
 0 0 -5 0 
 0 0 0 -1 
 
The first and last columns of V are the same, thus we must switch to jordan: 
 
V = 
 
 -0.0833 2.8333 -6.0000 -2.7500 
 0.3333 1.4167 0 -1.5000 
 0 0 0 1.0000 
 -0.5000 0 0 0.5000 
 
 
 
85 
 
 
J = 
 
 -5 0 0 0 
 0 1 0 0 
 0 0 -1 1 
 0 0 0 -1 
 
 
16. Show that the rotation matrix R of Problem 7 is orthogonal. Then, verify that each of its 
eigenvalues has an absolute value of 1. 
Solution 
It is readily observed that 
 1
cos sin 0
sin cos 0
0 0 1
T
θ θ
θ θ−
− 
 = = 
  
R R 
Therefore, R is orthogonal. The three eigenvalues of R are the roots of 0λ− =R I : 
 1, cos sinjθ θ± 
Each of the three clearly has an absolute value of 1. 
 
17. Prove that if k const= and matrix A has eigenvalues 1, ... , nλ λ with corresponding 
eigenvectors 1, ... , nv v , then the eigenvalues of kA are 1, ... , nk kλ λ with eigenvectors 
1, ... , nv v . 
Solution 
Multiply both sides of λ=Av v by k to get ( ) ( )k kλ=A v v . This is the eigenvalue problem for 
kA so that the eigenvalues are 1, ... , nk kλ λ and the corresponding eigenvectors are 1, ... , nv v . 
 
In Problems 18 through 20, decide whether the statement is true or false. 
 
18. If 3n ≥ , then every n n× matrix with any repeated eigenvalues is defective. 
 Solution 
False 
 
19. The rank of a singular 4 4× matrix is 3. 
Solution 
False 
 
20. If all eigenvalues of a matrix are real, then the matrix is symmetric. 
Solution 
False 
 
 
 
 
 
86 
 
Problem Set 4.1 
In Problems 1 through 8, assuming general initial conditions, express the system model in 
(a) Configuration form. 
(b) Standard, second-order matrix form. 
 
1. 
1 1
1 1 1 22 3
2 /31 1
2 2 1 22 3
( ) 0 
( ) t
x x x x
x x x x e−
 + + − =

+ − − =
 
 
 
Solution 
(a) The generalized coordinates are 1 1q x= and 2 2q x= . Then 
 
 
2
1 1 1 2 1 1 2 1 23
2 /31 1
2 2 1 2 2 1 2 1 22 3
2 ( ) ( , , , , ) 
( ) ( , , , , )t
q q q q f q q q q t
q q q q e f q q q q t−
 = − − − =

= − + − + =
   
   
 
 
(b) The second-order matrix form is obtained as 
 
 
1 11
1 1 13 32
1 1 1 2 /3
2 2 22 3 3
01 00
00 1 t
x x x
x x x e−
   −          + + =           −              
 
 
 
 
2. 
21
1 1 1 23
3
2 1 25
2( )
2( ) 0 
tx x x x e
x x x
− + + − =

− − =
 
 
Solution 
(a) The second equation is algebraic, hence may be used to eliminate one of the two variables: 
Inserting 10
2 113x x= into the first equation, we find 261
1 1 13 13
tx x x e−+ + =  . Therefore, there is 
only one generalized coordinate 1 1q x= and the configuration form is derived as 
218
1 1 1 1 1133 3 ( , , )tq q q e f q q t−= − − + =   . 
 
(b) The second-order matrix form happens to be scalar ( 1n = ) 
 
 [ ] 261
13 13 , 1 , , , tx e−   = = = = =   M C K x f 
 
 
3. 1 1 1 1 1 2 2 1 1
2 2 2 2 1 2
( ) ( )
( ) ( ) 
m x k x cx k x x F t
m x k x x F t
+ + − − =
 + − =
 

; Mechanical system in Figure 4.3 
Solution 
(a) The generalized coordinates are 1 1q x= and 2 2q x= . Then 
 
 
 
87 
 
 
[ ]
[ ]
1 1 1 1 2 2 1 1 1 1 2 1 2
1
2 2 2 1 2 2 1 2 1 2
2
1 ( ) ( ) ( , , , , ) 
1 ( ) ( ) ( , , , , ) 
q k q cq k q q F t f q q q q t
m
q k q q F t f q q q q t
m
 = − − + − + =

 = − − + =

   
  
 
 
 
 Figure 4.2 Mechanical system in Problem 3. 
 
 
(b) The second-order matrix form is obtained as 
 
 1 1 1 1 2 2 1 1
2 2 2 2 2 2 2
0 0
0 0 0
m x x k k k x Fc
m x x k k x F
+ −            
+ + =            −            
 
 
 
 
 
 
4. 1 1 1 1 1 1 2 2 1 2 2 1
2 2 2 2 1 2 2 1
( ) ( ) ( )
( ) ( ) 0 
m x c x k x k x x c x x F t
m x k x x c x x
+ + − − − − =
 + − + − =
   
  
; Mechanical system in Figure 4.3 
 
 Figure 4.3 Mechanical system in Problem 4. 
Solution 
(a) The generalized coordinates are 1 1q x= and 2 2q x= . Then 
 
 
[ ]
[ ]
1 1 1 1 1 2 2 1 2 2 1 1 1 2 1 2
1
2 2 2 1 2 2 1 2 1 2 1 2
2
1 ( ) ( ) ( ) ( , , , , ) 
1 ( ) ( ) ( , , , , ) 
q c q k q k q q c q q F t f q q q q t
m
q k q q c q q f q q q q t
m
 = − − + − + − + =

 = − − − − =

     
    
 
 
k1
x2x1
m1
F t1( )
m2
F t2( )
1x 2x
1k 2k
1m 2m
1c
( )F t
2c
 
 
88 
 
(b) The second-order matrix form is obtained as 
 
 1 1 1 2 2 1 1 2 2 1
2 2 2 2 2 2 2 2
0 ( )
0 0
m x c c c x k k k x F t
m x c c x k k x
+ − + −             
+ + =            − −             
 
 
 
 
 
5. 
/23
1 1 1 22
3
2 2 1 22
( ) sin
( ) 0 
te tθ θ θ θ
θ θ θ θ
− + + − =

+ − − =
 
 
 
Solution 
(a) The generalized coordinates are 1 1q θ= and 2 2q θ= . Then 
 
 
/23
1 1 1 2 1 1 2 1 22
3
2 2 1 2 2 1 2 1 22
( ) sin ( , , , , )
( ) ( , , , , ) 
tq q q q e t f q q q q t
q q q q f q q q q t
− = − − − + =

= − + − =
  
   
 
 
(b) The second-order matrix form is obtained as 
 
 
3 3
11 1 2 2
3 3
22 2 2 2
1 0 1 0 sin
0 1 0 1 0
te tθθ θ
θθ θ
−       −       + + =            −              
 
 
 
 
 
6. 
1
1 1 2 1 2 1 12
2 1
2 3 2 2 1 2 13 2
2
3 3 3 2 3 23
( ) 2 ( ) ( ) 
( ) ( ) 2 ( ) 0 
2 ( ) ( ) 
mx kx k x x c x x F t
mx k x x k x x c x x
mx kx k x x cx F t
 + − − − − =
 − − + − + − =
 + + − + =
  
  
 
 
Solution 
(a) The generalized coordinates are 1 1q x= , 2 2q x= , and 3 3q x= . Then 
 
 
1
1 1 2 1 2 1 1 1 1 2 3 1 2 32
2 1
2 3 2 2 1 2 1 2 1 2 3 1 2 33 2
2
3 3 3 2 3 23
1 ( ) 2 ( ) ( , , , , , , ) 
1 ( ) ( ) 2 ( ) ( , , , , , , ) 
1 ( )
2
q kq k q q c q q F f q q q q q q t
m
q k q q k q q c q q f q q q q q q t
m
q kq k q q cq F
m
 = − + − + − + = 
 = − − − − − = 
 = − − − − + = 
     
     
  3 1 2 3 1 2 3( , , , , , , ) f q q q q q q t








  
 
 
(b) The second-order matrix form is obtained as 
 
 
3 1
1 1 1 12 2
71 2
2 2 22 6 3
52
3 3 3 23 3
0 0 2 2 0 0
0 0 2 2 0 0
0 0 2 0 0 0
m x c c x k k x F
m x c c x k k k x
m x c x k k x F
 − −          
           + − + − − =           
            −           
 
 
 
 
 
 
 
89 
 
7. 
1 1
1 1 2 1 2 1 12 2
1
2 3 2 2 1 2 1 3 2 22
3 3 3 2 3 2
2 ( ) ( ) ( ) 
2 ( ) ( ) ( ) ( ) ( ) 
2 ( ) ( ) 0 
mx kx k x x c x x F t
mx k x x k x x c x x c x x F t
mx kx k x x c x x
+ − − − − =
− − + − + − − − =
+ + − + − =
  
    
   





 
Solution 
(a) The generalized coordinates are 1 1q x= , 2 2q x= , and 3 3q x= . Then 
 
 
1
1 1 2 1 2 1 1 1 1 2 3 1 2 32
1
2 3 2 2 1 2 1 3 2 2 2 1 2 3 1 2 32
3 3 3
2 2 ( ) ( ) ( , , , , , , ) 
1 ( ) ( ) ( ) ( ) ( , , , , , , )
2
1 2 (
q kq k q q c q q F f q q q q q q t
m
q k q q k q q c q q c q q F f q q q q q q t
m
q kq k q q
m
 = − + − + − + = 
 = − − − − − + − + = 
= − − −
     
       
 [ ]2 3 2 3 1 2 3 1 2 3) ( ) ( , , , , , , ) c q q f q q q q q q t






 − − =
    
 
 
(b) The second-order matrix form is obtained as 
 
 
1 1 1
1 1 1 12 2 2
31
2 2 2 22 2
3 3 3
0 0 0 3 0 ( )
0 2 0 2 ( )
0 0 0 0 3 0
m x c c x k k x F t
m x c c c x k k k x F t
m x c c x k k x
   − −        
           + − − + − − =            
            − −           
 
 
 
 
 
 
8. 
1
4 ( )
2 2
x x F t
xϕ ϕ ϕ
 = − +

= + −
 
  
 
Solution 
(a) The generalized coordinates are 1q x= and 2q ϕ= . Using the expression for x given by the 
first equation in the second equation, we have 
 
 
1
4
1
4
 
2 2
x x F
x Fϕ ϕ ϕ
 = − +
 = + + −
 
  
 
 
The configuration form is then found as 
 
 
1
1 1 1 1 2 1 24
1 1 1
2 2 2 1 2 1 2 1 22 8 2
( , , , , ) 
( , , , , )
q q F f q q q q t
q q q q F f q q q q t
 = − + =
 = + + − =
   
    
 
 
(b) The second-order matrix form is obtained as 
 
 
1
4
1
4
01 0 0 0
20 2 0 1
x x x F
Fϕ ϕ ϕ
            
+ + =           − − − −             
 
 
 
 
 
 
90 
 
Problem Set 4.2 
In Problems 1 through 8, find a suitable set of state variables, derive the state-variable equations, 
and form the state equation. 
1. 2 /3 , , 0tmx bx e m b const−+ = = >  
Solution 
The ODE is second-order in x , therefore two initial conditions, (0)x and (0)x , are needed. 
Thus, there are two state variables: 1x x= , 2x x=  . The state-variable equations are derived as 
 
 
1 2
2 /31
2 2
 
t
m
x x
x bx e−
=
  = − +  


 
The state equation is then formed as 
 
 1 2 /3
1
2
0 1 0
 , , , , 
0
t
b
m m
x x
u u e
x x
−      
= + = = = = =       −      
x Ax B x A B

 
 
 
2. 1 1
3 2 2 cosx x x x t+ + + =   
Solution 
The ODE is third-order in x , therefore three initial conditions, (0)x , (0)x , (0)x , are needed, and 
there are three state variables: 1x x= , 2x x=  , 3x x=  . The state-variable equations are then 
derived as 
 
 
1 2
2 3
1
3 3 2 12
 
 
3 2 cos
x x
x x
x x x x t
 =
 =
  = − − − +  



 
 
The state equation is subsequently formed as 
 
 
1
2
3
3 2
0 1 0 0
 , , 0 0 1 , 0 , cos
6 3 3
x
u x u t
x
    
    = + = = = =     
     − − −     
x Ax B x A B 
 
3. 
1 2
1 1 1 2 14 3
2
2 2 2 1 23
( ) ( ) 
2 ( ) ( ) 
x x x x F t
x x x x F t
 + + − =
 + + − =
 

 
Solution 
The first ODE is second-order in 1x , hence two initial conditions, 1(0)x , 1(0)x , are needed. The 
second one is first-order in 2x so that only 2 (0)x is needed. Therefore, there are a total of three 
state variables: 1 1x x= , 2 2x x= , 3 1x x=  . As a result, 
 
 
91 
 
 
 
1 3
1 2
2 2 2 1 22 3
1 2
3 3 1 2 14 3
 
( ) ( ) 
( ) ( ) 
x x
x x x x F t
x x x x F t
 =
  = − − − +  
 = − − − +



 
 
Noting that 1
2
F
F
 
=  
 
u , the state equation is obtained as 
 
1
151 1
3 2 2 1 2 3 6 2
22 2 1
3 3 3 4
0 0 1 0 0
 , , 0 , 0 , 
1 0
x
F
x
F
x
× ×
    
     = + = = − = =             − −     
x Ax B u x A B u 
 
4. 
2
5
1
2
2 
2 ( ) 
x x x y
y y x F t
 + + =
 + + =
 

 
Solution 
The first ODE is second-order in x , hence two initial conditions, (0)x , (0)x , are needed. The 
second one is first-order in y so that only (0)y is needed. Therefore, there are a total of three 
state variables: 1x x= , 2x y= , 3x x=  . The state-variable equations are 
 
 
1 3
1 1
2 2 12 4
2
3 3 1 25
 
( )
2 
x x
x x x F t
x x x x
 =
  = − − +  
 = − − +



 
The state equation is then formed as 
 
 
1
1 1 1
2 2 8 2
2
3 5
0 0 1 0
 , , 0 , , ( )
1 2 0
x
u x u F t
x
    
    = + = = − − = =     
     − −     
x Ax B x A B 
 
5. 1 1 1 2
1
2 1 23
2 2( ) ( )
2( ) 0 
x x x x F t
x x x
+ + − =
 − − =
 
 
Solution 
Because the second equation is algebraic, variables 1x and 2x are linearly dependent, hence 
cannot be selected as state variables. Solve the second equation for 2x to find 6
2 17x x= and 
insert into the first equation to obtain 2
1 1 172 ( )x x x F t+ + =  . We now have a second-order ODE 
in 1x and can proceed as usual. There are two state variables here: 1 1x x= and 2 1x x=  . Note 
that 2x is the standard notation for a state variable and should not be confused with 2x in the 
original system. Consequently, 
 
 
92 
 
 
 1 2
2
2 2 17
 
2 ( )
x x
x x x F t
=
 = − − +


 
The state equation is 
 1
2
1 7
0 1 0
 , , , , ( )
2 1
x
u u F t
x
    
= + = = = =     − −     
x Ax B x A B

 
 
6. 
/31
1 1 1 22
1
2 1 22
( )
 , 
( ) 0 
tz z k z z e
k const
z k z z
− + + − = =
− − =
 
 
Solution 
Because the second equation is algebraic, variables 1z and 2z are linearly dependent, hence 
cannot be selected as state variables. Solve the second equation for 2z to find 2 12
k
kz z+= and 
insert into the first equation to obtain /3
1 1 12
tk
kz z z e−++ + =  . We now have a second-order ODE 
in 1z and can proceed as usual. There are two state variables here: 1 1x z= and 21x z=  . Then, 
 
 1 2
/3
2 2 12
 
tk
k
x x
x x x e−+
=

= − − +


 
The state equation is 
 
 1 /3
1 2
0 1 0
 , , , , 
1 1
t
k
k
z
u u e
z
−
+
    
= + = = = =     − −     
x Ax B x A B

 
 
 
7. 1 1 2 1 2 1
2 1 2 1 2 2
2 4 2 ( )
2 2 ( )
x x x x x F t
x x x x x F t
+ − + − =
 − + − + =
  
  
 
Solution 
The two second-order ODEs require a total of four initial conditions for complete solution, hence 
there are four state variables. We choose them as 1 1x x= , 2 2x x= , 3 1x x=  , and 4 2x x=  . Then, 
 
 [ ]
1 3
2 4
1
3 3 4 1 2 12
4 3 4 1 2 2
 
 
4 2 ( )
2 2 ( ) 
x x
x x
x x x x x F t
x x x x x F t
=
 =
 = − + − + +
 = − + − +




 
 
Noting that 1
2
F
F
 
=  
 
u , the state equation is obtained as 
 
 
 
93 
 
 
1
2 1
4 2 2 1 1 1 1
3 22 2 2
4
0 0 1 0 0 0
( )0 0 0 1 0 0
 , , , , 
( )2 1 0
1 1 2 2 0 1
x
x F t
x F t
x
× ×
     
            = + = = = =      − −         − −    
x Ax B u x A B u 
 
8. 
2 2 2
1 1 1 2 1 2 2
1 22 2 2
1 2 1 2 2 2 1
( ) 0
 , , , , ,
( ) 0
mL mgL kL kL
m k g L L const
mL mgL kL kL
θ θ θ
θ θ θ
 + + − = =
+ + − =


 
Solution 
The two second-order ODEs require a total of four initial conditions for complete solution, hence 
there are four state variables. We choose them as 1 1x θ= , 2 2x θ= , 3 1x θ=  , and 4 2x θ=  . Then, 
 
 ( )
( )
1 3
2 4
2 2
3 1 2 1 2 22
1
2 2
4 1 2 2 2 12
1
 
 
1
1
x x
x x
x mgL kL x kL x
mL
x mgL kL x kL x
mL
=
 =
  = − + +  

  = − + +  




 
 
The state equation is obtained as 
 
 
1
2 2
1 2 22
2 2
1 13
2 2
4 2 1 2
2 2
1 1
0 0 1 0
0 0 0 1
0 0 , , 
0 0
x
mgL kL kLx
mL mLx
x kL mgL kL
mL mL
 
 
   
   +  − = = = 
  
   +   −
  
x Ax x A 
 
In Problems 9 and 10, derive the state-variable equations (in vector form) for the given nonlinear 
system model. 
 
9. 2 2 /31
32 2 tx x x e−+ + =  
Solution 
The ODE is second-order in x , therefore two initial conditions, (0)x and (0)x , are needed, and 
there are two state variables: 1x x= , 2x x=  . The state-variable equations are derived as 
 
 1 2
2 2 /31 1
2 2 16 2
 
t
x x
x x x e−
=

= − − +


 
 
 
94 
 
 
In vector form, 
 ( , , )t=x f x u 
where 
 21 2 /31
22 1
2 2 16
 , , t
xx
u e
x x x u
−     = = =   
− − +    
x f 
 
10. 
31
1 1 1 23
2 1 2sin 
x x x x
x x t
 + =

= +
  

 
Solution 
The first ODE is second-order in 1x , hence two initial conditions, 1(0)x , 1(0)x , are needed. The 
second one is first-order in 2x so that only 2 (0)x is needed. Therefore, there are a total of three 
state variables: 1 1x x= , 2 2x x= , 3 1x x=  . Then, 
 
 
1 3
2 1
31
3 3 3 23
 
2sin 
x x
x x t
x x x x
 =
 = +
 = − +



 
 
 
In vector form, 
 ( , , )t=x f x u 
where 
 
1 3
2 1
313 3 3 23
 , 2sin , 
x x
x u t x u
x x x x
  
    = = = +   
   − +    
x f 
 
Problems 11 through 14 are concerned with the stability of systems. A linear dynamic system is 
stable if the homogeneous solution of its mathematical model, subjected to the prescribed initial 
conditions, decays. More practically, a linear system is stable if the eigenvalues of its state 
matrix all have negative real parts, that is, they all lie in the left half-plane. 
 
11. Decide whether the system in Problem 1 is stable. 
Solution 
The state matrix is formed as 
 
0 1
0 b
m
 
=  − 
A 
 
The eigenvalues of A are 0, b
m− . Since one of these is always 0, we conclude that the system is 
unstable regardless of the values of m and b . 
 
 
95 
 
 
 
12. Decide whether the system in Problem 5 is stable. 
Solution 
The state matrix is 
 2
7
0 1
2
 
=  − − 
A 
 
Since its eigenvalues are 0.1548, 1.8452− − , the system is stable. 
 
 
13. Decide whether the system in Problem 4 is stable. 
Solution 
The state matrix is 
 1 1
2 8
2
5
0 0 1
0
1 2
 
 = − − 
 − − 
A 
Since the eigenvalues are 0.1620 1.1397 , 0.8490j± − , the system is unstable. 
 
 
14. Find the range of values of b for which a system described by ( 1) ( )y b y y F t− − − =  is 
stable. 
Solution 
Selection of 1x y= and 2x y=  as state variables leads to the state matrix 
 
 
2eigenvalues0 1 1 ( 1) 4
 ( )
1 1 2
b b
b
l
− ± − + 
= ⇒ = − 
A A 
 
However, since 2( 1) 4 1b b− + > − for all values of b , at least one of the two eigenvalues will 
always have a positive real part. Therefore, the system is unstable for all values of b . 
 
 
 
 
In Problems 15 through 18, find the state-space form of the mathematical model 
 
15. 
3
1 1 1 22
3
2 2 1 22
3 ( ) 0 
2 ( ) ( )
x x x x
x x x x F t
 + + − =
 + − − =
 

 , outputs are 1x and 1x . 
Solution 
Choosing state variables 1 1x x= , 2 2x x= , 3 1x x=  , the state-variable equations are obtained as 
 
 
 
96 
 
 
1 3 1
5 3 3 51
2 2 1 22 2 2 4 4
3 33 3 32 23 3 1 22 2
 0 0 1 0
 0 1 , , 
3 03 
x x x
x x x F u x u F
xx x x x
 =           = − + + ⇒ = − + = =      
     − −= − − +     
x x x

 

 
 
The outputs are 1x and 1x so that the output equation is written as 
 
1
1
2
3 2 3
3
1 0 0 0
0 0 1 0
x
x
x u
x
x×
 
      = = +      
      
 
y 
 
16. 
3
1 1 2 1 14
3
2 2 14
( ) 2 ( )
( ) 0 
z z z z z F t
z z z
 + − + + =
 + − =
 

 , outputs are 1z and 2z . 
Solution 
With state variables 1 1x z= , 2 2x z= , 3 1x z=  , we find 
 
 
1 3 1
3 3 3 3
2 2 1 24 4 4 4
7 3 7 3
3 1 2 3 34 4 4 4
 0 0 1 0
 0 0 , , ( )
2 2 1
x x x
x x x u x u F t
x x x x F x
  =   
     = − + ⇒ = − + = =     
     = − + − + − −      
x x x

 

 
 
The outputs are 1z and 2z , thus 
 
1
1
2
2 2 3
3
1 0 0 0
0 1 0 0
x
x
x u
x
x×
 
      = = +      
      
 
y 
 
17. 
1 1 3 2 1 2 1
2 2 1 2 1
3 1 3
2 9( ) 0.8( ) 2( ) ( )
0.8( ) 2( ) 0 
2 0.45( ) 
x x x x x x x F t
x x x x x
x x x
+ − − − − − =
 + − + − =
 = −
  
   , outputs are 2x and 2x . 
Solution 
The third equation is algebraic. Solving for 3x yields 0.45
3 12.45x x= . Using this, eliminate 3x in the 
first equation, while the second equation remains as is. Then 
 
 1 1 2 1 2
2 2 1 2 1
2 9.35 0.8( ) 2 ( )
0.8( ) 2( ) 0 
x x x x x F t
x x x x x
+ − − − =
 + − + − =
  
  
 
 
Select state variables as 1 1x x= , 2 2x x= , 3 1x x=  , 4 2x x=  , noting that 3x is not the same as 3x 
in the original model. The state equation is formed as 
 
 
 
97 
 
 
1 1
2 2
3 1
4 2
0 0 1 0 0
0 0 0 1 0
 , , ( )
4.675 1 0.4 0.4 1
2 2 0.8 0.8 0
x x
x x
u u F t
x x
x x
      
      
      = + = = =      − −           − −       
x x x


 
 
The output equation is 
 
1
2 2
4 32 4
4
0 1 0 0 0
0 0 0 1 0
x
x x
u
x x
x
×
 
       = = +      
      
  
y 
 
18. 
32 1
1 1 3 2 1 2 15 5 2
3 1
2 2 1 2 15 2
3
3 1 35
( ) ( ) ( ) 0
( ) ( ) ( ) 
( ) 
x x x x x x x
x x x x x F t
x x x
 + − − − − − =
 + − + − =
 = −
  
  

, outputs are 2x and 3x . 
Solution 
There are five state variables:1 1x x= , 2 2x x= , 3 3x x= , 4 1x x=  , 5 2x x=  . The state-variable 
equations are then obtained as 
 
 
1 4
2 5
3 3
3 1 35 5
9 3 31 2
4 1 2 3 4 510 2 5 5 5
3 3 1 1
5 5 4 25 5 2 2
 
 
 
x x
x x
x x x
x x x x x x
x x x x
=
=
= −
= − + + − +
= − + − +




 1 ( ) x F t






+
 
 
Consequently, the state equation u= +x Ax B is formed as 
 
 3 3
5 5
9 3 31 2
10 2 5 5 5
3 31 1
2 2 5 5 5 15 5
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 , , ( )0
0
0 1
u F t
××
   
   
   
 −  = = =
   − −   
   − −   
A B 
 
System outputs are 2x and 3x , hence the output equation is 
 
 
 
98 
 
 
1
2
2
3
3 2 5 2 1
4
5
0 1 0 0 0 0
0 0 1 0 0 0
x
x
x
x u
x
x
x
× ×
 
 
       = = +      
      
 
  
y 
 
19. A dynamic system is governed by 4 8 3 ( )y y y f t+ + =  , where f and y are the system input 
and output, respectively. Derive the state-space form of the decoupled system. 
Solution 
Selecting the state variables as 1x y= and 2x y=  , the state equation is formed as 
 
 1
3 1
24 4
0 1 0
 , , ( )
2
x y
u u u f t
x y
       
= + = + = = =      − −      
x x Ax B x

 
 
The output equation is 
 [ ] 1
1
2
1 0 0
x
y x u Du
x
 
= = + ⋅ = + 
 
Cx 
 
 
 31
2 2( ) ,l = − −A , 
2 2
1 3
− − 
=  
 
V . The transformed (via x = Vx ) state-space form is 
 
 
[ ]
11
82
3 1
82
Decoupled
0
 
0
2 2 0 
u
y u
−   −
= +   −       



= − − + ⋅
x x
x

 

 
 
20. A dynamic system model is derived as 
 
 1 1 1 2
1
2 1 23
4 3 ( )
2 0 
x x x x f t
x x x
− − + =
 + + =
 

 
 
 Where the input is f and the outputs are 2x and 1x . 
(a) Find the state-space form. 
(b) Derive the state-space form of the decoupled system. 
Solution 
(a) There are three state variables 1 1x x= , 2 2x x= , 3 1x x=  , which lead to 
 
 
 
99 
 
 
[ ]
1 3
1
2 1 23
1
3 3 1 24
 
2 
3
x x
x x x
x x x x f
 =
 = − −
 = + − +



 
 
The resulting state equation is u= +x Ax B where 
 
 
1
1
23
3 1 1 1
34 4 4 4
0 0 1 0
2 0 , 0 , ( ) , 
x
u f t x
x
     
     = − − = = =     
    −     
A B x 
The output equation is 
 
1
2
2
3 2 3 2 1
3
0 1 0 0
0 0 1 0
x
x
x u
x
x× ×
 
      = = +      
      
 
y 
(b) 
% Define state-space form matrices A,B,C,D 
>> A = [0 0 1;-1/3 -2 0;3/4 -1/4 1/4]; 
>> B = [0;0;1/4]; C = [0 1 0;0 0 1]; D = [0;0]; 
>> [V,Dtilda] = eig(A); % Solve eigenvalue problem for A 
>> Btilda = V\B; Ctilda = C*V; % Define transformed matrices B,C 
>> dec_sys = ss(Dtilda,Btilda,Ctilda,D) 
 
dec_sys = 
 
 a = 
 x1 x2 x3 
 x1 1.016 0 0 
 x2 0 -0.7885 0 
 x3 0 0 -1.977 % Decoupled 
 
 b = 
 u1 
 x1 -0.1996 
 x2 -0.184 
 x3 -0.02369 
 
 c = 
 x1 x2 x3 
 y1 0.07732 -0.2112 0.9887 
 y2 -0.7104 -0.6052 0.1338 
 
 d = 
 u1 
 y1 0 
 y2 0 
 Continuous-time state-space model. 
 
Problem Set 4.3 
In Problems 1 through 8, find all possible input-output equations. 
 
 
100 
 
 
1. 
1
1 1 1 23
1
2 1 23
( ) ( ) 
( ) 0 
x x x x f t
x x x
 + + − =
 − − =
 

 , ( )f t = input, 1 2,x x = outputs 
Solution 
Since there are two outputs and one input, we expect two I/O equations. Assuming zero initial 
conditions, Laplace transformation of the governing equations yields 
 
 
2 1 1
1 23 3
1 1
1 23 3
( ) ( ) ( ) ( ) 
 ( ) ( ) ( ) 0
s s X s X s F s
X s s X s
 + + − =

− + + =
 
 
Since both 1x and 2x are the outputs, we solve the above system once for 1( )X s and a second 
time for 2 ( )X s by means of Cramer’s rule: 
 
 
1
3
1 1 cross multiply
3 3 23
1 13 22 4 21 1
3 33 3
1 1
3 3
( )
0 ( ) ( )
( ) (3 4 2 ) ( ) (3 1) ( )
F s
s s F s
X s s s s X s s F s
s s ss s
s
−
+ +
= = ⇒ + + = +
+ ++ + −
− +
 
 
 
2 1
3
1 1 cross multiply
3 3 23
2 23 22 4 21 1
3 33 3
1 1
3 3
( )
0 ( )
( ) (3 4 2 ) ( ) ( )
s s F s
F s
X s s s s X s F s
s s ss s
s
+ +
−
= = ⇒ + + =
+ ++ + −
− +
 
 
Interpretation of these two equations in time domain results in the two desired I/O equations: 
 
 1 1 13 4 2 3x x x f f+ + = +   , 2 2 23 4 2x x x f+ + =   
 
 
 
2. 
3
1 1 1 2 15
3
2 2 1 2 25
2 ( ) ( ) 
3 ( ) ( )
x x x x f t
x x x x f t
 + + − =
 + − − =
 
 
, 1 2( ), ( )f t f t = inputs, 2x = output 
Solution 
Two I/O equations are expected. Laplace transformation of the original system leads to 
 
 
2 3 3
1 2 15 5
23 3
1 2 25 5
( 2 ) ( ) ( ) ( ) 
( ) (3 ) ( ) ( )
s s X s X s F s
X s s s X s F s
 + + − =

− + + + =
 
 
 
 
101 
 
Since 2x is the output, we solve the above system for 2 ( )X s via Cramer’s rule: 
 
 
2 3
15
23 3 3
25 5 5
2 2 12 3 3
5 5
23 3
5 5
2 ( )
( ) 2
( ) ( ) ( )
( ) ( )2
3
s s F s
F s s s
X s F s F s
s ss s
s s
+ +
− + +
= = +
∆ ∆+ + −
− + +
 
 
where 4 3 2 922
5 5( ) 3 7s s s s s∆ = + + + . When 1f is the input, we set 2 0f = so that 2 ( ) 0F s = in the 
above equation, and 
 
 
3
4 3 252 9 322
2 15 5 5
1
( ) (3 7 ) ( ) ( )
( ) ( )
X s s s s s X s F s
F s s
= ⇒ + + + =
∆
 
 
When 2f is the input, we set 1 0f = so that 1( ) 0F s = , and 
 
 
2 3
4 3 2 252 9 322
2 25 5 5
2
2( ) (3 7 ) ( ) ( 2 ) ( )
( ) ( )
s sX s s s s s X s s s F s
F s s
+ +
= ⇒ + + + = + +
∆
 
 
Transforming the data back to the time domain, the two I/O equations are 
 
 
(4) 9 322
2 2 2 12 5 5 53 7x x x x f+ + + =   , (4) 9 322
2 2 2 2 2 22 5 5 53 7 2x x x x f f f+ + + = + + 
   
 
 
3. 
1
1 1 1 2 13
1
2 1 2 23
( ) ( )
( ) ( ) 
x x x x f t
x x x f t
 + + − =
 − − =
 

, 1( )f t , 2 ( )f t = inputs, 1x = output 
Solution 
Two input-output equations are expected. Laplace transformation of the original system leads to 
 
 
2 1 1
1 2 13 3
21 1
1 2 23 3
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
s s X s X s F s
X s s X s F s
 + + − =

− + + =
 
 
Since 1x is the output, we solve the above system for 1( )X s via Cramer’s rule: 
 
 
1
1 3
2 21 1 1
2 1 23 3 3
1 2 1 1
3 3
21 1
3 3
( )
( ) ( ) ( ) ( )
( )
( ) ( )
F s
F s s s F s F s
X s
s ss s
s
−
+ +
= = +
∆ ∆+ + −
− +
 
 
 
102 
 
 
where 4 3 22 1
3 3( )s s s s s∆ = + + + . When 1f is the input, we set 2 0f = so that 2 ( ) 0F s = in the 
above equation, and 
 
 
2 1
4 3 2 231 2 1 1
1 13 3 3
1
( ) ( ) ( ) ( ) ( )
( ) ( )
sX s s s s s X s s F s
F s s
+
= ⇒ + + + = +
∆
 
 
When 2f is the input, we set 1 0f = so that 1( ) 0F s = , and 
 
 
1
4 3 231 2 1 1
1 23 3 3
2
( ) ( ) ( ) ( )
( ) ( )
X s s s s s X s F s
F s s
= ⇒ + + + =
∆
 
 
Transforming the data back to the time domain, the two I/O equations are 
 
 
(4) 2 1 1
1 1 1 1 11 3 3 3x x x x f f+ + + = +   , (4) 2 1 1
1 1 1 21 3 3 3x x x x f+ + + =   
 
 
 
 
4. 1 1 1 2
2 2 1 2
2( ) 0 
2( ) ( )
x x x x
x x x x f t
+ + − =
 + − − =
 
 
 , ( )f t = input, 1x = output 
Solution 
One I/O equation is expected. Laplace transformation of the original system leads to 
 
 
2
1 2
2
1 2
( 2) ( ) 2 ( ) 0 
2 ( ) ( 2) ( ) ( )
s s X s X s
X s s s X s F s
 + + − =

− + + + =
 
 
Since 1x is the output, we solve the above system for 1( )X s via Cramer’s rule: 
 
 
2
1 4 3 22
2
0 2
( ) 2 2 ( )( )
2 5 42 2
2 2
F s s s F sX s
s s s ss s
s s
−
+ +
= =
+ + ++ + −
− + +
 
 
The I/O equation is therefore obtained as 
 
 (4)
1 1 11 2 5 4 2x x x x f+ + + =  103 
 
 
5. 
2 1
1 1 1 23 3
1
2 1 23
( ) ( ) 
( ) 0 
u tθ θ θ θ
θ θ θ
 + + − =
 − − =
 
, ( )u t = input, 2θ = output 
Solution 
The second equation is algebraic. Solving for 1θ in the second equation, we have 1 24θ θ= . 
Inserting into the first equation yields the I/O equation as 8
2 2 234 ( )u tθ θ θ+ + =  . 
 
6. 1 1 1 2 1
2 1 2 2
3( ) ( ) 
3( ) ( ) 
x x x x f t
x x x f t
+ + − =
 − − =
 

, 1( )f t , 2 ( )f t = inputs, 1x , 2x = outputs 
Solution 
Four input-output equations are expected. Laplace transformation of the original system leads to 
 
 
2
1 2 1
2
1 2 2
( 3) ( ) 3 ( ) ( ) 
3 ( ) ( 3) ( ) ( ) 
s s X s X s F s
X s s X s F s
 + + − =

− + + =
 
 
Solving the above system for 1( )X s via Cramer’s rule, we find 
 
 
1
2 2
2 2 1
1 4 3 2 4 3 22
2
( ) 3
( ) 3 3 ( ) ( 3) ( )( )
6 3 6 33 3
3 3
F s
F s s F s s F sX s
s s s s s s s ss s
s
−
+ +
= = +
+ + + + + ++ + −
− +
 (a) 
 
Solving the system for 2 ( )X s via Cramer’s rule, we find 
 
 
2
1
2
2 1 2
2 4 3 2 4 3 22
2
3 ( )
3 ( ) 3 ( ) ( 3) ( )( )
6 3 6 33 3
3 3
s s F s
F s F s s s F sX s
s s s s s s s ss s
s
+ +
− + +
= = +
+ + + + + ++ + −
− +
 (b) 
 
When 1f is the input, we set 2 ( ) 0F s = in Eqns. (a) and (b), and two of the I/O equations are 
obtained as 
 
 
2
(4)1
1 1 1 1 1 114 3 2
(4)1
2 2 2 2 124 3 2
( 3) ( )( ) 6 3 3
6 3
3 ( )( ) 6 3 3
6 3
s F sX s x x x x f f
s s s s
F sX s x x x x f
s s s s
+
= ⇒ + + + = +
+ + +
= ⇒ + + + =
+ + +

  
  
 
 
 
104 
 
 
When 2f is the input, we set 1( ) 0F s = in Eqns. (a) and (b), and the other two I/O equations are 
obtained as 
 
 
(4)2
1 1 1 1 214 3 2
2
(4)2
2 2 2 2 2 2 224 3 2
3 ( )( ) 6 3 3
6 3
( 3) ( )( ) 6 3 3
6 3
F sX s x x x x f
s s s s
s s F sX s x x x x f f f
s s s s
= ⇒ + + + =
+ + +
+ +
= ⇒ + + + = + +
+ + +
  
 
  
 
 
 
 
7. 
1 1
1 1 1 2 12 3
2 1 2
2( ) ( )
2( ) 0 
q q q q q v t
q q q
 + + − + =

− − =
 

 , ( )v t = input, 1q , 2q = outputs 
Solution 
There are two I/O equations. Proceeding as always, 
 
 
2 71
1 22 3
1 2
( ) ( ) 2 ( ) ( )
2 ( ) ( 2) ( ) 0 
s s Q s Q s V s
Q s s Q s
 + + − =

− + + =
 
 
Solve for 1( )Q s and 2 ( )Q s separately: 
 
 3 21 11 2
1 12 3 33 22 1 11 271
2 3 32 3
( ) 2
0 2 ( 2) ( )( ) ( 2 ) ( ) ( 2) ( )
22
2 2
V s
s s V sQ s s s s Q s s V s
s s ss s
s
−
+ +
= = ⇒ + + + = +
+ + ++ + −
− +
 
 
 
 
2 71
2 3
3 21 11 2
2 22 3 33 22 1 11 271
2 3 32 3
( )
2 0 2 ( )( ) ( 2 ) ( ) 2 ( )
22
2 2
s s V s
V sQ s s s s Q s V s
s s ss s
s
+ +
−
= = ⇒ + + + =
+ + ++ + −
− +
 
 
In time-domain, the two I/O equations are described by 
 
 1 11 2 1 11 2
1 1 1 1 2 2 2 22 3 3 2 3 32 2 , 2 2q q q q v v q q q q v+ + + = + + + + =       
 
 
 
 
105 
 
8. 
1
1 1 3 2 1 2 12
1
2 2 1 2 12
3 1 3
( ) ( ) ( )
( ) 0 
 
x x x x x x x f t
x x x x x
x x x
 + − − − − − =
 + − + − =
 = −
  
  

, ( )f t = input, 2x , 3x = outputs 
Solution 
Two input-output equations are expected. Laplace transformation of the original system leads to 
 
 
2 3 1
1 2 32 2
21 1
1 22 2
1 3
( ) ( ) ( ) ( ) ( ) ( ) 
( ) ( ) ( ) ( ) 0 
 ( ) ( 1) ( ) 0 
s s X s s X s X s F s
s X s s s X s
X s s X s
 + + − + − =
 − + + + + =
 − + + =
 
 
Solving for 2 ( )X s and 3( )X s via Cramer’s rule: 
 
 
2 3
2
1
2
1
2
2 5 4 3 22 13 1
22 2
21 1
2 2
( ) 1
( ) 0 0
1 0 1 ( )( 1) ( )
( )
3 4 2( ) 1
( ) 0
1 0 1
s s F s
s
s s s F s
X s
s s s s ss s s
s s s
s
+ + −
− +
− + + +
= =
+ + + ++ + − + −
− + + +
− +
 
 
 
2 3 1
2 2
21 1
2 2
2 1
2
3 5 4 3 22 13 1
22 2
21 1
2 2
( ) ( )
( ) 0
1 0 0 ( ) ( )
( )
3 4 2( ) 1
( ) 0
1 0 1
s s s F s
s s s
s s F s
X s
s s s s ss s s
s s s
s
+ + − +
− + + +
− + +
= =
+ + + ++ + − + −
− + + +
− +
 
 
In time-domain, the two I/O equations are described by 
 
 
(5) (4) 31 1
2 2 22 2 2 2 2
(5) (4) 1 1
3 3 33 3 2 2
3 4 2
3 4 2 
x x x x x f f f
x x x x x f f f
+ + + + = + +
+ + + + = + +
 
  
 
  
 
 
 
9. A mechanical system model is derived as ( )mx bx kx f t+ + =  where 2m = , 1
2b = , 5k = , 
and applied force 2
5( ) tf t e−= , all in consistent physical units. The system is subjected to 
zero initial conditions. Assuming x is the output, find the transfer function. 
 
 
106 
 
Solution 
 The transfer function is independent of the specific nature of the input, and is found as 
 
 2 2 1
2
( ) 1 1
( ) 2 5
X s
F s ms bs k s s
= =
+ + + +
 
 
10. In Problem 9, find the transfer function if x is the output. 
Solution 
The transfer function is independent of the specific nature of the input. It is readily seen that 
 
 2 2 1
2
( ) 1 1
( ) 2 5
X s
F s ms bs k s s
= =
+ + + +
 
 
Noting that { } ( ) (0) ( )x sX s x sX s= − =L due to zero initial conditions, the transfer function is 
found as 
 { }
{ } 2 1
2
( )
( ) 2 5
x sX s s
f F s s s
= =
+ +
L
L
 
 
 
 
In Problems 11 through 14, a system model, as well as its inputs and outputs are provided. Find 
the appropriate transfer function or matrix. 
 
11. 1 1 1 2
2 2 2 1
0.4 9( ) ( )
0.6 0.8 9( ) 0 
x x x x f t
x x x x
+ + − =
 + + − =
 
 
 , ( )f t = input, 1x = output 
Solution 
Laplace transformation yields 
 
 
2
1 2
2
1 2
( 0.4 9) ( ) 9 ( ) ( ) 
9 ( ) (0.6 0.8 9) ( ) 0
s s X s X s F s
X s s s X s
 + + − =

− + + + =
 
 
Solving for 1( )X s results in 
 
 
2 2
1 4 3 22
2
( ) 9
0 0.6 0.8 9 0.6 0.8 9( ) ( )
0.6 1.04 14.72 10.80.4 9 9
9 0.6 0.8 9
F s
s s s sX s F s
s s s ss s
s s
−
+ + + +
= =
+ + ++ + −
− + +
 
 
Since the system is SISO, the only transfer function is 
 
 
 
107 
 
 
2
1
4 3 2
( ) 0.6 0.8 9
( ) 0.6 1.04 14.72 10.8
X s s s
F s s s s s
+ +
=
+ + +
 
 
 
12. 
1
1 1 1 2 12
1
2 1 22
( ) ( )
( ) 0 
q q q q q v t
q q q
 + + − + =
 − − =
 

 , ( )v t = input, 1q = output 
Solution 
Laplace transformation of the original system leads to 
 
 
2 3 1
1 22 2
1 1
1 22 2
( ) ( ) ( ) ( )
( ) ( ) ( ) 0 
s s Q s Q s V s
Q s s Q s
 + + − =

− + + =
 
 
Solve this system of algebraic equations for 1( )Q s via Cramer’s rule: 
 
 
1
2
1 1
2 2
1 3 22 3 13 1
2 22 2
1 1
2 2
( )
0 ( ) ( )
( )
2
V s
s s V s
Q s
s s ss s
s
−
+ +
= =
+ + ++ + −
− +
 
Since the system is SISO, there is only a single transfer function 
 
 
1
1 2
3 23 1
2 2
( )
( ) 2
sQ s
V s s s s
+
=
+ + +
 
 
 
13. 1 1 2 1 2 1 1
2 2 1 2 1 2
( ) ( )
2 ( ) ( ) 
x x x x x x f
x x x x x f
+ − − − − =
 + − + − =
  
  
 , 1( )f t , 2 ( )f t = inputs, 1x , 2x = outputs 
Solution 
We expect a 2 2× transfer matrix with the following structure: 
 2 1
2 1
1 1
1 20 011 12
21 22 2 2 2 2
1 20 0
( ) ( )
( ) ( )( ) ( )
( )
( ) ( ) ( ) ( )
( ) ( )
F F
F F
X s X s
F s F sG s G s
s
G s G s X s X s
F s F s
= =
×
= =
 
 
  
= =   
   
 
 
G 
To find the four transfer functions listed above, we take the Laplace transform of the governing 
equations, with zero initial conditions, 
 
 
2
1 2 1
2
1 2 2
( 2) ( ) ( 1) ( ) ( ) 
( 1) ( ) (2 1) ( ) ( ) 
s s X s s X s F s
s X s s s X s F s
 + + − + =

− + + + + =
 
 
 
108 
 
 
Solving for 1( )X s , we find 
 
 
1
2 2
2
1 1 2
( ) ( 1)
( ) 2 1 2 1 1( ) ( ) ( )
( ) ( ) ( )
F s s
F s s s s s sX s F s F s
s s s
− +
+ + + + +
= = +
∆ ∆ ∆
 (a) 
where 
 
2
4 3 2
2
2 ( 1)
( ) 2 3 5 1
( 1) 2 1
s s s
s s s s s
s s s
+ + − +
∆ = = + + + +
− + + +
 
 
Solving for 2 ( )X s , yields 
 
 
2
1
2
2
2 2 1
2 ( )
( 1) ( ) 2 1( ) ( ) ( )
( ) ( ) ( )
s s F s
s F s s s sX s F s F s
s s s
+ +
− + + + +
== +
∆ ∆ ∆
 (b) 
 
where ( )s∆ is as before. The four transfer functions are then determined as follows. 
 
 
2 1
2
1 1
1 20 0
( ) ( )2 1 1 , 
( ) ( ) ( ) ( )F F
X s X ss s s
F s s F s s= =
+ + +
= =
∆ ∆
 
 
 
2 1
2
2 2
1 20 0
( ) ( )1 2 , 
( ) ( ) ( ) ( )F F
X s X ss s s
F s s F s s= =
+ + +
= =
∆ ∆
 
 
Ultimately, the transfer matrix is formed as 
 
 
2
2
2 1 1
( ) ( )
( )
1 2
( ) ( )
s s s
s s
s
s s s
s s
 + + +
 ∆ ∆ =  + + + 
∆ ∆  
G 
 
 
14. 
1 1 3 2 1 1
2 2 1 2
3 1 3
2( ) ( )
2( ) ( ) 
 
x x x x x f t
x x x f t
x x x
+ − − − =
 + − =
 = −
  
  

 , 1( )f t , 2 ( )f t = inputs, 1x , 2x = outputs 
Solution 
Taking the Laplace transform, we find 
 
 
 
109 
 
 
2
1 2 3 1
2
1 2 2
1 3
( 2 1) ( ) 2 ( ) ( ) ( )
2 ( ) ( 2 ) ( ) ( ) 
( ) ( 1) ( ) 0 
s s X s sX s X s F s
sX s s s X s F s
X s s X s
 + + − − =
− + + =
− + + =
 
 
Solving separately for 1( )X s and 2 ( )X s , yields 
 
 
1
2
2
2
1 1 25 4 3 2 5 4 3 22
2
( ) 2 1
( ) 2 0
0 0 1 ( 1)( 2 ) 2 ( 1)( ) ( ) ( )
5 5 2 5 5 22 1 2 1
2 2 0
1 0 1
F s s
F s s s
s s s s s sX s F s F s
s s s s s s s ss s s
s s s
s
− −
+
+ + + +
= = +
+ + + + + ++ + − −
− +
− +
 
 
 
 
2
1
2
2 3 2
2 1 25 4 3 2 5 4 3 22
2
2 1 ( ) 1
2 ( ) 0
1 0 1 2 2 3 3( ) ( ) ( )
5 5 2 5 5 22 1 2 1
2 2 0
1 0 1
s s F s
s F s
s s s s s sX s F s F s
s s s s s s s ss s s
s s s
s
+ + −
−
− + + + +
= = +
+ + + + + ++ + − −
− +
− +
 
 
Note that we do not cancel terms involving s from the numerator and denominator in each 
fraction because valuable information about the system will be lost otherwise. The entries of the 
2 2× transfer matrix ( )sG are finally formed as 
 
 
2 1
2
1 1
11 125 4 3 2 5 4 3 2
1 20 0
( ) ( )( 1)( 2 ) 2 ( 1)( ) , ( )
( ) ( )5 5 2 5 5 2F F
X s X ss s s s sG s G s
F s F ss s s s s s s s= =
+ + +
= = = =
+ + + + + +
 
 
 
 
2 1
2 3 2
2 2
21 225 4 3 2 5 4 3 2
1 20 0
( ) ( )2 3 3( ) , ( )
( ) ( )5 5 2 5 5 2F F
X s X ss s s s sG s G s
F s F ss s s s s s s s= =
+ + +
= = = =
+ + + + + +
 
 
 
15. The governing equation for an electric circuit is derived as 
 
 
110 
 
 
0
1 ( ) ( )
tdiL Ri i t dt v t
dt C
+ + =∫ 
 where L , R , and C are the inductance, resistance, and capacitance, all constants, ( )i t is the 
current and ( )v t is the applied voltage. Initial conditions are assumed to be zero. If ( )v t and 
( )i t are the system input and output, respectively, find the transfer function. 
Solution 
Laplace transformation of the governing equation yields 
 
 
Transfer function
2
1 ( ) 1( ) ( ) 1( ) 1
I s CsLs R I s V s
Cs V s LCs RCsLs R
Cs
 + + = ⇒ = =  + +  + +
 
 
 
16. Electric charge q and electric current i are related via /i dq dt= . In Problem 15, find the 
transfer function if ( )v t and ( )q t are the system input and output, respectively. 
 Solution 
 Noting /i dq dt= , the governing equation can be rewritten as 1 ( )Lq Rq q v t
C
+ + =  . With zero 
initial conditions, Laplace transformation yields 
 
 
Transfer function
2
22
1 ( ) 1( ) ( ) 1( ) 1
Q s CLs Rs Q s V s
C V s LCs RCsLs Rs
C
 + + = ⇒ = =  + +  + +
 
 
 
17. The I/O equation for a dynamic system is given as 3 2 4 ( )x x x f t+ + =  where f and x 
denote the input and output, respectively. 
(a) Find the system’s transfer function. 
(b) Assuming ( )f t is the unit-impulse, find the expression for ( )X s using (a). 
(c) Find the steady-state value ssx via the final-value theorem. 
Solution 
(a) 2
( ) 1
( ) 3 2 4
X s
F s s s
=
+ +
 
(b) Knowing ( ) 1F s = , from (a) we find 2
1( )
3 2 4
X s
s s
=
+ +
. 
(c) Since the poles of ( )X s are 0.3333 1.1055j− ± , the final-value theorem applies, and 
 
 { } 20 0
lim ( ) lim 0
3 2 4ss s s
sx sX s
s s→ →
 = = = 
+ + 
 
 
 
18. The mathematical model of a dynamic system is given as 
 
 
111 
 
 
 
1
1 1 1 22
1
2 2 1 22
2 ( ) ( )
3 ( ) 0 
x x x x f t
x x x x
 + + − =
 + − − =
 
 
 
 
 where f is the input and 1x and 2x are the outputs. 
(a) Find the appropriate transfer functions. 
(b) Assuming ( )f t is the unit-impulse, find the expressions for 1( )X s and 2 ( )X s using (a). 
(c) Find the steady-state values of 1x and 2x using the final-value theorem. 
Solution 
(a) Laplace transformation of the model yields 
 
 
2 1 1
1 22 2
21 1
1 22 2
( 2 ) ( ) ( ) ( ) 
 ( ) (3 ) ( ) 0
s s X s X s F s
X s s s X s
 + + − =

− + + + =
 
 
Solving for 1( )X s and 2 ( )X s yields 
 
 
1
2
2 21 1
2 2
1 4 3 22 31 1
22 2
21 1
2 2
( )
0 3 3
( ) ( )
3 7 42
3
F s
s s s s
X s F s
s s s ss s
s s
−
+ + + +
= =
+ + ++ + −
− + +
 
 
 
2 1
2
1 1
2 2
2 4 3 22 31 1
22 2
21 1
2 2
2 ( )
0
( ) ( )
3 7 42
3
s s F s
X s F s
s s s ss s
s s
+ +
−
= =
+ + ++ + −
− + +
 
 
The two transfer functions are 
 
 
2 1
1 2
4 3 2 3
2
3( )
( ) 3 7 4
s sX s
F s s s s s
+ +
=
+ + +
 , 
1
2 2
4 3 2 3
2
( )
( ) 3 7 4
X s
F s s s s s
=
+ + +
 
 
(b) Knowing ( ) 1F s = , from (a) we find 
 
 
2 1
2
1 4 3 2 3
2
3
( )
3 7 4
s s
X s
s s s s
+ +
=
+ + +
 , 
1
2
2 4 3 2 3
2
( )
3 7 4
X s
s s s s
=
+ + +
 
 
(c) Poles of 1( )X s and 2 ( )X s are located at the origin (simple pole) and in the left half-plane, 
 
 
112 
 
thus the final-value theorem applies, and 
 
 { }
2 1
2
1 1 3 2 30 0
2
3 1lim ( ) lim
33 7 4ss s s
s s
x sX s
s s s→ →
 + + = = = 
+ + +  
 
 { }
1
2
2 2 3 2 30 0
2
1lim ( ) lim
33 7 4ss s s
x sX s
s s s→ →
  = = = 
+ + +  
 
 
 
19. The state-space representation of a system model is described as 
 
 
u
y Du
= +
 = +
x Ax B
Cx

 
 where 
 [ ]1
2
0 1 0
 , , , 0 1 , 0 , 
2 1 1
x
D u f
x
     
= = = = = =     − −    
x A B C 
 Find 
(a) The I/O equation. 
(b) The transfer function. 
Solution 
(a) The state-variable equations are 
 1 2
2 1 2
 
2
x x
x x x f
=
 = − − +


 
 
Take the Laplace transform and solve for 2 ( )X s , because 2x is the output, to find 
 
 1 2
2 2
1 2
( ) ( ) 0 ( ) ( )
2 ( ) ( 1) ( ) ( ) 2
sX s X s sF sX s
X s s X s F s s s
− =
⇒ = + + = + +
 
 
This yields the I/O equation as 2 2 22x x x f+ + =   . 
 
(b) The transfer function is 2
2
( )
( ) 2
X s s
F s s s
=
+ +
. 
 
 
20. The state-space representation of a system model is described as 
 
 
u
u
= +
 = +
x Ax B
y Cx D

 
 where 
 
 
113 
 
 1
1
2 2
0 1 0 1 0 0
 , , , , , 
3 1 0 1 0
x
u f
x
        
= = = = = =         − −         
x A B C D 
 
 Find the transfer matrix. 
Solution 
The state-variable equations are 
 1 2
1
2 1 22
 
3
x x
x x x f
=
 = − − +


 
Laplace transformation yields 
 1 2
1
1 22
( ) ( ) 0 
( ) ( 3) ( ) ( )
sX s X s
X s s X s F s
− =
 + + =
 
 
The output equation indicates that 1x and 2x are the outputs. Solving the above system for 
1( )X s and 2 ( )X s , we find 
 1 2 1
2
1
2
0 1
( ) 3 1( ) ( )
1 3
3
F s s
X s F s
s s s
s
−
+
= =
− + +
+
 
 
 
1
2
2 2 1
2
1
2
0
( )
( ) ( )
1 3
3
s
F s sX s F s
s s s
s
= =
− + +
+
 
Then, 
 1 2
11 212 21 1
2 2
( ) ( )1( ) , ( )
( ) ( )3 3
X s X s sG s G s
F s F ss s s s
= = = =
+ + + +
 
Finally, 
 
2 1
2
2 1
2
1
3
( )
3
s s
s
s
s s
 
 + + =  
 
+ +  
G 
 
 
Problem Set 4.4 
In Problem 1 through 6, find the state-space form directly from the input-output equation. 
 
 
 
114 
 
1. 1
2 y y y u+ + =   
Solution 
Multiply by 2 to obtain 2 2 2y y y u+ + =   . Comparing with Eq. (4.14) we have 2n = , 1 2a = , 
2 2a = , 0 0b = , 1 2b = , 2 0b = . By Eq. (4.17), 
 
 1
2
0 1 0
 ,, , 
2 2 1
x
u u
x
     
= = = =     − −    
x A B 
 
By Eq. (4.18) we have [ ]0 2 , 0D= =C . The state-space form is therefore obtained as 
 
 
[ ]
0 1 0
2 2 1
0 2 
u
y
    
= +    − −    


 =
x x
x

 
 
2. 3 2 3 2y y y u u+ + = +   
Solution 
Divide by 3 to rewrite in standard form, as 1 2 2
3 3 3y y y u u+ + = +   . Comparing with Eq. (4.14) we 
find 3n = , 1 0a = , 1
2 3a = , 2
3 3a = , 0 0b = , 1 1b = , 2 0b = , 2
3 3b = . By Eq. (4.17), 
 
 
1
2
2 1
3 3 3
0 1 0 0
 , 0 0 1 , 0 , 
0 1
x
x u u
x
    
    = = = =     
     − −     
x A B 
 
By Eq. (4.18) we have 2
3 0 1 , 0D = = C . The state-space form is therefore obtained as 
 
 2 1
3 3
2
3
0 1 0 0
0 0 1 0
0 1
0 1 
u
y
    
    = +    
    − −    


  =  
x x
x

 
 
3. 2
5 2 2y y y y u u+ + + = +     
Solution 
Compare with Eq. (4.14): 3n = , 2
1 5a = , 2 1a = , 3 2a = , 0 1b = , 1 0b = , 2 2b = , 3 0b = . Then, 
Eqns. (4.17) and (4.18) yield 
 
 
115 
 
 
 2
5
2
5
0 1 0 0
0 0 1 0
2 1 1
2 1 1 
u
y u
    
    = +    
    − − −    


  = − − + ⋅ 
x x
x

 
 
4. 32 1
3 2 3y y y u u u+ + = + +    
Solution 
Multiply by 3
2 to get 3 9 3 31
2 4 2 2 2y y y u u u+ + = + +    and compare with Eq. (4.14) to deduce 2n = , 
3
1 2a = , 9
2 4a = , 1
0 2b = , 3
1 2b = , 3
2 2b = . The state equation is then obtained as 
 
 9 3
4 2
0 1 0
1
u
   
= +   − −   
x x 
 
Equation (4.18) yields 3 3
8 4 =  C and 1
2D = so that the output equation is 3 3 1
8 4 2y u = + ⋅  x . 
The state-space form is 
 
 
9 3
4 2
3 3 1
8 4 2
0 1 0
1
 
u
y u
    
= +    − −   


  = + ⋅ 
x x
x

 
 
 
5. (4)2 2 2y y y y u u u+ + + = + +    
Solution 
Divide by 2 to get (4) 1 1 1 1
2 2 2 2y y y y u u u+ + + = + +    with Eq. (4.14): 4n = , 1 0a = , 1
2 2a = , 
3 1a = , 1
4 2a = , 0 0b = , 1
1 2b = , 2 1b = , 3 0b = , 1
4 2b = . Then, Eqns. (4.17) and (4.18) yield 
 
 1 1
2 2
1 1
2 2
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
1 0 1
0 1 
u
y
    
    
    = +         − − −    


 =  
x x
x

 
 
 
 
116 
 
6. (4) 1
4 4y y y y u+ + + =  
Solution 
Compare with Eq. (4.14): 4n = , 1 0a = , 2 1a = , 1
3 4a = , 4 1a = , 0 0b = , 1 0b = , 2 0b = , 3 0b = , 
4 4b = . The state-space form is then formed as 
 
 
[ ]
1
4
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
1 1 0 1
4 0 0 0 
u
y
    
    
    = +    
    − − −    


=
x x
x

 
 
 
 
7. Derive the state-space representation in controller canonical form as described by Eqns. 
(4.19) and (4.20). 
Solution 
Choosing the state variables as ( 1) ( 2)
1 2 1, , ... , , n n
n nx v x v x v x v− −
−= = = = , the state-variable 
equations are formed as 
 
 
1 1 1 2 2
2 1
1 2
1
 
 
... 
 
 
n n
n n
n n
x a x a x a x u
x x
x x
x x
− −
−
= − − + ⋅⋅⋅ − +
=
=
=



 
 
 
Thus, the corresponding state equation is u= +x Ax B where 
 
 
1 2 2 1... 1
1 0 ... 0 0 0 0
 , , ... ... ... ... ... ... ...
0 0 ... 1 0 0 0
0 0 ... 0 1 0 0
n n na a a a a
u u
− −− − − − −   
   
   
   = = =
   
   
      
A B 
 
The output equation is written as 
 
 
( ) ( 1)
0 1 1
0 1 1 1 1 1
 
 
n n
n n
n n n n
y b v b v b v b v
b x b x b x b x
−
−
− −
= + + ⋅⋅⋅ + +
= + + ⋅⋅⋅ + +


 
 
 
 
117 
 
Since the output equation cannot contain 1x , we substitute for 1x using the last relation in the 
state-variable equations above. The result is 
 
 0 1 1 2 2 1 1 1 1
0 1 1 1 0 2 2 2 0 0
( ) 
 ( ) ( ) ( )
n n n n n n
n n n
y b a x a x a x u b x b x b x
b a b x b a b x b a b x b u
− −= − − − ⋅⋅⋅ − + + + ⋅⋅⋅ + +
= − + + − + + ⋅⋅⋅ + − + +
 
 
Finally, the output equation is expressed as y Du= +Cx where 
 
 [ ]0 1 1 0 2 2 0 01... , n n nb a b b a b b a b D b×= − + − + − + =C 
 
 
8. A system’s I/O equation is provided as 
 
 (4)5 1 1 1
6 6 2 3y y y y u u u+ + + = + +    
 
(a) Find the state-space representation in controller canonical form. 
(b) Confirm the results of (a) in MATLAB. 
Solution 
(a) Multiply by 6
5 to get (4) 6 3 6 61 2
5 5 5 5 5 5y y y y u u u+ + + = + +    . Comparing with Eq. (4.14), 
 
 4n = , 6
1 5a = , 1
2 5a = , 3 0a = , 3
4 5a = , 0 0b = , 6
1 5b = , 2 0b = , 6
3 5b = , 2
4 5b = 
 
Selecting the state variables as 1x y= , 2x y=  , 3x y=  and 4x y= , Eqns. (4.19) and (4.20) yield 
 
 
6 31
5 5 5
6 6 2
5 5 5
10
01 0 0 0
 , , 0 , 0
00 1 0 0
00 0 1 0
D
 − − −  
   
     = = = =    
   
  
A B C 
 
(b) The transfer function is directly obtained from the I/O equation, as 
 
 
3 1
3
4 3 25 1 1
6 6 2
( )
( )
s sY s
U s s s s
+ +
=
+ + +
 
 
>> Num = [1 0 1 1/3]; Den = [5/6 1 1/6 0 1/2]; 
>> [A,B,C,D] = tf2ss(Num,Den) 
 
A = 
 -1.2000 -0.2000 0 -0.6000 
 1.0000 0 0 0 
 0 1.0000 0 0 
 0 0 1.0000 0 
 
 
118 
 
 
B = 
 1 
 0 
 0 
 0 
C = 
 1.2000 0 1.2000 0.4000 
 
D = 
 0 
 
 
 
In Problems 9 through 12, given the transfer function ( ) / ( )Y s U s , find 
(a) The input-output equation. 
(b) The state-space form directly from the input-output equation in (a). 
 
9. 
3
4
3 1
3 1
s
s s
+
+ +
 
Solution 
(a) Using the given transfer function, 
 
 
3
34 31
3 43 1
3
( ) ( 1) ( ) ( ) ( )
( ) 1
sY s s s Y s s U s
U s s s
+
= ⇒ + + = +
+ +
 
 
Transforming back to time domain, we find 31
3 4y y y u u+ + = +   . 
 
(b) Compare with Eq. (4.14), and subsequently use Eqns. (4.17) and (4.18) to find 
 
 1
3
3
4
0 1 0 0
0 0 1 0
1 0 1
1 0 0 
u
y u
    
    = +    
    − −    


  = + ⋅ 
x x
x

 
10. 
21
5
2 3 1
5 10
1s
s s
+
+ +
 
Solution 
(a) Using the given transfer function, 
 
 
21
2 25 3 1 1
5 10 52 3 1
5 10
1( ) ( ) ( ) ( 1) ( )
( )
sY s s s Y s s U s
U s s s
+
= ⇒ + + = +
+ +
 
 
 
119 
 
 
Transforming back to time domain, we have 3 1 1
5 10 5y y y u u+ + = +   . 
 
(b) Compare with Eq. (4.14), and use Eqns. (4.17) and (4.18) to find 
 
 
31
10 5
49 3 1
50 25 5
0 1 0
1
u
y u
    
= +    − −   


  = − + ⋅ 
x x
x

 
 
 
11. 
3
3 22
3
2 1
1
s s
s s s
+ +
+ + +
 
Solution 
(a) Using the given transfer function, 
 
 
3
3 2 32
33 22
3
( ) 2 1 ( 1) ( ) ( 2 1) ( )
( ) 1
Y s s s s s s Y s s s U s
U s s s s
+ +
= ⇒ + + + = + +
+ + +
 
 
Transforming back to time domain, we have 2
3 2y y y y u u u+ + + = + +     . 
 
(b) Multiply by 3
2 to get 3 3 3 3 3
2 2 2 2 23y y y y u u u+ + + = + +     . Compare with Eq. (4.14), and use 
Eqns. (4.17) and (4.18) to find 
 
 3 3 3
2 2 2
3 3 9 3
4 4 4 2
0 1 0 0
0 0 1 0
1
 
u
y u
    
    = +    
    − − −   


  = − − + ⋅ 
x x
x

 
12. 
3
4 2
2
2 3 2
s s
s s s
+
+ + +
 
Solution 
(a) Using the given transfer function, 
 
 
3
4 2 3
4 2
( ) 2 (2 3 2) ( ) ( 2 ) ( )
( ) 2 3 2
Y s s s s s s Y s s s U s
U s s s s
+
= ⇒ + + + = +
+ + +
 
 
Transforming back to time domain, we have (4)2 3 2 2y y y y u u+ + + = +   . 
 
 
120 
 
 
(b) Divide by 2, compare with Eq. (4.14), then use Eqns. (4.17) and (4.18) to find 
 
 3 1
2 2
1
2
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
1 0 1
1 0 0 0 
u
y u
    
    
   = +         − − −    


 = + ⋅  
x x
x

 
 
13. The state-variable equations and the output equation for a dynamic system are given as 
 
 1 2 1
1 22
2 2 1
 
 , 
x x
y x x
x x x u
=
= + = − − +


 
 
Find the transfer function (or matrix) by determining the Laplace transforms of 1x and 2x in the 
state-variable equations and using them in the Laplace transform of the output equation. 
Solution 
Laplace transformation of state-variable equations yields 
 
 1 2
1 2
( ) ( ) 0 
( ) ( 1) ( ) ( )
sX s X s
X s s X s U s
− =
 + + =
 
Using Cramer’s rule, 
 1 22 2
1( ) ( ) , ( ) ( )
1 1
sX s U s X s U s
s s s s
= =
+ + + +
 
 
Insert into the Laplace transform of the output equation: 
 
 
1
21
1 22 2 2( ) ( ) ( ) ( ) ( )
1 1
sY s X s X s U s U s
s s s s
= + = +
+ + + +
 
 
Finally, the transfer function is found as 
 
1
2
2
( )( )
( ) 1
sY sG s
U s s s
+
= =
+ +
 
 
14. Repeat Problem 13 for 
 
 1 2 1
1 1
2 2 1 22 2
 
 , 
x x x
x x x u x
=  
=  = − − +  
y


 
 
 
121 
 
Solution 
Laplace transformation of state-variable equations yields 
 
 1 2
1 1
1 22 2
( ) ( ) 0 
( ) ( ) ( ) ( )
sX s X s
X s s X s U s
− =
 + + =
 
Using Cramer’s rule, 
 
1 1
2 2
1 22 21 1
2 2
( ) ( ) , ( ) ( )
1 1
s
X s U s X s U s
s s s s
= =
+ + + +
 
 
Insert into the Laplace transform of the output equation: 
 
 
1
2
2 1
2
1
2
2 1
2
( )
1
( )
( )
1
U s
s s
s
s
U s
s s
 
 
+ + =  
 
 + + 
Y 
The transfer matrix is then obtained as 
 
1
2
2 1
2
1
2
2 1
2
1
( )
1
s s
s
s
s s
 
 
+ + =  
 
 + + 
G 
 
 
In Problems 15 through 19, the matrices in the state-space form of a system model are given. 
(a) Find the transfer function (or transfer matrix) using Eq. (4.21). 
(b) Verify the result of (a) using the ss2tf command. 
 
15. 1
1 2
7
0 1 0
 , , 1 , 0
2 2
D
   
 = = = =     − −   
A B C 
Solution 
(a) The system is SISO, hence there is only one transfer function. Since ( )sG is 1 1× , it is 
simply denoted by ( )G s . With 0D = , Eq. (4.21) reduces to 1( ) ( )G s s −= −C I A B . Then, 
 
 1 1
2 2 2 2 1
7
7 1 7 01 7(2 1) 2 1( ) ( ) 1
14 7 27 14 7 14 2
s s sG s s
ss s s s s s
− +    + + = − = = =     −+ + + + + +   
C I A B 
(b) 
>> A = [0 1;-2 -1/7]; B = [0;2]; C = [1/2 1]; D = 0; [n,d] = ss2tf(A,B,C,D) 
 
n = 
 0 2 1 % Num = 2s+1; 
d = 
 
 
122 
 
 1.0000 0.1429 2.0000 % Den = s2+(1/7)s+2 
 
16. [ ] 1
2 3
3
00 1
 , , 0 1 , 
1 2
D
  
= = = =  − −   
A B C 
Solution 
 (a) The system is SISO, hence there is only one transfer function. Since ( )sG is 1 1× , it is 
simply denoted by ( )G s . Eq. (4.21) yields 
 [ ]
2
1
22 2
3
02 11 1 4 1( ) ( ) 0 1
1 32 1 3( 2 1)
s s sG s s D
ss s s s
−  +  + +
= − + = + =  −+ + + +   
C I A B 
(b) 
>> A = [0 1 ;-1 -2]; B = [0;2/3]; C = [0 1]; D = 1/3; 
>> [n,d] = ss2tf(A,B,C,D) 
 
n = 
 0.3333 1.3333 0.3333 
 
d = 
 1 2 1 
 
 
17. 2 1
2 1 1
3 3 2
0 1 0 0
1 0 0
0 0 1 , 0 , , 
0 0 1
1
×
   
    = = = =    
    − − −   
A B C D 0 
Solution 
(a) The output equation is 2 3 3 1 2 1u× × ×= +y C x D , hence there are two outputs and one input. 
Therefore, ( )sG is 2 1× . By Eq. (4.21), 
 
 
1
2
3 2
12
2
3 2 2
( ) ( )
3 3 1 3( 1) 3 0
1 0 0 1 2 3 ( 1) 3 0
0 0 1 3 3 2
2 ( 2) 3
13 
2(3 3 2)
s s
s s s
s s s
s s s
s s s
s s s s
−= − +
   + + +
    = − +    + + +     − − +   
 
=  
+ + +  
G C I A B D
 
(b) 
>> A = [0 1 0;0 0 1;-2/3 -1/3 -1]; B = [0;0;1/2]; C = [1 0 0;0 0 1]; 
>> D = [0;0]; [n,d] = ss2tf(A,B,C,D) 
 
n = 
 0 0 0 0.5000 
 0 0.5000 0 0 
 
d = 
 1.0000 1.0000 0.3333 0.6667 
 
 
123 
 
 
 
18. 
1
2
1
4
0 1 0 0 0
0 1 0 0
0 0 1 , 1 0 , , 
0 0 1 0 0
2 1 2 0
  
    = = = =           − − −   
A B C D 
Solution 
(a) The output equation is 2 3 3 1 2 2 2 1× × × ×= +y C x D u , hence there are two outputs and two inputs. 
Therefore, ( )sG is 2 2× . By Eq. (4.21), 
 
 
1
2
3 2 2 3 2
1
2
3 2 2 3 2
1
42
3 2 2 3 2
3 2
3
( ) ( )
2 1 1 1
2 2 1 2 2 0 0
0 1 0 02 1 0
0 0 1 0 02 2 1 2 2 0
2 1
2 2 1 2 2
4 5 2
2(
 
s s
s s
s s s s s s s
s s
s s s s s s s
s s
s s s s s s s
s s s
s
−= − +
 + +
 
+ + + + + + +   
    −  = +      + + + + + + +        − −
 
+ + + + + + + 
+ + +
+
=
G C I A B D
2 3 2
2
3 2 3 2
2 2) 4( 2 2)
( 2)
2 2 4( 2 2)
s
s s s s s
s s
s s s s s s
 
 
+ + + + + 
 − + 
 + + + + + +  
(b) 
>> A = [0 1 0;0 0 1;-2 -1 -2]; B = [0 0;1 0;0 1/4]; C = [0 1 0;0 0 1]; 
>> D = [1/2 0;0 0]; 
>> [n1,d1] = ss2tf(A,B,C,D,1) % Input 1 
 
n1 = 
 0.5000 2.0000 2.5000 1.0000 % 0.5s3+2s2+2.5s+1 
 0 0 -1.0000 -2.0000 % -(s+2) 
 
d1 = 
 1.0000 2.0000 1.0000 2.0000 
 
 
>> [n2,d2] = ss2tf(A,B,C,D,2) % Input 2 
 
n2 = 
 0 0 0.2500 0 % 0.25s 
 0 0.2500 0 0 % 0.25s2 
 
d2 = 
 1.0000 2.0000 1.0000 2.0000 
 
 
 
 
 
124 
 
19. 
0 1 0 0 0
1 0 0 0 0
0 0 1 , 1 0 , , 
0 1 0 0 1
1 0 1 0 2
   
      = = = =             − −   
A B C D 
Solution 
(a) The output equation is 2 3 3 1 2 2 2 1× × × ×= +y C x D u , hence there are two outputs and two inputs. 
Therefore, ( )sG is 2 2× . By Eq. (4.21), 
 
 
1
3 2
2
3 2 3 2
3 2
3 2 3 2
( ) ( )
( 1) 1 1 0 0
1 0 0 0 01 1 ( 1) 1 0
0 1 0 0 11 0 21
1 2
1 1 
( 1) 2 1
1 1
s s
s s s
s s s
s s
s s
s
s s s s
s s s s s
s s s s
−= − +
 + +  
      = − + +      + +      − −   
+ 
 + + + + =
 + + + +
 + + + + 
G C I A B D
 
(b) 
>> A = [0 1 0;0 0 1;-1 0 -1]; B = [0 0;1 0;0 2]; C = [1 0 0;0 1 0]; 
>> D = [0 0;0 1]; 
>> [n1,d1] = ss2tf(A,B,C,D,1) % Input 1 
 
n1 = 
 
 0 0 1 1 % s+1 
 0 1 1 0 % s2+s 
 
d1 = 
 
 1.0000 1.0000 -0.0000 1.0000 
 
>> [n2,d2] = ss2tf(A,B,C,D,2) % Input 2 
 
n1 = 
 
 0 0 0 2.0000 % 2 
 1.0000 1.0000 2.0000 1.0000 % s3+s2+2s+1 
 
d1 = 
 
 1.0000 1.0000 -0.0000 1.0000 
 
 
20. A dynamic system model has state variables 1x and 2x and input u . The two I/O equations 
are derived as 
 1
1 1 1 2 2 232 , 2 2x x x u u x x x u+ + = + + + =     
 Find the state-space representation for this model and subsequently the transfer matrix. 
 
 
125 
 
Solution 
Using Eqns. (4.17) and (4.18), the first I/O equation generates 
 
 1
3
0 1 0
 , , 1 , 0
2 1 1
D     = = = =     − −   
A B C 
The second I/O equation gives 
 
 [ ]0 1 0
 , , 2 0 , 0
2 1 1
D   
= = = =   − −   
A B C 
 
As expected, A and B are the same in both cases. This implies that the system model state-
space form is 
 
u
u
= +
 = +
x Ax B
y Cx D

 
where 
 
1
30 1 0 01
 , , , 
2 1 1 02 0
      
= = = =      − −      
A B C D 
 
The transfer matrix is therefore 2 1× and obtained as 
 
 
1
21
3
2
2
( ) ( )
3 1
1 1 01 1 3( 2) 
2 12 0 2 2
2
s s
s
s s s
ss s
s s
−= −
+ 
 +      + + = =     −  + +     
 + + 
G C I A B
 
The two individual transfer functions in ( )sG are clearly valid as observed by the given I/O 
equations. 
 
 
 
 
 
126 
 
Problem Set 4.5 
1. Reduce the block diagram in Figure 4.25 by moving the constant gain block K to the right of 
the summing junction. Subsequently, find the transfer function ( ) / ( )Y s U sand 
generate a figure that can be imported into a document. 
Integrated signal
amplified by 2
Sine Wave
Amplitude = 0.8
Frequency = 1.5 r/s
1
s
Integrator
Scope
1
Out1
2
Gain
0 1 2 3 4 5 6 7 8 9 10
-1
-0.5
0
0.5
1
1.5
2
2.5
t
Original signal
Output
 
 
10 
 
 
 Figure 1.23 Problem 20. 
 
Solution 
 
 Figure Review1No20 
 
 
 
21 Build the Simulink model shown in Figure 1.24. Select the signal generator as a square 
wave with amplitude 1 and frequency 0.5 rad/sec . To flip the gain block, right click on it, 
then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum 
(Commonly Used Blocks) allows control over the list of desired signs. Run simulation and 
generate a figure that can be imported into a document. 
 
 Figure 1.24 Problem 21. 
 
 
SumSignal generator
1
Out1
Scope
1
s
Integrator
5
Feedback gain
Square wave
Amplitude = 0.5
Frequency = 0.5 r/s
0 1 2 3 4 5 6 7 8 9 10
-0.1
-0.05
0
0.05
0.1
0.15
t
Sum
Signal generator
1
Out1
Scope
1
s
Integrator
2
Feedback gain
Square wave
Amplitude = 1
Frequency = 0.5 r/s
 
 
11 
 
Solution 
 
 Figure Review1No21 
 
22 Build the Simulink model shown in Figure 1.25. To flip the gain block, right click on it, 
then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum 
allows control over the list of desired signs. Run simulation and generate the response plot. 
 
 Figure 1.25 Problem 22. 
 
Solution 
 
 
 Figure Review1No22 
 
0 1 2 3 4 5 6 7 8 9 10
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
t
Sum
1
Out1
Scope
1
s
Integrator1
2
Feedback gain
Sine Wave
Amplitude = 1
Frequency = 1.25 r/s
1
s
Integrator 2
0 1 2 3 4 5 6 7 8 9 10
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
t
 
 
12 
 
 
23 Build the Simulink model shown in Figure 1.26. Transfer function represents the ratio of the 
Laplace transforms of the output and the input signals associated with a block; see Chapter 4. 
The Transfer Fcn block is located in the Continuous library. Run simulation and generate 
the response plot. 
 
 Figure 1.26 Problem 23. 
 
Solution 
 
 
 Figure Review1No23 
 
24 Build the Simulink model shown in Figure 1.27. The Step signal has a step time of 1 and a 
final value of 1, that is, it assumes a value of 1 immediately after 1t = and a value of 0 
otherwise. The Transfer Fcn block is located in the Continuous library. Run simulation 
and generate the response plot. 
 
 
 Figure 1.27 Problem 24. 
Sum
1
Out1
Scope
1
2s+1
Transfer Fcn
5
Gain
Sine Wave
Amplitude = 1.5
Frequency = 1.25 r/s
0 1 2 3 4 5 6 7 8 9 10
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
t
1
Out1
Scope
s
s +2s+22
Transfer Fcn
4
Gain 2
2.5
Gain1
1
s
Integrator
Step
 
 
13 
 
 
Solution 
 
 Figure Review1No24 
 
25 Build and simulate a Simulink model that integrates a sawtooth signal (amplitude 1, 
frequency 1 rad/sec) and displays the result, along with the sawtooth signal itself. 
 
Solution 
 
 Figure Model in Problem 25 
 
 
 Figure Review1No25 
0 1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t
1
s
Integrator
Scope
1
Out1
Sawtooth
Amplitude = 1
Frequency = 1 r/s
0 1 2 3 4 5 6 7 8 9 10
-5
-4
-3
-2
-1
0
1
2
3
4
5
t
Integrated
signal 
Sawtooth
 
 
14 
 
 
26 Build the model shown in Figure 1.28, which is the Simulink model of the RLC circuit 
considered in this chapter and is equivalent to the Simscape model presented in Figure 1.19. 
Perform the simulation to confirm that both models yield the same response. 
 
 Figure 1.28 Problem 26. 
Solution 
Simulation generates the same response curve as in Figure 1.21. 
 
 
 Figure Review1No26 
 
 
27 Build a Simscape model for an RL circuit ( 5 R = Ω , 0.5 HL = ) driven by an applied voltage 
represented by a pulse (amplitude 0.6 for 1 3t. 
 
 
 Figure 4.25 Problem 1. 
 
Solution 
Based on the strategy in Figure 4.15, the block diagram reduces to what is shown in Figure PS4-
5 No1. This is a negative feedback loop, hence 
 
 
( )
( ) 11
Y s KG KG
HU s GHKG
K
= =
++
 
 
 
 Figure PS4-5 No1 
 
 
2. Consider the block diagram in Figure 4.26. Use the block diagram reduction steps listed 
below to find ( ) / ( )Y s U s . 
(a) Replace the single loop containing 1( )G s and ( )H s with a single equivalent block. 
(b) Move the gain block K to the right of the summing junction. 
(c) Replace the resulting loop with a single equivalent block. 
 
 
 Figure 4.26 Problem 2. 
 
Solution 
Details are shown in Figure PS4-5No2. 
 
( )Y s+( )U s ( )G s
( )H s
K
−
Σ
( )Y s+( )U s G
/H K
K
−
Σ
( )Y s+( )U s 1( )G s
( )H s
K −Σ 2 ( )G s
−
 
 
127 
 
 
 Figure PS4-5 No2 
 
3. For the block diagram in Figure 4.27, find the transfer function ( ) / ( )Y s U s using 
(a) Block diagram reduction techniques. 
(b) Mason’s rule. 
 
 
 Figure 4.27 Problem 3. 
 
Solution 
(a) The parallel connection involving 2G and 3G is replaced with a single block 2 3G G+ . This 
block, together with 1G and 4G form a series combination, hence replaced with 1 2 3 4( )G G G G+ . 
This new block and H are in a parallel connection. The final block is 1 2 3 4( )G G G G H+ + . 
(b) There are no loops and three forward paths with gains H , 1 2 4G G G and 1 3 4G G G . All paths 
are also coupled, thus the special case applies and yields 
 
 1 2 4 1 3 4
( )
( )
Y s H G G G G G G
U s
= + + 
 
This agrees with the result of (a). 
 
4. Using Mason’s rule, find ( ) / ( )Y s U s for the block diagram in Figure 4.16, Example 4.21. 
Solution 
There is one loop with gain 1 1G H− and two forward paths with gains 1 2G G K and 1 2G H K . 
Since all paths are coupled, special Mason’s rule gives 
( )Y s+( )U s 1
11
G
HG+
K Σ 2G
−
( )Y s
+( )U s 1 2
11
G G
HG+
KΣ
−
1/ K
( )Y s( )U s 1 2
1 1 21
KG G
HG G G+ +
( )a
( )b
( )c
( )Y s( )U s 1( )G s
( )H s
2 ( )G s
3( )G s
+
+
4 ( )G s
+ +
Σ Σ
 
 
128 
 
 
 1 2 1 2 1 2 2
1 1 1 1
( )( )
( ) 1 1
G G K G H K G K G HY s
U s G H G H
+ +
= =
+ +
 
 
5. Find ( ) / ( )Y s U s in Figure 4.28 using Mason’s rule. 
Solution 
There are three loops with gains 1 1G H , 1 2 3KG G H− and 2 2G H− , as well as one forward path with 
gain 1 2KG G . Since the three loops and the forward path are coupled, the special case of Mason’s 
rule applies. Therefore, 
 
 1 2
1 2 3 1 1 2 2
( )
( ) 1
KG GY s
U s KG G H G H G H
=
+ − +
 
 
 
6. Use block diagram reduction techniques listed below to find the overall transfer function for 
the diagram in Figure 4.28. 
(a) Move the summing junction of the negative feedback loop containing 2 ( )H s outside of the 
psitive loop containing 1( )H s . 
(b) In the ensuing diagram, replace the loop containing 1( )H s with a single block. 
(c) Similarly, replace the two remaining loops with single equivalent blocks to obtain one single 
block whose input is ( )U s and whose output is ( )Y s . 
 
 
 Figure 4.28 Problems 5 and 6. 
Solution 
Details are shown in Figure PS4-5No6. The overall transfer function is readily obtained as 
 
 1 2
1 2 3 1 1 2 2
( )
( ) 1
KG GY s
U s KG G H G H G H
=
+ − +
 
( )Y s+( )U s 1 ( )G s
1( )H s
2 ( )G s+
−
2 ( )H s
+
3( )H s
−
Σ Σ ΣK
+
 
 
129 
 
 
 Figure PS4-5No6 
 
7. In Figure 4.29, find the transfer function ( ) / ( )Y s U s using 
(a) Block diagram reduction steps. 
(b) Mason’s rule. 
Solution 
(a) The negative feedback loop is replaced with a single block of 1
1 11
G
G H+
. This, together with 
Σ ( )Y s+( )U s 1( )G s
1( )H s
2 ( )G sΣ+
+
2
1
( )
( )
H s
G s
Σ+
−
3( )H s
−
( )a
Σ ( )Y s+( )U s 2 ( )G s
2
1
( )
( )
H s
G s
Σ+
3( )H s
−
( )b
1
1 1
( )
1 ( ) ( )
G s
G s H s−
Σ ( )Y s
+( )U s
3( )H s
−
( )c
1 2
1 1 2 2
( ) ( )
1 ( ) ( ) ( ) ( )
G s G s
G s H s G s H s− +
( )Y s( )U s
( )d
1 2
1 1 2 2 1 2 3
( ) ( )
1 ( ) ( ) ( ) ( ) ( ) ( ) ( )
KG s G s
G s H s G s H s KG s G s H s− + +
K
K
−
K
 
 
130 
 
K and 2G form a series connection, hence replaced with 1 2
1 11
KG G
G H+
. This block and 2H are in a 
parallel connection, thus the transfer function is 
 
 1 2
2
1 11
KG GY H
U G H
= +
+
 
 
(b) Since not all paths are coupled, general case of Mason’s rule will be applied. There are two 
forward paths and one negative feedback loop. The quantities in Eq. (4.25) are: 
 
 1 2 2 1 2 1 1 , , 1F H F KG G D G H= = = + 
 
 1D = same as D when restricted to the bottom portion 1 11 G H= + 
 
 2D = same as D when restricted to the top portion 1= 
Then 
 
 2 1 1 1 2 1 2
2
1 1 1 1
(1 )( )
( ) 1 1
H G H KG G KG GY s H
U s G H G H
+ +
= = +
+ +
 
 
 
 
 Figure 4.29 Problem 7. 
 
 
 
8. Find the overall transfer function in Figure 4.30 using Mason’s rule. 
Solution 
There are two loops with gains 2 2G H and 2 3 3G G H− , and one forward path with gain 1 2 3G G G . 
All paths are coupled hence Mason’s rule (special case) applies: 
 
 1 2 3
2 2 2 3 3
( )
( ) 1
G G GY s
U s G H G G H
=
− +
 
 
 
( )Y s
+
( )U s
K
2 ( )H s
+
+
−
Σ
Σ 1 ( )G s 2 ( )G s
1( )H s
 
 
131 
 
 
 Figure 4.30 Problem 8. 
 
9. For the block diagram in Figure 4.31, find ( ) / ( )Y s U s using 
(a) Block diagram reduction. 
(b) Mason’s rule. 
 
 
 Figure 4.31 Problem 9. 
Solution 
(a) The negative loop on the left is replaced with 1
1 11
G
G H+
. The negative loop containing 2G 
and 2H is replaced with 2
2 21
G
G H+
. This block and 3G are in a series and part of a negative 
feedback, which is replaced with 
 
 
2 3
2 32 2
2 3 2 2 2 3
2 2
1
11
1
G G
G GG H
G G G H G G
G H
+
=
+ ++
+
 
This block and 1
1 11
G
G H+
 are in a series, hence 
 
 1 2 3
1 1 2 2 2 3
( )
( ) (1 )(1 )
G G GY s
U s G H G H G G
=
+ + + 
 
 
 
(b) General case of Mason’s rule applies because the loop on the left is not coupled with the 
other two. There is one forward path and three negative feedback loops. The quantities in Eq. 
(4.25) are: 
 
 
 1 1 2 3 1 1 2 2 2 3 1 1 2 2 1 1 2 3
Single-loop gains Gain product of
non-touching two-loops
 , 1F G G G D G H G H G G G H G H G H G G= = + + + + +
 
 
 
( )Y s
+
( )U s
−
Σ 2 ( )G s 3 ( )G s
2 ( )H s
3 ( )H s
1 ( )G s
+
( )Y s+( )U s
−
+
−Σ Σ1 ( )G s 2 ( )G s
2 ( )H s
3 ( )G s
1( )H s
−
 
 
132 
 
 1D = same as D when restricted to the portion not touching the forward path 1= 
 
Therefore, 
 1 2 3
1 1 2 2 2 3 1 1 2 2 1 1 2 3
( )
( ) 1
G G GY s
U s G H G H G G G H G H G H G G
=
+ + + + +
 
 
Of course, this agrees with the result of (a). 
 
 
10. Consider the block diagram in Figure 4.32 where ( )V s represents an external disturbance. 
(a) Find the transfer function ( ) / ( )Y s U s by setting ( ) 0V s = . 
(b) Find ( ) / ( )Y s V s by setting ( ) 0U s = . 
Solution 
(a) When ( ) 0V s = , the forward path going from ( )U s to ( )Y s has a gain of 1 2G G , while the two 
loops have gains 2 2G H− and 1 2 1G G H− . Using the special case of Mason’s rule, 
 
 1 2
2 2 1 2 1
( )
( ) 1
G GY s
U s G H G G H
=
+ +
 
 
(b) When ( ) 0U s = , the forward path goes from ( )V s to ( )Y s and has gain 2G . The two loops 
have gains 2 2G H− and 1 2 1G G H− . Mason’s rule (special case) yields 
 
 2
2 2 1 2 1
( )
( ) 1
GY s
V s G H G G H
=
+ +
 
 
 
 
 Figure 4.32 Problem 10. 
 
 
11. The state-variable equations and output equation of a system are given as 
 
 1 2
1 21
2 1 22
 
 , 2
2
x x
y x x
x x x u
=
= + = − − +


 
( )Y s
+( )U s
−
+
−
+
( )V s
External disturbance
Σ 1 ( )G s 2 ( )G s
1( )H s
2 ( )H s
Σ
 
 
133 
 
 where u and y are the input and output, respectively. 
(a) Construct the block diagram. 
(b) Find the transfer function ( ) / ( )Y s U s directly from the blockdiagram. 
(c) Find the transfer function ( ) / ( )Y s U s from the state-space form and compare with (b). 
Solution 
(a) Details are shown in Figure PS4-5No11. 
 
 
 Figure PS4-5No11 
 
(b) There are two forward paths with gains 2
4
s
 and 2
s
, and two loops with gains 1
s
− and 2
1
2s
− . 
All paths are coupled and by the special case of Mason’s rule we have 
 
2
2
2
4 2
( ) 4 8
1 1( ) 2 2 11
2
Y s sss
U s s s
s s
+ +
= =
+ ++ +
 
(c) In state-space form, we have 
 [ ]1
2
0 1 0
 , , 2 1
1 2
   
= = =   − −   
A B C 
Therefore, 
 1
2
( ) 4( 2)( )
( ) 2 2 1
Y s ss
U s s s
− +
= − =
+ +
C I A B 
This agrees with the result of (b). 
 
12. The state-space representation of a system model is given as 
 
u
y Du
= +
 = +
x Ax B
Cx

 
with 
 1 1
2 2
2 3
0 1 0
 , , , 1 , 0 , 
2 3
x
D u u
x
    
 = = = = = =       − −     
x A B C 
(a) Contruct the block diagram. 
(b) Find the transfer function ( ) / ( )Y s U s directly from the block diagram. 
(c) Find the transfer function ( ) / ( )Y s U s from the state-space form and compare with (b). 
Solution 
(a) The state-space form reveals the following information: 
2x
( )Y s( )U s Σ
+
−
−
1
s
1
s
2 Σ
+
1
2
+
2x
1x
1
12 x
2x
2 2x
 
 
134 
 
 1 2 1
1 22 2
2 1 23
 
 , 
2 3
x x
y x x
x x x u
=
= + = − − +


 
 
Proceeding as always, the block diagram is constructed and shown in Figure PS4-5No12. 
 
 
 Figure PS4-5No12 
 
(b) Two forward paths with gains 2
3
s
 and 3
2s
, and two loops with gains 2
3s
− and 2
2
s
− . All 
paths are coupled and by the special case of Mason’s rule we have 
 
 
2
2
2
3 3
( ) 3(3 6)2
2 2( ) 2(3 2 6)1
3
Y s sss
U s s s
s s
+ +
= =
+ ++ +
 
(c) In state-space form, we have 
 
 1
2 2
3
0 1 0
 , , 1
2 3
   
 = = =     − −   
A B C 
Therefore, 
 1
2
( ) 9( 2)( )
( ) 2(3 2 6)
Y s ss
U s s s
− +
= − =
+ +
C I A B 
 
This agrees with the result of (b). 
 
 
In problems 13 through 16, the I/O equation of a dynamic system is provided. 
(a) Find the state-space model. 
(b) Construct the block diagram based on the information in (a). 
(c) Determine the overall transfer function directly from the block diagram in (b). 
 
13. 1
33 2y y y u+ + =   
Solution 
(a) With 1x y= and 2x y=  , we find 
 
3 ( )Y s( )U s Σ
+
−
−
1
s
1
s Σ
+
+
1x
12x
2x
2
3
2
2x
2
23 x
1
2
 
 
135 
 
 
1 2
21 1
2 1 23 3
 
 , 
2
x x
y x
x x x u
= =  = − − +  


 
 
Therefore, the matrices in the state-space representation are 
 
 [ ]2 1 1
3 3 9
0 1 0
 , , 0 1 , 0D
   
= = = =   − −   
A B C 
 
(b) The block diagram is constructed as shown in Figure PS4-5No13. 
 
 
 Figure PS4-5No13 
(c) Noting that 2x is the output, there is one forward path with gain 1
9s
, and two loops with 
gains 1
3s
− and 2
2
3s
− . All paths are coupled and by the special case of Mason’s rule we have 
 2
2
1
( ) 9
1 2( ) 3(3 2)1
3 3
Y s ss
U s s s
s s
= =
+ ++ +
 
 
14. 2 1
3 32 2y y y u u+ + = +   
Solution 
(a) With 1x y= and 2x y=  , we find 
 1 2 1
1 23 3
2 1 22
 
 , 2
( 2 )
x x
y x x
x x x u
=
= + = − − +


 
(b) The block diagram can then be constructed as shown in Figure PS4-5No14. 
 
 
 Figure PS4-5No14 
1
9
( )U s Σ
+
−
−
1
s
1
s
1x2x
2x
1
3
2
3
2y x=
( )Y s( )U s Σ
+
−
−
1
s
1
s
Σ+
+
1x2x
3
23
2
2x
3
2
1
3
 
 
136 
 
(c) Two forward paths with gains 1
2s
 and 2
3
s
, and two loops with gains 3
s
− and 2
3
2s
− . All 
paths are coupled and by the special case of Mason’s rule we have 
 
2
2
2
1 3
( ) 62
3 3( ) 2 6 31
2
Y s ss s
U s s s
s s
+ +
= =
+ ++ +
 
 
15. 2 2y y y u u+ + = +    
Solution 
(a) With 1x y= , 2x y=  and 3x y=  , we find (Section 4.4) 
 1 1 1
4 4 2
1 1
2 2
0 1 0 0
0 0 1 , 0 , 1 , 
0 1
D
   
     = = = − − =    
   − −   
A B C 
The above can then be expressed as 
 
1 2
1 1 1
2 3 1 2 34 4 2
1 1
3 3 12 2
 
 , 
x x
x x y x x x u
x x x u
 =
 = = − + − +
 = − − +



 
(b) The block diagram can then be constructed as shown in Figure PS4-5No15. 
 
 
 Figure PS4-5No15 
(c) Because not all paths are coupled, the general case of Mason’s rule applies as follows: Four 
forward paths with gains 1
1
2
F = , 2
1
4
F
s
−
= , 3 3
1
4
F
s
−
= and 4 2
1F
s
= , and two loops with gains 
1
2s
− and 3
1
2s
− . 
 1 2 3 43
1 11 , , 1
2 2
D D D D D D
s s
= + + = = = = 
 
4 2 31
2 3 41 2
3 2 3 2
( ) 1 4 1 2
( ) 2 2(2 1) 2 1
i ii F D D F F FY s s s s s
U s D D s s s s
= + + + − + − +
= = = + =
+ + + +
∑ 
( )Y s( )U s Σ
+
−
−
1
s
1
s Σ
+
−
3x
3x
1
2
1
s
2x 1x3x
1
2
1
4
−
1
4
1
2
+
Direct link from input to output
 
 
137 
 
 
16. 1 1
2 22y y y u u+ + = +   
Solution 
(a) With 1x y= , 2x y=  and 3x y=  , we find (Section 4.4) 
 1 1
4 2
1
2
0 1 0 0
0 0 1 , 0 , 0 0 , 
2 0 1
D
   
     = = = − =    
   − −   
A B C 
The above can then be expressed as 
 
1 2
1 1
2 3 24 2
1
3 2 12
 
 , 
2
x x
x x y x u
x x x u
 =
 = = − +
 = − − +



 
(b) The block diagram can then be constructed as shown in Figure PS4-5No16. 
 
 
 Figure PS4-5No16 
(c) Because not all paths are coupled, general case of Mason’s rule applies as follows: There are 
two forward paths with gains 1
1
2
F = and 2 2
1
4
F
s
−
= , and two loops with gains 2
1
2s
− and 3
2
s
− . 
 1 22 3
1 21 , , 1
2
D D D D
s s
= + + = = 
 
2 31 221 2
3
2 3
1
( ) 1 24
1 2( ) 2 2 41
2
i ii F D D FY s ss
U s D D s s
s s
=
−+ +
= = = + =
+ ++ +
∑ 
 
17. A system is described by its transfer function 
 
1
3
2
( )
( ) (2 2)
sY s
U s s s s
+
=
+ +
 
(a) Find the I/O equation. 
(b) Find the state-space model from the I/O equation. 
(c) Build a block diagram based on the information in (b). Find the transfer function directly 
from the block diagram and compare with the given transfer function. 
Solution 
(a) The input-output equation is derived as 1
32 2y y y u u+ + = +    . 
( )Y s
( )U s Σ
+
−
−
1
s
1
s
3x
1
2
1
s
2x 1x3x
2
1
4 Σ
+−
1
2
Direct link from input to output
 
 
138 
 
(b) With 1x y= , 2x y=  and 3x y=  , we find 
 
1 2
1 1
2 3 1 26 2
1
3 3 22
 
 , 
( 2 )
x x
x x y x x
x x x u
 =
 = = +
 = − − +



 
(c) The block diagram can then be constructed as shown in Figure PS4-5No17. There are two 
forward paths with gains 2
1
2s
 and 3
1
6s
, and two loops with gains 1
2s
− and 2
1
s
− . All paths are 
coupled and by the special case of Mason’s rule we have 
 
2 3
2
2
1 1
( ) 3 12 6
1 1( ) 3 (2 2)1
2
Y s ss s
U s s s s
s s
+ +
= =
+ ++ +
 
This agrees with the original transfer function. 
 
 
 Figure PS4-5No17 
 
18. Repeat Problem 17 for 
 
2 1
3
2
( )
( ) 2 2
sY s
U s s s
+
=
+ +
 
Solution 
(a) The input-output equation is derived as 1
32 2y y y u u+ + = +   . 
(b) With 1x y= and 2x y=  , we find (Section 4.4) 
 1 1 1
1 3 4 2
2
0 1 0
 , , , 
1 1
D
     = = = − − =     − −   
A B C 
The above can then be expressed as 
 1 2 1 1 1
1 21 3 4 2
2 1 22
 
 , 
x x
y x x u
x x x u
=
= − − + = − − +


 
(c) Because not all paths are coupled, the general case of Mason’s rule applies as follows: 
Three forward paths with gains 1
1
2
F = , 2 2
1
3
F
s
= − and 3
1
4
F
s
= − , and two loops with gains 1
2s
− 
and 2
1
s
− . 
( )Y s( )U s Σ
+
−
−
1
s
1
s Σ
+3x
1
2
1
s
2x 1x3x 1
6
+
1
2
 
 
139 
 
 1 2 32
1 11 , , 1
2
D D D D D
s s
= + + = = = 
 
3 2 11 22 31 32
2
2
1 1
() 1 43
1 1( ) 2 2 21
2
i ii F D sD F FY s ss
U s D D s s
s s
=
− − ++ +
= = = + =
+ ++ +
∑ 
This agrees with the original transfer function. 
 
 
 Figure PS4-5No18 
 
 
19. The block diagram representation of a system model is presented in Figure 4.33, where 1x 
and 2x denote the two state variables. 
(a) Derive the state-space form directly from the block diagram. 
(b) Find the transfer function directly from the block diagram. 
 
 
 Figure 4.33 Problem 19. 
Solution 
(a) The left and right summing junctions respectively yield 
 
1
2 2 13
1
1 1 22
( ) ( ) 3 ( ) 2 ( )
( ) ( ) ( ) 
sX s X s X s U s
sX s X s X s
 = − − +
 = − +
 
The output is clearly 1( ) ( )Y s X s= . Therefore, 
 
1
1 1 22
11
2 1 23
 
 , 
3 2
x x x
y x
x x x u
 = − + = = − − +


 
The state-space form is then obtained as 
 
u
y Du
= +
 = +
x Ax B
Cx

 
with 
( )Y s( )U s Σ
+
−
−
1
s
1
s Σ
−
1
2
2x 1x2x 1
3
−
1
4
1
2
+
Direct link from input to output
( )Y s( )U s
− −
2 1( )X s2 ( )X s+ Σ
1
s
1
s
+ Σ
1
3
3
1
2
−
 
 
140 
 
 [ ]
1
1 2
1
2 3
1 0
 , , , 1 0 , 0 , 
3 2
x
D u u
x
 −   
= = = = = =    − −      
x A B C 
(b) To find the transfer function, we must use the general case of Mason’s rule. The pertinent 
quantities are 
 
 1 12 2 2
Non-touchingSingle-loop gains two-loops
2 1 1 3 1 , 1 , 1
3 2 6
F D D
s ss s s
= = + + + + =

 
 
 The transfer function is then found as 
 
 
2
1 1
2
2 2
2 1( ) 12
1 1 3 1( ) 6 5 191
3 2 6
F DY s s
U s D s s
s s s s
⋅
= = =
+ ++ + + +
 
 
20. Repeat Problem 19 for the block diagram in Figure 4.34, where 1x , 2x and 3x denote the 
state variables. 
 
 
 Figure 4.34 Problem 20. 
Solution 
(a) The left summing junction yields 2 1
3 3 13 2( ) ( ) ( ) ( )sX s X s X s U s= − − + . The next two 
integrators give 2 3( ) ( )sX s X s= and 1 2( ) ( )sX s X s= . The right summing junction yields 
1
1 3 3( ) ( ) 2 ( ) ( )Y s X s X s U s= − + . Therefore, 
 
1 2
1
2 3 1 3 3
2 1
3 3 13 2
 
 , 2
x x
x x y x x u
x x x u
 =
 = = − +
 = − − +



 
 
The state-space form is then obtained as 
 
u
y Du
= +
 = +
x Ax B
Cx

 
with 
( )Y s( )U s
−
−
2
1( )X s3( )X s+ Σ
1
s
1
s + Σ
2
3
1
2
1
3
−
1
s
2 ( )X s
+
 
 
141 
 
 [ ]
1
1
2 3
1 2
3 2 3
0 1 0 0
 , 0 0 1 , 0 , 1 0 2 , , 
0 1
x
x D u u
x
    
    = = = = − = =     
     − −     
x A B C 
(b) To find the transfer function, we must use the general case of Mason’s rule. The pertinent 
quantities are 
 
 1 2 3 1 2 33 3
1 2 1 2 1 , , , 1 , , 1 , 1
3 3 2
F F F D D D D D
s ss s
= = − = = + + = = = 
The transfer function is then found as 
 
 
3 21 32 33
3 2
3
2 1
( ) 1 6 32 21
2 1( ) 3 3(6 4 3)1
3 2
D F FY s s ss s
U s D s s
s s
− ++ + − +
= = + =
+ ++ +
 
 
Problem Set 4.6 
In Problems 1 through 10, the mathematical model of a nonlinear system is provided. Using the 
procedure outlined in this section, find the operating point and derive the linearized model. 
1. 3 1
22 ( ) , ( ) unit-step , (0) , (0) 0x x x u t u t x x+ + = = = =   
Solution 
The 4-step procedure is followed as listed below: 
Step 1: The operating point satisfies 3 1x = so that 1x = . 
Step 2: The nonlinearity 3( )f x x= is linearized about 1x = as 
 
 3 2
1
( ) (1) 3 1 3
x
f x x f x x x
=
 = ≅ + ∆ = + ∆  
 
Step 3: Substitute 1x x= + ∆ and the linear approximation into the original equation: 
 
 (1 ) 2(1 ) [1 3 ] ( ) 2 3 0x x x u t x x x+ ∆ + + ∆ + + ∆ = ⇒ ∆ + ∆ + ∆ = 
  
 
Step 4: Adjust the initial conditions: 
 
 1
2(0) (0) 1 , (0) (0) 0x x x x∆ = − = − ∆ = =  
 
Therefore, the linearized model is obtained as 
 
 
1
22 3 0 , (0) , (0) 0x x x x x∆ + ∆ + ∆ = ∆ = − ∆ =   
 
 
2. 31 1
3 3 sin , (0) 1 , (0) 1x x t x x+ = + = =  
 
 
142 
 
Solution 
The 4-step procedure is followed as listed below: 
Step 1: The operating point satisfies 31 1
3 3x = so that 1x = . 
Step 2: The nonlinearity 3( )f x x= is linearized about 1x = as 
 
 3 2
1
( ) (1) 3 1 3
x
f x x f x x x
=
 = ≅ + ∆ = + ∆  
 
Step 3: Substitute 1x x= + ∆ and the linear approximation into the original equation: 
 
 1 1
3 3(1 ) [1 3 ] sin sinx x t x x t+ ∆ + + ∆ = + ⇒ ∆ + ∆ =
 
 
Step 4: Adjust the initial conditions: 
 
 (0) (0) 1 0 , (0) (0) 1x x x x∆ = − = ∆ = =  
 
Therefore, the linearized model is obtained as 
 
 sin , (0) 0 , (0) 1x x t x x∆ + ∆ = ∆ = ∆ =  
 
 
3. 1 1 1
2 3 23 cos , (0) , (0) 0x x x x t x x+ + = + = − =   
Solution 
We follow the 4-step procedure outlined in the section. First note that 
 
 
2 Differentiate
2
3 if 0 6 if 0
( ) 3 ( )
6 if 03 if 0
x x x x
f x x x f x
x xx x
 ≥ ≥ ′= = ⇒ = − leads to 2 1
33x = so that 1
3x = ± . But since 0x > , the only valid solution is 
1
3x = . Therefore, the operating point is 1
3x = . 
 
Step 2: Linearizing ( ) 3f x x x= about the operating point, we find 
 
 [ ] 1
3
1 1
3 3( ) 3 ( ) 6 2xf x x x f x x x
=
= ≅ + ⋅∆ = + ∆ 
 
Step 3: Insert 1
3x x= ∆ + and the linear approximation into the original equation: 
 
 
 
143 
 
 1 1 1 1
2 3 3 22 cos 2 cosx x x t x x x t ∆ + ∆ + + ∆ = + ⇒ ∆ + ∆ + ∆ =     
 
Step 4: Initial conditions are adjusted: 51
3 6(0) (0)x x∆ = − = − , (0) (0) 0x x∆ = =  . Therefore, the 
linearized model is 
 
 51
2 62 cos , (0) , (0) 0x x x t x x∆ + ∆ + ∆ = ∆ = − ∆ =   
 
 
4. 1 1
8 2sin( ) , (0) 0 , (0) 1x x x x t x x+ + = + = =   
Solution 
First note that 
 
 
 
3
2
3
2
 if 0 if 0 
( ) ( )
 if 0 if 0
x xx x x
f x x x f x
x xx x x
 ≥≥ ′= = ⇒ = 
− leads to 1
8x x = , thus 1
4x = . Thus, the operating point is 1
4x = . 
 
Step 2: Linearization of ( )f x x x= about 1
4x = yields 
 
 
1
4
3 31 1
8 2 8 4( )
x
f x x x x
=
 ≅ + ∆ = + ∆  
 
Step 3: Substitute 1
4x x= + ∆ and the linear approximation just obtained, into the original model: 
 
 3 31 1 1 1
8 4 8 2 4 2sin( ) sin( )x x x t x x x t ∆ + ∆ + + ∆ = + ⇒ ∆ + ∆ + ∆ =     
 
Step 4: Adjust the initial conditions: 1 1
4 4(0) (0)x x∆ = − = − , (0) (0) 1x x∆ = =  . Therefore, the 
linearized model is described as 
 
 3 1 1
4 2 4 sin( ) , (0) , (0) 1x x x t x x∆ + ∆ + ∆ = ∆ = − ∆ =   
 
5. /32 1
3 3
3 if 0
( ) 1 , (0) 1 , (0) , ( )
3 if 0
t x x
x x f x e x x f x
x x
−
 ≥+ + = + = = = 
− ≤
   
 
 
144 
 
Solution 
Step 1: The operating point satisfies ( ) 1f x = . 
 
Case (1): 0x leads to 3 1x = , thus 1
9x = . Thus, the operating point is 1
9x = . 
 
Step 2: Linearization of ( )f x about 1
9x = yields 
 
 ( )
1
9
91
9 2
3( ) 1
2 x
f x f x x
x =
≅ + ∆ = + ∆ 
 
Step 3: Substitute 1
9x x= + ∆ and the linear approximation just obtained, into the original model: 
 
 /3 /39 92 2
3 2 3 21 1 t tx x x e x x x e− −∆ + ∆ + + ∆ = + ⇒ ∆ + ∆ + ∆ =    
 
Step 4: Adjust the initial conditions: 81
9 9(0) (0)x x∆ = − = , 1
3(0) (0)x x∆ = =  . Therefore, the 
linearized model is described as 
 
 /39 82 1
3 2 9 3 , (0) , (0)tx x x e x x−∆ + ∆ + ∆ = ∆ = ∆ =   
 
 
6. 1
3 ( ) 3 sin , (0) 0 , (0) 1x x f x t x x+ + = + = = −   , 
/21
2
/21
2
(1 ) if 0
( )
(1 ) if 0
x
x
e x
f x
e x
−
−
 − ≥= 
− −1: The operating point x satisfies ( ) 3f x = . Due to the nature of ( )f x , we need to inspect 
two cases: 
 
Case (1): 0x > leads to /21
2 (1 ) 3xe−− = , which has no solution because /2 0xe− > . 
Case (2): 0x leads to 2
1 1 2 0x x+ + = , which has no real solution. 
Case (2): 0x > sys='Problem11'; 
>> load_system(sys); 
>> opspec = operspec(sys); 
 
% Specify the properties of the first state variable. 
>> opspec.States(1).SteadyState = 1; 
>> opspec.States(1).x = 0; % Initial value 
>> opspec.States(1).Min = 0; % Minimum value 
 
% Do the same for the second state variable 
>> opspec.States(2).SteadyState = 1; 
>> opspec.States(2).x = 1; 
>> opspec.States(2).Min= 0; 
 
% Find the operating point based on the specifications listed above. 
>> [op,opreport] = findop(sys,opspec); 
 
 Operating Point Search Report: 
--------------------------------- 
 
 Operating Report for the Model Problem11. 
 (Time-Varying Components Evaluated at time t=0) 
 
Operating point specifications were successfully met. 
States: 
---------- 
(1.) Problem11/Integrator 1 
 x: 0 dx: 0 (0) 
(2.) Problem11/Integrator 2 
 x: 1 dx: 0 (0) 
 
Inputs: None 
---------- 
 
 
 
150 
 
Outputs: 
---------- 
(1.) Problem11/Out1 = x1 
 y: 1 [-Inf Inf] 
 
 
% Get the linearization annotations 
>> IO=getlinio(sys); 
 
% Extract the linear state-space model 
>> LIN = linearize('Problem11',op,IO) 
 
 
a = 
 Integrator 1 Integrator 2 
 Integrator 1 -1 -2 
 Integrator 2 1 0 % Controller canonical form 
 
b = 
 Sum1 
 Integrator 1 1 
 Integrator 2 0 
 
c = 
 Integrator 1 Integrator 2 
 x1 0 1 
 
d = 
 Sum1 
 x1 0 
 
Continuous-time state-space model. 
 
Note that 
o the state matrix is in the controller canonical form, and 
o the input in the linear model is comprised of the time-varying portion of the input in the 
original nonlinear system. In the current case, the input for the nonlinear system is 
1 sin t+ . This implies that the input for the linear model is simply sin t . 
 
(b) The operating point is (1,0) and the linearized model is 
 2 sinx x x t∆ + ∆ + ∆ =  
To derive the controller canonical form, choose the state variables as 1x x∆ = ∆ and 2x x∆ = ∆ , 
and the state-variable equations for the linearized model are obtained as 
 1 1 2
2 1
2 sin
 
x x x t
x x
∆ = −∆ − ∆ +
∆ = ∆


 
In vector form, 
 
 
u
y Du
= +
 = +
x A x B
C x
∆ ∆
∆
 (b) 
where 
 
 
151 
 
 [ ]1
2
1 2 1
 , , , 0 1 , 0 , sin
1 0 0
x
D u t
x
∆ − −     
= = = = = =     ∆     
x A B C∆ 
This confirms the set of results realized in Simulink. 
 
 
12. 3 ( ) , ( ) unit-step , (0) 1 , (0) 1x x x u t u t x x+ + = = = =   
Solution 
(a) The Simulink model is built as shown in Figure PS4-6No12 and saved as 'Problem12'. 
Employing the procedure outlined in this section, we find the operating point and the state-space 
model as follows. 
 
>> sys='Problem12'; 
>> load_system(sys); 
>> opspec = operspec(sys); 
>> opspec.States(1).SteadyState = 1; 
>> opspec.States(1).x = 1; 
>> opspec.States(1).Min = 0; 
>> opspec.States(2).SteadyState = 1; 
>> opspec.States(2).x = 1; 
>> opspec.States(2).Min = 0; 
>> [op,opreport] = findop(sys,opspec); 
 
Operating Point Search Report: 
--------------------------------- 
 
 Operating Report for the Model Problem12. 
 (Time-Varying Components Evaluated at time t=0) 
 
Operating point specifications were successfully met. 
States: 
---------- 
(1.) Problem12/Integrator 1 
 x: 0 dx: 9.93e-09 (0) 
(2.) Problem12/Integrator 2 
 x: 1 dx: 0 (0) 
 
Inputs: None 
---------- 
 
Outputs: 
---------- 
(1.) Problem12/Out1 = x1 
 y: 1 [-Inf Inf] 
 
>> IO=getlinio(sys); 
>> LIN = linearize('Problem12',op,IO) 
 
LIN = 
LIN = 
 
 a = 
 Integrator 1 Integrator 2 
 Integrator 1 -1e-10 -1 
 
 
152 
 
 Integrator 2 1 0 
 
 b = 
 Empty matrix: 2-by-0 
 
 c = 
 Integrator 1 Integrator 2 
 x1 0 1 
 
 d = 
 Empty matrix: 1-by-0 
 
Continuous-time state-space model. 
 
 
 
 Figure PS4-6No12 
 
 
(b) The 4-step procedure is followed as listed below: 
Step 1: The operating point satisfies 1x = . 
Step 2: The nonlinearity 3( )f x x=  is linearized about 0x = as 
 
 2
0
( ) (0) 3 0
x
f x f x x
=
 ≅ + ∆ = 

  
 
 
Step 3: Substitute 1x x= + ∆ and the linear approximation into the original equation: 
 
 (1 ) 0 [1 ] ( ) 0x x u t x x+ ∆ + + + ∆ = ⇒ ∆ +∆ =
 
 
Letting 1x x= ∆ and 2x x= ∆ , the above reduces to 
 
 1 2
2 1 
x x
x x
= −
 =


 
 
The matrices in the state-space form are then identified as 
 
IC: x1(0) = 1IC: x2(0) = 1
1
Out1 = x1
Scope
1
s
Integrator 2
1
s
Integrator 1
u^3
Nonlinear element
Unit step
at t = 0 x2 x1
 
 
153 
 
 [ ]0 1 0
 , , 0 1 , 0
1 0 0
D
−   
= = = =   
   
A B C 
 
This agrees with (a). 
 
 
 
13. 1 1 1
2 2 2cos( ) , (0) , (0) 0x x x x t x x+ + = + = =    
Solution 
(a) The Simulink model is built as shown in Figure PS4-6No13 and saved as 'Problem13'. 
Noting that ( )1
2cos 2 sin 2t t π= + , we generate cos 2t using the Sine Wave block from the 
Sources library by setting the amplitude to 1, frequency to 2 rad/sec , and phase to 1
2π . 
Employing the procedure outlined in this section, we find the operating point and the state-space 
model as follows. 
 
>> sys='Problem13'; 
>> load_system(sys); 
>> opspec = operspec(sys); 
>> opspec.States(1).SteadyState = 1; 
>> opspec.States(1).x = 1/2; 
>> opspec.States(1).Min = 0; 
>> opspec.States(2).SteadyState = 1; 
>> opspec.States(2).x = 0; 
>> opspec.States(2).Min = 0; 
>> [op,opreport] = findop(sys,opspec); 
 
 Operating Point Search Report: 
--------------------------------- 
 
 Operating Report for the Model Problem13. 
 (Time-Varying Components Evaluated at time t=0) 
 
Operating point specifications were successfully met. 
States: 
---------- 
(1.) Problem13/Integrator 1 
 x: 0 dx: 1.06e-09 (0) 
(2.) Problem13/Integrator 2 
 x: 1.5 dx: 0 (0) 
 
Inputs: None 
---------- 
 
Outputs: 
---------- 
(1.) Problem13/Out1 = x1 
 y: 1.5 [-Inf Inf] 
 
>> IO=getlinio(sys); 
>> LIN = linearize('Problem13',op,IO) 
 
 
 
154 
 
LIN = 
 
 a = 
 Integrator 1 Integrator 2 
 Integrator 1 -1e-05 -1 
 Integrator 2 1 0 
 
 b = 
 Sum1 
 Integrator 1 1 
 Integrator 2 0 
 
 c = 
 Integrator 1 Integrator 2 
 x1 0 1 
 
 d = 
 Sum1 
 x1 0 
 
Continuous-time state-space model. 
 
 
 Figure PS4-6No13 
 
 
(b) The 4-step procedure is followed as listed below: 
Step 1: The operating point satisfies 1
2x = . 
Step 2: The nonlinearity ( )f x x x=   is linearized about 0x = as 
 
 [ ] 0( ) (0) 2 0xf x f x x
=
≅ + ∆ =

  
 
 
Step 3: Substitute 1
2x x= + ∆ and the linear approximation into the original equation: 
 
 1 1 1 1 1
2 2 2 2 2( ) 0 [ ] cos( ) cos( )x x t x x t+ ∆ + + + ∆ = + ⇒ ∆ + ∆ =
 
 
Letting 1x x= ∆ and 2x x= ∆ , the above reduces to 
 
IC: x1(0) = 1/2IC: x2(0) = 0
1
Out1 = x1
cos(t/2)
Scope
1
s
Integrator 2
1
s
Integrator 1
1/2
Constant
u*abs(u)
Nonlinear element
x2 x1
 
 
155 
 
 
1
1 2 2
2 1
cos( )
 
x x t
x x
 = − +

=


 
 
The matrices in the state-space form are then identified as 
 
 [ ] 1
2
0 1 1
 , , 0 1 , 0 , cos( )
1 0 0
D u t
−   
= = = = =   
   
A B C 
 
This agrees with (a). 
 
 
14. 3 1
82 sin , (0) 1 , (0) 1x x t x x+ = + = =  
(a) The Simulink model is built as shown in Figure PS4-6No14 and saved as 'Problem14'. 
Employing the procedure outlined in this section, we find the operating point and the state-space 
model as follows. 
>> sys='Problem14'; 
>> load_system(sys); 
>> opspec = operspec(sys); 
>> opspec.States(1).SteadyState = 1; 
>> opspec.States(1).x = 1; 
>> opspec.States(1).Min = 0; 
>> opspec.States(2).SteadyState = 1; 
>> opspec.States(2).x = 1; 
>> opspec.States(2).Min = 0; 
>> [op,opreport] = findop(sys,opspec); 
 
Operating Point Search Report: 
--------------------------------- 
 
 Operating Reportfor the Model Problem14. 
 (Time-Varying Components Evaluated at time t=0) 
 
Operating point specifications were successfully met. 
States: 
---------- 
(1.) Problem14/Integrator 1 
 x: 0 dx: -2.33e-09 (0) 
(2.) Problem14/Integrator 2 
 x: 0.5 dx: 0 (0) 
 
Inputs: None 
---------- 
 
Outputs: 
---------- 
(1.) Problem14/Out1 = x1 
 y: 0.5 [-Inf Inf] 
 
>> IO=getlinio(sys); 
>> LIN = linearize('Problem14',op,IO) 
 
 
 
156 
 
LIN = 
 
 a = 
 Integrator 1 Integrator 2 
 Integrator 1 0 -0.375 
 Integrator 2 1 0 
 
 b = 
 Sum1 
 Integrator 1 0.5 
 Integrator 2 0 
 
 c = 
 Integrator 1 Integrator 2 
 x1 0 1 
 
 d = 
 Sum1 
 x1 0 
 
Continuous-time state-space model. 
 
 
 Figure PS4-6No14 
 
(b) The 4-step procedure is followed as listed below: 
Step 1: The operating point satisfies 1
2x = . 
Step 2: The nonlinearity 3( )f x x= is linearized about 1
2x = as 
 
 
1
2
2 31 1
2 8 4( ) ( ) 3
x
f x f x x x
=
 ≅ + ∆ = + ∆  
 
Step 3: Substitute 1
2x x= + ∆ and the linear approximation into the original equation: 
 
 3 31 1 1
2 8 4 8 42( ) [ ] sin 2 sinx x t x x t+ ∆ + + ∆ = + ⇒ ∆ + ∆ =
 
 
Letting 1x x= ∆ and 2x x= ∆ , the above reduces to 
 
IC: x1(0) = 1IC: x2(0) = 1
1
Out1 = x1
sin(t)
Scope
1
s
Integrator 2
1
s
Integrator 1
1/8
Constant
u^3
Nonlinear element
1/2
Gain
x2 x1
 
 
157 
 
 
31
1 22 4
2 1
( sin )
 
x x t
x x
 = − +

=


 
 
The matrices in the state-space form are then identified as 
 
 [ ]
3 1
280
 , , 0 1 , 0 , sin
01 0
D u t
   −
= = = = =   
  
A B C 
 
This agrees with (a). 
 
Review Problems 
1. The mathematical model of a dynamic system is derived as 
 1 1 1 2
1
2 1 24
3 0 
2 ( )
x x x x
x x x f t
+ + − =
 − − =
 

 
(a) If ( )f t is the input, and 1x and 1x are the outputs, obtain the state-space form. 
(b) Determine if the system is stable (Problem Set 4.2) by examining the state-space form. 
(c) Find the transfer matrix directly from the given model. 
(d) Confirm the result of (c) using ss2tf. 
Solution 
(a) There are three state variables: 1 1x x= , 2 2x x= , 3 1x x=  . The state-variable equations are 
 
 
1 3
1 1
2 1 22 4
3 3 1 2
 
3 
x x
x x x f
x x x x
=

 = + +  
 = − − +



 
 
The state equation is then formed as 
 
 
1
1 1 1
2 2 8 2
3
0 0 1 0
 , , 0 , , ( )
1 3 1 0
x
u x u f t
x
     
     = + = = = =     
     − −    
x Ax B x A B 
 
The output equation is 
 
 
1 0 0 0
 , , 
0 0 1 0
u    
= + = =   
   
y Cx D C D 
(b) 
>> A = [0 0 1;1/2 1/8 0;-1 3 -1]; eig(A) 
 
ans = 
 
 
 
158 
 
 0.7640 + 0.0000i 
 -0.8195 + 1.2065i 
 -0.8195 - 1.2065i 
 
System is unstable because one of the eigenvalues of the state matrix lies in the right half-plane. 
 
(c) Laplace transformation yields 
 
 
2
1 2
1
1 24
( 1) 3 0 
(2 ) ( )
s s X X
X s X F s
 + + − =

− + − =
 
Solve for 1X : 
 
 
1
4 1
1 3 2 3 22 7 7 13
4 4 4
1
4
0 3
2 3 12 
2 8 7 7 131 3
1 2
F s XFX
Fs s s s s ss s
s
−
−
= = ⇒ =
+ + − + + −+ + −
− −
 
 
The transfer function corresponding to x is 
 
 1
3 2
12
8 7 7 13
sX s
F s s s
=
+ + −
 
 
Therefore, the transfer matrix is 
 
 
3 2
3 2
12
8 7 7 13( )
12
8 7 7 13
s s ss
s
s s s
 
  + + −=  
 
 + + − 
G 
 
(d) 
 
>> A = [0 0 1;1/2 1/8 0;-1 3 -1]; B = [0;1/2;0]; 
>> C = [1 0 0;0 0 1]; D = [0;0]; 
>> [NUM,DEN] = ss2tf(A,B,C,D,1) % option '1' may be omitted 
 
NUM = 
 
 0 0 0 1.5000 
 0 0 1.5000 0 
 
 
DEN = 
 
 1.0000 0.8750 0.8750 -1.6250 
 
 
 
159 
 
2. A system is described by its governing equations 
 
1
1 1 2 12
2 2 1
2( ) ( )
2( ) 0 
x x x x f t
x x x
 + − − =

+ − =


 
 where f is the input, while 2x and 2x are the outputs. 
(a) Obtain the state-space form. 
(b) Find the transfer matrix directly from the state-space form. 
(c) Find the transfer matrix using the governing equations, and compare with (b). 
Solution 
(a) There are four state variables: 1 1x x= , 2 2x x= , 3 1x x=  , 4 2x x=  . The state-variable 
equations are 
 
 
1 3
2 4
3 1 2
4 1 2
 
 
6 4 2
2 2 
x x
x x
x x x f
x x x
=
 =
 = − + +
 = −




 
 
The state equation is then formed as 
 
 
1
2
3
4
0 0 1 0 0
0 0 0 1 0
 , , , , ( )
6 4 0 0 2
2 2 0 0 0
x
x
u u f t
x
x
     
     
     = + = = = =     −       −    
x Ax B x A B 
 
The output equation is 
 
 2
2 4 4 1
4
0 1 0 0 0
 , , 
0 0 0 1 0
x
u
x × ×
     
= = + = =     
    
y C x D C D 
(b) 
>> A = [0 0 1 0;0 0 0 1;-6 4 0 0;2 -2 0 0]; B = [0;0;2;0]; 
>> C = [0 1 0 0;0 0 0 1]; D = [0;0]; 
>> [nG, dG] = ss2tf(A,B,C,D) 
 
nG = 
 
 0 0 0 0 4 
 0 0 0 4 0 
 
 
dG = 
 
 1.0000 -0.0000 8.0000 -0.0000 4.0000 
 
 
 
 
160 
 
The transfer matrix is 2 1× because there are two outputs and one input: 
 
 
4 2
4 2
4
8 4( )
4
8 4
s ss
s
s s
 
  + +=  
 
 + + 
G 
 
(c) Laplace transformation yields 
 
 
21
1 22
2
1 2
( 3) 2
2 ( 2) 0 
s X X F
X s X
 + − =

− + + =
 
 
Solve for 2X : 
 
 
21
2
2
2 4 2 4 22 1 11
2 22
2
3
2 0 2 2 
4 2 4 23 2
2 2
s F
XFX
Fs s s ss
s
+
−
= = ⇒ =
+ + + ++ −
− +
 
 
Since the other output is 2x , the second transfer function is simply 
 
 2
4 21
2
2
4 2
sX s
F s s
=
+ +
 
 
Results clearly agree with (b). 
 
 
 
 
3. A dynamic system with input f and output x is modeled as 
 3
4 ( ) , 0x x kx f t k const+ + = = >  
(a) Find the state-space form, and determine the value(s) of k for which the system is stable. 
(b) Find the transfer function directly from the state-space form. 
(c) Find the transfer function directly from the given model, and compare with (b). 
Solution 
(a) There are two state variables: 1x x= , 2x x=  . The state-variable equations are 
 
 1 2
4
2 2 13
 
( )
x x
x x kx f
=
 = − − +


 
 
 
 
161 
 
The state equation is then formed as 
 
 1
4 4 4
2 3 3 3
0 1 0
 , , , , ( )
x
u u f t
kx
    
= + = = = =     − −     
x Ax B x A B 
 
The output is 2y x x= = , thus 
 
 [ ] , 0 1 , 0y Du D= + = =Cx C 
 
The eigenvalues of the state matrix are 2 4 4
3 9 3 kl = − ± − . Since 0k > , 4 4 2
9 3 3k− . 
 
(b) Since the system is SISO, there is one transfer function: 
 
 [ ]
1
1
4 4 4 23
3 3 3 4
1 0
( ) ( ) 0 1
s sG s s
k s s s k
−
− −   
= − = =   + + +   
C I A B 
 
(c) Laplace transformation of the model yields 
 
 23
4 23
4
( ) 1( ) ( ) ( ) 
( )
X ss s k X s F s
F s s s k
+ + = ⇒ =
+ +
 
 
Since x is the output, the transfer function is 
 
 23
4
( )
( )
sX s s
F s s s k
=
+ +
 
4. The I/O equation of a system is derived as 
 1 1
2 32y y y u u u+ + = + +    
(a) Find the state-space representation. 
(b) Find the transfer function from the state-space form. 
Solution 
(a) State-space form is found as 
 
 1
2 1 1
2 4
0 1 0 0
 , , 0 0 1 , 0 , 
0 1
x
u u u
x
   
     = + = = = =          − −   
x Ax B x A B 
 
 1 1 1 1
4 6 8 2 , , Du D = + = − = y Cx C 
 
 
 
162 
 
(b) Using the state-space form, 
 
 
1
3
1 1 1 1
4 6 8 3 2
1 1
2 4
1 0 0
1 6 2 6( ) ( ) 0 1 0
2 12 3 60 1
s
s sG s s D s
s ss
−
−
 −  
+ +    = − + = − − + =     + +   +   
C I A B 
 
 
5. A system’s transferfunction is given as 
 
2
3 21
3
( ) (2 1)
( ) 2
Y s s s
U s s s s
+
=
+ + +
 
(a) Find the state-space form. 
(b) Find the transfer function from (a). 
Solution 
(a) The I/O equation is 1
3 2 2y y y y u u+ + + = +    .The state-space form is obtained as 
 
 
1
2
3
0 1 0 0
 , , 0 0 1 , 0 , 
6 3 3 1
x
u x u u
x
     
     = + = = = =     
     − − −    
x Ax B x A B 
 
 
 [ ] , 36 18 15 , 6Du D= + = − − − =y Cx C 
 
 
(b) Using the state-space form, 
 
 
[ ]
1
1
2
3 2
( ) ( )
1 0 0
3 (2 1) 36 18 15 0 1 0 6
3 3 66 3 3 1
G s s D
s
s ss
s s ss
−
−
= − +
−   
+   = − − − − + =    + + +   +   
C I A B
 
 
6. Find the input-output equation of a SISO system whose state-space form is given as 
 
u
y Du
= +
 = +
x Ax B
Cx

 
 where 
 2 1
31 1 3 3
4 2 2
0 1 0
 , , , , 0u u D
   
 = = = = =     − −   
A B C 
Solution 
We will first find the transfer function, as 
 
 
 
163 
 
 
1
1 2 1
31 13 3 2
4 2 2
1 0 2( 2)( ) ( )
4 2 1
s sG s s D
s s s
−
− −    + = − + = =     + + +   
C I A B 
 
As a result, the I/O equation is 4 2 2 4y y y u u+ + = +   . 
 
 
7. Repeat Problem 6 for 
 [ ]0 1 0
 , , , 2 1 , 2
1 4 3
u u D   
= = = = − =   − −   
A B C 
Solution 
We will first find the transfer function, as 
 
 [ ]
1 2
1
2
1 0 2 5 8( ) ( ) 2 1 2
1 4 3 4 1
s s sG s s D
s s s
−
− −    + +
= − + = − + =   + + +   
C I A B 
 
As a result, the I/O equation is 4 2 5 8y y y u u u+ + = + +    . 
 
 
8. Find all possible input-output equations for a system with state-space form 
 
u
u
= +
 = +
x Ax B
y Cx D

 
 where 
 11
32
00 1 1 0 0
 , , , , 
1 0 2 0
u u
      
= = = = =      − −       
A B C D 
Solution 
We will first find the transfer matrix, as 
 
 
2
3
2
1
11 4
32 3
2
011 0 0 2 2 1( ) ( )
10 2 0
2 2 1
s s ss s
s s
s s
−
 
 −       + + = − + = + =     +          
+ + 
G C I A B D 
 
The system has one input u and two outputs 1y and 2y . As a result, the two I/O equations are 
obtained as 
 
 2 4
1 1 1 2 2 23 32 2 , 2 2y y y u y y y u+ + = + + =     
 
 
 
 
 
 
164 
 
In Problems 9 through 12, given the block diagram representation of the system, find the overall 
transfer function. 
 
9. Figure 4.38 
 
 Figure 4.38 Problem 9. 
 
Solution 
Based on Figure Reviwe4No9, we write 
 
 2
1 3 4( ) ( ) [ ( ) ( )]s X s k U s k sX s k X s= − + 
 
which yields 
 
 1
2
3 4
( )
( )
kX s
U s s k s k
=
+ +
 
 
Since 2( ) ( )Y s k X s= , the transfer function is found as 
 
 1 2
2
3 4
( )
( )
k kY s
U s s k s k
=
+ +
 
 
 
 
 
 
 Figure Review4No9 
 
 
 
10. Figure 4.39 
( )Y s
+( )U s
−
Σ
1
s1k
+
Σ
+
1
s 2k
3k
4k
( )Y s
+( )U s
−
Σ
1
s1k
+
Σ
+
1
s 2k
3k
4k
2 ( )s X s ( )sX s ( )X s
 
 
165 
 
 
 
 Figure 4.39 Problem 10. 
 
Solution 
There are three forward paths and two loops: 
 
 Forward path gain Loop gain 
 1 4F G= 1 1G H− 
 2 2 3F H G= 3 3G H− 
 3 1 2 3F G G G= 
 
In the general case of Mason’s rule, we have 
 
 [ ]1 1 3 3 1 3 1 31D G H G H G G H H= − − − + 
 
 1 Same as when restricted to portion not touching 1st pathD D D= = 
 2 1 1Same as when restricted to portion not touching 2nd path 1D D G H= = + 
 3 Same as when restricted to portion not touching 3rd path 1D D= = 
 
Then, Mason’s rule yields 
 
 
[ ]4 1 1 3 3 1 3 1 3 1 1 2 3 1 2 3
1 1 3 3 1 3 1 3
1 (1 )( )
( ) 1
G G H G H G G H H G H H G G G GY s
U s G H G H G G H H
+ + + + + +
=
+ + +
 
 
 
 
 
 
 
11. Figure 4.40 
 
 
( )Y s+( )U s
++
−
+ +
−
1 ( )G s 2 ( )G s
1( )H s
2 ( )H s
Σ Σ
3( )H s
3 ( )G s
4 ( )G s
Σ
 
 
166 
 
 
 Figure 4.40 Problem 11. 
 
Solution 
There are two forward paths and four loops, as shown in Figure Review4No11: 
 
 Forward path Gain Loop Gain 
 12456 1 1 2 3F G G G= 2482 1 1G H− 
 1236 2 4F G= 5675 3 3G H 
 4594 2 2G H 
 236759482 4 3 2 1G H H H− 
 
 
 
 
 
 Figure Review4No11 
 
 
In the general case of Mason’s rule, we have 
 
 [ ]1 1 2 2 3 3 4 3 2 1 1 1 3 31D G H G H G H G H H H G H G H= − − + + − − 
 
 1 Same as when restricted to portion not touching 1st path 1D D= = 
 2 2 2Same as when restricted to portion not touching 2nd path 1D D G H= = − 
 
Then, Mason’s rule yields 
 
( )Y s+( )U s +
−
+ +
1 ( )G s 2 ( )G s
1( )H s
2 ( )H s
Σ Σ
3 ( )H s
3 ( )G s
4 ( )G s
Σ
+
+ +
Σ
( )Y s+( )U s +
−
+ +
1 ( )G s 2 ( )G s
1( )H s
2 ( )H s
Σ Σ
3 ( )H s
3 ( )G s
4 ( )G s
Σ
+
+ +
Σ
1
2
3
4
5
6
78
9
 
 
167 
 
 
[ ]
1 2 3 2 2 4
1 1 2 2 3 3 4 3 2 1 1 1 3 3
(1 )( )
( ) 1
G G G G H GY s
U s G H G H G H G H H H G H G H
+ −
=
− − + + − −
 
 
 
 
12. Figure 4.41 
 
 
 Figure 4.41 Problem 12. 
 
Solution 
There are two forward paths and two loops: 
 
 Forward path gain Loop gain 
 1 4F G= 1 1G H− 
 2 1 2 3F kG G G= 2 2G H− 
 
 
In the general case of Mason’s rule, we have 
 
 [ ]1 1 2 2 1 1 2 21D G H G H G H G H= − − − + 
 
 1 Same as when restricted to portion not touching 1st pathD D D= = 
 2 Same as when restricted to portion not touching 2nd path 1D D= = 
 
Then, Mason’s rule yields 
 
 
[ ]4 1 1 2 2 1 1 2 2 1 2 3
1 1 2 2 1 1 2 2
1( )
( ) 1
G G H G H G H G H kG G GY s
U s G H G H G H G H
+ + + +
=
+ + +
 
 
 
 
In Problems 13 and 14, given the block diagram, find the system’s I/O equation. 
 
13. Figure 4.42 
 
( )Y s+( )U s +
−
+ +
−
1 ( )G s k
1( )H s 2 ( )H s
Σ Σ 3 ( )G s2 ( )G s
4 ( )G s
Σ
 
 
168 
 
 
 
 Figure 4.42 Problem 13. 
 
 
Solution 
Because not all paths are coupled, the general case of Mason’s rule applies as follows: There are 
three forward paths with gains 1
1
3
F = , 2 2
4
9
F
s
= and 3 3
1
3
F
s
−
= , and two loops with gains 2
2
3s
− 
and 3
1
s
− . 
 1 2 32 3
2 11 , , 1, 1
3
D D D D D
s s
= + + = = = 
 
3 31 2 32 31 3
3
2 3
4 1
( ) 1 29 3
2 1( ) 3 3 2 31
3
i ii F D D F FY s s ss s
U s D D s s
s s
=
−+ + +
= = = + =
+ ++ +
∑ 
Subsequently, the I/O equation is formed as 3 2 3 2y y y u u+ + = +    . 
 
 
 
14. Figure 4.43 
 
 
 Figure 4.43 Problem 14. 
( )Y s
+( )U s
−
+ +
1
s
2
3
Σ
Σ
−
−
1
s
1
s
4
9
1
3
1
3
( )Y s+( )U s
−
+ +1
s
1
2
Σ Σ
−
−
1
s
1
s
1
4
1
2
1
4 −
1
2
 
 
169 
 
 
 
 
Solution 
Because not all paths are coupled, the general case of Mason’s rule applies as follows: Four 
forward paths with gains 1
1
2
F = , 2
1
4
F
s
−
= , 3 3
1
4
F
s
−
= and 4 2
1F
s
= , and two loops with gains 
1
2s
− and 3
1
2s
− . 
 1 2 3 43
1 11 , , 1
2 2
D D D D D D
s s
= + + = = = = 
 
4 2 31
2 3 41 2
3 2 3 2
( ) 1 4 1 2
( ) 2 2(2 1) 2 1
i ii F D D F F FY s s s s s
U s D D s s s s
= + + + − + − +
= = = + =
+ + + +
∑ 
Therefore, the I/O equation is formed as 2 2y y y u u+ + = +    . 
 
 
In Problems 15 and 16, given the state-space representation of the system model, construct the 
appropriate block diagram, and directly use it to find the transfer matrix. 
 
15. 
u
y Du
= +
 = +
x Ax B
Cx

 
where 
 [ ]
1
2
1 1
3 2 2
0 1 0 0
 , 0 0 1 , 0 , 1 2 0 , 0 , 
1 1
x
x D u u
x
    
    = = = = = =     
     − − −     
x A B C 
Solution 
The block diagram is shown in Figure Review4No15. 
 
 
 
 Figure Review4No15 
 
 
 
 
170 
 
There are two forward paths with gains 3
1
s
 and 2
2
s
, and three loops with gains 1
2s
− , 2
1
s
− , and 
3
1
2s
− . All paths are coupled. By Mason’s rule, we have 
 
 
3 2
3 2
2 3
1 2
( ) 2(2 1)
1 1 1( ) 2 2 11
2 2
Y s ss s
U s s s s
s s s
+ +
= =
+ + ++ + +
 
 
Note that in the above transfer function, the numerator and denominator have 2 1s + in common 
which can algebraically be cancelled to yield 2
2
1s +
. However, this– known as pole-zero 
cancellation – must be avoided because important information about the system may be lost. 
 
 
 
 
16. 
u
u
= +
 = +
x Ax B
y Cx D

 
where 
 1
2
0 1 0 1 0 0
 , , , , , 
1 3 1 0 2 0
x
u u
x
         
= = = = = =         − −        
x A B C D 
Solution 
The block diagram is shown in Figure Review4No16. 
 
 
 
 
 Figure Review4No16 
 
 
There are two outputs and one input, hence the transfer matrix ( )sG is 2 1× . To find 1( )
( )
Y s
U s
, we 
note that there is one forward path with gain 2
1
s
 and two feedback loops with gains 3
s
− , 2
1
s
− . 
 
 
171 
 
By Mason’s rule, 
 
2
1
11 2
2
1
( ) 1( ) 3 1( ) 3 11
Y s sG s
U s s s
s s
= = =
+ ++ +
 
 
To find 2 ( )
( )
Y s
U s
, we note that there is one forward path with gain 2
s
 and two feedback loops with 
gains 3
s
− , 2
1
s
− . By Mason’s rule, 
 2
21 2
2
2
( ) 2( ) 3 1( ) 3 11
Y s ssG s
U s s s
s s
= = =
+ ++ +
 
 
Therefore, the transfer matrix is 11
21
( )
( )
( )
G s
s
G s
 
=  
 
G . 
 
 
 
 
17. A system is atable if the poles of the overall transfer function lie in the left half-plane; these 
poles are the same as the eigenvalues of the state matrix. Consider the block diagram 
representation of a system shown in Figure 4.44, where 0k > is a parameter. Determine the 
range of values of k for which the system is stable. 
 
 
 Figure 4.44 Problem 17. 
Solution 
The overall transfer function is formed by Mason’s rule, as 
 
 
1
6
2
1
( ) 6 (2 1)( ) 1 3( ) ( 2) 2 31
2 1 (2 1)
Y s s sG s ksU s k s s
s s s
+= = =
+ + + ++ +
+ +
 
 
The poles of ( )G s are 1,2 1 1 3( 2)s k= − ± − + . Since 0k > , both poles are complex with 
negative real parts. This implies that the system is stable for all 0k > . 
( )Y s+( )U s
−
1
2 1s +
1ks +
Σ
−
1
3
3
1
2
1
s
 
 
172 
 
 
 
 
18. Repeat Problem 17 for the block diagram in Figure 4.45, where 0k > is a parameter. 
 
 
 Figure 4.45 Problem 18. 
Solution 
The overall transfer function is formed by Mason’s rule, as 
 
 2
2 4
( ) 4 2( 1) 1( ) 2( ) 2( ) 3 (1 2 ) 21
1 ( 1)
Y s ss s sG s s k kU s s k s k
s s s
+
++ += = =
+ + + ++ +
+ +
 
 
The poles of ( )G s are 
2
1,2
(1 2 ) (1 2 ) 24
6
k k k
s
− + ± + −
= . But 2(1 2 ) 24 1 2k k k+ − . Therefore, both poles lie in the left half-plane for all values of 0k > . This implies that 
the system is stable for all 0k > . 
 
 
 
19. Derive the linearized model for the nonlinear system described by 
 11 1 1 2
22 1 2
(0) 1 1 sin
 , 
(0) 0 1 
xx x x x t
xx x x
= − = + − +
 == − − −


 
Solution 
Step 1: The operating point ( )1 2,x x satisfies 
 
Add
1 1 2
1 1 1
1 2
0 1
 2 0
0 1 
x x x
x x x
x x
 = + −
⇒ − − =
= − − −
 
Case (1): 1 0x > leads to 2
1 1 1 12 ( 2)( 1) 0x x x x− − = − + = , hence 1 2, 1x = − . Since 1 0x > , then 
we have 1 2x = so that 2 3x = − . Therefore, the operating point is (2, 3)− . 
Case (2): 1 0xI
= = = 
 
Figure PS5-1 No4 
 
5. Determine the elastic potential energy of the system shown in Figure 5.15 if the 10-kg block moves downward a 
distance of 0.05 m. Assume that the block is originally in static equilibrium. 
 
Figure 5.15 Problem 5. 
Solution 
Since the block is originally in static equilibrium, the spring is compressed and the deformation is 
10(9.81) 0.20 m
500
mgx
k
∆ = = = 
If the block moves a distance of 0.05 m, the total deformation of the spring is 0.25 m. Thus, the elastic potential energy 
is 
2 21 1
2 2 (500)(0.25 ) 15.63 JeV kx= = = 
 
6. If the disk in Figure 5.16 rotates in the clockwise direction by 10°, determine the elastic potential energy of the 
system. Assume that the springs are originally undeformed. 
 
Figure 5.16 Problem 6. 
 
178 
 
 
Solution 
If the disk rotates in the clockwise direction by 10°, the two torsional springs will be twisted in the opposite directions. 
The angular deformation of each spring is 10°. The total elastic potential energy is 
( ) ( )
2
21 1
2 2
102 θ 2 (400) 12.18 J
180eV K π = = = 
 
 
 
7. Determine the equivalent spring constant for the system shown in Figure 5.17. 
 
Figure 5.17 Problem 7. 
Solution 
For the three springs in parallel, we have eq1 2 3 4k k k k= + + . They are connected with another two springs in series. 
Thus, the equivalent spring constant for the system is 
eq 1 eq1 5 1 2 3 4 5
1 1 1 1 1 1 1
k k k k k k k k k
= + + = + +
+ +
 
1 2 3 4 5
eq
1 5 2 3 4 1 5
( )
( )( )
k k k k k
k
k k k k k k k
+ +
=
+ + + +
 
 
8. Determine the equivalent spring constant for the system shown in Figure 5.18. 
 
Figure 5.18 Problem 8. 
Solution 
For the two springs in series, 
eq1 1 2
1 1 1
k k k
= + or 1 2
eq1
1 2
k kk
k k
=
+
 
For the two springs in parallel, we have eq2 3 4k k k= + . eq1k and eq2k are connected in parallel. Thus, the equivalent 
spring constant for the system is 
1 2
eq eq1 eq2 3 4
1 2
k kk k k k k
k k
= + = + +
+
 
9. Derive the spring constant expression for the axially loaded bar shown in Figure 5.19. Assume that the 
cross-sectional area is A and the modulus of elasticity of the material is E. 
 
Figure 5.19 Problem 9. 
Solution 
The force-deflection relation of an axially loaded bar is Lx f
EA
= . Therefore, eq
f EAk
x L
= = . 
179 
 
 
10. The uniform circular shaft in Figure 5.20 acts as a torsional spring. Assume that the elastic shear modulus of the 
material is G. Derive the equivalent spring constant corresponding to a pair of torques applied at two free ends. 
 
Figure 5.20 Problem 10. 
Solution 
For a shaft, we have 
θ τL
GJ
= 
where the polar moment of inertia for a cylinder is 41
32 πJ d= . Thus, 
4
32θ τ
π
L
G d
= 
The equivalent spring stiffness is 
4
eq
τ π=
θ 32
G dK
L
= 
 
11. A rod is made of two uniform sections, as shown in Figure 5.21. The two sections are made of the same material, 
and the modulus of elasticity of the rod material is E. The areas for the two sections are A1 and A2, respectively. 
Derive the equivalent spring constant corresponding to a tensile force applied at the free end. 
 
Figure 5.21 Problem 11. 
Solution 
The rod with two uniform sections can be treated as two axial springs in series. Using the result obtained in Problem 9, 
the equivalent spring stiffness for each section is 
i
i
i
EAk
L
= 
where 1, 2i = . Thus, the equivalent spring stiffness for the rod is 
1 2
eq 1 2 1 2
1 1 1 L L
k k k EA EA
= + = + or 1 2
eq
1 2 2 1
EA Ak
L A L A
=
+
 
 
12. Derive the spring constant expression of the fixed-fixed beam in Figure 5.22. The Young’s modulus of the 
material is E and the moment of inertia of cross-sectional area is I. Assume that the force f and the deflection x are 
at the center of the beam. 
 
Figure 5.22 Problem 12. 
Solution 
For a fixed-fixed beam with a load at midspan, the force-deflection relation is 
3
192
Lx f
EI
= . Therefore, the equivalent 
spring stiffness is eq 3
192f EIk
x L
= = . 
 
180 
 
 
13. Derive the spring constant expression of the simply supported beam in Figure 5.23. The Young’s modulus of the 
material is E and the moment of inertia of cross-sectional area is I. Assume that the force f and the deflection x are 
at the center of the beam. 
 
Figure 5.23 Problem 13. 
Solution 
For a simply supported beam with a load at midspan, the force-deflection relation is 
3
48
Lx f
EI
= . Therefore, the 
equivalent spring stiffness is eq 3
48f EIk
x L
= = . 
 
14. Derive the spring constant expression of the simply supported beam in Figure 5.24. The Young’s modulus of the 
material is E and the moment of inertia of cross-sectional area is I. Assume that the applied load f is anywhere 
between the supports. 
 
Figure 5.24 Problem 14. 
Solution 
For a simply supported beam with a load applied anywhere between supports, the force-deflection relation is 
2 2
3
a bx f
EIL
= 
where L is the length of the beam and L = a+b. Therefore, the equivalent spring stiffness is 
eq 2 2
3 ( )f EI a bk
x a b
+
= = 
 
15. Determine the equivalent damping coefficient for the system shown in Figure 5.25. 
 
Figure 5.25 Problem 15. 
Solution 
For the two dampers in series, 
eq1 1 2
1 1 1
b b b
= + or 1 2
eq1
1 2
b bb
b b
=
+
 
eq1b and 3b are connected in parallel. Thus, the equivalent damping coefficient for the system is 
1 2
eq eq1 3 3
1 2
b bb b b b
b b
= + = +
+
 
 
16. Determine the equivalent damping coefficient for the system shown in Figure 5.26. 
 
Figure 5.26 Problem 16. 
181 
 
 
Solution 
For the two dampers in parallel, 
eq1 1 2b b b= + 
eq1b and 3b are connected in series. Thus, the equivalent damping coefficient for the system is 
eq eq1 3 1 2 3
1 1 1 1 1
b b b b b b
= + = +
+
 or 1 3 2 3
eq
1 2 3
b b b b
b
b b b
+
=
+ +
 
 
Problem Set 5.2 
1. For the system shown in Figure 5.38, the input is the force f and the output is the displacement x of the mass. 
 a. Draw the necessary free-body diagram and derive the differential equation of motion. 
b. Using the differential equation obtained in Part (a), determine the transfer function. Assume initial conditions 
x(0) = 0 and (0) 0x = . 
 c. Using the differential equation obtained in Part (a), determine the state-space representation. 
 
Figure 5.38 Problem 1. 
Solution 
a. The free-body diagram is shown in the figure below. Due to gravity, the spring is compressed by stδ when the 
mass is in static equilibrium and stmg kδ= . Choosing the static equilibrium as the origin of the coordinate x and 
applying Newton’s second law in the x-direction gives 
( )st:x f mg k x bx mxδ+ ↓ + − + − =  or ( )mx bx kx f t+ + =  
 
Figure PS5-2 No1 
b. Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 
2
( ) 1
( )
X s
F s ms bs k
=
+ +
 
c. The state, the input, and the output are 
 1
2
, ,
x x
u f y x
x x
   
= = = =   
  
x

 
1 1
2 2
0 1 0
1
x x
uk bx x
m m m
         = +      − −      


 
[ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
 
2. Repeat Problem 1 for the system shown in Figure 5.39. 
182 
 
 
 
Figure 5.39 Problem 2. 
Solution 
a. The free-body diagram is shown in the figure below. Due to gravity, the spring is stretched by stδ when the mass 
is in static equilibrium and stsin 30mg kδ=a . Choosing the static equilibrium as the origin of the coordinate x 
and applying Newton’s second law in the x-direction gives 
+: ( )sin 30 stf mg k x bx mxδ+ − + − =a
  
( )mx bx kx f t+ + =  
 
Figure PS5-2 No2 
b. Refer to the solution to Part (b) in Problem 1. 
c. Refer to the solution to Part (c) in Problem 1. 
 
3. Repeat Problem 1 for the system shown in Figure 5.40. 
 
Figure 5.40 Problem 3. 
Solution 
a. The free-body diagram is shown in the figure below. Due to gravity, the spring is stretched by stδ when the mass 
is in static equilibrium andstmg kδ= . Choosing the static equilibrium as the origin of the coordinate x and 
applying Newton’s second law in the x-direction gives 
( )st: δx f mg k x bx mx+ ↓ + − + − =  
( )mx bx kx f t+ + =  
 
Figure PS5-2 No3 
b. Refer to the solution to Part (b) in Problem 1. 
183 
 
 
c. Refer to the solution to Part (c) in Problem 1. 
4. Repeat Problem 1 for the system shown in Figure 5.41. 
 
Figure 5.41 Problem 4. 
Solution 
a. The free-body diagram is shown in the figure below. Choosing the static equilibrium as the origin of the 
coordinate x and applying Newton’s second law in the x-direction gives 
:x f kx bx mx+ → − − =  
( )mx bx kx f t+ + =  
 
Figure PS5-2 No4 
b. Refer to the solution to Part (b) in Problem 1. 
c. Refer to the solution to Part (c) in Problem 1. 
 
5. For Problems 1 through 4, use Simulink to construct block diagrams to find the displacement output x(t) of 
the system subjected to an applied force f(t) = 10u(t), where u(t) is the unit-step function. The parameter values are 
m = 1 kg, b = 2 N·s/m, and k = 5 N/m. Assume zero initial conditions. 
Solution 
Since the systems in Problem 1 through 4 have the same mathematical model, we can use the same Simulink block 
diagram to find the displacement output. 
 
Figure PS5-2 No5a 
 
184 
 
 
Figure PS5-2 No5b 
6. The system shown in Figure 5.42 simulates a machine supported by rubbers, which are approximated as four 
same spring-damper units. The input is the force f and the output is the displacement x of the mass. The parameter 
values are m = 500 kg, b = 250 N·s/m, and k = 200,000 N/m. 
 a. Draw the necessary free-body diagram and derive the differential equation of motion. 
 b. Determine the transfer function form. Assume zero initial conditions. 
 c. Determine the state-space representation. 
 d. Find the transfer function from the state-space form and compare with the result obtained in Part (b). 
 
Figure 5.42 Problem 6. 
Solution 
a. The free-body diagram is shown in the figure below. Note that four same spring-damper units are connected in 
parallel. Therefore, the equivalent spring stiffness is 4k and the equivalent damping coefficient is 4b. Let us 
choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled by the 
weight. Applying Newton’s second law in the x-direction gives 
: 4 4x f kx bx mx+ ↑ − − =  
4 4 ( )mx bx kx f t+ + =  or 500 1000 800000 ( )x x x f t+ + =  
 
Figure PS5-2 No6 
b. Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 
2
( ) 1
( ) 500 1000 800000
X s
F s s s
=
+ +
 
c. The state, the input, and the output are 
 1
2
, ,
x x
u f y x
x x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
1 1
2 2
00 1
11600 2
500
x x
u
x x
       = +      − −       


, [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
d. Using the results obtained in Part (c), we can derive the transfer function 
[ ]
1
2 2
( ) ( ) ( )
( ) ( )
02 11 1 5001 0 116002 1600 2 1600
500
X s Y s s
F s U s
s
ss s s s
−= = − +
 +   = =   −+ + + +   
C I A B D
 
185 
 
 
 which is the same as the one obtained in Part (b). 
7. Repeat Problem 6 for the system shown in Figure 5.43. 
 
Figure 5.43 Problem 7. 
Solution 
a. The free-body diagram is shown in the figure below. Choosing the static equilibrium as the origin of the 
coordinate x and applying Newton’s second law in the x-direction gives 
: 2 2x f kx bx mx+ → − − =  
2 2 ( )mx bx kx f t+ + =  or 20 250 800 ( )x x x f t+ + =  
 
Figure PS5-2 No7 
b. Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 
2
( ) 1
( ) 20 250 800
X s
F s s s
=
+ +
 
c. The state, the input, and the output are 
 1
2
, ,
x x
u f y x
x x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
1 1
2 2
0 1 0
40 12.5 0.05
x x
u
x x
      
= +      − −      


, [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
d. Using the results obtained in Part (c), we can derive the transfer function 
[ ]
[ ]
1
1
2 2
1 0( ) ( ) ( ) 1 0
40 12.5 0.05( ) ( )
12.5 1 01 0.051 0
40 0.0512.5 40 12.5 40
sX s Y s s
sF s U s
s
ss s s s
−
− −   
= = − + =    +   
+   
= =   −+ + + +   
C I A B D
 
which is the same as the one obtained in Part (b). 
 
8. The system shown in Figure 5.44 simulates a vehicle traveling on a rough road. The input is the displacement z. 
 a. Draw the necessary free-body diagram and derive the differential equation of motion. 
b. Assuming zero initial conditions, determine the transfer function for two different cases of output: (1) 
displacement x and (2) velocity x . 
 c. Determine the state-space representation for two different cases of output: (1) displacement x and (2) velocity 
x . 
 
186 
 
 
Figure 5.44 Problem 8. 
Solution 
a. We choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled 
by the weight. Assume that 0x z> > . Applying Newton’s second law in the x-direction gives 
( ) ( ):x k x z b x z mx+ ↑ − − − − =  
mx bx kx bz kz+ + = +   
 
 
Figure PS5-2 No8 
b. Assuming zero initial conditions and taking Laplace transform of the above differential equation gives 
( ) ( )2 ( ) ( )ms bs k X s bs k Z s+ + = + 
If the displacement x is the output, the transfer function is 
2
( )
( )
X s bs k
Z s ms bs k
+
=
+ +
 
If the velocity x is the output, the transfer function is 
2
2
( ) ( )
( ) ( )
V s sX s bs ks
Z s Z s ms bs k
+
= =
+ +
 
c. Rewrite the transfer function ( ) ( )X s Z s as 
2
( ) ( ) ( ) 1
( ) ( ) ( )
X s X s W s b ks b kZ s W s Z s m m s s
m m
 = = + 
  + +
 
where 
( )
( )
X s b ks
W s m m
= + 
2
( ) 1
( )
W s
b kZ s s s
m m
=
+ +
 
Thus in the time-domain, we have 
b kx w w
m m
= + 
b kw w w z
m m
= − − +  
Select the state and the input as [ ] , Tw w u z= =x  . The state-space equations with the displacement x as the 
output are 
1 1
2 2
0 1 0
1
x x
uk bx x
m m
       = +      − −      


 1
2
0
xk by u
xm m
  = + ⋅     
 
If the velocity x is the output, following the same procedure gives the output equation 
2
1
2 2
2
xbk b k by u
xm m m m
  
= − − + +  
  
 
187 
 
 
9. Repeat Problem 8 for the system shown in Figure 5.45, where the cam and follower impart a displacement z to the 
lower end of the system. 
 
Figure 5.45 Problem 9. 
Solution 
a. We choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled 
by the weight Assume that 0x z> > . The free-body diagram is shown in the figure below. Applying Newton’s 
second law in the x-direction gives 
( )1 2:x k x bx k x z mx+ ↑ − − − − =  
( )1 2 2mx bx k k x k z+ + + =  
 
Figure PS5-2 No9 
b. Assuming zero initial conditions and taking Laplace transform of the above differential equation give 
( )2
1 2 2( ) ( )ms bs k k X s k Z s+ + + = 
If the displacement x is the output, the transfer function is 
2
2
1 2
( )
( )
kX s
Z s ms bs k k
=
+ + +
 
If the velocity x is the output, the transfer function is 
2
2
1 2
( ) ( )
( ) ( )
k sV s sX s
Z s Z s ms bs k k
= =
+ + +
 
c. The state and the input are [ ] , Tx x u z= =x  . The state-space equations with the displacement x as the output is 
1 1
1 2 2
2 2
0 1 0x x
uk k kbx x
m m m
         = ++      − −      


 
[ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
If the velocity x is the output, the output equation is 
[ ] 1
2
0 1 0
x
y u
x
 
= + ⋅ 
 
 
 
188 
 
 
10. For the system shown in Figure 5.46, the input is the force f and the outputs)
(2 )
j j
j
+ − +
−
 
Solution 
( ) ( )
( )
3 21.2490 2.03443 2 7.8158
9.6706
4 4 1.85440.4636
10 5 (1 3 ) ( 1 2 ) 50 10 2 10 6.1344 1.5391
(2 ) 255 
j j j
j
jj
e ej j e e j
j ee
−−
+ − +
= = = = − −
−
 
 
16. 
3
1 4
j
j
 
 + 
 
Solution 
( )
3 /2
5.5482
31.3258
1 0.0106 0.0096
1 4 17 1717
j
j
j
j e e j
j e
π−
− 
= = = + + 
 
 
 
 
In Problems 17 through 21, find all possible values for each expression. 
 
17. 1/3( 1 2 )j− + 
Solution 
1/3 1/3 2.0344 2 2.0344 2( 1 2 ) ( 5) cos sin , 0,1, 2
3 3
k kj j kπ π+ + − + = + =  
 
 
The three values are obtained as 
 
 1.0183 0.8203 , 1.2196 0.4717 , 0.2013 1.2921j j j+ − + − 
 
 
18. 1/4( 1)− 
Solution 
The goal is to find 4w z= where 1z = − . Noting that 1z = − is located on the negative real 
axis, one unit from the origin, we have 1r = and θ π= , hence 1 jz e π= − = . Then, 
 4 4 2 21 1 cos sin , 0,1, 2,3
4 4
k kj kπ π π π+ + − = + = 
 
 
Therefore, the four roots have magnitude 1 and phases 3 5 71
4 4 4 4, , , π π π π . The roots are 
 
 
20 
 
 2 2
2 2(1 ), ( 1 )j j± − ± 
 
 
19. 1/4( 3 )j− 
Solution 
 
1 1
1/4 1/4 6 62 2
( 3 ) (2) cos sin , 0,1, 2,3
4 4
k k
j j k
π π π π − + − +
− = + = 
 
 
 
The four values are then found as 
 
 1.1790 0.1552 , 0.1552 1.1790 , 1.1790 0.1552 , 0.1552 1.1790j j j j− + − + − − 
 
 
20. 1 2j+ 
Solution 
The goal is to find w z= where 1 2z j= + . Since 0.95531 2 3 jz j e= + = , we have 
 
 1/4 0.9553 2 0.9553 21 2 3 cos sin , 0,1
2 2
k kj j kπ π+ + + = + = 
 
 
 
Therefore, the two values are 1.1688 0.6050 , 1.1688 0.6050j j+ − − , located on the circle of 
radius 1/43 and centered at the origin. 
 
 
21. 2/32
3( )j− + 
Solution 
We first rewrite the expression as 
 
 ( ) ( ) ( )
1/321/3 1/32/3 2 2.1588 4.317613 132 2
3 3 3 9( ) ( ) j jj j e e − + = − + = = 
 
 
 
Proceeding as always, the three values are formed as 
 
 ( )1/34.3176 1/313 13
9 9
4.3176 2 4.3176 2( ) cos sin , 0,1, 2
3 3
j k ke j kπ π+ + = + = 
 
 
 
This yields the three values 0.1483 1.1206 , 1.0447 0.4319 , 0.8963 0.6888j j j+ − − − . 
 
 
22. Find all possible roots of 3 (1 ) 0jz j− + = . 
 
 
21 
 
Solution 
The equation is rewritten as 3 1 1jz j
j
+
= = − , hence the three values of 1/3(1 )j− are sought. 
Noting /41 2 jj e π−− = , 
 
1 1
1/3 1/6 4 42 2
(1 ) 2 cos sin , 0,1, 2
3 3
k k
j j k
π π π π − + − +
− = + = 
 
 
 
This gives 1.0842 0.2905 , 0.2905 1.0842 , 0.7937 0.7937z j j j= − − + − − . 
 
 
Problem Set 2.2 
In Problems 1 through 4, solve the linear, first-order initial-value problem. 
 
1. 1
3 cos 2 , (0) 1x x t x+ = = 
Solution 
Since 1
3( )g t = , we have 1
3( )h t t= and 
 
 
/3 /3 /3 /39 1
37 3
/39 1
37 3
( ) cos 2 (2sin 2 cos 2 )
 (2sin 2 cos 2 )
t t t t
t
x t e e tdt c e e t t c
t t ce
− −
−
   = + = + +  
= + +
∫ 
 
Using the initial condition 34
37c = so that /39 341
37 3 37( ) (2sin 2 cos 2 ) tx t t t e−= + + . 
 
 
2. 1 1 1
2 2 3 , (0)x tx t x+ = = 
Solution 
Writing the ODE in standard form yields ( ) 2g t t= so that 2( )h t t= and 
 
 
2 2 2 2 21 1
2 2( ) t t t t tx t e e t dt c e e c ce− − −   = + = + = +   ∫ 
 
By the initial condition 1
6c = − and the solution is obtained as 
21 1
2 6( ) tx t e−= − . 
 
 
3. ( 1) 2 , (0) 1t u tu t u− + = = 
Solution 
Writing the ODE in standard form yields ( )
1
tg t
t
=
−
 and 2( )
1
tf t
t
=
−
 so that 
 
 
 
22 
 
 ( ) ln( 1)
1
th t dt t t
t
= = + −∫
−
 
and 
 ln( 1) ln( 1) ln( 1)2( ) 2( 1) 2
1 1
t
t t t t t t tt ceu t e e dt c e t e c
t t
−
− − − + − − − −   = + = − + = +   − − ∫ 
By the initial condition we have 1c = , hence ( ) 2
1
teu t
t
−
= +
−
. 
 
4. 1
2(1 )cos , (0)u u t u= − = 
Solution 
Writing the ODE in standard form yields ( ) cos ( )g t t f t= = so that ( ) sinh t t= and 
 
 sin sin sin sin sin( ) cos 1t t t t tu t e e tdt c e e c ce− − −   = + = + = +  ∫ 
 
By the initial condition we have 1
2c = − , hence sin1
2( ) 1 tu t e−= − . 
 
 
In Problems 5 through 10, solve the linear, second-order initial-value problem. 
 
5. 4 4 , (0) 1 , (0) 1tx x x e x x−+ + = = =   
Solution 
Characteristic equation 2 4 4 0λ λ+ + = yields 2, 2λ = − − and 2
1 2( ) t
hx c c t e−= + . Based on 
( ) tf t e−= we pick t
px Ke−= , no special case, and insert into the ODE to get 1K = hence 
t
px e−= . A general solution is 2
1 2( ) ( ) t tx t c c t e e− −= + + . Initial conditions yield 1 20, 2c c= = , 
and the solution is 2( ) 2 t tx t te e− −= + . 
 
 
6. 4 , (0) 0 , (0) 1x x t x x+ = = =  
Solution 
Characteristic values are 2 jλ = ± , hence 1 2cos 2 sin 2hx c t c t= + . Pick px At B= + , no special 
case, and insert into the original ODE to find 1
4 , 0A B= = so that a general solution is 
1
1 2 4( ) cos 2 sin 2x t c t c t t= + + . Using the initial conditions, we find 1 0c = , 3
2 8c = and thus 
3 1
8 4( ) sin 2x t t t= + . 
 
 
7. sin , (0) 1 , (0) 0x x t x x+ = = =  
Solution 
Characteristic values are jλ = ± so that 1 2cos sinhx c t c t= + . Based on the nature of ( )f t we 
pick ( cos sin )px t A t B t= + , special case, and insert into the ODE to get 1
2A = − , 0B = hence 
 
 
23 
 
1
2 cospx t t= − . A general solution is 1
1 2 2( ) cos sin cosx t c t c t t t= + − . By initial conditions, we 
have 1
1 2 21,c c= = , and the solution is 1 1
2 2( ) cos sin cosx t t t t t= + − . 
 
8. 34 3 2 , (0) 0 , (0) 1tx x x e x x−+ + = = = −   
Solution 
Characteristic values are 1, 3λ = − − so that 3
1 2
t t
hx c e c e− −= + . Because 3te− coincides with an 
independent solution we pick 3t
px Kte−= and insert into the ODE to get 1K = − , hence 
3t
px te−= − . A general solution is 3 3
1 2( ) t t tx t c e c e te− − −= + − . By initial conditions, 1 20c c= = 
and the solution is 3( ) tx t te−= − . 
 
 
9. 6 7 2 65cos , (0) 0 , (0) 5u u u t u u+ + = = =   
Solution 
Characteristic values are 2 1
3 2,λ = − − so that /2 2 /3
1 2
t t
hu c e c e− −= + . Based on the nature of ( )f t 
we pick cos sinpu A t B t= + , no special case, insert into ODE to find 4A = − , 7B = . Therefore, 
4cos 7sinpu t t= − + and a general solution is /2 2 /3
1 2( ) 4cos 7sint tu t c e c e t t− −= + − + . By initial 
conditions, 1 24, 0c c= = , and the solution is /2( ) 4 4cos 7sintu t e t t−= − + . 
 
 
10. 4 4 5 0 , (0) 0 , (0) 1u u u u u+ + = = =   
Solution 
Characteristic values are 1
2 jλ = − ± so that /2
1 2( cos sin )t
hu u e c t c t−= = + . Using the initial 
conditions, we have 1 20, 1c c= = , and the solution is /2( ) sintu t e t−= . 
 
 
In Problems 11 through 14, write the expression in the form sin( )D tω φ+ . 
 
11. 1
3cos sint t+ 
Solution 
Write 1
3cos sin sin( ) sin cos cos sint t D t D t D tφ φ φ+ = + = + and compare the two sides to find 
 
 
1st quadrant10 /3
1
3
sin 1 sin 0
 tan 3 1.2490 rad
cos cos 0
DD
D
φ φ
φ φ
φ φ
== >
⇒ ⇒ = ⇒ =
= >
 
 
Therefore, 1
3
10cos sin sin( 1.2490)
3
t t t+ = + . 
 
 
 
24 
 
12. 1
2 cos3 sin 3t t− 
Solution 
Write 1
2 cos3 sin 3 sin(3 ) sin 3 cos cos3 sint t D t D t D tφ φ φ− = + = + . Comparing the two sides, 
 
 
1 2nd quadrant5 /2
12
2
sin 0sin 
 tan 2.6779 rad
cos 0cos 1
DD
D
φφ
φ φ
φφ
= >=
⇒ ⇒ = − ⇒ =
⇒ ⇒ = − ⇒ =
are the displacements x1 and x2 of the 
masses. 
 a. Draw the necessary free-body diagrams and derive the differential equations of motion. 
 b. Write the differential equations of motion in the second-order matrix form. 
 c. Using the differential equations obtained in Part (a), determine the state-space representation. 
 
Figure 5.46 Problem 10. 
Solution 
a. We choose the displacements of the two masses 1x and 2x as the generalized coordinates. The static equilibrium 
positions of 1m and 2m are set as the coordinate origins. Assume that 2 1 0x x> > . Applying Newton’s second 
law in the x-direction gives 
: x xx F ma+ ↓ ∑ = 
Mass 1: ( ) ( )1 1 2 2 1 2 1 1 1k x k x x b x x m x− + − + − =   
Mass 2: ( ) ( )2 2 1 2 1 2 2f k x x b x x m x− − − − =   
Rearranging the equations, we have 
( )1 1 1 2 1 2 1 2 2 0m x bx bx k k x k x+ − + + − =   
2 2 1 2 2 1 2 2m x bx bx k x k x f− + − + =   
 
Figure PS5-2 No10 
b. The differential equations can be expressed in second-order matrix form as 
1 1 1 1 2 2 1
2 2 2 2 2 2
0 0
0
m x x k k k xb b
m x x k k xb b f
+ −−            
+ + =            −−            
 
 
 
c. The state, the input, and the output are 
189 
 
 
 
1 1
2 2 1
3 1 2
4 2
, ,
x x
x x x
u f
x x x
x x
   
        = = = =     
    
      
x y


 
The state-space representation in matrix form is 
1 1
2 21 2 2
1 1 1 13 3
4 42 2
2
2 2 2 2
0 0 1 0
0
0 0 0 1
0
0
1
x x
x xk k k b b
u
m m m mx x
x xk k b b
m
m m m m
 
              +   − −  = +     
      
           − −   
 




 
1
2
3
4
1 0 0 0 0
0 1 0 0 0
x
x
u
x
x
 
     = +    
    
  
y 
 
11. Repeat Problem 10 for the system shown in Figure 5.47. 
 
Figure 5.47 Problem 11. 
Solution 
a. We choose the displacements of the two masses 1x and 2x as the generalized coordinates. The static equilibrium 
positions of 1m and 2m are set as the coordinate origins. Assume that 1 2 0x x> > . Applying Newton’s second 
law in the x-direction gives 
+: x xF ma∑ = 
Mass 1: ( ) ( )1 1 2 1 1 2 1 1f k x x b x x m x− − − − =   
Mass 2: ( ) ( )1 1 2 1 1 2 2 2 2 2 2 2k x x b x x k x b x m x− + − − − =    
Rearranging the equations, we have 
1 1 1 1 1 2 1 1 1 2m x b x b x k x k x f+ − + − =   
2 2 1 1 1 2 2 1 1 1 2 2( ) ( ) 0m x b x b b x k x k k x− + + − + + =   
 
Figure PS5-2 No11 
b. The differential equations can be expressed in second-order matrix form as 
1 1 1 1 1 1 1 1
2 2 1 1 2 2 1 1 2 2
0
0 0
m x b b x k k x f
m x b b b x k k k x
− −             
+ + =            − + − +             
 
 
 
190 
 
 
c. The state, the input, and the output are 
 
1 1
2 2 1
3 1 2
4 2
, ,
x x
x x x
u f
x x x
x x
   
        = = = =     
    
      
x y


 
The state-space representation in matrix form is 
1 1
2 21 1 1 1
1 1 1 13 3
1
4 41 1 2 1 1 2
2 2 2 2
0 0 1 0
0
0 0 0 1
0
1
0
x x
x xk k b b
u
m m m mx x
m
x xk k k b b b
m m m m
 
                 − −  = +     
      
      + +     − −   
 




 
1
2
3
4
1 0 0 0 0
0 1 0 0 0
x
x
u
x
x
 
     = +    
    
  
y 
 
12. Repeat Problem 10 for the system shown in Figure 5.48. 
 
Figure 5.48 Problem 12. 
Solution 
a. We choose the displacements of the two masses 1x and 2x as the generalized coordinates. The static equilibrium 
positions of 1m and 2m are set as the coordinate origins. Assume that 2 1 0x x> > . Applying Newton’s second 
law in the x-direction gives 
: x xx F ma+ → ∑ = 
Mass 1: ( )1 1 1 1 2 2 1 1 1k x b x k x x m x− − + − =  
Mass 2: ( )2 2 1 2 2f k x x m x− − =  
Rearranging the equations, we have 
( )1 1 1 1 1 2 1 2 2 0m x b x k k x k x+ + + − =  
2 2 2 1 2 2m x k x k x f− + = 
 
Figure PS5-2 No12 
b. The differential equations can be expressed in second-order matrix form as 
1 1 1 1 2 2 11
2 2 2 2 2 2
0 0 0
0 0 0
m x x k k k xb
m x x k k x f
+ −           
+ + =            −            
 
 
 
c. The state, the input, and the output are 
 
1 1
2 2 1
3 1 2
4 2
, ,
x x
x x x
u f
x x x
x x
   
        = = = =     
    
      
x y


 
The state-space representation in matrix form is 
191 
 
 
1 1
2 21 2 2 1
1 1 13 3
4 42 2
2
2 2
0 0 1 0
0
0 0 0 1
0
0 0
1
0 0
x x
x xk k k b
u
m m mx x
x xk k
m
m m
 
              +   − −= +      
     
        −     
 




 
1
2
3
4
1 0 0 0 0
0 1 0 0 0
x
x
u
x
x
 
     = +    
    
  
y 
 
13. For Problems 10 through 12, use MATLAB commands to define the systems in the state-space form and 
then convert to the transfer function form. Assume that the displacements of the two masses, x1 and x2, are the 
outputs, and all initial conditions are zero. The masses are m1 = 5 kg and m2 = 15 kg. The spring constants are k1 = 
7.5 kN/m and k2 = 15 kN/m. The viscous damping coefficients are b1 = b2 = b = 250 N·s/m. 
Solution 
For the system in Problem 10, the MATLAB session is 
% define the system parameters 
m1 = 5; 
m2 = 15; 
b1 = 250; 
b2 = 250; 
b = 250; 
k1 = 7500; 
k2 = 15000; 
% define the system in the state-space form 
A = [0 0 1 0; 
 0 0 0 1; 
 -(k1+k2)/m1 k2/m1 -b/m1 b/m1; 
 k2/m2 -k2/m2 b/m2 -b/m2]; 
B = [0; 0; 0; 1/m2]; 
C = [1 0 0 0; 0 1 0 0]; 
D = zeros(2,1); 
% convert to the transfer function form 
sys_ss = ss(A,B,C,D); 
sys_tf = tf(sys_ss); 
 
The command tf returns two transfer functions 
1
4 3 2 4 6
2
2
4 3 2 4 6
( ) 3.333 200
( ) 66.67 5500 2.5 10 1.5 10
( ) 0.06667 3.333 300
( ) 66.67 5500 2.5 10 1.5 10
X s s
F s s s s s
X s s s
F s s s s s
+
=
+ + + × + ×
+ +
=
+ + + × + ×
 
 
For the systems in Problems 11 through 12, we only need to modify the state-space matrices. 
% define the system in Problem 11 in the state-space form 
A = [0 0 1 0; 
 0 0 0 1; 
 -k1/m1 k1/m1 -b1/m1 b1/m1; 
 k1/m2 -(k1+k2)/m2 b1/m2 -(b1+b2)/m2]; 
B = [0; 0; 1/m1; 0]; 
C = [1 0 0 0; 0 1 0 0]; 
D = zeros(2,1); 
 
The command tf returns two transfer functions for the system in Problem 11 
192 
 
 
2
1
4 3 2 4 6
2
4 3 2 4 6
( ) 0.2 6.667 300
( ) 83.33 3833 7.5 10 1.5 10
( ) 3.333 100
( ) 83.33 3833 7.5 10 1.5 10
X s s s
F s s s s s
X s s
F s s s s s
+ +
=
+ + + × + ×
+
=
+ + + × + ×
 
 
% define the system in Problem 12 in the state-space form 
A = [0 0 1 0; 
 0 0 0 1; 
 -(k1+k2)/m1 k2/m1 -b1/m1 0; 
 k2/m2 -k2/m2 0 0]; 
B = [0; 0; 0; 1/m2]; 
C = [1 0 0 0; 0 1 0 0]; 
D = zeros(2,1); 
The command tf returns two transfer functions for the system in Problem 11 
1
4 3 2 4 6
2
2
4 3 2 4 6
( ) 200
( ) 50 5500 5 10 1.5 10
( ) 0.06667 3.333 300
( ) 50 5500 5 10 1.5 10
X s
F s s s s s
X s s s
F s s s s s
=
+ + + × + ×
+ +
=
+ + + × + ×
 
 
14. For the system in Figure 5.49, the inputs are the forces f1 and f2 applied to the masses and the outputs are the 
displacements x1 and x2 of the masses. 
 a. Draw the necessary free-body diagrams and derive the differential equations of motion. 
 b. Write the differential equations of motion in the second-order matrix form. 
 c. Using the differential equations obtained in Part (a), determine the state-space representation. 
 
Figure 5.49 Problem 14. 
Solution 
a. We choose the displacements of the two masses 1x and 2x as the generalized coordinates. The static equilibrium 
positions of 1m and 2m are set as the coordinate origins. Assume that 2 1 0x x> > . Applying Newton’s second 
law in the x-direction gives 
: x xx F ma+ → ∑ = 
Mass 1: ( ) ( )1 1 1 2 2 1 2 2 1 1 1f k x kx x b x x m x− + − + − =   
Mass 2: ( ) ( )2 3 2 2 2 1 2 2 1 2 2f k x k x x b x x m x− − − − − =   
Rearranging the equations, we have 
1 1 2 1 2 2 1 2 1 2 2 1( )m x b x b x k k x k x f+ − + + − =   
2 2 2 1 2 2 2 1 2 3 2 2( )m x b x b x k x k k x f− + − + + =   
 
Figure PS5-2 No14 
193 
 
 
b. The differential equations can be expressed in second-order matrix form as 
1 2 21 1 2 2 1 1 1
2 2 32 2 2 2 2 2 2
0
0
k k km x b b x x f
k k km x b b x x f
+ −−             
+ + =            − +−            
 
 
 
c. The state, the input, and the output are 
 
1 1
2 2 1 1
3 1 2 2
4 2
, ,
x x
x x f x
x x f x
x x
   
          = = = =       
      
      
x u y


 
The state-space representation in matrix form is 
1 1
2 2 11 2 2 2 2
1 1 1 1 13 3 2
4 42 32 2 2
22 2 2 2
0 0 1 0 0 0
0 0 0 1 0 0
1 0
10
x x
x x uk k k b b
m m m m mx x u
x xk kk b b
mm m m m
   
               +     − −= +       
       
      +   − −   
    




 
1
2 1
3 2
4
1 0 0 0 0 0
0 1 0 0 0 0
x
x u
x u
x
 
       = +      
      
  
y 
 
15. Repeat Problem 14 for the system shown in Figure 5.50. 
 
Figure 5.50 Problem 15. 
Solution 
a. We choose the displacements of the two masses 1x and 2x as the generalized coordinates. The static equilibrium 
positions of 1m and 2m are set as the coordinate origins. Assume that 2 1 0x x> > . 
 
 
Figure PS5-2 No15 
 
Applying Newton’s second law in the x-direction gives 
: x xx F ma+ → ∑ = 
Mass 1: 1 1 1 1 1 2 2 1 1 1( )f k x b x k x x m x− − + − =  
Mass 2: 2 2 2 1 3 2 2 2( )f k x x k x m x− − − =  
Rearranging the equations, we have 
1 1 1 1 1 2 1 2 2 1( )m x b x k k x k x f+ + + − =  
2 2 2 1 2 3 2 2( )m x k x k k x f− + + = 
b. The differential equations can be expressed in second-order matrix form as 
1 2 21 1 1 1 11
2 2 32 2 2 2 2
0 0
0 0 0
k k km x x x fb
k k km x x x f
+ −           
+ + =            − +           
 
 
 
c. The state, the input, and the output are 
194 
 
 
 
1 1
2 2 1 1
3 1 2 2
4 2
, ,
x x
x x f x
x x f x
x x
   
          = = = =       
      
      
x u y


 
The state-space representation in matrix form is 
1 1
2 2 11 2 2 1
1 1 1 13 3 2
4 42 32
22 2
0 0 1 0 0 0
0 0 0 1 0 0
10 0
100 0
x x
x x uk k k b
m m m mx x u
x xk kk
mm m
   
               +     − −= +       
       
      +   −   
    




 
1
2 1
3 2
4
1 0 0 0 0 0
0 1 0 0 0 0
x
x u
x u
x
 
       = +      
      
  
y 
 
16. For Problems 14 and 15, use MATLAB commands to define the systems in the state-space form and then 
convert to the transfer function form. Assume that the displacements of the two masses, x1 and x2, are the outputs, 
and all initial conditions are zero. The masses are m1 = 5 kg and m2 = 15 kg. The spring constants are k1 = 7.5 
kN/m, k2 = 15 kN/m, and k3 = 30 kN/m. The viscous damping coefficients are b1 = b2 = 250 N·s/m. 
Solution 
For the system in Problem 14, the MATLAB session is 
% define the system parameters 
m1 = 5; 
m2 = 15; 
b1 = 250; 
b2 = 250; 
k1 = 7500; 
k2 = 15000; 
k3 = 30000; 
% define the system in the state-space form 
A = [0 0 1 0; 0 0 0 1; 
 -(k1+k2)/m1 k2/m1 -b2/m1 b2/m1; 
 k2/m2 -(k2+k3)/m2 b2/m2 -b2/m2]; 
B = [0 0; 0 0; 1/m1 0; 0 1/m2]; 
C = [1 0 0 0; 0 1 0 0]; 
D = zeros(2,2); 
% convert to the transfer function form 
sys_ss = ss(A,B,C,D); 
sys_tf = tf(sys_ss); 
The command tf returns four transfer functions 
2
1
2
1 2
4 3 2 5 7
2 2
1 2
1
0.2 3.333 600 3.333 200( ) ( )
( ) ( ) 3.333 200 0.06667 3.333 300
( ) ( ) 66.67 7500 1.25 10 1.05 10
( ) ( )
s s sX s X s
F s F s s s s
X s X s s s s s
F s F s
 + + +     + + +    =
  + + + × + ×
 
 
 
For the systems in Problem 15, we only need to modify the state-space matrices. 
% define the system in Problem 15 in the state-space form 
A = [0 0 1 0; 
 0 0 0 1; 
 -(k1+k2)/m1 k2/m1 -b2/m1 b2/m1; 
 k2/m2 -(k2+k3)/m2 b2/m2 -b2/m2]; 
B = [0 0; 0 0; 1/m1 0; 0 1/m2]; 
C = [1 0 0 0; 0 1 0 0]; 
D = zeros(2,2); 
195 
 
 
The command tf returns four transfer functions for the system in Problem 15 
1
2
1
2
1 2
4 3 2 5 7
2 2
1 2
0.2 600 200( ) ( )
( ) ( ) 200 0.06667 3.333 300
( ) ( ) 50 7500 1.5 10 1.05 10
( ) ( )
sX s X s
F s F s s s
X s X s s s s s
F s F s
 +     + +    =
  + + + × + ×
 
 
 
 
17. For the system in Figure 5.51, the input is the force f and the outputs are the displacement x1 of the mass and the 
displacement x2 of the massless junction A. 
a. Draw the necessary free-body diagrams and derive the differential equations of motion. Determine the 
number of degrees of freedom and the order of the system. 
 b. Write the differential equations of motion in the second-order matrix form. 
 c. Using the differential equation obtained in Part (a), determine the state-space representation. 
 
Figure 5.51 Problem 17. 
Solution 
a. One massless junction is included in this system. We choose the displacements of the mass and the junction A as 
the generalized coordinates, which are denoted by 1x and 2x . Thus, this is a two-degree-of-freedom system. 
Assume that 2 1 0x x> > . 
 
Figure PS5-2 No17 
Applying Newton's second law gives 
Mass: : x xx F ma+ → ∑ = 
 1 1 1 1 2 2 1 2 2 1 1( ) ( )k x b x k x x b x x mx− − + − + − =    
Massless junction: : 0x xx F ma+ → ∑ = = 
A: 2 2 1 2 2 1( ) ( ) 0k x x b x x f− − − − + =  
The equations can be rearranged as follows. Thus, this is a third-order system. 
1 1 2 1 2 2 1 2 1 2 2( ) ( ) 0mx b b x b x k k x k x+ + − + + − =   
 2 2 2 1 2 1 2 2b x b x k x k x f− − + =  
b. The differential equations can be expressed in second-order matrix form as 
1 1 2 2 1 1 2 2 11
2 2 2 2 2 2 2
0 0
0 0
x b b b x k k k xm
x b b x k k x f
+ − + −           
+ + =            − −            
 
 
 
c. The state, the input, and the output are 
 
1 1
1
2 2
2
3 1
, ,
x x
x
x x u f
x
x x
   
    = = = =     
      
x y

 
The state-space representation in matrix form is 
1 1
2 2 2 2 2 2 2
3 1 1 3
0 0 1 0
1 1
0 1
x x
x k b k b x b u
x k m b m x m
      
      = − +      
      − −      



 
1
2
3
1 0 0 0
0 1 0 0
x
x u
x
 
    = +    
    
 
y 
196 
 
 
18. Repeat Problem 17 for the system in Figure 5.52, the input is the displacement z and the outputs are the 
displacements x1 and x2. 
 
Figure 5.52 Problem 18. 
Solution 
a. Assume that 1 2 0x x z> > > . 
 
Figure PS5-2 No18 
Applying Newton's second law give 
Mass: : x xx F ma+ ↑ ∑ = 
 ( ) ( )1 1 2 1 2 1k x x b x x mx− − − − =   
Massless junction A : : 0x xx F ma+ ↑ ∑ = = 
( ) ( ) ( )1 1 2 1 2 2 2 0k x x b x x k x z− + − − − =  
The equations can be rearranged as 
1 1 2 1 1 1 2 0mx bx bx k x k x+ − + − =   
 2 1 1 1 1 2 2 2( )bx bx k x k k x k z− − + + =  
b. The differential equations can be expressed in second-order matrix form as 
1 1 1 1 1
2 2 1 1 2 2 2
00
0 0
x x k k xm b b
x x k k k x k zb b
−−           
+ + =            − +−            
 
 
 
c. The state, the input, and the output are 
 
1 1
1
2 2
2
3 1
, ,
x x
x
x x u z
x
x x
   
    = = = =     
      
x y

 
The state-space representation in matrix form is 
1 1
2 1 1 2 2 2
3 2 3 2
0 0 1 0
( ) 1
0 0
x x
x k b k k b x k b u
x k m x k m
      
      = − + +      
      −      



 
12
3
1 0 0 0
0 1 0 0
x
x u
x
 
    = +    
    
 
y 
 
197 
 
 
Problem Set 5.3 
1. Consider the rotational system shown in Figure 5.61. The system consists of a massless shaft and a uniform thin 
disk of mass m and radius r. The disk is constrained to rotate about a fixed longitudinal axis along the shaft. The 
shaft is equivalent to a torsional spring of stiffness K. Draw the necessary free-body diagram and derive the 
differential equation of motion. 
 
Figure 5.61 Problem 1. 
Solution 
The free-body diagram of the disk is shown below. Applying the moment equation to the fixed point O gives 
+↶: O OαM I∑ = 
Oτ θ θ θK B I− − =  
where the mass moment of inertia for a thin disk about its center of gravity is 21
O 2I mr= . Thus, we have 
21
2 θ θ θ τmr B K+ + =  
 
Figure PS5-3 No1 
 
2. Repeat Problem 1 for the system shown in Figure 5.62. 
 
Figure 5.62 Problem 2. 
Solution 
The free-body diagram of the disk is shown below. Applying the moment equation to the fixed point O gives 
+↶: O OαM I∑ = 
Oτ 2 θ 2 θ θK B I− − =  
where the mass moment of inertia for a thin disk about its center of gravity is 21
O 2I mr= . Thus, we have 
21
2 θ 2 θ+2 θ τmr B K+ =  
 
Figure PS5-3 No2 
 
3. Consider the torsional mass–spring system in Figure 5.63. The mass moments of inertia of the two disks are I1 and 
I2, respectively. The massless torsional springs represent the elasticity of the shafts. 
198 
 
 
 a. Draw the necessary free-body diagrams and derive the differential equations of motion. Provide the 
equations in the second-order matrix form. 
 b. Determine the transfer functions Θ1(s)/T(s) and Θ2(s)/T(s). All the initial conditions are assumed to be zero. 
 c. Determine the state-space representation with the angular displacements θ1 and θ2 as the outputs. 
 
Figure 5.63 Problem 3. 
Solution 
a. We choose the angular displacements θ1 and θ2 as the generalized coordinates. Assume that 1 2θ θ 0> > . The 
free-body diagrams are shown in the figure below. Applying the moment equation about the fixed points O1 and 
O2, respectively, gives 
+↶: O OαM I∑ = 
11 1 2 12 1( )K K Iτ − θ θ − θ = θ−  
22 2 21( )K Iθ − θ = θ 
Rearranging the equations results in 
1 1 1 2 1 2 2( )I K K Kθ + + θ − θ = τ 
2 2 2 1 2 2 0I K Kθ − θ + θ = 
which can be written in the second-order matrix form 
1 1 2 2 11
2 2 2 22
0 θ τθ
0 θ 0θ
I K K K
I K K
+ −          + =        −         


 
 
Figure PS5-3 No3 
 
b. All the initial conditions are assumed to be zero. Taking Laplace transform of the above differential equations 
gives 
 
2
1 1 2 2 1
2
22 2 2
( ) ( )
.
( ) 0
I s K K K s T s
sK I s K
 + + − Θ   
=     Θ− +    
 
Using Cramer’s rule, we can solve for 1( ) ( )s sΘ Τ and 2 ( ) ( )s sΘ Τ , 
2 2
1 2 2 2 2
2 2 2 4 2
1 1 2 2 2 2 1 2 1 2 2 1 2 2 1 2
2 2 2
2 2 2 4 2
1 1 2 2 2 2 1 2 1 2 2 1 2 2 1 2
( )
,
T( ) ( )( ) ( )
( )
.
T( ) ( )( ) ( )
s I s K I s K
s I s K K I s K K I I s I K I K I K s K K
s K K
s I s K K I s K K I I s I K I K I K s K K
Θ + +
= =
+ + + − + + + +
Θ
= =
+ + + − + + + +
 
 
c. The state, the input, and the output are 
 
1 1
2 2 1
3 1 2
4 2
θ
θ θ
, τ,
θ θ
θ
x
x
u
x
x
   
        = = = =     
    
      
x y


 
The state-space representation in matrix form is 
199 
 
 
1 1
2 21 2 2
1 13 3
1
4 42 2
2 2
0 0 1 0
0
0 0 0 1
0
0 0 1
0 0 0
x x
x xK K K
u
I Ix x
I
x xK K
I I
 
              +   −  = +     
      
           −   
 




 
1
2
3
4
1 0 0 0 0
0 1 0 0 0
x
x
u
x
x
 
     = +    
    
  
y 
 
4. Repeat Problem 3 for the system shown in Figure 5.64, where the torsional viscous damper represents the fluid 
coupling. The input is the angular displacement φ at the end of the shaft. 
 
Figure 5.64 Problem 4. 
Solution 
a. Assume that 1 2θ θ 0φ> > > . The free-body diagrams are shown in the figure below. 
 
 
Figure PS5-3 No4 
 
Applying the moment equation about the fixed points O1 and O2, respectively, gives 
+↶: O OαM I∑ = 
( )1 1 1 1 3 1 2 1 1θ θ θ θ θK B K I− − − − =  
( ) ( )3 1 2 2 2 2 2θ θ θ θK K Iφ− − − =  
Rearranging the equations results in 
( )1 1 1 1 1 3 1 3 2θ θ θ θ 0I B K K K+ + + − =  
( )2 2 3 1 2 3 2 2θ θ θI K K K K φ− + + = 
which can be written in the second-order matrix form 
1 3 31 111 1
3 2 32 2 22 2
0 θ 00θ θ
0 θ0 0θ θ
K K KI B
K K KI K φ
+ −           
+ + =            − +           
 
 
 
b. All the initial conditions are assumed to be zero. Taking Laplace transform of the above differential equations 
gives 
 
2
11 1 1 3 3
2
2 23 2 2 3
( ) 0
( ) ( )
sI s B s K K K
s K sK I s K K
Θ + + + −    
=     Θ Φ− + +     
 
Using Cramer’s rule, we can solve for 1( ) ( )s sΘ Φ and 2 ( ) ( )s sΘ Φ , 
200 
 
 
2 31( )
( ) ( )
K Ks
s s
Θ
=
Φ ∆
 
( )2
2 1 1 1 32 ( )
( ) ( )
K I s B s K Ks
s s
+ + +Θ
=
Φ ∆
 
where 
( ) ( )2 2 2
1 1 1 3 2 2 3 3( )s I s B s K K I s K K K∆ = + + + + + − 
c. The state, the input, and the output are 
 
1 1
2 2 1
3 1 2
4 2
θ
θ θ
, ,
θ θ
θ
x
x
u
x
x
φ
   
        = = = =     
    
      
x y


 
The state-space representation in matrix form is 
1 1
2 21 3 3 1
1 1 13 3
2
4 43 2 3
2
2 2
0 0 1 0
0
0 0 0 1
0
0 0
0 0
x x
x xK K K B
u
I I Ix x
K
x xK K K
I
I I
 
              +   − −= +      
     
     +   −     
 




 
1
2
3
4
1 0 0 0 0
0 1 0 0 0
x
x
u
x
x
 
     = +    
    
  
y 
 
5. Consider the pendulum system shown in Figure 5.65. The system consists of a bob of mass m and a uniform rod of 
mass M and length L. The pendulum pivots at the joint O. Draw the necessary free-body diagram and derive the 
differential equation of motion. Assume small angles for θ. 
 
Figure 5.65 Problem 5. 
Solution 
For the pendulum system, the free-body diagram is shown in the figure below, where Rx and Ry are the x and y 
components of the reaction force at the joint O, respectively. Note that the system rotates about a fixed axis through the 
point O. 
 
Figure PS5-3 No5 
Applying moment equation about the fixed point O gives 
O O: M I+ = α∑ 
1
O2 sin · sin ·L Mg L mg B I− θ − θ − θ = θ  
The system consists of a point mass and a slender rod. The total mass moment of inertia about point O is 
201 
 
 
O O _ mass O _ rodI I I= + 
where 2
O _ massI mL= and IO_rod can be obtained using the parallel-axis theorem, 
( )22 2 21 1 1
O _ rod C _ rod 12 2 3I I Md ML M L ML= + = + = 
Then ( ) 21
O 3I m M L= + . Substituting IO into the equation and rearranging it in the input–output differential equation 
form, we obtain 
( ) ( )21 1
3 2 sin 0m M L B m M gL+ θ + θ + + θ =  
For small angles θ, sin θ ≈ θ. The equation of motion becomes 
( ) ( )21 1
3 2 0m M L B m M gL+ θ + θ + + θ =  
 
6. Repeat Problem 5 for the systems shown in Figure 5.66, where (a) the mass of the rod is neglected and (b) no bob 
is attached to the rod. 
 
Figure 5.66 Problem 6. 
Solution 
a. The free-body diagram of the system in (a) is shown below, where the mass of the rod is neglected. Applying 
moment equation about the fixed point O gives 
O O: M I+ = α∑ 
Osin ·L mg B I− θ − θ = θ  
 where the mass moment of inertia about point O is 2
OI mL= . Therefore, the differential equation of motion is 
2 sin 0mL B mgLθ + θ + θ =  
 For small angles θ, sin θ ≈ θ. The equation of motion becomes 
2 0mL B mgLθ + θ + θ =  
 
Figure PS5-3 No6a 
b. The free-body diagram of the system in (b) is shown below, where no bob is attached to the rod. Applying 
moment equation about the fixed pointO gives 
O O: M I+ = α∑ 
1
O2 sin ·L Mg B I− θ − θ = θ  
202 
 
 
 where the mass moment of inertia about point O can be obtained using the parallel-axis theorem, 
( )22 2 21 1 1
O C 12 2 3I I Md ML M L ML= + = + = 
 Thus, the differential equation of motion is 
21 1
3 2 sin 0ML B MgLθ + θ + θ =  
 For small angles θ, sin θ ≈ θ. The equation of motion becomes 
21 1
3 2 0ML B MgLθ + θ + θ =  
 
Figure PS5-3 No6b 
 
7. The system shown in Figure 5.67 consists of a uniform rod of mass M and length L and a translational spring of 
stiffness k at the rod’s left tip. The friction at the joint O is modeled as a damper with coefficient of torsional 
viscous damping B. The input is the force f and the output is the angle θ. The position θ = 0 corresponds to the 
static equilibrium position when f = 0. 
 
Figure 5.67 Problem 7. 
 a. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. 
 b. Using the linearized differential equation obtained in Part (a), determine the transfer function Θ(s)/F(s). 
Assume that the initial conditions are θ(0) = 0 and θ(0) 0= . 
 c. Using the differential equation obtained in Part (a), determine the state-space representation. 
Solution 
a. At equilibrium, we have 
+↷: O 0M∑ = 
1
st2 0L kδ⋅ = 
st 0δ = 
Applying the moment equation to the fixed point O gives 
+↷: O OαM I∑ = 
1 1
O2 2cosθ cosθ θ θkL f L f B I− ⋅ + ⋅ − =  
where, for small angular motions, the spring force can be approximated as 1
2( θ)kf k L= . 
Note that the mass moment of inertia about point O is 21
O C 12I I ML= = . 
Thus, the differential equation of motion for small angles θ is 
2 21 1 1
12 4 2θ+ θ+ θ=ML B kL Lf  
203 
 
 
 
Figure PS5-3 No7 
b. Taking Laplace transform of the above linearized differential equation gives 
2 2 2
( ) 6
( ) 12 3
s L
F s ML s Bs kL
Θ
=
+ +
 
c. The state, the input, and the output are 
 1
2
θ
, , θ
θ
x
u f y
x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
1 1
2 22
0 1 0
3 12 6
x x
uk Bx x
M ML ML
         = +      − −      


 [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
 
8. Repeat Problem 7 for the system shown in Figure 5.68. When θ = 0 and f = 0, the springs are at their free lengths. 
 
Figure 5.68 Problem 8. 
Solution 
a. Applying the moment equation to the fixed point O gives 
+↷: O OαM I∑ = 
1 1
O2 2cosθ sin θ sin θ cosθ (2 ) θkL f L mg L Mg L f I⋅ + ⋅ + ⋅ − ⋅ =  
where, for small angular motions, the spring force can be approximated as 
1
2( θ)kf k L= 
 
Figure PS5-3 No8 
Note that the mass moment of inertia about point O is 
2 21
O O _ mass O _ rod 3I I I mL ML++ == 
Thus, the differential equation of motion for small angles θ is 
2 21 1 1
3 2 2( ) θ+ θ ( ) θ=m M L kL m M gL fL+ − + 
b. Taking Laplace transform of the above linearized differential equation gives 
21 1 1
3 2 2
( ) 1
( ) ( ) ( )
s
F s m M Ls kL m M g
Θ
=
+ + − +
 
204 
 
 
c. The state, the input, and the output are 
 1
2
θ
, , θ
θ
x
u f y
x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
1 11 1
2 2
2 2 11
33
0 1 0
( ) 1
0
( )( )
x x
ukL m M g
x x
m M Lm M L
   
      = +− + +           + +   


 
[ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
 
9. Example 5.4 Part (d) shows how one can represent a linear system in Simulink based on the differential 
equation of the system. A linear system can also be represented in transfer function or state-space form. The 
corresponding blocks in Simulink are Transfer Fcn and State-Space, respectively. Consider Problem 7 
and construct a Simulink block diagram to find the output θ(t) of the system, which is represented using (a) the 
linearized differential equation of motion, (b) the transfer function, and (c) the state-space form obtained in 
Problem 7. The parameter values are M = 0.8 kg, L = 0.6 m, k = 100 N/m, B = 0.4 N·s/m, and g = 9.81 m/s2. The 
input force f is the unit-impulse function, which has a magnitude of 20 N and a time duration of 0.01 s. 
Solution 
The linearized differential equation of motion obtained in Problem 7 is 
0.024θ+0.4θ+9θ=0.3 f  
The transfer function obtained in Problem 7 is 
2
( ) 3.6
( ) 0.288 4.8 108
s
F s s s
Θ
=
+ +
 
The state-space equations obtained in Problem 7 is 
1 1
2 2
0 1 0
375 16.67 12.5
x x
u
x x
      
= +      − −      


 [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
The Simulink block diagrams constructed based on the differential equation, the transfer function, the state-space form 
are shown below. Run simulations and the same response in Plot d is obtained. 
 
 
Figure PS5-3 No9a 
 
 
Figure PS5-3 No9b 
 
 
Figure PS5-3 No9c 
205 
 
 
 
Figure PS5-3 No9d 
 
10. Repeat Problem 9 using the linearized differential equation of motion, the transfer function, and the 
state-space form obtained in Problem 8. The parameter values are m = 0.2 kg, M = 0.8 kg, L = 0.6 m, k = 100 N/m, 
and g = 9.81 m/s2. The input force f is the unit-impulse function, which has a magnitude of 20 N and a time 
duration of 0.01 s. 
Solution 
The linearized differential equation of motion obtained in Problem 8 is 
0.168θ+14.47θ=0.6f 
The transfer function obtained in Problem 8 is 
2
( ) 1
( ) 0.28 24.114
s
F s s
Θ
=
+
 
The state-space equations obtained in Problem 7 is 
1 1
2 2
0 1 0
86.12 0 3.57
x x
u
x x
      
= +      −      


 [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
The Simulink block diagrams constructed based on the differential equation, the transfer function, the state-space form 
are similar to those in Problem 9. Run simulations and the same response as shown in the figure below is obtained. 
 
Figure PS5-3 No10 
 
11. Consider the system shown in Figure 5.69, in which the motion of the rod is small angular rotation. When θ = 0, 
the springs are at their free lengths. 
 a. Determine the mass moment of inertia of the rod about point O. Assume that a > b. 
 b. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. 
0 0.2 0.4 0.6 0.8 1
Time (s)
-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
 (r
ad
)
0 0.5 1 1.5 2
Time (s)
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
 (r
ad
)
206 
 
 
 
Figure 5.69 Problem 11. 
Solution 
a. Applying the parallel axis theorem gives the mass moment of inertia of the rod about point O 
( )22 2 2 21 1 1
O C 12 2 3( ) ( ) ( )I I Md M a b M a a b M a b ab= + = + + − + = + − 
b. Applying the moment equation to the fixed point O gives 
+↶: O OαM I∑ = 
1
1 2 O2cosθ ( ( ))sin θ cosθ θk ka f a a b Mg b f I− ⋅ + − + ⋅ − ⋅ =  
where sin θ θ≈ , cosθ 1≈ , and the spring forces are 1 1 θkf k a= and 2 2 θkf k b= . 
Thus, the differential equation of motion for small angles θ is 
2 2 2 21 1
1 23 2( )θ+ θ θ ( ) θ=0M a b ab a k b k a b Mg+ − + − − 
 
Figure PS5-3 No11 
12. Consider the system shown in Figure 5.70, in which a lever arm has a spring–damper combination on the other 
side. When θ = 0, the system is in static equilibrium. 
 a. Assuming that the lever arm can be approximated as a uniform slender rod, determine the mass moment of 
inertia of the rod about point O. 
 b. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. 
 
Figure 5.70 Problem 12. 
Solution 
a. Applying the parallel axis theorem gives the mass moment of inertia of the rod about point O 
( )22 2 271 1 1
O C 12 2 4 48I I Md ML M L L ML= + = + − = 
b. At equilibrium, we have 
207 
 
 
+↷: O 0M∑ = 
31
st4 4 0L Mg L kδ⋅ − ⋅ = 
st3Mg kδ= 
 
Figure PS5-3 No12 
Applying the moment equation to the fixed point O gives 
+↷: O OαM I∑ = 
3 3 1
O4 4 4cosθ cosθ cosθ θk bL f L f L Mg I− ⋅ − ⋅ + ⋅=  
where sin θ θ≈ , cosθ 1≈ , the spring force 
3
st4( θ δ )kf k L= + 
and the damping force 
3
4( θ)bf b L=  
Thus, the differential equation of motion for small angles θ is 
2 2 27 9 9 3 1
st48 16 16 4 4θ θ θ δ =0ML bL kL k L MgL+ + + −  
Substituting the static equilibrium condition gives 
2 2 27 9 9
48 16 16θ θ θ=0ML bL kL+ +  
 
13. Consider the two-degree-of-freedom system shown in Figure 5.71, in which two simple inverted pendulums are 
connected by a translational spring of stiffness k. Each pendulum consists of a point mass m concentrated at the tip 
of a massless rope of length L. θ1 and θ2 are the angular displacements of the pendulums. When θ1 = 0 and θ2 = 0, 
the spring is at its free length. 
 a. Draw the necessary free-body diagrams and derive the differential equations of motion. Assume small angles 
for θ1 and θ2. Provide the equations in the second-order matrix form. 
 b. Using the differential equations obtained in Part (a), determine the state-space representation with the angular 
velocities 1θ and 2θ as the outputs. 
 
Figure 5.71 Problem 13. 
Solution 
a. Assume that 1 2θ θ 0> > . The free-body diagrams are shown in the figure below. Applying the moment equation 
about the fixed points O1 and O2, respectively, gives 
+↷: O OαM I∑ = 
11 1 O 1sin θ cosθ θkL mg L f I− ⋅ − ⋅ =  
22 2 O 2sin θ cosθ θkL mg L f I− ⋅ + ⋅ =  
where 
1 2
2
O OI I mL= = . Assuming small angles, we have cosθ ≈ 1 and 
208 
 
 
( ) ( )1 2 1 2sinθ sin θ θ θkf kL kL= − ≈ − 
Rearranging the equations results in 
2 2
1 1 2 1θ (θ θ ) θ 0mL kL mgL+ − + = 
2 2
2 1 2 2θ (θ θ ) θ 0mL kL mgL− − + = 
which can be written in the second-order matrix form 
2 2 2
11
2 2 2
22
θ 0θ0
θ 0θ0
mL kL mgL kL
mL kL kL mgL
    + −      + =        − +        


 
 
Figure PS5-3 No13 
b. The state and the output are 
 
1 1
2 2 1
3 1 2
4 2
θ
θ θ
,
θ θ
θ
x
x
x
x
   
          = = =     
     
      
x y




 
The state-space representation in matrix form is 
1 1
2 2
3 3
4 4
0 0 1 0
0 0 0 1
0 0
0 0
x x
x xmg kL k
x xmL m
x xk mg kL
m mL
 
    
    
    − −=    
    
   − −    
  




 
1
2
3
4
0 0 1 0
0 0 0 1
x
x
x
x
 
    =   
   
  
y 
 
14. Repeat Problem 13 for the system shown in Figure 5.72, in which each pendulum is a uniform slender rod of mass 
m and length L. The inputs are the torques τ1 and τ2 applied to the pivots O1 and O2, respectively. When θ1 = 0, θ2 
= 0, τ1 = 0, and τ2 = 0, the spring is at its free length. 
 
Figure 5.72 Problem 14. 
Solution 
a. Assume that 1 2θ θ 0> > . The free-body diagrams are shown in the figure below. Applying the moment equation 
about the fixed points O1 and O2, respectively, gives 
+↶: O OαM I∑ = 
1
1 2
1 1 1 O 12 3τ sin θ cosθ θkL mg L f I+ ⋅ − ⋅ =  
2
1 2
2 2 2 O 22 3τ sin θ cosθ θkL mg L f I+ ⋅ + ⋅ =  
209 
 
 
where 
1 2
21
O O 3I I mL= = . Assuming small angles, we have cosθ ≈ 1 and 
( )( ) ( )( )2 2
1 2 1 23 3sinθ sin θ θ θkf k L k L= − ≈ − 
Rearranging the equations results in 
2 21 4 1
1 1 2 1 13 9 2θ (θ θ ) θ τmL kL mgL+ − − = 
2 21 4 1
2 1 2 2 23 9 2θ (θ θ ) θ τmL kL mgL− − − = 
which can be written in the second-order matrix form 
2 2 21 4 1 4
1 113 9 2 9
2 2 21 4 4 1
2 223 9 9 2
θ τ0 θ
θ τ0 θ
mL kL mgL kL
mL kL kL mgL
     − −      + =        − −         


 
 
Figure PS5-3 No14 
b. The state, the input, and the output are 
 
1 1
2 2 1 1
3 1 2 2
4 2
θ
θ τ θ
, ,
θ τ θ
θ
x
x
x
x
   
         = = = =       
      
      
x u y




 
The state-space representation in matrix form is 
1 1
2 2 1
2
3 3 2
4 4
2
0 00 0 1 0
0 00 0 0 1
39 8 4 00 0
6 3
4 9 8 30 0 0
3 6
x x
x x umg kL k
x x umL m mL
x xk mg kL
m mL mL
  
     
             −= +               
   −      
     




 
1
2 1
3 2
4
0 0 1 0 0 0
0 0 0 1 0 0
x
x u
x u
x
 
       = +      
      
  
y 
 
15. Consider the system shown in Figure 5.73, in which a uniform sphere of mass m and radius r rolls along an 
inclined plane of 30°. A translational spring of stiffness k is attached to the sphere. Assuming that there is no 
slipping between the sphere and the surface, draw the necessary free-body diagram and derive the differential 
equation of motion. 
 
Figure 5.73 Problem 15. 
Solution 
The free-body diagram of the system is shown in the figure below, where the normal force N and the friction force f are 
reaction forces at the contact point. Assuming that the sphere rolls down the incline, the spring is in compression and fk 
is the spring force. When the sphere is at the static equilibrium position, we have fk = kδst, where δst is the static 
deformation of the spring. Then, 
+ ↶: IC 0M =∑ 
210 
 
 
stsin 30 · · 0r mg r k° − δ = 
st0.5mg k= δ 
We choose the static equilibrium position as the origin. Because of no slipping, the contact point is the IC. Applying 
the moment equation to the IC gives 
+ ↶: IC ICM I= α∑ 
st ICsin 30 · · ( )r mg r k x I° − + δ = θ 
where 2 2 2 272
IC C 5 5I I md mr mr mr= + = + = . Introducing the static equilibrium condition, 0.5 mg = kδst, and the 
assumption of no slipping, x = rθ, we obtain the differential equation of motion 
2 27
5 0mr krθ + θ = 
 
Figure PS5-3 No15 
 
16. Consider the system shown in Figure 5.74. A uniform solid cylinder of mass m, radius R, and length L is fitted 
with a frictionless axle along the cylinder’s long axis. A spring of stiffness k is attached to a bracket connected to 
the axle. Assume that the cylinder rolls without slipping on a horizontal surface. Draw the necessary free-body 
diagram and derive the differential equation of motion. 
 
Figure 5.74 Problem 16. 
Solution 
The free-body diagram of the system is shown, where the normal force N and the friction force f are reaction forces at 
the contact point. When the cylinder is at the static equilibrium position, δst = 0, where δst is the static deformation of 
the spring. We choose the static equilibrium position as the origin. Because the cylinder rolls without slipping, the 
contact point is the IC. Applying the moment equation to the IC gives 
+ ↷: IC ICM I= α∑ 
IC·R kx I− = θ 
where 2 2 2 231
IC C 2 2I I md mR mR mR= + = + = . Introducing the assumption of no slipping, x = Rθ, we obtain the 
differential equation of motion 
2 23
2 0mR kRθ + θ = 
 
Figure PS5-3 No16 
211 
 
 
Problem Set 5.4 
1. For the pulley system in Example 5.14, draw the free-body diagram and kinematic diagram, and derive the 
equation of motion using the force/moment approach. 
 
Figure 5.82 A pulley system. 
Solution 
The free-body diagram and the kinematic diagram are shown below. 
 
Figure PS5-4 No1 
 
At static equilibrium, we have 
O: 0M+ =∑ 
st· · 0r mg r k− δ = or stmg k= δ 
where δst is the static deformation of the spring. Assume that the block and the pulley are displaced in their positive 
directions. The spring force is fk = k(x + δst). 
For the block (translation only), applying the force equation in the x direction gives 
C: x xx F ma
mg T mx
+ ↓ =
− =
∑

 
For the pulley (rotation only), applying the moment equation about the fixed point O gives 
O O
st O
:
· · ( )
M I
r T r k x I
+ = α
− + δ = θ
∑

 
Combining the two equations and eliminating the unknown T results in 
st O( ) ( )r mg mx rk x I− − + δ = θ 
Introducing the geometric constraint, x = rθ, and the static equilibrium condition, mg = kδst, the equation becomes 
2 2
O( ) 0I mr kr+ θ + θ = 
 
2. The double pulley system shown in Figure 5.89 has an inner radius of r1 and an outer radius of r2. The mass 
moment of inertia of thepulley about point O is IO. A translational spring of stiffness k and a block of mass m are 
suspended by cables wrapped around the pulley as shown. Draw the free-body diagram and kinematic diagram, 
and derive the equation of motion using the force/moment approach. 
212 
 
 
 
Figure 5.89 Problem 2. 
Solution 
The free-body diagram and the kinematic diagram are shown below. 
At static equilibrium, we have stδkf k= and 
+↶: O 0M∑ = 
1 2 stδ 0r mg r k⋅ − ⋅ = 
where stδ is the static deformation of the spring. 
Choose the equilibrium as the origin. Assume that the block and the double pulley are displaced in their positive 
directions. Note that 1θx r= and the spring force ( )2 stθ δkf k r= + . 
 
Figure PS5-4 No2 
 
For the block (translation only), applying the force equation in the x-direction gives 
: x Cxx F ma+ ↓ ∑ = 
1θmg T mx mr− = = 
 
For the double pulley (rotation only), applying the moment equation about the fixed point O yields 
+↶: O OαM I∑ = 
( )1 2 2 st Oθ δ θr T r k r I⋅ − ⋅ + =  
Combining the two equations and eliminating the unknown T results in 
( ) ( )1 1 2 2 st Oθ θ δ θr mg mr r k r I− − + =  
Introducing the static equilibrium condition, 1 2 stδr mg r k= , the equation becomes 
( )2 2
O 1 2θ θ 0I mr kr+ + = 
 
3. Consider the mechanical system shown in Figure 5.90. A disk-shaft system is connected to a block of mass m 
through a translational spring of stiffness k. The elasticity of the shaft and the fluid coupling are modeled as a 
torsional spring of stiffness K and a torsional viscous damper of damping coefficient B, respectively. The radius 
of the disk is r and its mass moment of inertia about point O is IO. Assume that the friction between the block and 
the horizontal surface cannot be ignored and is modeled as a translational viscous damper of damping coefficient 
b. The input to the system is the force f. Draw the necessary free-body diagram and the kinematic diagram, and 
derive the equations of motion. 
213 
 
 
 
Figure 5.90 Problem 3. 
Solution 
This is a mixed system, where the mass block undergoes translational motion along x-direction and the motion of the 
disk is pure rotation about point O that is fixed. Choose the displacement of the block x and the angular displacement 
of the disk θ as the generalized coordinates. Assuming that the block and the disk are displaced in their positive 
directions and rθ > x > 0, the spring is in tension and the magnitude of the spring force is fk = k(rθ − x). 
 
The free-body diagram and the kinematic diagram of the system are shown in the figure blow, where W is the weight 
of the disk and N is the normal force supporting the block. 
 
Figure PS5-4 No3 
 
For the block (translation only), applying the force equation in the x direction gives 
C:
( )
x xx F ma
k r x bx mx
+ ← =
θ − − =
∑
 
 
For the disk (rotation only), applying the moment equation about the fixed point O gives 
+ ↷: O OM I= α∑ 
O· ( ) ·K B r k r x r f I− θ − θ − θ − + = θ  
Rearranging the two equations gives 
0mx bx kx kr+ + − θ = 
 2
O ( )I B K kr krx rfθ + θ + + θ − =  
 
4. Consider the mechanical system shown in Figure 5.91, where the motion of the rod is small angular rotation. 
When θ = 0 and f = 0, the deformation of each spring is zero and the system is at static equilibrium. Assume that 
the friction between the block of mass m1 and the horizontal surface cannot be ignored and is modeled as a 
translational viscous damper of damping coefficient b. 
 a. Assuming that a > c > 0, determine the mass moment of inertia of the rod about the pivot point O. 
 b. Draw the necessary free-body diagram and the kinematic diagram, and derive the equations of motion for 
small angles. 
214 
 
 
 
Figure 5.91 Problem 4. 
Solution 
a. The mass moment of inertia of the rod about point O can be found by applying the parallel axis theorem 
( )22 2 2 21 1 1
O C 12 2 3( ) ( ) ( )I I Md M a c M a a c M a c ac= + = + + − + = + − 
b. Choose the displacements of the blocks, x1 and x2, and the angular displacement of the rod θ as the generalized 
coordinates. Assume that the blocks and the rod are displaced in their positive directions, x1 > aθ > 0, and cθ > x2 
> 0. Both springs are in tension and the magnitudes are 
1 1 1( θ)kf k x a= − 
2 2 2( θ )kf k c x= − 
The free-body diagram and the kinematic diagram of the system are shown in the figure below. 
 
Figure PS5-4 No4 
For the blocks (translation only), applying the force equation in the x direction gives 
Mass 1: C: x xx F ma+ → =∑ 
1 1 1 1 1( )f k x a bx m x− − θ − =  
Mass 2: C: x xx F ma+ ← =∑ 
2 2 2 2( )k c x m xθ − =  
 For the rod (rotation only), applying the moment equation about the fixed point O gives 
+ ↷: O OM I= α∑ 
1
1 1 2 2 O2cos ( ) ( ( ))sin cos ( )a k x a a a c Mg c k c x Iθ⋅ − θ + − + θ⋅ − θ⋅ θ − = θ 
where sin θ θ≈ and cosθ 1≈ for small angles. Substituting the mass moment of inertia IO and rearranging the 
three equations give 
215 
 
 
1 1 1 1 1 1m x bx k x k a f+ + − θ =  
2 2 2 2 2 0m x k x k c+ − θ =
 2 2 2 21 1
1 2 1 1 2 23 2( ) ( ( ) ) 0M a c ac k a k c a c Mg k ax k cx+ − θ + + − − θ − − = 
 
5. Consider the mechanical system shown in Figure 5.92, in which a simple pendulum is pivoted on a cart of mass m 
and is constrained to rotate in a vertical plane. The pendulum consists of a point mass M concentrated at the tip of 
a massless rod of length L. Assume that the friction between the cart and the horizontal surface cannot be ignored. 
Denote the displacement of the cart as x and the angular displacement of the pendulum as θ. Draw the necessary 
free-body diagram and kinematic diagram, and derive the equations of motion for small angles. 
 
Figure 5.92 Problem 5. 
Solution 
The free-body diagram and the kinematic diagram are shown in the figure below. 
 
Figure PS5-4 No5 
Applying the force equation to the whole system along the x-direction gives 
( )
2
1
: x i Ci x
i
x F m a
=
+ → ∑ = ∑ 
2θ cosθ θ sin θbx kx mx Mx ML ML− − = + + ⋅ − ⋅ 
   
Applying the moment equation to the pendulum about the pivot point P results in 
+↷: _α
CP C eff mM I M∑ = + a 
sin θ cosθ θL Mg L Mx L ML⋅ = ⋅ + ⋅ 
 
Rearranging the two equations gives 
2( ) θ cosθ θ sin θ 0m M x ML ML bx kx+ + − + + = 
  
2cosθ θ sin θ 0MLx ML MgL+ − =
 
For small angular motions, we have cosθ 1≈ , sin θ θ≈ , 2θ θ 0≈ . Thus, 
( ) θ 0m M x ML bx kx+ + + + =
  
2θ θ 0MLx ML MgL+ − =
 
 
6. Consider the mechanical system shown in Figure 5.93. The inputs are the force f1 applied to the cart and the force 
f2 applied at the tip of the rod. The outputs are the displacement x of the cart and the angular displacement θ of the 
pendulum. 
216 
 
 
 a. Draw the free-body diagram and kinematic diagram, and derive the equations of motion for small angles. 
 b. Using the differential equation obtained in Part (a), determine the state-space representation. 
 
Figure 5.93 Problem 6. 
Solution 
a. The free-body diagram and the kinematic diagram are shown in the figure below. 
 
Figure PS5-4 No6 
Applying the force equation to the whole system along the x-direction gives 
( )
2
1
: x i Ci x
i
x F m a
=
+ → ∑ = ∑ 
2
1 2 2 2cosθ θ cosθ θ sin θL Lf kx f mx Mx M M− + = + + ⋅ − ⋅ 
  
Applying the moment equation to the pendulum about the pivot point P results in 
+↷: _α
CP C eff mM I M∑ = + a 
2 C2 2 2 2sin θ θ cosθ θL L L LMg Lf I Mx M⋅ + = + ⋅ + ⋅ 
 
where 21
C 12I ML= for a uniform thin rod. Rearranging the two equations gives 
21 1
1 22 2( ) θcosθ θ sin θ cosθm M x ML ML kx f f+ + − + = + 
 
21 1 1
22 3 2cosθ θ sin θMLx ML MgL Lf+ − =
 
For small angular motions, we have cosθ 1≈ , sin θ θ≈ , 2θ θ 0≈ . Thus, 
1
1 22( ) θm M x ML kx f f+ + + = +
 
21 1 1
22 3 2θ θMLx ML MgL Lf+ − =
 
b. The state, the input, and the output are specified as 
217 
 
 
1
2 1
32
4
θ
θ
θ
x x
x f x
x fx
x
   
          = = = =       
     
     
x u y


 
We then take the time derivative of each state variable. For the first two, 
 1 3x x x= =  
2 4θx x= = 
To find 3x and 4x , we rewrite the two linearized differential equations as 
 
1
1 22
2 11 1
2 22 3 θθ
f f kxm M ML x
Lf MgLML ML
+ −+    
=     +    


 
from which x and θ can be solved using Cramer’s rule, 
1 24 3 θ 4 2
4
kx Mg f fx
M m
− − + −
=
+
 
( ) ( )1 26 6 θ 6 6 12
θ
( 4 )
Mkx M m Mg Mf M m f
ML M m
+ + − + +
=
+
 
Thus, the state-space equation in matrix form is 
1 1
2 2 1
3 3 2
4 4
0 0 1 0 0 0
0 0 0 1 0 0
4 3 4 20 0
4 4 4 4
6 6( ) 6 6 120 0
( 4 ) ( 4 ) ( 4 ) ( 4 )
x x
x x fk Mg
x x fM m M m M m M m
k M m g M mx x
L M m L M m L M m ML M m
   
      
              = +− − −        + + + +        
   + +       −   + + + +   




 
1
2 1
3 2
4
1 0 0 0 0 0
0 1 0 0 0 0
x
x f
x f
x
 
       = +      
      
  
y 
 
7. For the mechanical system in Problem 2, use the energy method to derive the equation of motion. 
Solution 
The static equilibrium position and the deformed positions are shown in the figure below. At equilibrium, we have 
1 2 stδr mg r k= . Assuming that the block and the pulley move along the direction shown gives 
 2 2 2 2 21 1 1 1
block pulley O 1 O2 2 2 2θ θ θT T T mx I mr I= + = + = +  
 
( )21
1 2 st2θ θ δg eV V V mgr k r= + = − + + 
where the static equilibrium position is chosen as the datum for the gravitational potential energy. Because of the static 
equilibrium condition, the total potential energy becomes 
2 2 21 1
2 st2 2θ δV kr k= + 
The total energy of the system is 
2 2 2 2 2 21 1 1 1
1 O 2 st2 2 2 2θ θ θ δT V mr I kr k+ = + + +  
for which the time derivative is 
( ) 2 2
1 O 2θθ θθ θθd T V mr I kr
dt
+ = + +   
Applying the principle of conservation of energy gives 
( )2 2
1 O 2θ θ 0mr I kr+ + = 
218 
 
 
 
Figure PS5-4 No7 
 
8. For the mechanical system in Problem 11 of Problem Set 5.3, use the energy method to derive the equation of 
motion. 
Solution 
 
Figure PS5-4 No8 
The system is only subjected to the gravitational and spring forces, and it is a conservative system. 
When θ = 0, the springs are at their free length. Assuming that the rod rotates along the positive direction by a small 
angle θ gives 
 21
O2 θT I=  
2 21 1 1
1 22 2 2( ) cosθ ( θ) ( θ)g eV V V Mg a b k a k b= + = ⋅ − + + 
The total energy of the system is 
2 2 2 2 21 1 1 1
O 1 22 2 2 2θ ( )cosθ θ θT V I Mg a b k a k b+ = + − + + 
for which the time derivative is 
( ) 2 21
O 1 22θθ ( )sin θθ θθ θθd T V I Mg a b k a k b
dt
+ = + − + +    
Applying the principle of conservation of energy gives 
2 21
O 1 22θ ( )sin θ θ θ=0I Mg a b k a k b+ − + + 
where the mass moment of inertia of the rod about pivot O is 
2 21
O 3 ( )I M a b ab= + − 
 
9. Repeat Problem 13 of Problem Set 5.3 using Lagrange’s equations. 
Solution 
The system is only subjected to the gravitational and spring forces, and it is a conservative system. 
219 
 
 
 
Figure PS5-4 No9 
The kinetic energy of the system is 
2 2 2 21 1
ball1 ball2 1 22 2θ θT T T mL mL= + = +  
Assume that 1 2θ θ 0> > . Then, the spring is under elongation and the elastic potential energy is 
21
e 1 22 ( sin θ sin θ )V k L L= − 
Using the datum defined in the figure, we can obtain the potential energy 
 g 1 2 1 2cosθ cosθV mgh mgh mgL mgL= − − = − − 
The total potential energy is 
2 21
e g 1 2 1 22 (sin θ sin θ ) (cosθ cosθ )V V V kL mgL= + = − − + 
Apply Lagrange's equations 
0 1,2
i i i
d T T V i
dt q q q
 ∂ ∂ ∂
− + = = ∂ ∂ ∂ 
 
For 1i = , 1 1θq = , 1 1θq =  , 2
1
1
θ
θ
T mL∂
=
∂


 
2
1
1
θ
θ
d T mL
dt
 ∂
= 
∂ 


 
1
0
θ
T∂
=
∂
 
2
1 2 1 1
1
(sin θ sin θ )cosθ sin θ
θ
V kL mgL∂
= − +
∂
 
Substituting into Lagrange's equation results in 
2 2
1 1 2 1 1
1 11
θ (sin θ sin θ )cosθ sin θ 0
θ θθ
d T T V mL kL mgL
dt
 ∂ ∂ ∂
− + = + − + =  ∂ ∂∂ 


 
Similarly, for 2i = , 2 2θq = , 2 2θq =  , 2
2
2
θ
θ
T mL∂
=
∂


 
2
2
2
θ
θ
d T mL
dt
 ∂
= 
∂ 


 
2
0
θ
T∂
=
∂
 
2
1 2 2 2
2
(sin θ sin θ )( cosθ ) sin θ
θ
V kL mgL∂
= − − +
∂
 
Substituting into Lagrange's equation gives 
2 2
2 1 2 2 2
2 22
θ (sin θ sin θ )cosθ sin θ 0
θ θθ
d T T V mL kL mgL
dt
 ∂ ∂ ∂
− + = − − + =  ∂ ∂∂ 


 
For small motions, the two differential equations of motion after linearization are 
220 
 
 
2 2
1 1 2 1θ (θ θ ) θ 0mL kL mgL+ − + = 
2 2
2 1 2 2θ (θ θ ) θ 0mL kL mgL− − + = 
 
10. Repeat Problem 14 of Problem Set 5.3 using Lagrange’s equations. 
Solution 
Note that the system is subjected to external torques, τ1 and τ2, which are nonconservative. 
 
Figure PS5-4 No10 
The kinetic energy of the system is 
1 2
2 21 1
rod1 rod2 O 1 O 22 2θ θT T T I I= + = +  
where 
1 2
21
O O 3I I ML= = . Assume that 1 2θ θ 0> > . Then, the spring is under elongation and the elastic potential 
energy is 
21 2 2
e 1 22 3 3( sin θ sin θ )V k L L= − 
Using the datum defined in the figure, we can obtain the potential energy 
 1 1
g 1 2 1 22 2( cosθ ) ( cosθ )V mgh mgh mg L mg L= + = + 
The total potential energy is 
2 22 1
e g 1 2 1 29 2(sin θ sin θ ) (cosθ cosθ )V V V kL mgL= + = − + + 
Apply Lagrange's equations 
1,2i
i i i
d T T V Q i
dt q q q
 ∂ ∂ ∂
− + = = ∂ ∂ ∂ 
 
For 1i = , 1 1θq = , 1 1θq =  , 21
13
1
θ
θ
T mL∂
=
∂


 
21
13
1
θ
θ
d T mL
dt
 ∂
= 
∂ 


 
1
0
θ
T∂
=
∂
 
24 1
1 2 1 19 2
1
(sin θ sin θ )cosθ sin θ
θ
V kL mgL∂
= − −
∂
 
( ) ( ) ( ) ( )
2
1 1 1 2 2 1
1 1 11
j
j
j
Q
=
∂ ∂ ∂
= = τ θ + τ θ = τ
∂θ ∂θ ∂θ∑
r
F k k k k 
Substituting into Lagrange's equation results in 
2 21 4 1
1 1 2 1 1 13 9 2
1 11
θ (sin θ sin θ )cosθ sin θ τ
θ θθ
d T T V mL kL mgL
dt
 ∂ ∂ ∂
− + = + − − =  ∂ ∂∂ 


 
Similarly, for 2i = , 2 2θq = , 2 2θq =  , 21
23
2
θ
θ
T mL∂
=
∂


 
221 
 
 
21
23
2
θ
θ
d T mL
dt
 ∂
= 
∂ 


 
2
0
θ
T∂
=
∂
 
24 1
1 2 2 29 2
2
(sin θ sin θ )( cosθ ) sin θ
θ
V kL mgL∂
= − − −
∂
 
( ) ( ) ( ) ( )
2
2 1 1 2 2 2
2 2 21
j
j
j
Q
=
∂ ∂ ∂
= = τ θ + τ θ = τ
∂θ ∂θ ∂θ∑
r
F k k k k 
Substituting into Lagrange's equation gives 
2 21 4 1
2 1 2 2 2 23 9 2
2 22
θ (sin θ sin θ )cosθ sin θ τ
θ θθ
d T T V mL kL mgL
dt
 ∂ ∂ ∂
− + = − − − =  ∂ ∂∂ 


 
For small motions, the two differential equations of motion after linearization are 
2 21 4 1
1 1 2 1 13 9 2θ (θ θ ) θ τmL kL mgL+ − − = 
2 21 4 1
2 1 2 2 23 9 2θ (θ θ ) θ τmL kL mgL− − − = 
 
11. Repeat Problem 5 using Lagrange’s equations. 
Solution 
The system is subjected to a damping force, and it is a nonconservative system. 
 
Figure PS5-4 No11 
The kinetic energy of the system is 
2 21 1
block ball 2 2T T T mx Mv= + = + 
where 
2 2 2( θ) 2 ( θ)cosθv x L x L= + + 
  
Thus, we have 
2 2 21 1
2 2( ) θ θcosθT m M x ML MLx= + + + 
  
Using the datum defined in the figure, we can obtain the potential energy 
 2 21 1
e g 2 2 cosθV V V kx Mgh kx MgL= + = + = + 
We then apply Lagrange's equations 
1,2i
i i i
d T T V Q i
dt q q q
 ∂ ∂ ∂
− + = = ∂ ∂ ∂ 
 
For 1i = , 1q x= , 1q x=  , ( ) θcosθT m M x ML
x
∂
= + +
∂



 
( ) θcosθ θsinθ θd T m M x ML ML
dt x
∂  = + + − ⋅ ∂ 
  


 
0T
x
∂
=
∂
 
V kx
x
∂
=
∂
 
222 
 
 
( ) ( )1Q bx x bx
x x
∂ ∂
= = − = −
∂ ∂
rF i i  
Substituting into Lagrange's equation results in 
2( ) θ cosθ θ sin θd T T V m M x ML ML kx bx
dt x x x
∂ ∂ ∂  − + = + + − + = − ∂ ∂ ∂ 
 
  
Similarly, for 2i = , 2 θq = , 2 θq =  , 2θ cosθ
θ
T ML MLx∂
= +
∂



 
2θ cosθ sin θ θ
θ
d T ML MLx MLx
dt
∂  = + − ⋅ ∂ 
 
 

 
θsinθ
θ
T MLx∂
=
∂

 
sin θ
θ
V MgL∂
= −
∂
 
( ) ( )2 0Q bx x∂ ∂
= = − =
∂θ ∂θ
rF i i 
Substituting into Lagrange's equation gives 
2θ cosθ sin θ 0
θ θ θ
d T T V ML MLx MgL
dt
∂ ∂ ∂  − + = + − = ∂ ∂ ∂ 



 
For small motions, the equations of motion after linearization are 
( ) θ 0m M x ML bx kx+ + + + =
  
2θ θ 0ML MLx MgL+ − =
 
 
12. Repeat Problem 6 using Lagrange’s equations. 
Solution 
The system is subjected to two external forces, f1 and f2, which are nonconservative. 
 
Figure PS5-4 No12 
The kinetic energy of the system is 
21 1 1
cart rod
2
2
2
2 2 θC CT T T mx Mv I+ + += = 
 
where 21
12CI ML= and 
2 2
2 2( θcosθ) θsinθL L
Cv x= + +  
Thus, we have 
2 2 21 1 1
2 6 2( ) θ θ cosθT m M x ML MLx= + + + 
  
Using the datum defined in the figure, we can obtain the potential energy 
 2 21 1 1
e g 2 2 2 cosθV V V kx Mgh kx MgL= + = + = + 
We then apply Lagrange's equations 
223 
 
 
1, 2i
i i i
d T T V Q i
dt q q q
 ∂ ∂ ∂
− + = = ∂ ∂ ∂ 
 
For 1i = , 1q x= , 1q x=  , 1
2( ) θ cosθT m M x ML
x
∂
= + +
∂



 
1 1
2 2( ) θcosθ θsinθ θd T m M x ML ML
dt x
∂  = + + − ⋅ ∂ 
  


 
0T
x
∂
=
∂
 
V kx
x
∂
=
∂
 
( ) ( ) ( ) ( )
( ) ( )
2
1 1 2 2
1
1 2 2 1 2
cos sin sin cos
cos sin cos
j
j
j
Q f x f f x L L
x x x
f f f f f
=
∂ ∂ ∂
= = + θ + θ + θ − θ
∂ ∂ ∂
= + θ + θ = + θ
∑
r
F i i i j i i j
i i i j i
 
Substituting into Lagrange's equation results in 
21 1
12 2 2( ) θ cos coθ θ sin sθθd T T V m M x ML ML kx f
dt x x x
f∂ ∂ ∂  − + = + + − + = + ∂ ∂ ∂ 
 
 
Similarly, for 2i = , 2 θq = , 2 θq =  , 21 1
3 2θ cosθ
θ
T ML MLx∂
= +
∂



 
21 1 1
3 2 2θ cosθ sin θ θ
θ
d T ML MLx MLx
dt
∂  = + − ⋅ ∂ 
 
 

 
1
2 θsin θ
θ
T MLx∂
= −
∂

 
1
2 sin θ
θ
V MgL∂
= −
∂
 
( ) ( ) ( ) ( )
( ) ( )( ) ( )
2
2 1 2 2
1
2 2
1 2 2 2 2
cos sin sin cos
0 cos sin cos sin cos sin
j
j
j
Q f x f f x L L
f f f L L f L f L
=
∂ ∂ ∂
= = + θ + θ + θ − θ
∂θ ∂θ ∂θ
= + θ + θ θ + θ = θ + θ =
∑
r
F i i i j i i j
i i j i j
 
Substituting into Lagrange's equation gives 
21 1 1
23 2 2θ cosθ sin θ
θ θθ
d T T V ML MLx MgL f L
dt
∂ ∂ ∂  − + = + − =  ∂ ∂∂ 



 
For small motions, the equations of motion after linearization are 
1
1 22( ) θm M x ML kx f f+ + + = +
 
21 1 1
23 2 2θ θML MLx MgL f L+ − =
 
 
13. A robot arm consists of rigid links connected by joints allowing the relative motion of neighboring links. The 
dynamic model for a robot arm can be derived using Lagrange’s equations 
d , 1,2, , ,
d i
i ii
T T V i n
t
 ∂ ∂ ∂
− + = τ = …  ∂θ ∂θ∂θ 
 
 where θi is the angular displacement of the ith joint, τi is the torque applied to the ith joint, and n is the total 
number of joints. Consider a single-link planar robot arm as shown in Figure 5.94. Use Lagrange’s equations to 
derive the dynamic model of the robot arm. Assume that the motion of the robot arm is constrained in a vertical 
plane, and the joint angle varies between 0° and 360°. 
224 
 
 
 
Figure 5.94 Problem 13. 
Solution 
The kinetic energy of the robot arm is 
2 2 2 2 21 1 1 1
O2 2 3 6θ ( )θ θT I mL mL= = =   
Using the datum defined in the figure, we can obtain the potential energy 
 1
2 cosθgV V mgh mg L= = = ⋅ 
 
Figure PS5-4 No13 
We then apply Lagrange's equations 
d
d
T T V
t
∂ ∂ ∂  − + = τ  ∂θ ∂θ∂θ 
 
where 
 21
3 θ
θ
T mL∂
=
∂


 
21
3 θ
θ
d T mL
dt
∂  = ∂ 


 
0
θ
T∂
=
∂
 
1
2 sin θ
θ
V mgL∂
= −
∂
 
Substituting into Lagrange's equation results in 
21 1
3 2θ sin θ τ
θ θ θ
d T T V mL mgL
dt
∂ ∂ ∂  − + = − = ∂ ∂ ∂ 


 
 
14. Repeat Problem 13 for a two-link planar robot arm as shown in Figure 5.95. Assume that the motion of the robot 
arm is constrained in a horizontal plane, and the joint angles vary between 0° and 360°. 
 
Figure 5.95 Problem 14. 
Solution 
The motion of the robot arm is assumed to be constrained in a horizontal plane. Thus, the gravitational potential energy 
does not change ( θ 0iV∂ ∂ = ) and the Lagrange’s equations can be rewritten as 
d , 1,2, , ,
d i
ii
T T i n
t
 ∂ ∂
− = τ = … 
∂θ∂θ 
 
225 
 
 
The kinematic diagram is shown below. 
 
Figure PS5-4 No14 
The kinetic energy of the robot arm is 
link1 link2T T T= + 
Note that link 1 rotates about a fixed axis through the point O, and 
2 2 2 2 21 1 1 1
link1 1O 1 1 12 2 3 6θ ( )θ θT I mL mL= = =   
The link 2 undergoes general plane motion, and 
2 2 2 2 21 1 1 1 1
link2 2 2 2 2 1 22 2 2 2 12ω ( )(θ θ )C C CT mv I mv mL= + = + +  
where v2C is the velocity at the center of gravity of link 2. Denote the velocity at the end of link 1 as v1E and the relative 
velocity of the center of gravity of link 2 with respect to the end of link 1 as vr, we can obtain 
( ) ( ) ( ) ( ) ( )2 22 2 2 1 1
2 1 1 2 1 2 1 1 2 1 22 22 cos π θ θ θ θ 2 θ θ θ cosθC r E r Ev v v v v L L L L   = + − − = + + + +   
      
using the law of cosine. Thus, the total kinetic energy is 
( ) ( )
( ) ( )
22 2 2 2 2 2 2 21 1 1 1
1 1 2 1 1 1 2 2 1 26 2 4 24
22 2 2 2 22 1 1
1 1 2 1 1 2 23 6 2
θ θ θ θ θ θ θ cosθ (θ θ )
θ θ θ θ θ θ cosθ
T mL m L L L mL
mL mL mL
 = + + + + + + +  
= + + + +
        
     
 
We then apply Lagrange's equations. For 1i = , 
 ( ) ( )2 2 24 1 1
1 1 2 1 2 23 3 2
1
θ θ θ 2θ θ cosθ
θ
T mL mL mL∂
= + + + +
∂
    

 
( ) ( ) ( )2 2 2 24 1 1 1
1 1 2 1 2 2 1 2 2 23 3 2 2
1
θ θ θ 2θ θ cosθ 2θ θ sin θ θ
θ
d T mL mL mL mL
dt
 ∂
= + + + + − + ⋅ ∂ 
       

 
1
0
θ
T∂
=
∂
 
Substituting into Lagrange's equation results in 
( ) ( ) ( )
( )
1 1
2 2 2 24 1 1 1
1 1 2 1 2 2 1 2 2 23 3 2 2
2 2 2 25 1 1 1
2 1 2 2 1 2 2 2 13 3 2 2
θ θ
θ θ θ 2θ θ cosθ 2θ θ sin θ θ
( cosθ ) θ ( cosθ ) θ 2θ θ θ sin θ τ
d T T
dt
mL mL mL mL
mL mL mL
 ∂ ∂
− ∂ ∂ 
= + + + + − + ⋅
= + + + − + =

       
    
 
Similarly, for 2i = , 
( )2 21 1
1 2 1 23 2
2
θ θ θ cosθ
θ
T mL mL∂
= + +
∂
  

 
( )2 2 21 1 1
1 2 1 2 1 2 23 2 2
2
θ θ cos θ sin θ θ
θ
d T mL mL mL
dt
θ θ
 ∂
= + + − ⋅ ∂ 
   

 
( )2 21
1 1 2 22
2
θ θ θ sin θ
θ
T mL∂
= − +
∂
   
Substituting into Lagrange's equation gives 
226 
 
 
( ) ( )
2 2
2 2 2 2 21 1 1 1
1 2 1 2 1 2 2 1 1 2 23 2 2 2
2 2 2 21 1 1 1
2 1 2 1 2 23 2 3 2
θ θ
θ θ θ cosθ θ sin θ θ θ θ θ sin θ
( cosθ ) θ θ θ sin θ τ
d T T
dt
mL mL mL mL
mL mL mL
 ∂ ∂
− ∂ ∂ 
= + + − ⋅ + +
= + + + =

       
  
 
Thus, the two differential equations of motion are 
( )2 2 2 25 1 1 1
2 1 2 2 1 2 2 2 13 3 2 2( cosθ ) θ ( cosθ ) θ 2θ θ θ sin θ τmL mL mL+ + + − + =     
2 2 2 21 1 1 1
2 1 2 1 2 23 2 3 2( cosθ ) θ θ θ sin θ τmL mL mL+ + + =   
 
Problem Set 5.5 
1. Repeat Example 5.18, and determine a mathematical model for the simple one-degree-of-freedom system shown 
in Figure 5.96a in the form of a differential equation of motion in θ2. 
Solution 
As shown in Example 5.18, the differential equation of motion in 1θ is 
2
1
C1 C2 1 12
2
rI I
r
 
+ θ = τ  
 
 
Introducing the geometric constraint 2 1 1 2r rθ θ = , the above differential equation can be rewritten as 
2
1 2
C1 C2 2 12
12
r rI I
rr
   
+ θ = τ      
 
which can be simplified as 
2
2 2
C1 C2 2 12
11
r rI I
rr
 
+ θ = τ  
 
 
If the gears have negligible inertia or zero angular acceleration, and if the energy loss due to the friction between the 
gear teeth can be neglected, we have 
1 2 1
2 1 2
τ θ
τ θ
r
r
= = 
Thus, the dynamics seen from the output side is 
2
2
C1 C2 2 22
1
r I I
r
 
+ θ = τ  
 
 
 
2. Repeat Example 5.19, and determine a mathematical model for the single-link robot arm shown in Figure 5.97 in 
the form of a differential equation of motion in the load variable θ. 
Solution 
As shown inExample 5.19, the differential equation of motion in terms of the angular displacement of the motor is 
2 2
m m m m m( ) ( )I N I B N B+ θ + + θ = τ  
By the geometry of the gears, 
1
m 2
θ
θ
r N
r
= = 
the above differential equation can be rewritten in the load variable θ, 
2 2
m m m( ) ( )I N I B N B
N N
θ θ
+ + + = τ
 
 
which can be simplified as 
m m m2 2
1 1 1I I B B
NN N
   + θ + + θ = τ   
   
  
If the gears have negligible inertia or zero angular acceleration, and if the energy loss due to the friction between the 
227 
 
 
gear teeth can be neglected, we have 
m 1
m 2
τ θ
τ θ
r N
r
= = = 
Thus, the dynamics seen from the output side is 
m m2 2
1 1 1I I B B
NN N
   + θ + + θ = τ   
   
  
 
3. Consider the one-degree-of-freedom system shown in Figure 5.99. The system consists of two gears of mass 
moments of inertia I1 and I2 and radii r1 and r2, respectively. The applied torque on gear 1 is τ1. Assume that the 
gears are connected with flexible shafts, which can be approximated as two torsional springs of stiffnesses K1 and 
K2, respectively. 
 a. Draw the necessary free-body diagrams, and derive the differential equation of motion in θ1. 
 b. Using the differential equation obtained in Part (a), determine the transfer function Θ2(s)/T1(s). 
 c. Using the differential equation obtained in Part (a), determine the state-space representation with θ2 as the 
output. 
 
Figure 5.99 Problem 3. 
Solution 
a. The free-body diagram is shown in the figure below. 
 
Figure PS5-5 No3 
Applying the moment equation to each gear gives 
 +↶: O OαM I∑ = 
Gear 1: 1 1 1 1 1 1τ θ θK r F I− − =  
Gear 2: 2 2 2 2 2θ θK r F I− = −  
Combining the two equations and eliminating F yields 
1
1 1 1 2 2 2 2 1 1
2
θ ( θ θ ) θrK I K I
r
τ − − + =  
By the geometry of the gears, 2 1 2 1θ ( )θr r= . Substituting it into the previous differential equation gives 
2 2
1 1
1 2 1 1 2 1 12 2
2 2
θ θ τr rI I K K
r r
   
+ + + =   
   
 
228 
 
 
 
b. Assuming zero initial conditions and taking Laplace transform of the above differential equation, we have 
1
2 2
21 1 1
1 2 1 22 2
2 2
( ) 1
( )
s
s r rI I s K K
r r
Θ
=
Τ    
+ + +   
   
 
Applying the geometric constraint gives 
1
2 1 1 2
2 2
21 2 1 1 1
1 2 1 22 2
2 2
( ) ( )
( ) ( )
r
s r s r
s r s r rI I s K K
r r
Θ Θ
= =
Τ Τ    
+ + +   
   
 
c. The state, the input, and the output are 
 1 1
1 2
2 1
θ
, τ , θ
θ
x
u y
x
   
= = = =   
   
x

 
The state equation and the output equation in matrix form are 
1 12 2 2
2 1 1 2 2
2 22 2 2 2
2 1 1 2 2 1 1 2
0 1 0
0
x x
ur K r K rx x
r I r I r I r I
   
      = ++      −      + +   


 
11
22
0 0
xry u
xr
   
= + ⋅  
  
 
 
4. Consider the gear–train system shown in Figure 5.100. The system consists of a rotational cylinder and a pair of 
gears. The gear ratio is N = r1/r2. The applied torque on the cylinder is τa. Assume that the gears are connected with 
flexible shafts, which can be approximated as two torsional springs of stiffnesses K1 and K2, respectively. 
 a. Draw the necessary free-body diagrams, and derive the differential equations of motion. 
 b. Using the differential equation obtained in Part (a), determine the state-space representation. Use θa, θ1, ωa, 
and ω1 as the state variables, and use θ2 and ω2 as the output variables. 
 
Figure 5.100 Problem 4. 
Solution 
a. Assume that a 1θ θ 0> > . The free-body diagram is shown in the figure below. 
 
229 
 
 
Figure PS5-5 No4 
Applying the moment equation to each gear gives 
 +↶: O OαM I∑ = 
Gear 1: ( )a 1 a 1 aθ θ θK Iτ − − =  
Gear 2: ( )1 a 1 1 1 1θ θ θK r F I− − =  
Gear 3: 2 2 2 2 2θ θr F K I− + = −  
Combining the last two equations and eliminating F yields 
( ) 1
1 a 1 2 2 2 2 1 1
2
θ θ ( θ θ ) θ
rK I K I
r
− − + =  
By the geometry of the gears, we have ( )2 1 2 1 1θ θ θr r N= = . Substituting it into the previous equation gives 
( ) ( )2 2
1 2 1 1 a 1 2 1θ θ θ 0I N I K K N K+ − + + = 
The differential equations of motion for the system are 
a 1 a 1 1 aθ θ θ τI K K+ − = 
( ) ( )2 2
1 2 1 1 a 1 2 1θ θ θ 0I N I K K N K+ − + + = 
b. The state, the input, and the output are 
 
1 a
2 1 2
3 a 2
4 1
θ
θ θ
, τ,
ω ω
ω
x
x
u
x
x
   
        = = = =     
    
      
x y 
The state equation in matrix form is 
 
1 1
2 21 1
3 3
2
1 1 24 4
2 2
1 2 1 2
0 0 1 0
0
0 0 0 1
0
0 0 1
0 0 0
x x
x xK K u
x xI I
IK K N Kx x
I N I I N I
 
                 = +−      
     
     +   −     + + 




 
By the geometry of the gears, we have 2 1θ θN= and 2 1ω ωN= . Thus, the output equation in matrix form is 
 
1
2
3
4
0 0 0 0
0 0 0 0
x
xN
u
xN
x
 
     = +    
    
  
y 
 
5. A three-degree-of-freedom gear–train system is shown in Figure 5.101, which consists of four gears of moments 
of inertia I1, I2, I3, and I4. Gears 2 and 3 are meshed and their radii are r2 and r3, respectively. Gears 1 and 2 are 
connected by a relatively long shaft, and gears 3 and 4 are connected in the same way. The shafts are assumed to 
be flexible, and can be approximated by torsional springs. The applied torque and load torque are τa and τl on gear 
1 and gear 4, respectively. The gears are assumed to be rigid and have no backlash. Derive the differential 
equations of motion. 
 
230 
 
 
Figure 5.101 Problem 5. 
Solution 
Assume that 1 2θ θ 0> > and 3 4θ θ 0> > . The free-body diagram is shown in the figure below. 
 
Figure PS5-5 No5 
Applying the moment equation to each gear gives 
 +↶: O OαM I∑ = 
Gear 1: ( )a 1 1 2 1 1τ θ θ θK I− − =  
Gear 2: ( )1 1 2 2 2 2θ θ θK r F I− − =  
Gear 3: ( )3 2 3 4 3 3θ θ θr F K I− + − = −  
Gear 4: ( )2 3 4 4 4τ θ θ θl K I− − − = −  
Combining the equations of gears 2 and 3 and eliminating F yields 
( ) ( )2 2
1 1 2 3 3 2 3 4 2 2
3 3
θ θ θ θ θ θ
r rK I K I
r r
− − − − =  
By the geometry of the gears, we have ( )3 2 3 2θ θr r= . Substituting it into the previous differential equation gives 
2 2
2 2 2
2 3 2 1 1 1 2 2 2 42 2
33 3
θ θ θ θ 0
r r rI I K K K K
rr r
   
+ − + + − =   
   
 
Similarly, the equation of gear 4 becomes 
2
4 4 2 2 2 4
3
θ θ θ τl
rI K K
r
− + = 
Combining with the equation of gear 1 gives the differential equations of motion for the system 
1 1 1 1 1 2 aθ θ θ τI K K+ − = 
2 2
2 2 2
2 3 2 1 1 1 2 2 2 42 2
33 3
θ θ θ θ 0
r r rI I K K K K
rr r
   
+ − + + − =   
   
 
2
4 4 2 2 2 4
3
θ θ θ τl
rI K K
r
− + = 
 
6. Repeat Problem 5. Assume that the shaft connecting gears 1 and 2 is relatively short and rigid. 
Solution 
Assuming that the shaft connecting gears 1 and 2 is relatively short and rigid, we have 1 2θ θ= . Assume that 
3 4θ θ 0> > . The free-body diagram is shown in the figure below. Applying the moment equation to each gear gives 
 +↶: O OαM I∑ = 
Gears 1 and 2: a 2 1 2 2τ ( + )θr F I I− =  
231 
 
 
Gear 2: ( )1 1 2 2 2 2θ θ θK r F I− − =  
Gear 3: ( )3 2 3 4 3 3θ θ θr F K I− + − = −  
Gear 4: ( )2 3 4 4 4τ θ θ θl K I− − − = −  
Combining the first two equations and eliminating F yields 
( ) ( )2 2
a 3 3 2 3 4 1 2 2
3 3
τ θ θ θ θ
r rI K I I
r r
− − − = +  
By the geometry of the gears, 3 2 3 2θ ( )θr r= . Substituting it into the above equation and the equation of gear 4 gives 
2 2
2 2 2
1 2 3 2 2 2 2 4 a2 2
33 3
θ θ θ τ
r r rI I I K K
rr r
 
+ + + − = 
 
 
2
4 4 2 2 2 4
3
θ θ θ τl
rI K K
r
− + = 
 
Figure PS5-5 No6 
 
Problem Set 5.6 
(Note: All MATLAB figures are given at the end of Problem Set 5.6.) 
1. Consider the mass–spring–damper system in Problem 6 of Problem Set 5.2. Assume that the force actingon 
the mass block is a unit-impulse function with a magnitude of 100 N and a duration of 0.1 sec. The parameter 
values are m = 500 kg, b = 250 N·s/m, and k = 200,000 N/m. 
a. Build a Simulink model based on the differential equation of motion of the system and find the displacement 
output x(t). 
b. Build a Simscape model of the physical system and find the displacement output x(t). 
Solution 
The differential equation of the system is 
4 4mx bx kx f+ + =  or 1 ( 4 4 )mx f bx kx= − −  
 
2. Repeat Problem 1 for the mass–spring–damper system shown in Figure 5.117, in which the origin of the 
coordinate x is set at equilibrium. Assume x(0) = 0.1 m and (0) 0x = m/s. The parameter values are m = 20 kg, b 
= 125 N⋅s/m, and k = 400 N/m. 
 
 
Figure 5.117 Problem 2. 
Solution 
The differential equation of the system is 
232 
 
 
2 0mx bx kx+ + =  or 1 ( 2 )mx bx kx= − −  
 
3. Consider the two-degree-of-freedom mass–spring system shown in Figure 5.118. The parameter values are 
m1 = m2 = 5 kg, k1 = 2000 N/m, and k2 = 4000 N/m. Assume that initially x(0) = [0 0]T and [ ](0) 1 0 T=x . 
 
Figure 5.118 Problem 3. 
a. Build a Simulink model based on the differential equations of motion of the system and find the displacement 
outputs x1(t) and x2(t). 
b. Build a Simscape model of the physical system and find the displacement outputs x1(t) and x2(t). 
Solution 
The differential equations of motion of the system is 
1 1 1 2 1 2 2
2 2 2 1 2 2
( ) 0
0
m x k k x k x
m x k x k x
+ + − =
− + =


 
or 
[ ]
[ ]
1
2
1
1 1 2 1 2 2
1
2 2 1 2 2
( )m
m
x k k x k x
x k x k x
= − + +
= −


 
 
4. Repeat Problem 3 for the two-degree-of-freedom quarter-car model in Example 5.5. Assume that the surface 
of the road can be approximated as a sine wave z = Z0sin(2πvt/L), where Z0 = 0.01 m, L = 10 m, and the speed v = 
20 km/h. If the car moves at a speed of 100 km/h, rerun the simulations and compare the results with those 
obtained in the case of 20 km/h. Ignore the control force f for both cases. 
Solution 
The differential equations of motion of the system is 
1 1 1 1 1 2 1 1 1 2
2 2 1 1 1 2 1 1 1 2 2 2
0
( )
m x b x b x k x k x
m x b x b x k x k k x k z
+ − + − =
− + − + + =
  
  
 
or 
[ ]
[ ]
1
2
1
1 1 1 1 2 1 1 1 2
1
2 1 1 1 2 1 1 1 2 2 2( )
m
m
x b x b x k x k x
x b x b x k x k k x k z
= − + − +
= − + − + +
  
  
 
 
5. Consider the disk–shaft system in Problem 2 of Problem Set 5.3. The system is approximated as a 
single-degree-of-freedom rotational mass–spring system, where m = 10 kg, r = 0.05 m, B = 1 N⋅m⋅s/rad, and K = 
1000 N⋅m/rad. 
a. Assume that a torque τ = 50u(t) N⋅m is acting on the disk, which is initially at rest. Build a Simscape model of 
the physical system and find the angular displacement output θ(t). 
b. Assuming that the external torque is τ = 0 and the initial angular displacement is θ(0) = 0.1 rad, find the 
angular displacement output θ(t). 
Solution 
The differential equations of motion of the system is 
21
2 θ 2 θ 2 θ τmr B K+ + =  
or ( ) ( )2
2θ τ 2 θ 2 θ 80 τ 2θ 2000θ
mr
B K= − − = − −   
 
233 
 
 
6. Consider the pendulum-bob system in Problem 5 of Problem Set 5.3. The parameter values are m = 0.1 kg, M 
= 1.2 kg, L = 0.6 m, and B = 0.25 N⋅s/m. The initial angular displacement is θ(0) = 0.1 rad and the initial angular 
velocity is θ 0.1 rad/s= . 
a. Build a Simulink block diagram based on the nonlinear mathematical model of the system and find the 
angular displacement output θ(t). 
b. Build a Simscape model of the nonlinear physical system and find the angular displacement output θ(t). 
Solution 
The nonlinear mathematical model of the system is 
( ) ( )21 1
3 2 sin 0m M L B m M gL+ θ + θ + + θ =  
0.18 0.25 4.12sin 0θ + θ + θ =  
or 
( )1
0.18 0.25 4.12sinθ = − θ − θ  
 
234 
 
 
MATLAB Figures 
Problem 1 
 
 
Figure PS5-6 No1a Simulink block diagram. 
 
 
Figure PS5-6 No1b Simscape block diagram. 
 
 
Figure PS5-6 No1c 
 
S PS
Simulink-PS
Converter1
Signal 1
Group 1
Signal Builder1
R
S
C
R
C
S
Ideal Force Source
f(x) = 0
Solver
Configuration
R
C
V
PP
V
C
R
Ideal Translational
Motion Sensor
Mass
Mechanical
Translational
Reference2 Mechanical
Translational
Reference1
Mechanical
Translational
Reference
SPS
PS-Simulink
Converter
Scope
R
CC
R R
CC
R Translational
 Spring
R
CC
R
r1
R
CC
RR
CC
RTranslational
Damper
R
CC
R R
CC
R
r3
R
CC
R
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
D
is
pl
ac
em
en
t x
 (m
)
10 -4
235 
 
 
Problem 2 
 
 
Figure PS5-6 No2a Simulink block diagram. 
 
 
Figure PS5-6 No2b Simscape block diagram. 
 
 
Figure PS5-6 No2c 
 
R CCR
Translational 
Damper
f(x) = 0
Solver
Configuration
R
C
V
PP
V
C
R
Ideal Translational
Motion Sensor
Mass
SPS
PS-Simulink
Converter
Mechanical
Translational
Reference
Mechanical
Translational
Reference1RCC R
Translational
 Spring
RCC R
Translational
 Spring1
Scope
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (s)
-0.02
0
0.02
0.04
0.06
0.08
0.1
D
is
pl
ac
em
en
t x
 (m
)
236 
 
 
Problem 3 
 
 
Figure PS5-6 No3a Simulink block diagram. 
 
 
Figure PS5-6 No3b Simscape block diagram. 
x2x1
RCC R
Translational 
Spring k2
RCC R
Translational 
Spring k1
f(x) = 0
Solver
Configuration
Scope1Scope
S
P
SPS-Simulink
Converter1S
P
SPS-Simulink
Converter
Mechanical
Translational
Reference2
Mechanical
Translational
Reference1
Mechanical
Translational
Reference
Mass m2Mass m1
R
C V PPVC
R Ideal Translational
Motion Sensor1
R
C V PPVC
R Ideal Translational
Motion Sensor
237 
 
 
 
Figure PS5-6 No3c 
 
Problem 4 
 
 
Figure PS5-6 No4a Simulink block diagram. 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (s)
-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
D
is
pl
ac
em
en
t x
1
 (m
)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (s)
-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
D
is
pl
ac
em
en
t x
2
 (m
)
238 
 
 
 
Figure PS5-6 No4b Simscape block diagram. 
 
 
Figure PS5-6 No4c: v = 20 km/h 
 
 
Figure PS5-6 No4d: v = 100 km/h 
 
x2
x1
R
CC
R
Translational Spring 2
R
CC
R
Translational Spring 1
R
CC
R
Translational Damper
f(x) = 0
Solver
Configuration
S PS
Simulink-PS
Converter1
Scope1
Scope
SPS
PS-Simulink
Converter1
SPS
PS-Simulink
Converter
Mechanical
Translational
Reference2
Mechanical
Translational
Reference1
Mechanical
Translational
Reference
Mass 2
Mass 1
R
S C
R
CS
Ideal Translational
Velocity Source
R
C
V
PP
V
C
R
Ideal Translational
Motion Sensor1
R
C
V
PP
V
C
R
Ideal Translational
Motion Sensor
Z0*2*pi*v/L*cos(2*pi*v/L*u)
Fcn
velocity z_dot
(Z0 = 0.01 m
L = 10 m
v = 20 or 100 km/h)
Clock
0 1 2 3 4 5 6 7 8 9 10
Time (s)
-0.025
-0.02
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
0.02
D
is
pl
ac
em
en
t x
1
 (m
)
0 1 2 3 4 5 6 7 8 9 10
Time (s)
-0.02
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
D
is
pl
ac
em
en
t x
2
 (m
)
0 1 2 3 4 5 6 7 8 9 10
Time (s)
-6
-4
-2
0
2
4
6
D
is
pl
ac
em
en
t x
1
 (m
)
10
-3
0 1 2 3 4 5 6 7 8 9 10
Time (s)
-8
-6
-4
-2
0
2
4
6
8
D
is
pl
ac
em
en
t x
2
 (m
)
10
-3
239 
 
 
Problem 5 
 
Figure PS5-6 No5a Simscape block diagram. 
 
Set the external torque τ as 50 Nm and the initial angular displacement θ(0) as 0 rad. 
 
Figure PS5-6 No5b Angular displacement output. 
Set the external torque τ as 0 and the initial angular displacement θ(0) as 0.1 rad. 
 
Figure PS5-6 No5c Angular displacement output. 
Step1 Scope1
SPS
PS-Simulink
Converter
R
C
W
AA
W
C
R
Ideal Rotational
Motion Sensor
S PS
Simulink-PS
Converter
R
S
C
R
C
S
Ideal 
Torque Source
f(x) = 0
Solver
Configuration
Inertia
Mechanical
Rotational 
Reference
Mechanical
Rotational 
Reference1
RCC R
Rotational 
Spring
R CCRtω ω ω φ ω φ ω φ− = + = + . Comparison gives 
 
 
4th quadrant10
1
3
sin 1 sin 0
 tan 0.3218 rad
cos 3 cos 0
DD
D
φ φ
φ φ
φ φ
== − 
 
 
Therefore 3sin cos 10 sin( 0.3218)t t tω ω ω− = − . 
 
 
In Problems 15 and 16, write the expression in the form cos( )D tω φ+ . 
 
15. 3
4cos sint t− 
Solution 
Write 3
4cos sin cos( ) cos cos sin sint t D t D t D tφ φ φ− = + = − and compare the two sides to find 
 
 
5
43 1st quadrant
34
4
sin 0sin
 tan 0.6435 rad
cos 0cos 1
DD
D
φφ
φ φ
φφ
= >=
⇒ ⇒ = ⇒ =
>=
 
 
 
25 
 
 
Therefore, 3 5
4 4cos sin cos( 0.6435)t t t− = + . 
 
16. sin cost t− 
Solution 
Write sin cos cos( ) cos cos sin sint t D t D t D tφ φ φ− = + = − and compare the two sides to find 
 
 
3rd quadrant2
5
4
sin 1 sin 0
 tan 1 
cos 1 cos 0
DD
D
φ φ
φ φ π
φ φ
== − > syms a b t 
>> laplace(exp(a*t+b)) 
ans = 
-exp(b)/(a - s) 
 
 
2. cos( )tω φ+ , , constω φ = 
Solution 
(a) Using trigonometric expansion, we find 
 
 
2 2 2 2 2 2
{cos( )} {cos }cos {sin }sin
cos sin cos sin
t t t
s s
s s s
ω φ ω φ ω φ
ω φ ω φφ φ
ω ω ω
+ = −
−
= − =
+ + +
L L L
 
(b) 
 
>> syms w t p 
 
 
26 
 
>> laplace(cos(w*t+p)) 
 
ans = 
 
(s*cos(p) - w*sin(p))/(s^2 + w^2) 
 
3. sin( )tω φ− , , constω φ = 
Solution 
(a) Using trigonometric expansion, we find 
 
 
{ } { } { } { }
2 2 2 2 2 2
sin( ) sin cos cos sin sin cos cos sin
cos sin cos sin
t t t t t
s s
s s s
ω φ ω φ ω φ ω φ ω φ
ω ω φ φφ φ
ω ω ω
− = − = −
−
= − =
+ + +
L L L L
 
 
(b) 
 
>> syms w t p 
>> laplace(sin(w*t-p)) 
 
ans = 
 
(w*cos(p) - s*sin(p))/(s^2 + w^2) 
 
 
4. 3 1
2t − 
Solution 
(a) { }3 1
2 4
6 1
2
t
ss
− = −L 
(b) 
>> syms t 
>> laplace(t^3-1/2) 
ans = 
6/s^4 - 1/(2*s) 
 
 
5. 2sin t 
Solution 
(a) Noting that 2 1
2sin (1 cos 2 )t t= − , we find 
 
 { }
Simplify
2
2 2
1 1 1 2sin {1 cos 2 }
2 2 4 ( 4)
st t
s s s s
 = − = − = + + 
L L
 (b) 
 
>> syms t 
>> laplace(sin(t)^2) 
 
 
 
27 
 
ans = 
 
 2/(s*(s^2 + 4)) 
 
 
6. cost tω 
Solution 
(a) Following Eq. (2.16), 
 
 { }
2 2
2 2 2 2 2cos
( )
d s st t
ds s s
ωω
ω ω
− = − = + + 
L
 (b) 
 
>> syms t w 
>> simplify(laplace(t*cos(w*t))) 
 
ans = 
 
 (s^2 - w^2)/(s^2 + w^2)^2 
 
 
7. cosht t 
Solution 
(a) Comparing with Eq. (2.16), we have ( ) coshg t t= so that 
 
 { } { }
Simplify
2
1 1 1 1( ) cosh
2 2 1 1 1
t t sG s t e e
s s s
−  = = + = + = − + − 
L L 
Then, 
 { }
2
2 2 2
1cosh
1 ( 1)
d s st t
ds s s
+ = − = − − 
L
 (b) 
 
>> syms t 
>> simplify(laplace(t*cosh(t))) 
 
ans = 
 
(s^2 + 1)/(s^2 - 1)^2 
 
 
8. 2 1
2sin( )t t 
Solution 
(a) Following the general form of Eq. (2.16) with 1
2( ) sin( )g t t= and 2n = , 
 
 
 
28 
 
 { }
22 1 1
2 2 41
2 2 2 2 31 1
4 4
3
sin( )
( )
sdt t
ds s s
  −
= =  + + 
L
 
(b) 
>> G = sym('1/(2*(s^2+1/4))'); simplify(diff(G,2)) 
ans = 
 (16*(12*s^2 - 1))/(4*s^2 + 1)^3 
 
In Problems 9 through 12, 
(a) Express the signal in terms of unit-step functions. 
(b) Find the Laplace transform of the expression in (a) using the shift on t − axis. 
 
9. ( )g t in Figure 2.15 
Solution 
(a) ( ) ( 1) ( 2)g t u t u t= − − − 
(b) 
2 2
( )
s s s se e e eG s
s s s
− − − −−
= − = 
 
 
 Figure 2.15 Signal in Problem 9. 
 
 
10. ( )g t in Figure 2.16 
Solution 
(a) ( ) ( ) 2 ( 1) ( 2)g t u t u t u t= − + − − − 
(b) 
2 21 2 (1 )( )
s s se e eG s
s s
− − −− + − − −
= = 
 
( )g t
t
1
1
2
 
 
29 
 
 
 Figure 2.16 Signal in Problem 10 
 
 
11. 
0 if 0
( ) 1 if 0 1
0 if 1
t
g t t t
t

 
Solution 
(a) Construct ( )g t using the strategy shown in Figure PS2-3 No11, leading to 
 
 ( ) (1 ) ( ) (1 ) ( 1)g t t u t t u t= − − − − 
 
(b) Rewrite the expression obtained in (a) as ( ) ( ) ( ) ( 1) ( 1)g t u t tu t t u t= − + − − . Then 
 
 2 2
1 1( )
seG s
s s s
−
= − + 
 
 
 
 
 Figure PS2-3 No11 
 
 
( )g t
t
1
1
2
1−
 
 
30 
 
12. 
0 if 0
( ) if 0 1
1 if 1
t
g t t t
t

 
Solution 
(a) 
simplify
( ) ( ) ( 1) ( 1) ( ) ( 1) ( 1)g t tu t tu t u t tu t t u t= − − + − = − − − 
(b) 2 2 2
1 1( )
s se eG s
s s s
− −−
= − = 
 
 
 
In Problems 13 through 16, find the Laplace transform of each periodic function whose 
definition in one period is given. 
 
13. 
1
2
1
2
1 if 0
( )
1 if 1
t
h t
t
 > B = [1 0]; A = conv([1 1],[1 2 2]); 
>> [R,P,K] = residue(B,A) 
 
R = 
 0.5000 - 0.5000i 
 0.5000 + 0.5000i 
 -1.0000 + 0.0000i 
 
P = 
 -1.0000Rotational 
Spring1
RCC R
Rotational Damper
R CCR
Rotational Damper1
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
A
ng
ul
ar
 D
is
pl
ac
em
en
t 
 (r
ad
)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
A
ng
ul
ar
 D
is
pl
ac
em
en
t 
 (r
ad
)
240 
 
 
Problem 6 
 
 
Figure PS5-6 No6a Simulink block diagram. 
 
Rewrite the equation as 0.18 0.25 4.12sinθ + θ = − θ  , where 4.12sin− θ is considered as a torque applied to the 
system. 
 
Figure PS5-6 No6b Simscape block diagram. 
 
Run either of the simulations returns the curve shown below. 
 
Figure PS5-6 No6c 
Scope1
-4.12*sin(u)
Fcn
SPS
PS-Simulink
Converter
R
C
W
AA
W
C
R
Ideal Rotational
Motion Sensor
R
CC
R
Rotational Damper
S PS
Simulink-PS
Converter
R
S
C
R
C
S
Ideal 
Torque Source
f(x) = 0
Solver
Configuration
Inertia Mechanical
Rotational 
Reference1
Mechanical
Rotational 
Reference
Mechanical
Rotational 
Reference2
0 1 2 3 4 5 6 7 8 9 10
Time (s)
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
0.12
A
ng
ul
ar
 D
is
pl
ac
em
en
t 
 (r
ad
)
241 
 
 
Review Problems 
1. Determine the equivalent spring constant for the system shown in Figure 5.119. 
 
Figure 5.119 Problem 1. 
Solution 
The equivalent spring constant for the middle part is 
eq1 2 3
1 1 1
k k k
= + or 2 3
eq1
2 3
k k
k
k k
=
+
 
The equivalent spring constant for the system is 
1 2 1 3 2 3 2 4 3 4
eq 1 eq1 4
2 3
k k k k k k k k k k
k k k k
k k
+ + + +
= + + =
+
 
 
2. Determine the equivalent spring constant for the system shown in Figure 5.120. 
 
Figure 5.120 Problem 2. 
Solution 
The equivalent spring constant for the bottom part is 
eq1 2 3 4
1 1 1
k k k k
= +
+
 or 2 3 2 4
eq1
2 3 4
k k k kk
k k k
+
=
+ +
 
The equivalent spring constant for the system is 
1 2 1 3 1 4 2 3 2 4
eq 1 eq1
2 3 4
k k k k k k k k k kk k k
k k k
+ + + +
= + =
+ +
 
 
3. Consider the system shown in Figure 5.121, in which a mass–spring system is hung from the middle of a massless 
beam. Assume that the beam can be modeled as a spring and the equivalent stiffness at the midspan is 192EIA/L3, 
where E is the modulus of elasticity of beam material and IA is the area moment of inertia about the beam’s 
longitudinal axis. 
 
Figure 5.121 Problem 3. 
 a. Derive the differential equation of motion for the system. 
 b. Using the differential equation obtained in Part (a), determine the transfer function X(s)/F(s). Assume that the 
initial conditions are x(0) = 0 and (0) 0x = . 
242 
 
 
c. Using the differential equation obtained in Part (a), determine the state-space representation. Assume that the 
output is the displacement x of the mass. 
Solution 
a. The system is equivalent to the mass-spring system shown in the figure below, where the spring kb represents the 
flexibility of the beam and kb = 192EIA/L3. The equivalent spring stiffness of the system is 
3
eq b A
1 1 1 1
192
L
k k k EI k
= + = + 
or 
A
eq 3
A
192
192
EI kk
L k EI
=
+
 
Choose the static equilibrium as the origin of the coordinate x. The equation of motion for the system is 
eq ( )mx k x f t+ = 
 
Figure Review5 No3 
b. Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 
2
eq
( ) 1
( )
X s
F s ms k
=
+
 
c. The state, the input, and the output are 
 1
2
, ,
x x
u f y x
x x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
1 1
eq
2 2
0 1 0
1
0
x x
ukx x
mm
         = +      −       


 [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
 
4. An accelerometer attached to an object can be modeled as a mass–damper–spring system, as shown in Figure 
5.122. Denote the displacement of the mass relative to the object, the absolute displacement of the mass, and the 
absolute displacement of the object as x(t), y(t), and z(t), respectively, where x(t) = y(t) – z(t) and x(t) is measured 
electronically. 
a. Draw the necessary free-body diagram and derive the differential equation in terms of x(t). 
b. Using the differential equation obtained in Part (a), determine the transfer function X(s)/Z(s). Assume that the 
initial conditions are x(0) = 0 and (0) 0x = . 
c. Using the differential equation obtained in Part (a), determine the state-space representation. The input is the 
absolute displacement of the object z(t) and the output is the displacement of the mass relative to the object 
x(t). 
 
243 
 
 
 
Figure 5.122 Problem 4. 
Solution 
a. Choose the static equilibrium as the origin of the coordinate y and assume that 0y z> > . The free-body diagram 
of the mass is shown below. 
 
Figure Review5 No4 
Applying Newton’s second law in the y-direction gives 
( ) ( ):y k y z b y z my+ ↑ − − − − =  
Note that x(t) = y(t) – z(t). Thus, the equation can be rewritten as 
( )kx bx m x z− − = +   or mx bx kx mz+ + = −   
b. Assuming zero initial conditions and taking Laplace transform of the above differential equation gives 
2
2
( )
( )
X s ms
Z s ms bs k
−
=
+ +
 
c. Rewrite the transfer function ( ) ( )X s Z s as 
2
2
( ) ( ) ( ) 1
( ) ( ) ( )
X s X s W s s b kZ s W s Z s s s
m m
= = −
+ +
 
where 
2( )
( )
X s s
W s
= − 
2
( ) 1
( )
W s
b kZ s s s
m m
=
+ +
 
Thus in the time-domain, we have 
x w= −  
b kw w w z
m m
= − − +  
Select the state and the input as 
 1
2
,
x w
u z
x w
   
= = =   
  
x

 
The state equation is 
1 1
2 2
0 1 0
1
x x
uk bx x
m m
       = +      − −      


 
and the output equation is 
1
2
1
xk by u
xm m
  = − ⋅     
 
244 
 
 
5. Consider a quarter-car model shown in Figure 5.123, where m1 is the mass of the seats including passengers, m2 is 
the mass of one-fourth of the car body, and m3 is the mass of the wheel–tire–axle assembly. The spring k1 
represents the elasticity of the seat supports, k2 represents the elasticity of the suspension, and k3 represents the 
elasticity of the tire. z(t) is the displacement input due to the surface of the road. Draw the necessary free-body 
diagrams and derive the differential equations of motion. Write the differential equations of motion in the 
second-order matrix form. 
Solution 
Choose the displacements of the three masses 1x , 2x , and 3x as the generalized coordinates. The static equilibrium 
positions of 1m , 2m , and 3m are set as the coordinate origins. Assume that 1 2 3 0x x x z> > > > . The free-body 
diagram is shown below. 
 
 Figure 5.123 Problem 5. Figure Review5 No5 
 
Applying Newton’s second law in the x-direction gives 
: x xx F ma+ ↑ ∑ = 
Mass 1: ( ) ( )1 1 2 1 1 2 1 1k x x b x x m x− − − − =   
Mass 2: ( ) ( ) ( ) ( )1 1 2 1 1 2 2 2 3 2 2 3 2 2k x x b x x k x x b x x m x− + − − − − − =     
Mass 3: ( ) ( ) ( )2 2 3 2 2 3 3 3 3 3k x x b x x k x z m x− + − − − =   
Rearranging the equations into 
1 1 1 1 1 2 1 1 1 2 0m x b x b x k x k x+ − + − =   
( ) ( )2 2 1 1 1 2 2 2 3 1 1 1 2 2 2 3 0m x b x b b x b x k x k k x k x− + + − − + + − =    
( )3 3 2 2 2 3 2 2 2 3 3 3m x b x b x k x k k x k z− + − + + =   
The differential equations can be expressed in second-order matrix form as 
1 1 1 1 1 1 1 1
2 2 1 1 2 2 2 1 1 2 2 2
3 3 2 2 3 2 2 3 3 3
0 0 0 0 0
0 0 0
0 0 0 0
m x b b x k k x
m x b b b b x k k k k x
m x b b x k k k x k z
− −           
           + − + − + − + − =           
           − − +           
 
 
 
 
 
6. The system shown in Figure 5.124 consists of a uniform rod of mass m and length L and a translational spring of 
stiffness k at the rod’s tip. The friction at the joint O is modeled as a damper with coefficient of torsional viscousdamping B. The input is the force f and the output is the angle θ. The position θ = 0 corresponds to the static 
equilibrium position when f = 0. 
245 
 
 
 a. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. 
 b. Using the linearized differential equation obtained in Part (b), determine the transfer function Θ(s)/F(s). 
Assume that the initial conditions are θ(0) = 0 and θ(0) 0= . 
 c. Using the differential equation obtained in Part (b), determine the state-space representation. 
 
Figure 5.124 Problem 6. 
Solution 
a. At equilibrium, we have 
+↶: O 0M∑ = 
st 0kδ = 
st 0δ = 
Applying the moment equation to the fixed point O gives 
+↶: O OαM I∑ = 
1 1
O2 2cosθ sin θ θ cosθ θkL f L mg B L f I− ⋅ − ⋅ − + ⋅ =  
where, for small angular motions, the spring force can be approximated as 
θkf kL= 
Note that the mass moment of inertia about point O is 
2 2 2 21 1 1
O C 12 2 3( )I I md m mL L mL+ == + = 
Thus, the differential equation of motion for small angles θ is 
2 21 1 1
3 2 2θ θ θ θmL B kL mgL fL+ + + =  
 
Figure Review5 No6 
b. Taking Laplace transform of the above linearized differential equation gives 
2 2 2
( ) 3
( ) 2 6 6 3
s L
F s mL s Bs kL mgL
Θ
=
+ + +
 
c. The state, the input, and the output are 
 1
2
θ
, , θ
θ
x
u f y
x
   
= = = =   
  
x

 
The state equation and the output equation in matrix form are 
246 
 
 
1 1
2 22
0 1 0
6 3 3 3
2 2
x x
ukL mg Bx x
mL mLmL
         = +− −      −         


 [ ] 1
2
1 0 0
x
y u
x
 
= + ⋅ 
 
 
 
7. Consider the system shown in Figure 5.125. Assume that a cylinder of mass m rolls without slipping. Draw the 
necessary free-body diagram and derive the differential equation of motion for small angles θ. 
 
Figure 5.125 Problem 7. 
Solution 
The free-body diagram is shown in the figure below. Because the cylinder rolls without slipping, the contact point is 
the IC. 
 
Figure Review5 No7 
 
Choose the static equilibrium position as the origin. Applying the moment equation to the IC gives 
+↷: IC ICαM I∑ = 
( ) ( )1 2 ICθk ka r f a r f I− + ⋅ − + ⋅ =  
where for a cylinder, 
2 2 2 231
IC C 2 2I I md mr mr mr= + = + = 
For small angles θ , the spring forces are 1 1 θkf k a= and 2 2 θkf k a= . Thus, the differential equation of motion is 
( ) ( )23
1 22 θ θ 0mr a a r k k+ + + = 
 
8. The pulley of mass M shown in Figure 5.126 has a radius of r. The mass moment of inertia of the pulley about the 
point O is IO. A translational spring of stiffness k and a block of mass m are connected to the pulley as shown. 
Assume that the pulley rolls without slipping. Derive the equation of motion using (a) the force/moment 
approach, and (b) the energy approach. 
 
Figure 5.126 Problem 8. 
Solution 
The free-body diagram and the kinematic diagram are shown in the figure below. Because the pulley rolls without 
slipping, the contact point is the IC. At static equilibrium, we have stδkf k mg= = , where stδ is the static deformation 
of the spring. We choose the equilibrium as the origin. Assume that the block and the pulley are displaced in their 
positive directions. Note that θx r= and the spring force ( )stθ δkf k r= + . 
247 
 
 
 
Figure Review5 No8 
 
For the block (translation only), applying the force equation in the x-direction gives 
: x Cxx F ma+ ↓ ∑ = 
θmg T mx mr− = = 
 
For the pulley, applying the moment equation about the IC yields 
+↷: IC ICαM I∑ = 
( )st ICθ δ θr T r k r I⋅ − ⋅ + =  
where 2 2
IC O OI I Md I Mr= + = + . Combining the two equations and eliminating the unknown T results in 
( ) ( ) ( )2
st Oθ θ δ θr mg mr rk r I Mr− − + = +  
Introducing the static equilibrium condition, stδmg k= , the equation becomes 
( )2 2 2
O θ θ 0I Mr mr kr+ + + = 
 
9. Consider a half-car model shown in Figure 5.127, in which IC is the mass moment of inertia of the car body about 
the pitch axis, mb is the mass of the car body, mf is the mass the front wheel−tire−axle assembly, and mr is the mass 
of the rear wheel−tire−axle assembly. Each of the front and rear wheel−tire−axle assemblies is represented by a 
mass–spring–damper system. The input is the force f, and the car undergoes vertical and pitch motion. Derive the 
equations of motion using the force/moment approach. 
 
Figure 5.127 Problem 9. 
Solution 
Choose fx , rx , bx , and θ as the generalized coordinates. The static equilibrium positions of fm , rm , and bm are set 
as the coordinate origins. Assume that b fθ 0x a x− > > , b rθ 0x b x+ > > , and θ 0> . The free-body diagram and the 
kinematic diagram are shown in the figure below. 
 
248 
 
 
 
Figure Review5 No9 
 
Applying the force equations to fm , rm , and the rod bm gives 
: x Cxx F ma+ ↑ ∑ = 
sf b f sf b f f f f f( θ ) ( θ )k x a x b x a x k x m x− − + − − − =
   
sr b r sr b r r r r r( θ ) ( θ )k x b x b x b x k x m x+ − + + − − =
   
sf b f sf b f sr b r sr b r b b( θ ) ( θ ) ( θ ) ( θ )f k x a x b x a x k x b x b x b x m x− − − − − − − + − − + − = 
     
Applying the moment equation to the rod gives 
+↶: αC CM I∑ = 
sf b f sf b f sr b r sr b r C( θ ) cosθ ( θ ) cosθ ( θ ) cosθ ( θ ) cosθ cosθ θk x a x a b x a x a k x b x b b x b x b fc I− − + − − − + − − + − + =  
    
For small angular motions, we have cosθ 1≈ . The equations can be written in second-order matrix form 
f sf sf sff f
r sr sr srr r
b sf sr sf sr sf srb b
2 2
C sf sr sf sr sf sr
sf f sf sf
sr r
0 0 0 0
0 0 0 0
0 0 0
0 0 0 θ θ
0
0
m b b b ax x
m b b b bx x
m b b b b b a b bx x
I b a b b b a b b b a b b
k k k k a
k k k
−      
      − −      +
      − − + − +
      − − + +      
+ −
+ −
+
 
 
 
 
f
sr sr r
sf sr sf sr sf sr b
2 2
sf sr sf sr sf sr
0
0
θ
x
k b x
k k k k k a k b x f
k a k b k a k b k a k b fc
     
     −     =
     − − + − +
     − − + +     
 
 
10. Consider the mechanical system shown in Figure 5.128, where a uniform rod of mass m and length L is attached to 
a massless rigid link of equal length. Assume that the system is constrained to move in a vertical plane. Denote the 
angular displacement of the link as θ1, and the angular displacement of the rod as θ2. Derive the equations of 
motion for small angles using (a) the force/moment approach, and (b) the energy approach. 
 
Figure 5.128 Problem 10. 
Solution 
(a) The free-body diagram and the kinematic diagram are shown in the figure below. 
249 
 
 
 
Figure Review5 No10a 
 
Applying the force equations to the rod gives 
: x Cxx F ma+ → ∑ = 
2 21 1
1 1 1 2 2 1 1 2 22 2sin θ θ cosθ ( )θ cosθ θ sin θ ( )θ sin θT mL m L mL m L− = + − −    
: y Cyy F ma+ ↑ ∑ = 
2 21 1
1 1 1 2 2 1 1 2 22 2cosθ θ sin θ ( )θ sin θ θ cosθ ( )θ cosθT mg mL m L mL m L− = + + +    
Eliminating the unknown force T yields 
( ) ( )21 1
1 2 2 1 2 2 1 12 2θ θ cos θ θ θ sin θ θ sin θ 0mL mL mL mg+ − − − + =   
Applying the moment equation to the rod about the O1 results in 
+↶: 
1O C _α
Ceff mM I M∑ = + a 
( ) ( ) 21 1 1 1 1
2 2 2 1 1 2 1 1 22 2 2 2 2sin θ θ cos θ θ θ sin θ θ θ ( )θCL mg I L mL L mL L m L− ⋅ = + − ⋅ + − ⋅ + ⋅    
where 21
C 12I mL= . Rearranging the equation gives 
( ) ( ) 21 1 1 1
2 2 1 1 2 1 1 23 2 2 2θ cos θ θ θ sin θ θ θ sin θ 0mL mL mL mg+ − + − + =   
For small angular motions, we have 
1
1 2 12θ θ θ 0mL mL mg+ + =  
1 1 1
1 2 22 3 2θ θ θ 0mL mL mg+ + =  
(b) The system is only subjected to the gravitational forces, and it is a conservative system. The rod undergoes 
general plane motion, and 
( )2 2 2 2 21 1 1 1 1
22 2 2 2 12 θC CT mv I mv mLω= + = +  
where v is the velocity at the center of gravity of the rod, and 
( ) ( ) ( ) ( ) ( )2 22 1 1
2 1 2 1 2 12 2θ θ 2 θ θ cos θ θv L L L L= + + −    
using thelaw of cosine. Thus, the kinetic energy is 
( ) ( ) ( ) ( ) ( ) ( )
( )
2 2 2 21 1 1 1 1
2 1 2 1 2 1 22 2 2 2 12
2 2 2 2 21 1 1
1 2 1 2 2 12 6 2
θ θ 2 θ θ cos θ θ θ
θ θ θ θ cos θ θ
T m L L L L mL
mL mL mL
 = + + − +  
= + + −
    
   
 
Using the datum defined in the figure, we can obtain the potential energy 
 ( )1
1 22cosθ cosθgV V mgh mg L L= = − = − + 
 
Figure Review5 No10b 
250 
 
 
We then apply Lagrange's equations. For 1i = , 
 ( )2 21
1 2 2 12
1
θ θ cos θ θ
θ
T mL mL∂
= + −
∂
 

 
( ) ( ) ( )2 2 21 1
1 2 2 1 2 2 1 2 12 2
1
θ θ cos θ θ θ sin θ θ θ θ
θ
d T mL mL mL
dt
 ∂
= + − − − ⋅ − ∂ 
    

 
( ) ( )21
1 2 2 12
1
θ θ sin θ θ 1
θ
T mL∂
= − − ⋅ −
∂
  
1
1
sin θ
θ
V mgL∂
=
∂
 
Substituting into Lagrange's equation results in 
( ) ( )2 2 2 21 1
1 2 2 1 2 2 1 12 2
1 1 1
θ θ cos θ θ θ sin θ θ sin θ 0
θ θ θ
d T T V mL mL mL mgL
dt
 ∂ ∂ ∂
− + = + − − − + = ∂ ∂ ∂ 
  

 
Similarly, for 2i = , 
( )2 21 1
2 1 2 13 2
2
θ θ cos θ θ
θ
T mL mL∂
= + −
∂
 

 
( ) ( ) ( )2 2 21 1 1
2 1 2 1 1 2 1 2 13 2 2
2
θ θ cos θ θ θ sin θ θ θ θ
θ
d T mL mL mL
dt
 ∂
= + − − − ⋅ − ∂ 
    

 
( )21
1 2 2 12
2
θ θ sin θ θ
θ
T mL∂
= − −
∂
  
1
22
2
sin θ
θ
V mgL∂
=
∂
 
Substituting into Lagrange's equation gives 
( ) ( )2 2 2 21 1 1 1
2 1 2 1 1 2 1 23 2 2 2
2 2 2
θ θ cos θ θ θ sin θ θ sin θ 0
θ θ θ
d T T V mL mL mL mgL
dt
 ∂ ∂ ∂
− + = + − + − + = ∂ ∂ ∂ 
  

 
Thus, the two differential equations of motion are 
( ) ( )21 1
1 2 2 1 2 2 1 12 2θ θ cos θ θ θ sin θ θ sin θ 0mL mL mL mg+ − − − + =   
( ) ( )21 1 1 1
2 1 2 1 1 2 1 23 2 2 2θ θ cos θ θ θ sin θ θ sin θ 0mL mL mL mg+ − + − + =   
For small angular motions, we have 
1
1 2 12θ θ θ 0mL mL mg+ + =  
1 1 1
1 2 22 3 2θ θ θ 0mL mL mg+ + =  
 
11. Consider the mechanical system shown in Figure 5.129, in which a block of mass m is connected to a spring of 
stiffness k and it can slide without friction inside a vertical tube. Assume that the tube pivots at the joint O, and it 
can be approximated as a uniform rod of mass M and length L. The unstretched length of the spring is L/2. Denote 
the angular displacement of the tube as θ, and the relative displacement of the block with respect to the tube as x. 
Derive the nonlinear equations of motion using (a) the energy approach, and (b) the force/moment approach. 
 
Figure 5.129 Problem 11. 
 
251 
 
 
Solution 
(a) The system is a conservative system. The tube undergoes general plane motion, and 
( )2 2 21 1 1
tube O2 2 3 θT I MLω= =  
The block rotates together with the tube and at the same time slides in the tube, and the kinetic energy is 
2 2 2 21 1
block 2 2 ( θ )CT mv m x x= = +  
The total kinetic energy is 
2 2 2 2 21 1 1
6 2 2θ θT ML mx mx= + + 
 
Using the datum defined in the figure, we can obtain the potential energy 
 21
2 2 2cosθ cosθ ( )L L
g eV V V Mg mgx k x= + = − − + − 
 
Figure Review5 No11a 
 
We then apply Lagrange's equations. For 1i = , 
 2 21
3 θ θ
θ
T ML mx∂
= +
∂
 

 
2 21
3 θ θ 2
θ
d T ML mx mxx
dt
∂  = + + ∂ 
 


 
0
θ
T∂
=
∂
 
2 sinθ sin θ
θ
LV Mg mgx∂
= +
∂
 
Substituting into Lagrange's equation results in 
2 21
3 2θ θ 2 sinθ sin θ 0
θ θθ
Ld T T V ML mx mxx Mg mgx
dt
∂ ∂ ∂  − + = + + + + =  ∂ ∂∂ 
 


 
Similarly, for 2i = , 
T mx
x
∂
=
∂


 
d T mx
dt x
∂  = ∂ 


 
2θT mx
x
∂
=
∂
 
2cosθ ( )LV mg k x
x
∂
= − + −
∂
 
Substituting into Lagrange's equation gives 
2
2θ cosθ ( ) 0Ld T T V mx mx mg k x
dt x x x
∂ ∂ ∂  − + = − − + − = ∂ ∂ ∂ 



 
Thus, the two differential equations of motion are 
2 21
3 2θ θ 2 sinθ sin θ 0LML mx mxx Mg mgx+ + + + = 
 
2
2θ cosθ ( ) 0Lmx mx mg k x− − + − =
 
252 
 
 
(b) The free-body diagram and the kinematic diagram are shown in the figure below. Note that the tube and the block 
rotate together about pivot O, and at the same time the block moves relatively inside the tube. 
 
 
Figure Review5 No11b 
 
Applying the force equation to the block gives 
: x Cxx F ma+ → ∑ = 
xcosθkf mg ma− + = 
where the spring force 2( )L
kf k x= − . To find the acceleration ax, it is more convenient to use the polar coordinate 
system. The velocity components of the block in the polar coordinate system are θ θθv x u= 
 and x xv xu= 
 . Taking 
time derivatives gives 
θ θ θ θθ θ θv x u x u x u= + +   
  
 
 
x x xv xu xu= +  
 
  
Note that θ xθu u= − 

 and x θθu u= 

 . Thus, 
θ θ θ xθ θ θθv x u x u x u= + −   
   

 
x x θθv xu x u= +  


  
and the acceleration component along the x direction is 
x θθa x x= −  
 
Substituting it into the force equation gives 
2
2( ) cosθ θLk x mg mx mx− − + = − 
 
Applying the moment equation to the tube and the block about the O results in 
+↶: O C _α
Ceff mM I M∑ = +∑ ∑ a 
θ2 2 2sin θ sin θ θ θL L L
CMg x mg I M x ma− ⋅ − ⋅ = + ⋅ + ⋅  
where 21
C 12I mL= and the acceleration component of the block along the θ direction is 
θ θ θ θa x x x= + +  
  
Substituting them into the moment equation gives 
2 21 1
2 12 4sin θ sin θ θ θ ( θ θ θ)L Mg x mg ML ML xm x x x− ⋅ − ⋅ = + + + +    
  
Rearranging the above two equations gives 
2
2θ ( ) cosθ 0Lmx mx k x mg− + − − =
 
2 21 1
3 2θ 2 θ θ sin θ sin θ 0ML mxx mx MgL mgx+ + + + =  
 
 
12. A rack and pinion is a pair of gears that convert rotational motion into translation. As shown in Figure 5.130, a 
torque τ is applied to the shaft. The pinion rotates and causes the rack to translate. The mass moment of inertia of 
the pinion is I and the mass of the rack is m. Draw the free-body diagram and derive the differential equation of 
motion. 
253 
 
 
 
Figure 5.130 Problem 12. 
Solution 
The free-body diagram is shown in the figure below. 
 
Figure Review5 No12 
Applying the moment equation to the pinion gives 
+↷: O OαM I∑ = 
τ θrF I− =  
Applying the force equation to the track gives 
: x xx F ma+ ← ∑ = 
F mx=  
Combining the two equations and eliminating F yields 
θ τI mrx+ =
 
Introducing the geometric constrain θx r= gives the differential equation of motion 
( )2 θ τI mr+ = 
 
13. Consider the mass–spring–damper system shown in Problem 9 of Problem Set 5.2, in which the cam and 
follower impart a displacement z(t) in the form of a periodic sawtooth function (see Figure 5.131) to the lower end 
of the system. The values of the system parameters are m = 12 kg, b = 200 N⋅s/m, k1 = 4000 N/m, and k2 = 2000 
N/m. 
 a. Build a Simulink model based on the differential equation of motion of the system and find the displacement 
output x(t). 
 b. Build a Simscape model of the physical system and find the displacement output x(t). 
 
Figure 5.131 Problem 13. 
Solution 
The differential equation of motion of the system is 
( )1 2 2mx bx k k x k z+ + + =  
or ( )1
2 1 2mx k z bx k k x= − − +    
254 
 
 
 
Figure Review5 No13a Simulink block diagram. 
 
 
Figure Review5 No13b Simscape block diagram. 
 
 
Figure Review5 No13c 
 
14. Case Study 
Consider the ball and beam system [?] shown in Figure 5.132, in which I and Ib are the mass moments of inertia of 
the beam and the ball, respectively, m is the mass of the ball, and r is the radius of the ball. The beam is made to 
x
S PS
Simulink-PS
Converter
Signal 1
Group 1
Signal Builder
velocity
R
CC
R
Translational 
Damper
f(x) = 0
Solver
Configuration
R
C
V
PP
V
C
R
Ideal Translational
Motion Sensor
R
S C
R
CS
Ideal Translational
Velocity Source
Mass
SPS
PS-Simulink
Converter
Mechanical
Translational
Reference1
Mechanical
Translational
Reference
Mechanical
Translational
Reference2
R
CC
RTranslational
 Spring1
Scope
R
CC
R
Translational
 Spring
0 1 2 3 4 5 6
Time (s)
-2
0
2
4
68
10
D
is
pl
ac
em
en
t x
 (m
)
10 -3
255 
 
 
rotate in a vertical plane by applying a torque τ at the center of rotation and the ball is free to roll (with one 
degree-of-freedom) along the beam. It is assumed that the ball remains in contact with the beam and that the 
rolling occurs without slipping. The system parameters are I = 0.02 kg·m2, Ib = 2×10−6 kg·m2, m = 0.05 kg, and r = 
0.01 m. 
a. Choosing the beam angle θ and the ball position x as generalized coordinates, derive the nonlinear equations 
of motion of the system using the Lagrange’s equations. 
b. Choosing θ, θ , x, and x as state variables and τ as the input, find the state variable equations of the 
nonlinear system. 
c. Assuming small angular motion, find the state-space form for the linearized system (with θ and x as outputs). 
d. Build two Simulink block diagrams, one using the differential equations obtained in Part (a) and the 
other using the state-space form found in Part (c), compare the beam angle θ(t) and the ball displacement x(t) 
when the torque applied to the system is a pulse function, τ(t) = 1 Nm for 1 ≤ t ≤ 2 s. 
 
Figure 5.132 Problem 14. 
Solution 
a. The kinetic energy of the ball and beam system is 
beam ballT T T= + 
Note that the beam rotates about a fixed axis through the center, and 
21
beam 2 θT I=  
The motion of the ball is more complicated than that of the beam. As shown in the figure below, the ball rotates 
together with the beam about the pivot, and it rolls without slipping along the beam. The kinetic energy of the ball 
consists of three parts 
2 2 2 21 1 1
ball b b2 2 2( )θ + αT I mx mx I= + +  
Note that αr = x due to the geometric constraint. 
 
Figure Review5 No14a 
Thus, the total kinetic energy is 
2 2 2b1 1
b2 2 2( )θ ( )
I
T I I mx m x
r
= + + + +
 
The total potential energy is 
ball sin θV V mgx= = 
We then apply Lagrange's equations. For 1i = , 
 2
b( )θ
θ
T I I mx∂
= + +
∂


 
2
b( )θ 2 θ
θ
d T I I mx mxx
dt
∂  = + + + ∂ 
 


 
0
θ
T∂
=
∂
 
cosθ
θ
V mgx∂
=
∂
 
256 
 
 
( ) ( )1Q ∂ ∂
= = τ θ = τ
∂θ ∂θ
rF k k 
Substituting into Lagrange's equation results in 
2
b( )θ 2 θ cosθ τI I mx mxx mgx+ + + + = 
 
Similarly, for 2i = , 
b
2( )
IT m x
x r
∂
= +
∂


 
b
2( )
Id T m x
dt x r
∂  = + ∂ 


 
2θT mx
x
∂
=
∂
 
sin θ
θ
V mg∂
=
∂
 
2 0Q
x
∂
= =
∂
rF 
Substituting into Lagrange's equation gives 
2b
2( ) θ sin θ 0
I
m x mx mg
r
+ − + =
 
Thus, the two differential equations of motion are 
2
b( )θ 2 θ cosθ τI I mx mxx mgx+ + + + = 
 
2b
2( ) θ sin θ 0
I
m x mx mg
r
+ − + =
 
b. Writing the above nonlinear equations in the matrix form. 
2
b
2b
2
0 θ τ 2 θ cosθ
θ sin θ0
I I mx mxx mgx
I x mx mgm
r
 + +    − −  =     −+      
 



 
Applying the Cramer’s rule gives 
b
2
2 2
b
θ ( )(τ 2 θ cosθ)
( )( θ sin θ)
I
m mxx mgx
r
x I I mx mx mg
= + − −
= + + −
 



 
 Choose θ, θ , x, and x as state variables and τ as the input. The state variable equations are 
1 2
b
2 2 3 4 3 12
3 4
2 2
4 b 3 2 3 1
( )( 2 cos )
( )( sin )
x x
I
x m u mx x x mgx x
r
x x
x I I mx mx x mg x
=
= + − −
=
= + + −




 
c. Linearize the nonlinear equations about the equilibrium, θ ≈ 0 and x ≈ 0. Note that the nonlinear terms θ 0xx ≈ , 
2 0x ≈ , 2θ 0x ≈ , cosθ ≈ 1, and sinθ ≈ θ. The linearized model of the system is 
b
2
b
θ ( )(τ )
( )( θ)
I
m mgx
r
x I I mg
= + −
= + −


 
and the state-space form is 
257 
 
 
1 1
b b
2 22 2
3 3
4 4
b
0 1 0 0 0
0 0 ( ) 0
0 0 0 1 0
( ) 0 0 0 0
x x
I Ix xmg m m
ur rx x
x xmg I I
               − + +   = +      
      
         − +     




 
1
1 2
2 3
4
1 0 0 0 0
0 0 1 0 0
x
y x
u
y x
x
 
       = +      
      
  
 
d. The Simulink block diagram for the nonlinear system is shown below, in which two Fcn blocks are used to 
construct θ and x . According to the order of the input signals, the expressions in the two Fcn blocks are 
for θ : (m+Ib/r^2)*(u(5)-m*g*u(3)*cos(u(1))-2*m*u(3)*u(4)*u(2)) 
for x : (I+Ib+m*u(3)*u(3))*(-m*g*sin(u(1))+m*u(3)*u(2)*u(2)) 
 
Figure Review5 No14b 
The Simulink block diagram for the linearized system can be built in the same way, only with different 
expressions defined in the two Fcn blocks: 
for θ : (m+Ib/r^2)*(u(5)-m*g*u(3)) 
for x : (I+Ib)*(-m*g*u(1)) 
The nonlinear and linear responses of the beam angle and the ball displacement are shown in the figure below. For 
a small torque input, the nonlinear and linearized models give almost the same results. 
 
Figure Review5 No14c 
0 2 4 6 8 10
Time (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
B
ea
m
 a
ng
le
 
 (r
ad
)
Nonliear
Linear
0 2 4 6 8 10
Time (s)
-0.08
-0.07
-0.06
-0.05
-0.04
-0.03
-0.02
-0.01
0
ba
ll 
di
sp
la
ce
m
en
t x
 (m
)
Nonliear
Linear
258 
 
 
Problem Set 6.1 
1. Determine the equivalent resistance Req for the circuit shown in Figure 6.8. 
 
Figure 6.8 Problem 1. 
Solution 
The equivalent resistance for the parallel-connected resistors R1 and R3 is 
eq1 1 3
1 1 1
R R R
= + or 1 3
eq1
1 3
R RR
R R
=
+
 
They are connected with the resistor R2 in series. The equivalent resistance for the circuit is 
1 3 1 2 2 3 1 3
eq 2 eq1 2
1 3 1 3
R R R R R R R RR R R R
R R R R
+ +
= + = + =
+ +
 
 
2. Determine the equivalent resistance Req for the circuit shown in Figure 6.9. 
 
Figure 6.9 Problem 2. 
Solution 
The equivalent resistance for the series-connected resistors R2 and R3 is 
eq1 2 3R R R= + 
They are connected with the resistor R1 in parallel. The equivalent resistance for the circuit is 
eq eq1 1 2 3 1
1 1 1 1 1
R R R R R R
= + = +
+
 or 1 2 1 3
eq
1 2 3
R R R RR
R R R
+
=
+ +
 
 
3. Determine the equivalent resistance Req for the circuit shown in Figure 6.10. Assume that all resistors have the 
same resistance of R. 
 
Figure 6.10 Problem 3. 
259 
 
 
Solution 
The equivalent resistance for the parallel-connected resistors is 
eq1 2 3 4
1 1 1 1 1 1 1 3
R R R R R R R R
= + + = + + = or eq1 3
RR = 
They are connected with the resistor R1 and R5 in series. The equivalent resistance for the circuit is 
71
eq 1 eq1 5 3 3R R R R R R R R= + + = + + = 
 
4. Determine the equivalent resistance Req for the circuit shown in Figure 6.11. Assume that all resistors have the 
same resistance of R. 
 
Figure 6.11 Problem 4. 
Solution 
The resistors R3, R4, and R5 are connected in series and the equivalent resistance is 
eq1 3 4 5 3R R R R R R R R= + + = + + = 
They are connected with R1 and R2 in parallel. The equivalent resistance for the circuit is 
eq 1 2 eq1
1 1 1 1 1 1 1 7
3 3R R R R R R R R
= + + = + + = or eq2
3
7
R R= 
 
5. A potentiometer is a variable resistor with three terminals. Figure 6.12a shows a potentiometer connected to a 
voltage source. The two end terminals are labeled as 1 and 2, and the adjustable terminal is labeled as 3. The 
potentiometer acts as a voltage divider, and the total resistance is separated into two parts as shown in Figure 
6.12b, where R13 is the resistance between terminal 1 and terminal 3, and R32 is the resistance between terminal 3 
and terminal 2. Determine the relationship between the input voltage vi and the output voltage vo. 
 
Figure 6.12 Problem 5. (a) Potentiometer and (b) voltage divider. 
Solution 
Denoting the current in the circuit to be i, we obtain the input voltage 
( )i 13 32v i R R= + 
and the output voltage 
o 32v iR= 
Thus the relationship between the input voltage iv and the output voltage ov is 
o 32
i 13 32
v R
v R R
=
+
 
260 
 
 
6. The output voltage of a voltage divider is not fixed but varies according to the load. 
a. Find the output voltage vo in Figure 6.13 for two different values of load resistance: (1) RL = 10kΩ and (2) RL 
= 100 kΩ. 
b. Is it possible to get an output voltage vo of greater than 9 V? If yes, determine the minimum value of the 
 load resistance. 
 
Figure 6.13 Problem 6. 
Solution 
a. If L 10 kR = Ω , then the equivalent resistance for the parallel-connected resistors is 
eq 2 L
1 1 1 1 1 19
9 10 90R R R
= + = + = or eq
90 k
19
R = Ω 
 Refer to the result in Problem 5. The output voltage is 
( )eq
o i
1 eq
90 19 10 8.26 V
1 90 19
R
v v
R R
= = =
+ +
 
If L 100 kR = Ω , following the same procedure gives 
eq 2 L
1 1 1 1 1 109
9 100 900R R R
= + = + = or eq
900 k
109
R = Ω 
and 
( )eq
o i
1 eq
900 109 10 8.92 V
1 900 109
R
v v
R R
= = =
+ +
 
b. If the output voltage o 9 Vv > , then we have 
eq eq
i
1 eq eq
(10) 9 V
1
R R
v
R R R
= >
+ +
 
 Solving for Req gives eq 9 kR > Ω . However, 
eq 2 L L
1 1 1 1 1 1
9 9R R R R
= + = + > 
which implies that eq 9 kRform as follows: 
a1 1
a2 2
d /d
1d / 0
0
d
R R vi t i
vi t i
C
L
R R
         + =                 

 
Taking the Laplace transform of the differential equations gives 
a1
a2
( )
1 (
( )
( ))
R I s
I s
Ls R V s
sV sRs Rs
C
+      =      
+   
 
Using Cramer’s rule to solve for I2(s), we have 
2
2
2
a( ) ( )I s V s
L RRLs s
C C
Ls
=
+ +
 
Applying the voltage-current relation for a capacitor, ov idt C= ∫ gives 
o a2 2
1( ) ( ) ( )LsV s I s V s
Cs RLCs Ls R
= =
+ +
 
Taking the inverse Laplace transform gives the same input-output equation as the one obtained in Problem 3, 
o o o aRLCv Lv Rv Lv+ + =   
 
5. Consider the circuit shown in Figure 6.27. Use the node method to derive the input–output differential equation 
relating i and va. 
 
Figure 6.27 Problem 5. 
Solution 
Note that at node 1, R Ci i= , that is, 
a 1 1v v dvC
R dt
−
= 
1 1 a
1 1Cv v v
R R
+ = 
267 
 
 
 
Figure PS6-2 No5 
Assume that all the initial conditions are zero. Taking the Laplace transform of the above equation yields 
1
a
( ) 1
( ) 1
V s
V s RCs
=
+
 or 1 a
1( ) ( )
1
V s V s
RCs
=
+
 
Applying Kirchhoff’s current law gives 
 R L C Li i i i i= + = + 
1
a
1dvi C v dt
dt L
= + ∫ 
Taking the Laplace transform of the above equation yields 
1 a
1( ) ( ) ( )I s CsV s V s
Ls
= + 
Substituting V1(s) and rearranging the equation gives 
2
2
a
( ) 1
( )
I s LCs RCs
V s RLCs Ls
+ +
=
+
 
Taking the inverse Laplace transform gives the input-output equation 
2
a a a2
d i diRLC L LCv RCv v
dtdt
+ = + +  
 
6. Repeat Problem 5 using the loop method. 
Solution 
Assign loop currents as shown in the figure below. For loop 1, applying Kirchhoff’s voltage law gives 
R C av v v+ = 
 
Figure PS6-2 No6 
1 1 a
1Ri i dt v
C
+ =∫ 
1
1 a
1diR i v
dt C
+ =  
Assume that all the initial conditions are zero. Taking the Laplace transform of the above equation yields 
1
a
( )
( ) 1
I s Cs
V s RCs
=
+
 or 1 a( ) ( )
1
CsI s V s
RCs
=
+
 
 
268 
 
 
For loop 2, applying Kirchhoff’s voltage law gives 
L av v= 
a
2diL v
dt
= 
Taking the Laplace transform of the above equation yields 
2 a
1( ) ( )I s V s
Ls
= 
Note that i = i1 + i2. Combining with the above two equations gives 
2
a a2
1 1( ) ( ) ( )
1
Cs LCs RCsI s V s V s
RCs Ls RLCs Ls
+ + = + = + + 
 
Taking the inverse Laplace transform gives the same input-output equation obtained in Problem 5, 
2
a a a2
d i diRLC L LCv RCv v
dtdt
+ = ++  
 
7. Consider the circuit shown in Figure 6.28. Use the node method to derive the input–output differential equation 
relating vo and va. 
 
Figure 6.28 Problem 7. 
Solution 
Applying Kirchhoff’s current law to node 1 gives 
 L R C 0i i i+ − = 
Expressing the current through each element in terms of the node voltage, we have 
a 1 1
a 1
1 ( ) 0
v v dvv v dt C
L R dt
−
− + − =∫ 
Differentiating the above equation with respect to time results in 
a 1 a 1 1
1 1 1 1 0v v v v Cv
L L R R
− + − − =   
Note that v1 = vo. Rearranging the equation gives 
o o o a aRLCv Lv Rv Lv Rv+ + = +   
 
Figure PS6-2 No7 
269 
 
 
8. Repeat Problem 7 using the loop method. 
Solution 
Assign loop currents as shown in the figure below. 
 
Figure PS6-2 No8 
For loop 1, applying Kirchhoff’s voltage law gives 
L C av v v+ = 
1
1 2 a
1 ( )
diL i i dt v
dt C
++ =∫ 
2
1
1 2 a2
1 1d iL i i v
C Cdt
+ + =  
Similarly, applying Kirchhoff’s voltage law to loop 2 gives 
R C av v v+ = 
1 22 a
1 ( )Ri i i dt v
C
+ =+ ∫ 
1 2 a
2 1 1diR i i v
dt C C
+ + =  
The above two differential equations can be written in matrix form as follows: 
2 2
a1 1 1
2 2
a2 22
d /d d /d0 0 0 1 1
d /d 1d0 /0 1d 0
L
R
vi t i t iC C
vi t iC Ci t
             + + =            
            


 
Taking the Laplace transform of the differential equations gives 
1
a
2
a
2
1 1
( )
( )
( )
( )1 1
Ls I sC C
I s
sV s
sV sRs
C C
 
    
=     
  
+
 
+   

 
Using Cramer’s rule to solve for I1(s) and I2(s), we have 
a1 2( ) ( )I s V s
RLCs
R
s
C
L R
s
=
+ +
 
2 2 a
2
( ) ( )I s V s
RLCs
L
s
C
L R
s
=
+ +
 
Applying the voltage-current relation for a capacitor, 1o 2(1 ) ( )C i i tv d= +∫ , gives 
2o 1 a2
1( ) [ ( ) ( )] ( )V s I s I s V s
Cs RLCs Ls
R
R
Ls
= + =
+ +
+ 
Taking the inverse Laplace transform gives the same input-output equation obtained in Problem 7, 
o o o a aRLCv Lv Rv Lv Rv+ + = +   
 
9. Consider the circuit shown in Figure 6.29. Use the node method to derive the transfer function Vo(s)/ Va(s). 
Assume that all initial conditions are zero. 
270 
 
 
 
Figure 6.29 Problem 9. 
Solution 
All currents entering or leaving node 1 and node 2 are labeled as shown in the figure below. 
 
Figure PS6-2 No9 
At node 1, applying Kirchhoff’s current law gives 
 R1 Li i= 
( )a 1
1 2
1
1v v
v v dt
R L
−
= −∫ 
Differentiating the above equation with respect to time results in 
1 1 2 a
1 1
1 1 1 1v v v v
R L L R
+ − =  or 1 1 1 1 2 aLv R v R v Lv+ − =  
Applying Kirchhoff’s current law to node 2 and following the similar procedure give 
L R2 Ci i i= + 
2 2
1 2
2
1 ( )
v dvv v dt C
L R dt
− = +∫ 
Differentiating the above equation with respect to time results in 
2 2 2 1
2
1 1 1 0Cv v v v
R L L
+ + − =  or 2 2 2 2 2 2 1 0R LCv Lv R v R v+ + − =  
The above two differential equations can be written in matrix form as follows: 
 1 1 1 1 1 a
2 2 2 2 2 2
0 0 0
0 0 0
v v R R vL Lv
R LC v v R R vL
−            
+ + =            −            
  
 
 
Taking Laplace transform gives 
( )
( )
( )1 1 1 a
2
2 2 2 2 0
Ls R R V s LsV s
R R LCs Ls R V s
+ −        =    − + +      
 
Note that o 2v v= . Applying Cramer’s rule gives 
( )
( )
( )
( )
o 2 2
2
a a 2 1 2 1 2
V s V s R
V s V s R LCs Ls R R Cs R R
= =
+ + + +
 
 
10. Reconsider the circuit shown in Figure 6.29. Use the loop method to derive the transfer function Vo(s)/ Va(s). 
Assume that all initial conditions are zero. 
Solution 
Assign loop currents as shown in the figure below. 
271 
 
 
 
Figure PS6-2 No10 
For loop 1, applying Kirchhoff’s voltage law gives 
 R1 L R2 av v v v+ + = 
1
1 1 2 1 2 a( )
diR i L R i i v
dt
+ + − = 
Similarly, applying Kirchhoff’s voltage law to loop 2 gives 
C R2 0v v+ = 
2 2 2 1
1 ( ) 0i dt R i i
C
+ − =∫ 
2 1
2 2 2
1 0
di dii R R
C dt dt
+ − = 
The above two differential equations can be written in matrix form as follows: 
 
1
1 2 2
1 a
2 2 22
0
1 00
di R R RL i vdt
R R idi
Cdt
  + −          + =        −           
 
Taking Laplace transform gives 
( )
( )
( )1 1 2 2
1 a
22 2
1 0
L s R R R I s V s
I sR s R s
C
+ + −        =    − +       
 
Note that o C 2(1 )v v C i dt= = ∫ . Applying Cramer’s rule gives the same transfer function as the one obtained in 
Problem 9, 
( )
( )
( )
( )
o 2 2
2
a a 2 1 2 1 2
1V s I s R
V s Cs V s R LCs Ls R R Cs R R
= =
+ + + +
 
 
11. Consider the circuit shown in Figure 6.26. Determine a suitable set of state variables and obtain the state-space 
representation with vo as the output. 
Solution 
Refer to the figure shown in Problem 3. Note that the circuit has two energy storage elements, L and C. This implies 
that two states are needed, and they are 
 1 L 2 C,x i x v= = 
Their time derivatives are 
 L
1 L
1dix v
dt L
= = 
C
2 C
1dv
x i
dt C
= = 
where 
L C 2v v x= = 
a C
C R L L 2 1
1 1v v
i i i i u x x
R R R
−
= − = − = − − 
272 
 
 
Thus, the complete set of two state-variable equations is 
 1 2
1x x
L
= 
 2 2 1
1 1 1x u x x
RC RC C
= − − 
The output equation is 
 o C 2y v v x= = = 
The state equation and the output equation can be written in matrix form as follows:1 1
2 2
1 00
11 1
x xL u
x x
RCC RC
         = +          − −     


, [ ] 1
2
0 1 0
x
y u
x
 
= + ⋅ 
 
 
 
12. Repeat Problem 11 for the circuit shown in Figure 6.27. 
Solution 
Refer to the figure shown in Problem 5. Note that the circuit has two energy storage elements, C and L. This implies 
that two states are needed, and they are 
 1 C 2 L,x v x i= = 
Their time derivatives are 
 C
1 C
1dv
x i
dt C
= = 
L
2 L
1dix v
dt L
= = 
where 
a C
C R 1
1 1v v
i i u x
R R R
−
= = = − 
L av v u= = 
Thus, the complete set of two state-variable equations is 
 1 1
1 1x u x
RC RC
= − 
 2
1x u
L
= 
The output equation is 
 a C
R L L 1 2
1 1v v
y i i i i u x x
R R R
−
= = + = + = − + 
The state equation and the output equation can be written in matrix form as follows: 
1 1
2 2
11 0
10 0
x x RC uRC
x x
L
    −    = +                


 
1
2
1 11
x
y u
xR R
  = − + ⋅     
 
 
13. Repeat Problem 11 for the circuit shown in Figure 6.28. 
Solution 
Refer to the figure shown in Problem 7. Note that the circuit has two energy storage elements, L and C. This implies 
that two states are needed, and they are 
 1 L 2 C,x i x v= = 
Their time derivatives are 
273 
 
 
 L
1 L
1dix v
dt L
= = 
C
2 C
1dvx i
dt C
= = 
where 
L R a C 2v v v v u x= = − = − 
a C
C L R L 1 2
1 1v v
i i i i x u x
R R R
−
= + = + = + − 
Thus, the complete set of two state-variable equations is 
 1 2
1 1x u x
L L
= − 
 2 1 2
1 1 1x x u x
C RC RC
= + − 
The output equation is 
 C 2y v x= = 
The state equation and the output equation can be written in matrix form as follows: 
1 1
2 2
1 10
1 1 1
x xL L u
x x
C RC RC
   −      
= +      
      −      


 
[ ] 1
2
0 1 0
x
y u
x
 
= + ⋅ 
 
 
 
14. Repeat Problem 11 for the circuit shown in Figure 6.29. 
Solution 
Note that the circuit has two energy storage elements, L and C. This implies that two states are needed, and they are 
 1 L 2 C,x i x v= = 
Their time derivatives are 
 L
1 L
1dix v
dt L
= = 
C
2 C
1dvx i
dt C
= = 
where 
L a R1 C a 1 L C 1 1 2v v v v v R i v u R x x= − − = − − = − − 
C
C L R2 L 1 2
2 2
1vi i i i x x
R R
= − = − = − 
Thus, the complete set of two state-variable equations is 
 1
1 1 2
1 1Rx u x x
L L L
= − − 
 2 1 2
2
1 1x x x
C R C
= − 
The output equation is 
 o C 2y v v x= = = 
The state equation and the output equation can be written in matrix form as follows: 
1
1 1
2 2
2
1
1
1 1
0
R
x xL L uL
x x
C R C
 − −         = +         −    


, [ ] 1
2
0 1 0
x
y u
x
 
= + ⋅ 
 
 
274 
 
 
15. Repeat Problem 9 for the circuit shown in Figure 6.30. 
 
Figure 6.30 Problem 15. 
Solution 
Note that the circuit has two energy storage elements, L1 and L2. This implies that two states are needed, and they are 
 1 L1 2 L2,x i x i= = 
Their time derivatives are 
 L1
1 L1
1
1dix v
dt L
= = 
L2
1 L2
2
1dix v
dt L
= = 
where 
L1 a R1 R2 a 1 L1 2 L1 L2 1 2 1 2 2( ) ( )v v v v v R i R i i u R R x R x= − − = − − − = − + + 
L2 R2 2 L1 L2 2 1 2 2( )v v R i i R x R x= = − = − 
Thus, the complete set of two state-variable equations is 
 1 2 2
1 1 2
1 1 1
1 R R Rx u x x
L L L
+
= − + 
 2 2
2 1 2
2 2
R Rx x x
L L
= − 
The output equation is 
 o R2 2 L1 L2 2 1 2 2( )y v v R i i R x R x= = = − = − 
The state equation and the output equation can be written in matrix form as follows: 
1 2 2
1 11 1
1
2 22 2
2 2
1
0
R R R
L Lx x
L u
x xR R
L L
+ −         = +         −    
 


, [ ] 1
2 2
2
0
x
y R R u
x
 
= − + ⋅ 
 
 
 
16. Repeat Problem 9 for the circuit shown in Figure 6.31. 
 
Figure 6.31 Problem 16. 
Solution 
Note that the circuit has three independent energy storage elements, L1, L2, and C. This implies that three states are 
needed, and they are 
275 
 
 
1 L1 2 L2 3 C, ,x i x i x v= = = 
Their time derivatives are 
L1
1 L1
1
L2
2 L2
2
C
3 C
1
1
1
dix v
dt L
dix v
dt L
dv
x i
dt C
= =
= =
= =



 
where 
L1 a R1 a 1 L1 L2 1 1 2
L2 R1 R2 1 L1 L2 C 1 1 1 2 3
C
C L2 R 2 L2 2 3
2 2
( ) ( )
( )
1
v v v v R i i u R x x
v v v R i i v R x R x x
v
i i i i x x
R R
= − = − − = − −
= − = − − = − −
= − = − = −
 
Thus, the complete set of three state-variable equations is 
1 1
1 1 2
1 1 1
1 1
2 1 2 3
2 2 2
3 2 3
2
1
1
1 1
R Rx u x x
L L L
R Rx x x x
L L L
x x x
C R C
= − +
= − −
= −



 
The output equation is 
o C 3y v v x= = = 
The state equation and the output equation can be written in matrix form as follows: 
[ ]
1 1
1 1
11 1
1 1
2 2
2 2 2
3 3
2
1
2
3
0 1
1 0 , 
0
1 10
0 0 1 0
R R
L L Lx x
R Rx x u
L L L
x x
C R C
x
y x u
x
 
−   
             = − − +    
     
      
 −     
 
 = + ⋅ 
 
 



 
 
Problem Set 6.3 
1. The op-amp circuit shown in Figure 6.37 is a summing amplifier. Determine the relation between the input 
voltages v1, v2, and the output voltage vo. 
 
Figure 6.37 Problem 1. 
Solution 
Applying Kirchhoff’s current law to node 1 gives 
276 
 
 
 1 2 3 0i i i i−+ − − = 
Because the current drawn by the op-amp is very small, i.e., 0i− ≈ , we have 
1 2 3i i i+ = 
Using the voltage-current relation for each resistor yields 
o1 2
1 2 3
v vv v v v
R R R
−− − −− −
+ = 
where 0v v− +≈ ≈ . Thus, the relation between the input voltage vi and the output voltage vo is 
 o1 2
1 2 3
vv v
R R R
−
+ = or 3 3
o 1 2
1 2
R Rv v v
R R
 
= − + 
 
 
 
Figure PS6-3 No1 
 
2. The op-amp circuit shown in Figure 6.38 is a difference amplifier. Determine the relation between the input 
voltages v1, v2, and the output voltage vo. 
 
Figure 6.38 Problem 2. 
Solution 
Because the current drawn by the op-amp is very small, i.e., 0i i− +≈ ≈ , we have at node 1, 
1 2i i= 
o1
1 2
v vv v
R R
−− −−
= or 2 1 1 o 1 2( )R v R v R R v−+ = + 
and at node 2, 
3 4i i= 
2
1 2
v v v
R R
+ +−
= or 2 2 1 2( )R v R R v+= + 
where v v− +≈ . Eliminating v_ and v+ gives 
2 1 1 o 2 2R v R v R v+ = or 2
o 2 1
1
( )Rv v v
R
= − 
 
Figure PS6-3 No2 
277 
 
 
3. The op-amp circuit shown in Figure 6.39 is a noninverting amplifier. Determine the relation between the input 
voltage vi and the output voltage vo. 
 
Figure 6.39 Problem 3. 
Solution 
Because the current drawn by the op-amp is very small, i.e., 0i i− +≈ ≈ , we have 
R1 R 2i i= 
o
1 2
0 v vv
R R
−− −−
= 
where iv v v− +≈ = . Substituting it into the above equation and solving for vo gives 
( )o 2 1 i1v R R v= + 
 
4. Consider the op-amp integrator circuit shown in Figure 6.35. Derive the differential equation relating the input 
voltage vi and the output voltage vo. 
Solution 
Note that the current drawn by the op-amp is very small. Applying Kirchhoff’s current law gives 
R Ci i= 
( )i
o
v v dC v v
R dt
−
−
−
= − 
where 0v v− +≈ = . Thus, the differential equation for the op-amp integrator is 
i
o
v Cv
R
= −  or o i
1v v
RC
= − 
 
Figure PS6-3 No4 
 
5. Consider the op-amp circuit shown in Figure 6.40. Derive the differential equation relating the input voltage vi 
and the output voltage vo. 
 
Figure 6.40 Problem 5. 
Solution 
Note that the current drawn by the op-amp is very small. Applying Kirchhoff’s current law to node 1 gives 
278 
 
 
1 2i i= 
1 o1 ( )0 d v vv C
R dt
−−
= 
where 1v v v− += ≈ . Applying Kirchhoff’s current law to node 2 gives 
3 4i i= 
( )2i 2 0d vv v C
R dt
−−
= 
where 2 1v v v v+ −= ≈ = . 
 
Figure PS6-3 No5 
The above two equations can be rewritten in the s-domain with the assumption of zero initial conditions, 
1 o( 1) ( ) ( )RCs V s RCsV s+ = 
i 2( ) ( 1) ( )V s RCs V s= + 
Eliminating V1(s) and V2(s) gives 
i o( ) ( )V s RCsVs= 
Transforming from the s-domain to the time domain, we obtain the differential equation of the system 
o iRCv v= 
 
6. Repeat Problem 5 for the op-amp circuit shown in Figure 6.41. 
 
Figure 6.41 Problem 6. 
Solution 
Denote the currents entering and leaving the node as i1 and i2, respectively. We have 
i 1 1 1
1
1v v i dt R i
C−− = +∫ 
o 2 2 2
2
1v v i dt R i
C− − = +∫ 
where 0v v− +≈ = . Because the current drawn by the op-amp is very small, i.e., 0i i− +≈ ≈ , we have 1 2i i i= = . 
Therefore, the above two equations can be rewritten as 
i 1
1
1v idt R i
C
= +∫ 
o 2
2
1v idt R i
C
− = +∫ 
or in the s-domain with the assumption of zero initial conditions, 
1 i 1 1( ) (1 ) ( )C sV s R C s I s= + 
279 
 
 
2 o 2 2( ) (1 ) ( )C sV s R C s I s− = + 
Eliminating I(s) gives 
2 1 1 o 1 2 2 i(1 ) ( ) (1 ) ( )C R C s V s C R C s V s+ = − + 
Transforming from the s-domain to the time domain, we obtain the differential equation of the system 
1 1 2 o 2 o 2 1 2 i 1 iR C C v C v R C C v C v+ = − −  
 
Figure PS6-3 No6 
 
Problem Set 6.4 
1. Reconsider the armature-controlled motor in Figure 6.44. Equations 6.31 and 6.32 represent the dynamics of the 
system in terms of the variables ia and θ. 
a. Assuming the angle θ to be the output, draw a block diagram to represent the dynamics of the 
armature-controlled motor. 
b. Derive the transfer functions Θ(s)/Va(s) and Θ(s)/TL(s). All of the initial conditions are assumed to be zero. 
 c. Determine the state-space form. 
Solution 
a. Assuming the angle θ to be the output, the differential equations of the system are 
a
a a a e aθdiL R i K v
dt
+ + = 
t a Lθ θ τI B K i+ − = −  
All of the initial conditions are assumed to be zero. Taking the Laplace transform of both equations results in 
( ) ( ) ( ) ( )a a a a e aL sI s R I s K s s V s+ + Θ = 
2
t a L( ) ( ) ( ) ( )Is s Bs s K I s T sΘ + Θ − = − 
The block diagram is shown in the figure below. 
 
Figure PS6-4 No1 
b. The equations expressed in s-domain given in Part (a) can be rewritten in matrix form 
( )
( )
( )
( )
a a e a a
2
t L
L s R K s I s V s
K Is Bs s T s
+     
=    − + Θ −     
 
Applying Cramer’s rule gives 
( ) ( )
( ) ( ) ( )
( ) ( ) ( )a a t
a2 2
a a t e a a t e
L
L s R Ks T s V s
L s R Is Bs K K s L s R Is Bs K K s
− +
Θ = +
+ + + + + +
 
Thus, the transfer functions L( ) ( )s T sΘ and a( ) ( )s V sΘ are 
( )
( ) ( )
( ) ( )
a a
2
L a a t e
t
2
a a a t e
( )
( )
( )
( )
L s Rs
T s L s R Is Bs K K s
Ks
V s L s R Is Bs K K s
− +Θ
=
+ + +
Θ
=
+ + +
 
280 
 
 
c. Refer to the differential equations in Part (a). Let 
1 a 2 3θ θx i x x= = =  
and 
1 a 2 Lτu v u= = 
The state variable equations are 
( ) ( )a
1 a a a e a 1 e 3 1
a a
1 1θdix v R i K R x K x u
dt L L
= = − − = − − +
 
2 3θx x= = 
( ) ( )3 L t a t 1 3 2
1 1θ τ θx B K i K x Bx u
I I
= = − − + = − − 
 
Assuming the angle θ to be the output gives 
2θy x= = 
Thus, the state equation and the output equation in matrix form are 
a e
a a1 1 a
1
2 2
2
3 t 3
0
1 0
0 0 1 0 0
0 1
0
R K
L Lx x L
u
x x
u
x K x IB
I I
 − −               = +              −      −
  



, [ ] [ ]
1
1
2
2
3
0 1 0 0 0
x
u
y x
u
x
 
  = +   
  
 
 
 
2. Reconsider the field-controlled motor in Figure 6.47. Equations 6.44 and 6.45 represent the dynamics of the 
system in terms of the variables if and θ. 
a. Assuming the angle θ to be the output, draw a block diagram to represent the dynamics of the field-controlled 
motor. 
b. Derive the transfer functions Θ(s)/Vf(s) and Θ(s)/TL(s). All of the initial conditions are assumed to be zero. 
c. Determine the state-space form. 
Solution 
a. The differential equations of the field-controlled DC motor shown in Figure 6.47 are 
f
f f f f
d
d
iL R i v
t
+ = 
t f Lθ θ τI B K i+ − = −  
All of the initial conditions are assumed to be zero. Taking the Laplace transform of both equations results in 
f f f f f( ) ( ) ( )L sI s R I s V s+ = 
2
t f L( ) ( ) ( ) ( )Is s Bs s K I s T sΘ + Θ − = − 
The block diagram is shown in the figure below. 
 
Figure PS6-4 No2 
b. The equations expressed in s-domain given in Part (a) can be rewritten in matrix form 
( )
( )
( )
( )
f f f f
2
t L
0L s R I s V s
K Is Bs s T s
+     
=    − + Θ −     
 
Applying Cramer’s rule gives 
( ) ( )L f2 2
f f
1( ) ( ) ( )tKs T s V s
Is Bs L s R Is Bs
−
Θ = +
+ + +
 
Thus, the transfer functions L( ) ( )s T sΘ and f( ) ( )s V sΘ are 
281 
 
 
( ) ( )
2
L
t
2
f f f
( ) 1
( )
( )
( )
s
T s Is Bs
Ks
V s L s R Is Bs
Θ −
=
+
Θ
=
+ +
 
c. Refer to the differential equations in Part (a). Let 
1 f 2 3θ θx i x x= = =  
and 
1 f 2 Lτu v u= = 
The state variable equations are 
( ) ( )f
1 f f f f 1 1
f f
1 1dix v R i R x u
dt L L
= = − = − + 
2 3θx x= = 
( ) ( )3 L t f t 1 3 2
1 1θ τ θx B K i K x Bx u
I I
= = − − + = − − 
 
Assuming the angle θ to be the output gives 
2θy x= = 
Thus, the state equation and the output equation in matrix form are 
f
f
f1 1
1
2 2
2
3 t 3
1 00 0
0 0 1 0 0
0 10
R
LLx x
u
x x
u
x K xB
I I
I
 
   −              = +       
           −  
  −  
 



, [ ] [ ]
1
1
2
2
3
0 1 0 0 0
x
u
y x
u
x
 
  = +   
   
 
 
3. Consider the electromechanical system shown in Figure 6.49a. It consists of a cart of mass m moving without 
slipping on a ground track. The cart is equipped with an armature-controlled DC motor, which is coupled to a rack 
and pinion mechanism to convert the rotational motion to translation and to create the driving force f for the 
system. Figure 6.49b shows the equivalent electric circuit and the mechanical model of the DC motor, where r is 
the radius of the motor gear. The torque and the back emf constants of the motor are Kt and Ke, respectively. 
a. Determine the transfer function X(s)/Va(s). Assume that all initial conditions are zero. 
b. Determine the differential equation of the system relating the cart position x and the applied voltage va. 
 
Figure 6.49 Problem 3. 
Solution 
a. For the electrical circuit shown in Figure 6.49(b), applying Kirchhoff’s voltage law gives 
a
a a a b a 0diR i L e v
dt
+ + − = 
where b e eω θe K K= =  . Using the geometric constraint, θx r= , which implies that θ x r=  , we have 
a e
a a a a
di KL R i x v
dt r
+ + = 
Consider the translational motion of the cart as shown in Figure 6.49(a). Applying Newton’s second law along 
x-direction gives 
282 
 
 
f mx=  
where the driving force generated by the motor is 
tm
a
τ Kf i
r r
= = 
Thus, we have 
t
a 0K i mx
r
− = 
Rewriting the equations for the electrical and mechanical subsystems in the s-domain gives 
e
a a a a( ) ( ) ( ) ( )KL s R I s sX s V s
r
+ + = 
2t
a ( ) ( ) 0K I s ms X s
r
− = 
Applying Cramer’s rule gives 
t
2 2
a a a t e
( )
( ) ( )
K rX s
V s r L s R ms K K s
=
+ +
 
b. For the transfer function obtained in Part (a), taking inverse Laplapce transform yields 
2 2
a a t e t ar L mx r R mx K K x K rv+ + =   
 
4. Consider the single-link robot arm as shown in Figure 6.50a. It is driven by an armature-controlled DC motor 
through spur gears with a total gear ratio of N, and θ = Nθm. The mass moments of inertia of the motor and the load 
are Im and I, respectively. The coefficients of torsional viscous damping of the motor and the load are Bm and B, 
respectively. Figure 6.50b shows the equivalent electric circuit and the mechanical model of the DC motor. The 
torque and the back emf constants of the motor are Kt and Ke, respectively. 
a. Determine the transfer function Θ(s)/Va(s). Assume that all initial conditions are zero. 
b. Determine the differential equation relating the applied voltage va and the link angular displacement θ. 
 
Figure6.50 Problem 4. 
Solution 
a. The differential equation of the electrical subsystem of the motor is 
a
a a a b a 0diR i L e v
dt
+ + − = 
 where b e mθe K=  . By the geometry of the gears, 
mθ θN=  
 where N is the gear ratio, we have 
a e
a a a aθdi KR i L v
dt N
+ + = 
Refer to Example 5.19. The differential equation of the robot arm relating the angle mθ and the motor torque mτ 
is given by 
2 2
m m m m m( ) ( )I N I B N B+ θ + + θ = τ  
283 
 
 
where m t aτ K i= . By the geometry of the gears, mθ θN=  and mθ θN=  , the above equation can be rewritten as 
tm m
a22 0KI BI B
NNN
i   + θ + + θ   

−
 
=

  
Rewriting the equations for the electrical and mechanical subsystems in the s-domain gives 
e
a a a a( ) ( ) ( ) ( )KL s R I s s s V s
N
+ + Θ = 
( ) ( ) ( )2
m m t
2 2
a (( ) 0)I I s B sN B s KN N I s + + + Θ − = 
Applying Cramer’s rule gives 
( ) ( ) ( )2 2 2
a a a e
t
2
m m t
( )
( ) ( )
Ns
V s L s R N B N K N s
K
I I s B s K + + + + 
Θ
=
+
 
b. For the transfer function obtained in Part (a), taking inverse Laplapce transform yields 
( ) ( ) ( ) ( ) ( ) ( )2
m m m m t
2 2 2 2
a a a ea atθ θ θ θ+ θI I B I I RL N L B N R N B N NK N K vB K+ + + + + + + =     
 
5. A more complicated model of the armature-controlled motor is shown in Figure 6.51, in which the rotor is 
connected to an inertial load through a flexible and damped shaft. Km and Bm represent the torsional stiffness and 
the torsional viscous damping of the shaft, respectively. The mass moments of inertia of the motor and the load 
are Im and IL, respectively. Let m mω θ=  and L Lω θ=  . 
 
Figure 6.51 Problem 5. 
a. Assuming zero initial conditions, derive the transfer functions ΩL(s)/Va(s) and ΩL(s)/TL(s). 
b. Assuming the angular velocity ωL to be the output, draw a block diagram to represent the dynamics of the 
field-controlled motor. 
c. Determine the state-space form. 
Solution 
a. For the electrical part, applying Kirchhoff’s current law gives 
a
a a a e m a
d
d
θiR i L K v
t
+ + = 
where e m bθK e= . The mechanical part becomes a two-degree-of-freedom system. Assuming that θm > θL and 
applying the moment equation to the rotor and the load gives 
t a m m L m m L m m(θ θ ) (θ θ ) θK i K B I− − − − =   
( ) ( )L m m L m m L L L L L L Lτ θ θ θ θ θ θ θK B B K I− + − + − − − =    
 where t a mτK i = . Assuming all the initial conditions to be zero and taking the Laplace transform, we have 
a a a e m a) ( ) ( )( ( )L R Ks I s s s sV+ =+ Θ 
m m m m m m a
2
L t( ) ( ) ) ( ) ( )( 0I B K B Ks s s s s I sK+ + Θ − + Θ − = 
L m L m L m m m L
2
L+ ) ( ) ( ) ( ) ( )( s s s K sI B s sK sB K B+ + Θ = −Τ+ Θ − + 
Applying Cramer’s rule gives 
( ) ( ) ( ) ( )
2
a a m m m t e t m m
L L a
( )( ) ( ) ( )L s R I s B s K K K s K B s Ks T s V s
s s
+ + + + +
Θ = − +
∆ ∆
 
where 
( ) ( ) ( ) ( ) ( )22 2
a a m m m L m L m L m m
2
t e L m L m L( )
s L s R I s B s K I s B s B s K K B s K
K K s I s B s B s K K
 ∆ = + + + + + + + − + 
+ + + + +
 
284 
 
 
Thus, the transfer functions L L( ) T ( )s sΩ and L a( ) ( )s V sΩ are 
( )
( )
( )
( ) ( )
2 2
L L a a m m m t e
L L
( )( )
T T
s s s s L s R I s B s K K K s
s s s
Ω Θ + + + +
= = −
∆
 
( ) ( )
( )
L L t m m
a a
( )
( ) ( )
s s s sK B s K
V s V s s
Ω Θ +
= =
∆
 
b. A block diagram representing the dynamics of the armature-controlled motor is shown below. 
 
Figure PS6-4 No5 
c. Refer to the differential equations in Part (a). Let 
1 a 2 m 3 L 4 m 5 Lθ θ θ θx i x x x x= = = = =  
and 
1 a 2 Lτu v u= = 
The state variable equations are 
( ) ( )a a e m a
a
1 a 1 4
a
e 1
a
1 1θR i K R
di
x v x x u
d
K
t L L
= = − − = − − +
 
2 m 4θx x= =
 
3 L 5θx x= =
 
( )
4 m
m
t 1 m 2 m 3 m 4 m 5
m
t a m m L m m L(θ θ ) (θ θ )1θ
1
Kx
I
K x K x
i K B
K x B x B x
I
 = =  
= − + − +
− − − −  

 
( ) ( )
( ) ( )
5 L L
L
2
m m L m m L
m 2 m L 3 m 4 L
L
L L
m
L L
5
θ θ θ θ θ θ1θ τ
1
x
I
u K x K K x B x B B
B K
I
K
x
B = = − + 
= − + − + + − +
− + − − −
  
 


 
 Assuming the angular velocity Lω to be the output gives 
L L 5ω θy x= = = 
Thus, the state equation and the output equation in matrix form are 
285 
 
 
a e
a a a
1 1
2 2
3 3
4 4
t m m m m
5 5
m m m m m
m m L m m L
LL L L L
10 0 0 0
0 0 0 1 0 0 0
0 00 0 0 0 1
0 0
100
R K
L L L
x x
x x
x x
x x
K K K B B
x xI I I I I
K K K B B B
II I I I
   − −   
   
   
      
      
        = +     
     
     
− −        
  
  + + −− −  
  





1
2
u
u
  
 
 





[ ] [ ]
1
2
1
3
2
4
5
0 0 0 0 1 0 0
x
x
u
y x
u
x
x
 
 
    = +   
  
 
  
 
 
6. A more complicated model of the field-controlled motor is shown in Figure 6.52, in which the rotor is connected 
to an inertial load through a flexible and damped shaft. Km and Bm represent the torsional stiffness and the 
torsional viscous damping of the shaft, respectively. The mass moments of inertia of the motor and the load are Im 
and IL, respectively. 
a. Assuming zero initial conditions, derive the transfer functions ΩL(s)/Vf(s) and ΩL(s)/TL(s). 
b. Assuming the angular velocity ωL to be the output, draw a block diagram to represent the dynamics of the 
field-controlled motor. 
c. Determine the state-space form. 
 
Figure 6.52 Problem 6. 
Solution 
a. Applying Kirchhoff’s current law to the field circuit and the moment equation to the mechanical part gives 
 f
f f f f
diL R i v
dt
+ = 
( ) ( )m m m m L m m L m t fθ θ θ θ θ τI B K K i+ − + − = =   
L L L m L m m L m L m m Lθ ( )θ θ ( )θ θ τI B B B K K K+ + − + + − = −   
The above three equations are the system differential equations of the field-controlled DC motor shown in Figure 
6.52. Assuming all the initial conditions to be zero and taking the Laplace transform, we have 
 f f f f f( ) ( ) ( )L sI s R I s V s+ = 
286 
 
 
( ) ( )2
m m m m m m L t f( ) ( ) ( )I s B s K s B s K s K I s+ + Θ − + Θ = 
( ) ( ) ( ) ( ) ( )2
L L m L m L m m m LI s B s B s K K s B s K s T s+ + + + Θ − + Θ = − 
Applying Cramer’s rule gives 
( )
( ) ( )
( ) ( ) ( )
( )
2
f f m m m t m m
L L fT ( )
L s R I s B s K K B s K
s s V s
s s
− + + + +
Θ = +
∆ ∆
 
where 
( ) ( ) ( ) ( ) ( )22 2
f f m m m L L m L m m ms L s R I s B s K I s B s B s K K B s K ∆ = + + + + + + + − +  
Thus, the transfer functions L L( ) T ( )s sΩ and L f( ) ( )s V sΩ are 
( )
( )
( )
( )
( )( )
( )
2
f f m m mL L
L L
s L s R I s B s Ks s s
T s T s s
− + + +Ω Θ
= =
∆
 
( ) ( ) ( )
( )
L L t m m
f f( ) ( )
s s s K s B s K
V s V s s
Ω Θ +
= =
∆
 
b. A block diagram representing the dynamics of the field-controlled motor is shown below. 
theta_m_dot theta_m
theta_L_dot theta_L
i_f tau_mv_f
tau_L
1
Lf.s+Rf
1
s
1
s
1
s
1
s
1/IL
1/Im
Kt
KL
Bm
BL
Km
 
Figure PS6-4 No6 
c. Refer to the differential equations in Part (a). Let 
1 f 2 m 3 L 4 m 5 Lθ θ θ θx i x x x x= = = = =  
and 
1 f 2 Lτu v u= = 
The state variable equations are 
( ) ( )f
1 f f f f 1 1
f f
1 1dix v R i R x u
dt L L
= = − = − + 
2 m 4θx x= =
 
3 L 5θx x= =
 
( ) ( ) ( )4 m t f m m L m m L t 1 m 2 m 3 m 4 m 5
m m
1 1θ θ θ θ θx K i K B K x K x K x B x B x
I I
 = = − − − − = − + − + 
  
 
[ ]
5 L L m m L m L m m L m L
L
2 m 2 L m 3 m 4 L m 5
L
1θ τ θ ( )θ θ ( )θ
1 ( ) ( )
x K K K B B B
I
u K x K K x B x B B x
I
 = = − + − + + − + 
= − + − + + − +
  

 
287 
 
 
 Assuming the angular velocity Lω to be the output gives 
L L 5ω θy x= = = 
Thus, the state equation and the output equation in matrix form are 
f
f f
1 1
2 2
3 3
4 4
t m m m m
5 5
m m m m m
m L m m L m
LL L L L
10 0 0 0 0
0 0 0 1 0 0 0
0 00 0 0 0 1
0 0
100
R
L L
x x
x x
x x
x x
K K K B B
x xI I I II
K K K B B B
II I I I
   −   
   
   
      
      
         = +      
      
      
− −        
  
  + +
−− −  
  





1
2
u
u
 
 
 




 
[ ] [ ]
1
2
1
3
2
4
5
0 0 0 0 1 0 0
x
x
u
y x
u
x
x
 
 
    = +   
  
 
  
 
 
Problem Set 6.5 
1. Reconsider the RC circuit shown in Figure 6.24. Use the impedance method to determine the transfer function 
I(s)/Va(s) and the input–output differential equation relating vC and va. Assume that all the initial conditions are 
zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain. 
 
Figure PS6-5 No1 
Note 
 1Z R= 
 2
1( )Z s
Cs
= 
Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields 
 1 2 a( ) ( ) ( ) ( )Z I s Z s I s V s+ = 
Thus, the transfer function relating the input voltage va and the output current i is 
 
a 1 2
( ) 1
( ) ( ) 1
I s Cs
V s Z Z s RCs
= =
+ +
 
Note that the current is related to the output voltage by 
 C 2( ) ( ) ( )V s Z s I s= 
Thus, we have 
C 2
a a
( ) ( ) ( ) 1
( ) ( ) 1
V s Z s I s
V s V s RCs
= =
+
 
288 
 
 
By transforming C a( ) ( )V s V s from the s domain to the time domain with the assumption of zero initial conditions, we 
obtain the differential equation of the system 
 C C aRCv v v+ = 
 
2. Reconsider the RL circuit shown in Figure 6.25. Use the impedance method to determine the transfer function 
VL(s)/Va(s) and the input–output differential equation relating iL and va. Assume that all the initial conditions are 
zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the 
figure below, where 
 1Z R= 
 2 ( )Z s Ls= 
 
Figure PS6-5 No2 
Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields 
 1 L 2 L a( ) ( ) ( ) ( )Z I s Z s I s V s+ = 
Thus, the transfer function relating the input voltage va and the output current iL is 
L
a 1 2
( ) 1 1
( ) ( )
I s
V s Z Z s R Ls
= =
+ +
 
Note that the current is related to the output voltage by 
 ( ) ( ) ( )L 2 LV s Z s I s= 
Thus, we have 
L 2 L
a a
( ) ( )
( ) ( )
V s Z s I Ls
V s V s R Ls
= =
+
 
Also, by transforming L a( ) ( )I s V s from the s domain to the time domain with the assumption of zero initial 
conditions, we obtain the differential equation of the system 
 L
L a
diL Ri v
dt
+ = 
 
3. Reconsider the RLC circuit shown in Figure 6.26. Use the impedance method to determine the input–output 
differential equation relating vo and va. Assume that all the initial conditions are zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the 
figure below. 
 
Figure PS6-5 No3 
289 
 
 
Note that 
1Z R= 
 
2
1 1
( )
Cs
Z s Ls
= + or 2 2( )
1
LsZ s
LCs
=
+
 
Applying Kirchhoff’s voltage law in the equivalent impedance circuit yields 
 1 2 a( ) ( ) ( ) ( )Z I s Z s I s V s+ = 
Thus, the transfer function relating the input voltage va and the current i is 
2
2
a 1 2
( ) 1 1
( ) ( )
I s LCs
V s Z Z s RLCs Ls R
+
= =
+ + +
 
Note that the current is related to the output voltage by 
 o 2( ) ( ) ( )V s Z s I s= 
Thus, we have 
 o 2
2
a a
( ) ( ) ( )
( ) ( )
V s Z s I s Ls
V s V s RLCs Ls R
= =
+ +
 
By transforming o a( ) ( )V s V s from the s domain to the time domain with the assumption of zero initial conditions, we 
obtain the differential equation of the system 
 o o o aRLCv Lv Rv Lv+ + =   
 
4. Reconsider the RLC circuit shown in Figure 6.27. Use the impedance method to determine the input–output 
differential equation relating i and va. Assume that all the initial conditions are zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the 
figure below, where 
 1
1 1( ) RCsZ s R
Cs Cs
+
= + = 
 2 ( )Z s Ls= 
 
Figure PS6-5 No4 
Applying Kirchhoff’s current law to node 1 in the equivalent impedance circuit yields 
 R L( ) ( ) ( )I s I s I s= + 
where 
 ( ) a
R
1
( )V s
I s
Z
= 
( ) a
L
2
( )V s
I s
Z
= 
Thus, we have 
2
a a2
1 2
1 1 1( ) ( ) ( )LCs RCsI s V s V s
Z Z RLCs Ls
  + +
= + =  + 
 
which gives the transfer function 
290 
 
 
 
2
2
a
( ) 1
( )
I s LCs RCs
V s RLCs Ls
+ +
=
+
 
By transforming a( ) ( )I s V s from the s domain to the time domain with the assumption of zero initial conditions, we 
obtain the differential equation of the system 
 
2
a a a2
d i diRLC L LCv RCv v
dtdt
+ = + +  
 
5. Reconsider the RLC circuit shown in Figure 6.28. Use the impedance method to determine the input–output 
differential equation relating vo and va. Assume that all the initial conditions are zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the 
figure below, we have 
 
1
1 1 1
( )Z s Ls R
= + or 1( ) RLsZ s
Ls R
=
+
 
 2
1( )Z s
Cs
= 
Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields 
 1 2 a( ) ( ) ( ) ( ) ( )Z s I s Z s I s V s+ = 
Thus, we have 
2
2
a 1 2
( ) 1
( ) ( ) ( )
I s LCs RCs
V s Z s Z s RLCs Ls R
+
= =
+ + +
 
Note that the current is related to the output voltage by 
 o 2( ) ( ) ( )V s Z s I s= 
Thus, the transfer function relating the input voltage va and the output voltage vo, 
 o 2
2
a a
( ) ( ) ( )
( ) ( )
V s Z s I s Ls R
V s V s RLCs Ls R
+
= =
+ +
 
By transforming o a( ) ( )V s V s from the s domain to the time domain with the assumption of zero initial conditions, we 
obtain the differential equation of the system 
o o o a aRLCv Lv Rv Lv Rv+ + = +   
 
Figure PS6-5 No5 
 
6. Reconsider the RLC circuit shown in Figure 6.29. Use the impedance method to determine the input–output 
differential equation relating vo and va. Assume that all initial conditions are zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain, where 
 1 1( )Z s R Ls= + 
2 2
1 1
( )
Cs
Z s R
= + or 2
2
2
( )
1
RZ s
R Cs
=
+
 
291 
 
 
 
Figure PS6-5 No6 
Applying Kirchhoff’s voltage law yields 
 1 2 a( ) ( ) ( ) ( )Z s I s Z I s V s+ = 
Thus, we have 
2
2
2 1a 1 2 2 1 2
1( ) 1
( ) ( ) (( ))
CsI s
V s Z
R
R LCs Rs Z R s Rs C L R
+
=
+ + + +
=
+
 
Note that the output voltage is 
 o 2( ) ( ) ( )V s Z s I s= 
Thus, the transfer function relating the input voltage va and the output voltage vo is 
( )
( )
2
2
2 1 2a a 1
o 2
2(
( ) ( )
) )(
R
R LCs R R C
V s
L s R R
Z s I s
V s V s + + + +
= = 
Transforming o a( ) ( )V s V s from the s domain to the time domain with the assumption of zero initial conditions gives 
2 o 1 2 o 1 2 o 2 a( ) ( ) .R LCv R R C L v R R v R v+ + + + =  
 
7. Reconsider the RLC circuit shown in Figure 6.30. Use the impedance method to determine the input–output 
differential equation relating vo and va. Assume that all initial conditions are zero. 
Solution 
Replacing the passive elements with their impedance representations gives the circuit in the s domain, where 
 1 1 1( )Z s R L s= + 
2 2 2
1 1 1
( )Z s R L s
= + or 2 2
2
2 2
( )
R L sZ s
L s R
=
+
 
 
Figure PS6-5 No7 
Applying Kirchhoff’s voltage law yields 
 1 2 a( ) ( ) ( ) ( )Z s I s Z I s V s+ = 
Thus, we have 
2 2
a 1
2
2 1 2 2 1 2 1 22 1
( ) 1
( ) ( ) ( ) ( )
L s RI s
V s Z s Z s L L s R L s R L L s R R+ + + +
+
= =
+
 
Note that the output voltage is 
 o 2( ) ( ) ( )V s Z s I s= 
Thus, the transfer function relating the input voltage va and the output voltage vo is 
( )
( ) 2
2 1 2 2 1 2
o 2 2 2
a a 21 1
( ) ( )
( ( ))
V s Z s I s R L s
V s V s L L s R L s R L L s R R+ + + +
= = 
292 
 
 
Transforming from the s domain to the time domain with the assumption of zero initial conditions+ 1.0000i 
 -1.0000 - 1.0000i 
 -1.0000 + 0.0000i 
 
K = 
 [] 
 
 
 
20. 2 2
9 5
( 4 5)
s
s s s
+
+ +
 
Solution 
(a) Expand as 
 
 
2 2 2
2 2 2 2 2 2
9 5 ( 4 5) ( 4 5) ( )
( 4 5) 4 5 ( 4 5)
s A B Cs D As s s B s s s Cs D
ss s s s s s s s s
+ + + + + + + + +
= + + =
+ + + + + +
 
 
This leads to 
 
0
4 0
5 4 9
5 5
A C
A B D
A B
B
+ =
+ + =
+ =
=
 
 
Simultaneous solution gives 1, 1, 1, 5A B C D= = = − = − so that 
 
 
3 31 1
2 2 2 2
2 2 2 2 2
9 5 1 1 5 1 1
2 2( 4 5) 4 5
j js s
s s s j s js s s s s s s
− + − −+ − −
= + + = + + +
+ − + ++ + + +
 
 
(b) 
 
>> B = [9 5]; A = conv([1 0 0],[1 4 5]); 
>> [R,P,K] = residue(B,A) 
 
R = 
 -0.5000 + 1.5000i 
 -0.5000 - 1.5000i 
 1.0000 + 0.0000i 
 
 
34 
 
 1.0000 + 0.0000i 
 
P = 
 -2.0000 + 1.0000i 
 -2.0000 - 1.0000i 
 0.0000 + 0.0000i 
 0.0000 + 0.0000i 
 
K = 
 [] 
 
 
In Problems 21 through 26, find the inverse Laplace transform using the partial-fraction 
expansion method. 
 
21. 2
1
( 3)
s
s s
+
+
 
Solution 
Expand as 
 
 
2
92
1
32 2 2
2
9
0
1 ( 3) ( 3) 3 1 
3( 3) ( 3) 3 1
A C A
s A B C As s B s Cs A B B
s ss s s s s B C
+ = =
+ + + + +
= + + = ⇒ + = ⇒ =
++ +
= = −
 
 
Therefore, 
 
 
12 1 2
39 3 9 2 1 2
9 3 92 2
1 
3( 3)
ts t e
s ss s s
−
−−+
= + + ⇒ + −
++
L
 
 
 
22. 2
2 3
( 4)( 1)
s
s s
+
+ +
 
Solution 
Partial-fraction expansion leads to 
 
 
2
2 2 2
0
2 3 ( )( 1) ( 4) 2
1( 4)( 1) 4 ( 4)( 1) 4 3
A C
s As B C As B s C s A B
ss s s s s B C
+ =
− + + + + +
= + = ⇒ + =
++ + + + +
+ = −
 
 
Simultaneous solution gives 1, 1, 1A B C= = = − so that 
 
 
1
1
22 2 2 2
2 3 1 1 1 2 1 cos 2 sin 2
1 2 1( 4)( 1) 4 4 4
ts s s t t e
s ss s s s s
−
−− + − −
= + = + ⋅ + ⇒ + −
+ ++ + + + +
L
 
 
 
 
35 
 
 
23. 2 2
4 5
( 4 5)
s
s s s
+
+ +
 
Solution 
Forming partial fractions, 
 
 
3 2
2 2 2 2 2 2
4 5 ( ) ( 4 ) (4 5 ) 5
( 4 5) 4 5 ( 4 5)
s A B Cs D B C s A B D s A B s A
ss s s s s s s s s
+ + + + + + + + +
= + + =
+ + + + + +
 
 
Then 
 
 
solve
0 1
4 0 0
 
4 5 4 0
5 5 1
B C A
A B D B
A B C
A D
+ = =
+ + = =
⇒
+ = =
= = −
 
 
Therefore 
 
 
1
2
2 2 2 2 2
4 5 1 1 sin
( 4 5) ( 2) 1
ts t e t
s s s s s
−
−+
= − ⇒ −
+ + + +
L
 
 
 
24. 2
10
( 2 5)
s
s s s
+
+ +
 
Solution 
Using partial fractions, 
 
 
2
2 2 2
10 ( ) (2 ) 5
( 2 5) 2 5 ( 2 5)
s A Bs C A B s A C s A
ss s s s s s s s
+ + + + + +
= + =
+ + + + + +
 
 
Subsequently, 
 
 
solve
0 2
2 1 2
5 10 3
A B A
A C B
A C
+ = =
+ = ⇒ = −
= = −
 
 
Then 
 
 
1
2 2 2 2 2
10 2 2 3 2 2( 1) 1 1 2 2 cos 2 sin 2
2( 2 5) ( 1) 2 ( 1) 2
t ts s s e t e t
s ss s s s s
−
− −+ + + +
= − = − ⇒ − −
+ + + + + +
L
 
 
 
 
36 
 
 
 
25. 2
5 8
( 2)
s
s s
+
+
 
Solution 
Forming partial fractions, 
 
( ) ( )22
1 2 12 1 2 1
2 2 2 2
4 2 4( 2) ( 2)5 8
2( 2) ( 2) ( 2) ( 2)
A A s A A A s AA A A s A s A s ss A
s ss s s s s s s
+ + + + ++ + + ++
≡ + + = =
++ + + +
 
Then, 
 
1
2 1 2
1
0 2
4 2 5 1 
4 8 2
A A A
A A A A
A A
+ = =
 + + = ⇒ =
 = = −
 
Therefore, 
 
 
1
2 2
2 2
5 8 2 1 2 2 2
2( 2) ( 2)
t ts te e
s ss s s
−
− −+ −
≡ + + ⇒ + −
++ +
L
 
 
 
26. 
2
2
2 3 4
( 3)( 2 2)
s s
s s s
+ −
+ + +
 
Solution 
Partial-fraction expansion gives 
 
 
2 2
2 2 2
2 3 4 ( ) (2 3 ) 2 3
3( 3)( 2 2) 2 2 ( 3)( 2 2)
s s A Bs C A B s A B C s A C
ss s s s s s s s
+ − + + + + + + +
= + =
++ + + + + + + +
 
 
Comparing coefficients of like powers of s results in 
 
 
2 1
2 3 3 1
2 3 4 2
A B A
A B C B
A C C
+ = =
+ + = ⇒ =
+ = − = − 
Finally, 
 
 
2
2 2 2 2 2 2 2
2 3 4 1 2 1 1 13
3 3( 3)( 2 2) ( 1) 1 ( 1) 1 ( 1) 1
s s s s
s ss s s s s s
+ − − +
= + = + − ⋅
+ ++ + + + + + + + +
 
 
Term-by-term inversion yields 
 
 
37 
 
 
 3 cos 3 sint t te e t e t− − −+ − 
 
 
 
 
In Problems 27 through 32, 
(a) Solve the IVP. 
(b) Confirm the result of (a) in MATLAB. 
 
27. 2
3 2 ( ) ( 1), (0) 0x x u t u t x+ = − − = 
Solution 
(a) Taking the Laplace transform and using the zero initial condition, yields 
 
 
3 3
2 22
3
1( 2) ( ) ( )
( 3) ( 3)
s
ses X s X s e
s s s s s s
−
−+ = − ⇒ = −
+ +
 
 
Let 1( )
( 3)
G s
s s
=
+
 so that by partial-fraction expansion we have 31
3( ) (1 )tg t e−= − . Next, 
 
 { } { }
3 3
1 1 1 12 2 3 3
2 2( ) ( ) ( )
( 3) ( 3)
s sx t e G s e G s
s s s s
− − − − − −      = − = −   + +      
L L L L 
 
Using the shift on the t − axis for the second term, we find 3 3
2 2( ) ( ) ( 1) ( 1)x t g t g t u t= − − − . But 
( )g t was defined earlier, hence 
 
 3 3( 1)1 1
2 2( ) (1 ) 1 ( 1)t tx t e e u t− − − = − − − −  
(b) 
 
>> x = dsolve('(2/3)*Dx+2*x=heaviside(t)-heaviside(t-1)','x(0)=0'); 
 
>> pretty(x) 
 
 / heaviside(t) heaviside(t - 1) exp(3) 
-exp(-3 t) | ------------ - ----------------------- 
 \ 2 2 
 
 exp(3 t) heaviside(t) heaviside(t - 1) exp(3 t) \ 
 - --------------------- + ------------------------- | 
 2 2 / 
 
 
28. 2 ( 1) ( 2), (0) 0, (0) 0x x x u t u t x x+ + = − − − = =   
Solution 
 
 
38 
 
(a) Laplace transformation of the ODE and using the zero initial conditions, yields 
 
 
2
2 2
2 2
1 1( 1) ( ) ( )
( 1) ( 1)
s s
s se es X s X s e e
s s s s s
− −
− −−
+ = ⇒ = −
+ +
 
 
Let 2
1( )
( 1)
H s
s s
=
+
 so that by partial-fraction expansion we have ( ) 1 ( 1) th t t e−= − + . Next, 
 
 { } { }1 1 2 1 1 2
2 2
1 1( ) ( ) ( )
( 1) ( 1)
s s s sx t e e e H s e H s
s s s s
− − − − − − − −   
= − = −   
+ +   
L L L L 
 
By the shift on the t − axis, we find ( ) ( 1) ( 1) ( 2) ( 2)x t h t u t h t u t= − − − − − . Using the expression 
for ( )h t defined earlier, we have 
 
 ( 1) ( 2)( ) 1 ( 1) 1 ( 1) ( 2)t tx t te u t t e u t− − − −   = − − − − − −    
(b) 
>> x = simplify(dsolve('D2x+2*Dx+x=heaviside(t-1)-heaviside(t-2)','x(0)=0, 
Dx(0)=0')) 
 
x = 
 
heaviside(t - 1) - heaviside(t - 2) - heaviside(t - 2)*exp(2 - t) - 
t*heaviside(t - 1)*exp(1 - t) + t*heaviside(t - 2)*exp(2 - t) 
 
 
29. 21 1
4 2 , (0) 0 , (0)tx x e x x−+ = = =   
Solution 
(a) Laplace transformation of the ODE and using the initial conditions leads to 
 
 2 1
4 1
4
1 1 4( ) ( ) ( )
2 2 2 ( 2)( )
ss s X s X s
s s s s
+
+ − = ⇒ =
+ + +
 
 
Finally, using partial-fraction expansion, we find 
 
 1 2 /4
1
4
4 2 30( ) 4
2 ( 2)( ) 7 7
t tsx t e e
s s s
− − − + = = − + + +  
L 
(b) 
>> x = simplify(dsolve('D2x + (1/4)*Dx = exp(-2*t)','x(0)=0, Dx(0)=1/2')) 
 
x = 
 
(2*exp(-2*t))/7 - (30*exp(-t/4))/7 + 4 
 
 
 
39 
 
 
30. 1 1
2 2( 1) , (0) 0 , (0)x x x t x xδ+ + = − = =   
Solution 
(a) Laplace transform of the ODE and using the initial conditions, we find 
 
 
1
2 21 1
2 2 2 21 1
2 2
1( ) ( ) ( ) ( )s ss X s sX s X s e X s e
s s s s
− −− + + = ⇒ = +
+ + + +
 
 
Let 2 1
2
1( )H s
s s
=
+ +
 so that /2( ) 2 sin( / 2)th t e t−= . Also, 
 
 
{ } { }1 11 1
2 2
( 1)/2 /21
2
( ) ( ) ( ) ( 1) ( 1) ( )
 2 sin( ( 1)) ( 1) sin( / 2)
s
t t
x t H s e H s h t u t h t
e t u t e t
− − −
− − −
= + = − − +
= − − +
L L
 
 
 
(b) 
 
>> x = simplify(dsolve('D2x + Dx + (1/2)*x = dirac(t-1)','x(0)=0,Dx(0)=1/2')) 
 
 
x = 
 
exp(-t/2)*(sin(t/2) + 2*heaviside(t - 1)*exp(1/2)*sin(t/2 - 1/2)) 
 
 
 
31. 2
33 2 sin( ), (0) 0, (0) 3x x x t x x+ + = = =   
Solution 
(a) Taking the Laplace transform of the ODE and using the given initial conditions, yields 
 
 
22
2 3
2 24 4
9 9
3 2( 3 2) ( ) 3 ( )
( 1)( 2)( )
ss s X s X s
s s s s
+
+ + = + ⇒ =
+ + + +
 
 
By partial-fraction expansion we find 
 
 
2
2 24 4
9 9
3 2 3.4615 3.1500 0.3115 0.1615( )
1 2( 1)( 2)( )
s sX s
s ss s s s
+ − − +
= = + +
+ ++ + + +
 
Finally,gives 
 
 
8. Reconsider the op-amp circuit shown in Figure 6.41. Use the impedance method to determine the differential 
equation relating the input voltage vi and the output voltage vo. 
Solution 
Replacing the passive elements with their impedance representations gives the op-amp circuit in the s domain as 
shown in the figure below, we have 
 and 
 
Figure PS6-5 No8 
Because the current drawn by the op-amp is very small, applying Kirchhoff’s current law to node 1 yields 
 
 
where . Thus, we have 
 
which gives the transfer function relating the input voltage vi and the output voltage vo, 
 
Transforming from the s domain to the time domain with the assumption of zero initial conditions gives 
 
 
Problem Set 6.6 
(Note: All MATLAB figures are given at the end of Problem Set 6.6.) 
1. Consider the RL circuit shown in Figure 6.25 (Problem Set 6.2, Problem 2), in which R = 35 Ω and L = 10 H. 
When the switch is closed at 0 second, the circuit is driven by a 6V DC voltage source. Assume that all initial 
conditions are zero. 
 a. Build a Simscape model of the physical system and find the loop current iL(t) and the voltage across the 
inductor vL(t). 
 b. Build a Simulink model of the system based on the differential equation relating iL and va, and find the loop 
current iL(t). 
c. Build a Simulink model of the system based on the transfer function VL(s)/Va(s), and find the voltage across 
the inductor vL(t). 
Solution 
The differential equation relating iL and va is 
 or 
The transfer function VL(s)/Va(s) is 
 
293 
 
 
2. Consider the parallel RLC circuit shown in Example 6.2, in which R = 2 Ω, L = 1 H, and C = 0.5 F. The 
circuit is driven by a controlled current source ia(t) = 10u(t), where u(t) is a unit-step function. 
a. Build a Simscape model of the physical system and find the voltage across the capacitor vC(t) and the current 
through the inductor iL(t). 
 b. Refer to the results obtained in Example 6.2. Build a Simulink model of the system based on the transfer 
function VC(s)/Ia(s) and find the voltage across the capacitor vC(t). 
 c. Refer to the results obtained in Example 6.2. Build a Simulink model of the system based on the transfer 
function IL(s)/Ia(s) and find the current through the inductor iL(t). 
Solution 
The transfer function VC(s)/Ia(s) is 
The transfer function IL(s)/Ia(s) is 
 
3. A simple low-pass filter can be realized by an RLC circuit (see Figure 6.21 in Example 6.4), which passes 
signals with frequencies lower than a certain cutoff frequency and attenuates signals with frequencies higher than 
the cutoff frequency. Assume that the parameter values are R = 500 Ω, L = 100 mH, and C = 10 µF. The circuit is 
connected to an AC voltage source, which has an amplitude of 1 V and a varying frequency. 
a. Build a Simscape model of the physical system and find the output voltage vo(t) when the frequency of the 
input voltage is 500, 1000, and 2500 rad/s. 
b. Build a Simulink model of the system based on the transfer function Vo(s)/Va(s), and verify the results 
obtained in Part (a). 
Solution 
The transfer function Vo(s)/Va(s) is 
 
 
4. A simple band-pass filter can be realized by an RLC circuit (see Figure 6.73), which passes frequencies 
within a certain range and attenuates frequencies outside that range. Assume that the parameter values are R = 500 
Ω, L = 100 mH, and C = 10 µF. The circuit is connected to an AC voltage source, which has an amplitude of 1 V 
and a varying frequency. 
 a. Build a Simscape model of the physical system and find the output voltage vo(t) when the frequency of the 
input voltage is 1000, 800, and 1200 rad/s. 
b. Derive the transfer function Vo(s)/Va(s), build a Simulink model of the system based on this transfer function, 
and verify the results obtained in Part (a). 
 
Figure 6.73 Problem 4. 
Solution 
The transfer function Vo(s)/Va(s) is 
 
5. Consider the op-amp integrator in Figure 6.35. Assume that the parameter values are R = 1 MΩ and C = 1 
µF. Build a Simscape model of the op-amp circuit and find the output voltage vo(t) when the input voltage is vi = 
−0.1 V. 
 
294 
 
 
6. Consider the op-amp circuit shown in Figure 6.74, in which the parameter values are C1 = 0.8 µF, R1 = 10 
kΩ, C2 = 80 pF, and R2 = 100 kΩ. The circuit is connected to an AC voltage source, which has an amplitude of 1 
V and a frequency of 200 Hz. Build a Simscape model of the op-amp circuit and find the output voltage vo(t). 
 
Figure 6.74 Problem 6. 
295 
 
 
MATLAB Figures 
Problem 1 
vL(t)
iL(t)
+
V
--
+
V
Voltage Sensor
PSPS
Switch
f(x) = 0
Solver
Configuration
S PS
Simulink-PS
Converter
Scope1
Scope
+ --+
Resistor
SPS
PS-Simulink
Converter1
SPS
PS-Simulink
Converter
+-- +
Inductor
Electrical Reference
DC Voltage Source
+
I
--+
I
Current Sensor
Clock
 
Figure PS6-6 No1a 
 
 
Figure PS6-6 No1b 
 
 
Figure PS6-6 No1c 
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
In
du
ct
or
 C
ur
re
nt
 i
L
 (A
)
 
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
-1
0
1
2
3
4
5
6
In
du
ct
or
 V
ol
ta
ge
 v
L
 (A
)
 
Figure PS6-6 No1d 
296 
 
 
Problem 2 
vC(t)
iL(t)
+
V--
+
V
Voltage Sensor
Step
f(x) = 0
Solver
Configuration
S PS
Simulink-PS
Converter
Scope1
Scope
+
--
+
Resistor
SPS
PS-Simulink
Converter1
SPS
PS-Simulink
Converter+
--
+
Inductor
Electrical Reference
+
I--
+
I
Current Sensor
Controlled Current
Source
+
--
+
Capacitor
 
Figure PS6-6 No2a 
 
 
 
Figure PS6-6 No2b 
 
 
 
Figure PS6-6 No2c 
 
 
0 5 10 15
Time (s)
-4
-2
0
2
4
6
8
10
C
ap
ac
ito
r V
ol
ta
ge
 v
C
 (V
)
 
0 5 10 15
Time (s)
0
2
4
6
8
10
12
14
In
du
ct
or
 C
ur
re
nt
 i
L
 (A
)
 
Figure PS6-6 No2d 
297 
 
 
Problem 3 
vC(t)
+
V--
+
V
Voltage Sensor
f(x) = 0
Solver
Configuration
Scope
+
--
+
Resistor
SPS
PS-Simulink
Converter
+ --+
Inductor
Electrical Reference
+
--
+
CapacitorAC Voltage Source
 
Figure PS6-6 No3a 
 
 
Figure PS6-6 No3b 
 
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-1.5
-1
-0.5
0
0.5
1
1.5
2
O
ut
pu
t v
ol
ta
ge
 (V
)
500 rad/s
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-6
-4
-2
0
2
4
6
O
ut
pu
t v
ol
ta
ge
 (V
)
1000 rad/s
 
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
O
ut
pu
t V
ol
ta
ge
 (V
)
2000 rad/s
 
Figure PS6-6 No3c 
298 
 
 
Problem 4 
+
V--
+
V
Voltage Sensor
f(x) = 0
Solver
Configuration
Scope1
+ --+
Resistor
SPS
PS-Simulink
Converter1
+
--
+
Inductor
Electrical Reference
+
--
+
CapacitorAC Voltage Source
 
Figure PS6-6 No4a 
 
 
 
Figure PS6-6 No4b 
 
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-1
-0.5
0
0.5
1
O
ut
pu
t V
ol
ta
ge
 (V
)
1000 rad/s
 
 
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
O
ut
pu
t V
ol
ta
ge
 (V
)
800 rad/s
 
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time (s)
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
O
ut
pu
t V
ol
ta
ge
 (V
)
1200 rad/s
 
Figure PS6-6 No4c 
 
299 
 
 
Problem 5 
vo(t)
+
V--
+
V
Voltage Sensor
f(x) = 0
Solver
Configuration
Scope
+ --+
Resistor
SPS
PS-Simulink
Converter
+
--
+
Op-Amp
Electrical Reference
DC Voltage Source
+ --+
Capacitor
 
Figure PS6-6 No5a 
 
 
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
Time (s)
0
0.01
0.02
0.03
0.04
0.05
0.06
O
ut
pu
t V
ol
ta
ge
 (V
)
 
Figure PS6-6 No5b 
300 
 
 
Problem 6 
vo(t)
+
V--
+
V
Voltage Sensor
f(x) = 0
Solver
Configuration
Scope
+ --+
Resistor1
+ --+
Resistor
SPS
PS-Simulink
Converter
+
--
+
Op-Amp
Electrical Reference
+ --+
Capacitor1
+ --+
Capacitor
AC Voltage Source
 
Figure PS6-6 No6a 
 
0 0.01 0.02 0.032 2 2
3 3( ) 3.4615 3.1500 0.3115cos( ) 0.2423sin( )t tx t e e t t− −= − − + 
 
(b) 
 
 
40 
 
 
>> x=simplify(dsolve('D2x+3*Dx+2*x=sin(2*t/3)','x(0)=0,Dx(0)=3')) 
 
x = 
 
(45*exp(-t))/13 - (81*cos((2*t)/3))/260 - (63*exp(-2*t))/20 + 
(63*sin((2*t)/3))/260 
 
 
32. /23
42 , (0) 0, (0) 0tx x x e x x−+ + = = =   
Solution 
(a) Taking the Laplace transform of the ODE and using the zero initial conditions, yields 
 
 2 3
4 2 2 31 131 1
2 2 22 2 2
1 1 1 1 1( 2 ) ( ) ( )
( ) ( ) ( )
s s X s X s
s s ss s s
−
+ + = ⇒ = = + +
+ + ++ + +
 
 
Inversion yields /2 /2 3 /2( ) t t tx t e te e− − −= − + + . 
(b) 
 
>> x=simplify(dsolve('D2x+2*Dx+(3/4)*x=exp(-t/2)','x(0)=0,Dx(0)=0')) 
 
x = 
 
exp(-(3*t)/2)*(t*exp(t) - exp(t) + 1) 
 
 
In Problems 33 through 38 decide whether the final-value theorem is applicable, and if so, find 
ssx . 
33. 2( )
(4 1)( 3)
sX s
s s
=
+ +
 
Solution 
The poles are at 1
4 , 3j− ± , two of which are located on the imaginary axis. Therefore, the FVT 
does not apply. 
 
34. 
1
3
21
3
2
( )
( )( 1)
s
X s
s s s
+
=
+ + +
 
Solution 
The poles are at 31 1
3 2 2, j− − ± hence the FVT applies. 
 { }
1
3
210 0
3
(2 )
lim ( ) lim 0
( )( 1)ss s s
s s
x sX s
s s s→ →
 + = = = 
+ + +  
 
 
35. 3
2( )
( 1) ( 3)
sX s
s s
+
=
+ +
 
 
 
41 
 
Solution 
The poles are at 1, 1, 1, 3− − − − hence the FVT applies. 
 
 { } 30 0
( 2)lim ( ) lim 0
( 1) ( 3)ss s s
s sx sX s
s s→ →
 +
= = = 
+ + 
 
 
 
36. 2
1( )
( 3)( 2)
sX s
s s s
+
=
+ +
 
Solution 
The poles are at 0,0, 3, 2− − . Since the pole at the origin is repeated, the FVT does not apply. 
 
37. 
2
2
4 1( )
(4 4 2)
sX s
s s s
+
=
+ +
 
Solution 
The poles are at 1 1
2 20, j− ± hence the FVT applies. 
 { }
2
20 0
4 1 1lim ( ) lim
24 4 2ss s s
sx sX s
s s→ →
 + = = = 
+ +  
 
 
38. 
2
5
2 2
1
( )
( 2 2)
s
X s
s s s
+
=
+ +
 
Solution 
The poles are at 0, 1 j− ± hence the FVT applies. 
 { }
2
5
2 20 0
1 1lim ( ) lim
4( 2 2)ss s s
s
x sX s
s s→ →
 + = = = 
+ +  
 
 
 
In Problems 39 through 42, evaluate (0 )x + using the initial-value theorem. 
39. 
2
22
3
3 1( )
( 1)( 2)
sX s
s s
+
=
+ +
 
Solution 
 { }
2
22
3
(3 1) 27(0 ) lim ( ) lim
4( 1)( 2)s s
s sx sX s
s s
+
→∞ →∞
 + = = = 
+ +  
 
 
 
40. 
2
21
2
1( )
( 1)(9 6 2)
sX s
s s s
+
=
+ + +
 
Solution 
 
 
42 
 
 { }
2
21
2
( 1) 2(0 ) lim ( ) lim
9( 1)(9 6 2)s s
s sx sX s
s s s
+
→∞ →∞
 + = = = 
+ + +  
 
 
 
 
41. 
2
2
( 4)( )
( 1)( 2)( 3)
s sX s
s s s
+
=
+ + +
 
Solution 
 { }
2 2
2
( 4)(0 ) lim ( ) lim 0
( 1)( 2)( 3)s s
s sx sX s
s s s
+
→∞ →∞
 + = = = 
+ + +  
 
 
42. 
41
4
3 2
1
( )
(2 9)
s
X s
s s
+
=
+
 
Solution 
 { }
41
4
2 2
1 1(0 ) lim ( ) lim
8(2 9)s s
s
x sX s
s s
+
→∞ →∞
 + = = = 
+  
 
 
 
Review Problems 
In Problems 1 through 4, perform the operations and express the result in rectangular form. 
1. ( 1 2 )/(1 )j je − + + 
 
Solution 
Noting 
 1 2 1 1 3
1 1 2
j j j
j j
− + − +
× =
+ −
 
we have 
 ( 1 2 )/(1 ) (1 3 )/2 1/2 3 3
2 2(cos sin ) 0.1166 1.6446j j je e e j j− + + += = + = + 
 
2. 
8(0.15 0.85 )je − 
 
Solution 
We first write 1.39610.15 0.85 0.8631 jj e−− = so that 
 
 8 11.1690(0.15 0.85 ) 0.3081 0.0532 0.3034jj e j−− = = + 
Then 
 
8(0.15 0.85 ) 0.0532 0.3034 1.0065 0.3151j je e j− += = + 
 
 
 
43 
 
3. 
42 1
3 2
3 1
4 3
( )
(1 )( 1 )
j
j j
+
+ − +
 
Solution 
 
4 0.6435 4 2.57402 1
3 2
3 1
4 3
( ) (0.8333 ) 0.4823 0.4066 0.2593
(1 )( 1 ) 1.25 0.4167 1.25 0.4167 1.25 0.4167
j jj e e j
j j j j j
+ − +
= = =
+ − + − − − − − −
 
 
This yields 0.2306 0.2843j− . 
 
 
4. 
42
3
32
3
(1 )
( )
j
j
−
+
 
Solution 
 
4 0.5880 4 2.35202
5.30043 2
33 0.9828 3 2.94842
3
(1 ) (1.2019 ) 1.2019 1.2019
( ) (1.2019 )
j j
j
j j
j e e e j
j e e
− −
−−
= = = = +
+
 
 
 
5. Find all possible values of ( )4/31
33 j+ . 
Solution 
First, 
 ( ) ( )1/34/3 4 1/3 0.1107 4 0.4428 1/31 1
3 33 ((3 ) ) (3.0185 ) (83.0123 )j jj j e e+ = + = = 
 
The three values are generated by 
 
 1/3 0.4428 2 0.4428 283.0123 cos sin , 0,1, 2
3 3
k kj kπ π+ + + =  
 
 
which yields 
 
 4.3149 0.6413 , 2.7130 3.4160 , 1.6018 4.0575j j j+ − + − − 
 
6. Find all possible roots of the following equation: 
 
 4(2 ) 1 2 0j z j− + − = 
Solution 
The equation is re-written as 
 
 4 1 2 1 2 4 3
2 2 5
j j jz
j j
− + − + +
= × =
− −
 
 
This in turn is equivalent to 1/434
5 5( )z j= + . Therefore, the four values of z are generated by 
 
 
44 
 
 
 0.6435 2 0.6435 2cos sin , 0,1, 2,3
4 4
k kj kπ π+ +
+ = 
which gives 
 
 0.9871 0.1602 , 0.1602 0.9871 , 0.9871 0.1602 , 0.1602 0.9871j j j j+ − + − − − 
 
 
 
 
 
In Problems 7 through 10, solve the IVP. Do not use Laplace transformation. 
 
7. tan 1 , (0) 1y y t t y− = + = 
Solution 
By Eq. (2.12), 
 
 
[ ]ln cos ln cos ln costan ln cos , ( ) ( 1) ( 1)cos 
1 cos ( 1
cos
t t th tdt t y t e e t dt c e t t dt c
t t
t
− − = − = = + + = + +∫ ∫ ∫ 
= + +[ ])sin t c+
 
Application of the initial condition yields 0c = , hence 
 
 ( ) 1 ( 1) tany t t t= + + 
 
 
8. 1
2(1 )sin , ( ) 2y y t y π= − = 
 
Solution 
Rewrite the ODE as 
 
 (1 )sin sin ln 1 cos
1
dy dyy t tdt y t c
dt y
= − ⇒ = ⇒ − − = − +∫ ∫
−
 
 
This implies 
 cos if 1 0
1 , 
 if 1 0
c
t
c
e y
y ke k
e y
−
−
 − >
− = = 
− −= , we find 
 
 
3/2
3 /2
0
1( ) ( )
1
st
sG s e g t dt
e
−
−= ∫
−
 
 
The integral is evaluated as 
 
 
3/2 1 3/2
3 /2
2
0 0 1
1 2( ) ( 2 3) (1 ) ( )st st st s s se g t dt e dt e t dt e e e
s s
− − − − − −= + − + = − + −∫ ∫ ∫ 
 
Simplifying and substituting back, we find 
 
 
3 /2
2
3 /2
1 2 ( )
( )
1
s s
s
e e
s sG s
e
− −
−
+ −
=
−
 
 
 
 
( )g t
t
1
1
3
2
 
 
47 
 
 
 Figure 2.20 Problem 12. 
 
13. Find the response ( )x t of a system governed by 
 
 3 7 2 ( 1) , (0) 0, (0) 0x x x tu t x x+ + = − = =   
 
Solution 
Taking the Laplace transform of the ODE, using the zero ICs, we find 
 
 
( )
2
2 2 2 2
1 1(3 7 2) ( ) ( )
(3 7 2) (3 7 2)
H s
s s
se es s X s X s e
s s s s s s s s
− −
−
 
 + + = + ⇒ = + + + + + 
 
 
 
Inverse Laplace transforms of the two terms in ( )H s are 
 
 1 2 /39 71 1
2 20 5 42 2
1
(3 7 2)
t tt e e
s s s
− − − 
= − + − + + 
L 
 
 1 2 /331 1
10 5 22
1
(3 7 2)
t te e
s s s
− − − 
= − + + + 
L 
so that 
 
 { }1 2 /36 51 1
2 20 5 4( ) ( ) t th t H s t e e− − −= = + + −L 
 
Finally, by the shift on t − axis, we have 
 
 { }1 2( 1) ( 1)/36 51 1
2 20 5 4( ) ( ) ( 1) ( 1) ( 1) ( 1)s t tx t H s e h t u t t e e u t− − − − − − = = − − = − + + − − L 
 
 
 
14. Evaluate the convolution ( )t u t a∗ − . 
 
Solution 
By definition of convolution, 
( )g t
t
1
1
3
2
35
2
 
 
48 
 
 
 
0
( ) ( )( )
t
t u t a u a t dt t t∗ − = − −∫ 
 
Since 
0 if 
( )
1 if 
a
u a
a
t
t
t

, the above reduces to 
 
 21
2( ) ( ) ( )
t
a
t u t a t d t at t∗ − = − = −∫ 
 
 
15. Evaluate the convolution ( )( )g h t∗ where ( )g t is shown in Figure 2.21 and ( ) sinh t t= . 
 
Solution 
By definition of convolution, 
 
 
0
( )( ) ( )sin( )
t
g h t g t dt t t∗ = −∫ (*) 
 
The analytical description of ( )g t is 
 
 
1 if 0 1
( )
0 if 1
t t
g t
t
− > syms s 
>> ilaplace((4*s^3+8*s^2+2)/s^2/(4*s^2+1)) 
( )g t
t
1
1
 
 
50 
 
 
ans = 
 
2*t + cos(t/2) 
 
 
17. (a) Find 
 
 
2
1
2
(4 12 5)
2 (4 8 5)
se s s
s s s
−
−  + +
 + + 
L 
 
 (b) Confirm the result of (a) in MATLAB. 
Solution 
(a) Let 
 
2 Partial-fraction expansion
2 2
4 12 5 2 1( )
2 (4 8 5) 4( 1) 1 2
s sH s
s s s s s
+ +
= = +
+ + + +
 
so that 
 
 { }1 1 1
2 2( ) ( ) sin( )th t H s e t− −= = +L 
 
Shift on t − axis yields 
 
 { }
2
1 1 ( 1) 1 1
2 22
(4 12 5) ( ) sin( ( 1)) ( 1) ( 1)
2 (4 8 5)
s
s te s s e H s e t u t u t
s s s
−
− − − − − + +
= = − − + − + + 
L L 
 
(b) 
>> syms s 
>> ilaplace(exp(-s)*(4*s^2+12*s+5)/(2*s)/(4*s^2+8*s+5)) 
 
ans = 
 
heaviside(t - 1)/2 + heaviside(t - 1)*exp(1 - t)*sin(t/2 - 1/2) 
 
 
18. Consider 
 
2
2
3 1( )
( 1)
s sX s
s s
+ +
=
+
 
(a) Using the final-value theorem, if applicable, evaluate ssx . 
(b) Confirm the result of (a) by evaluating lim ( )
t
x t
→∞
. 
Solution 
(a) Poles of ( )X s are at 0, 1, 1− − so that the FVT is applicable: 
 
 
 
51 
 
 { }
2
20 0
3 1lim ( ) lim 1
( 1)ss s s
s sx sX s
s→ →
+ +
= = =
+
 
 
(b) Since ( ) 1tx t te−= + , we find lim ( ) 1
t
x t
→∞
= , confirming the result of (a). 
 
 
 
19. Consider 
 
2
2
2(25 30 113)( )
(25 10 226)
s sX s
s s s
− +
=
+ +
 
(a) Using the initial-value theorem, evaluate (0 )x + . 
(b) Confirm the result of (a) by evaluating 
0
lim ( )
t
x t
→
. 
Solution 
(a) 
 
 { }
2
20
2(25 30 113)(0 ) lim ( ) lim 2
25 10 226s s
s sx sX s
s s
+
→∞ →
− +
= = =
+ +
 
 
(b) Since 0.2( ) (cos3 sin 3 ) 1tx t e t t−= − + , we find 
0
lim ( ) 2
t
x t
→
= , confirming the result of (a). 
 
 
20. Consider 
 2
0.2 0.3( )
(3 2.5 0.5)
sX s
s s s
+
=
+ +
 
(a) Evaluate ssx using the final-value theorem. 
(b) Confirm the result of (a) by evaluating lim ( )
t
x t
→∞
 . 
Solution 
(a) 
 
 [ ]{ } 20 0
(0.2 0.3)lim ( ) lim 0
3 2.5 0.5ss s s
s sx s sX s
s s→ →
+
= = =
+ +
 
 
(b) Since /2 /37 34
5 5 5( ) t tx t e e− −= − + , we have /3 /27 2
15 5( ) t tx t e e− −= − and 
 
 { }/3 /27 2
15 50
lim ( ) lim 0t t
t x
x t e e− −
→ →∞
= − = 
 
 
 
 
 
 
52 
 
Problem Set 3.1 
In Problems 1 through 8, perform the indicated operations, if defined, for the vectors and 
matrices below. 
 
 
0 2 1 1
3 1 1 1
1 3 3 , , , 0
0 4 5 2
2 1 2 1
−   
−      = − = = =      −       −   
A B v w 
1. Tw A 
Solution 
[ ]2 3 1T = − − −w A 
 
 
2. BAw 
Solution 
1
8
 
=  − 
BAw 
 
3. ( )TAB v 
Solution 
5
( ) 9
31
T
 
 = − 
 − 
AB v 
 
 
4. ( )T−B vw A 
Solution 
5 7 3
( )
10 13 20
T  
− =  − 
B vw A 
 
 
5. T−Aw B v 
Solution 
2
5
9
T
 
 − =  
 
 
Aw B v 
 
 
 
 
 
 
53 
 
6. ( )T+A BA vw 
Solution 
Operation is undefined! A is 3 3× , but T+BA vw is 2 3× . 
 
7. Tv Bw 
Solution 
6T =v Bw 
 
8. 2( )T+A B B w 
Solution 
2
16
( ) 19
33
T
 
 + = − 
 − 
A B B w 
 
 
In Problems 9 through 12, find the rank by inspecting the row-echelon form of the matrix. 
 
9. 
3 2 0
1 2 4
4 4 3
− 
 = − 
 − − 
A 
Solution 
3 2 0 1 2 4 1 2 4 1 2 4
1 2 4 3 2 0 0 4 12 0 1 3 REF( ) rank( ) 3
4 4 3 4 4 3 0 4 13 0 0 1
− − − −       
       = − → − → → = ⇒ =       
       − − − −       
A A A
 
 
 
10. 
2 1 1
3 3 2
1 2 0
− 
 =  
 − 
A 
Solution 
1
3
2 1 1 1 2 0 1 2 0 1 2 0
3 3 2 3 3 2 0 9 2 0 9 2 REF( ) rank( ) 3
1 2 0 2 1 1 0 3 1 0 0
 − − − −     
      = → → → = ⇒ =      
      − −       
A A A 
 
 
 
 
 
 
54 
 
11. 
1 1 1 0
3 2 0 1
0 2 4 1
1 1 3 1
− − 
 − =
 
 
 
A 
Solution 
1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0
3 2 0 1 0 1 3 1 0 1 3 1 0 1 3 1
REF( )
0 2 4 1 0 2 4 1 0 0 2 1 0 0 2 1
1 1 3 1 0 2 4 1 0 0 2 1 0 0 0 0
− − − − − − − −       
       −       = → → → =
       − − − −
       − − −       
A A 
 
Therefore, rank( ) 3=A . 
 
 
12. 
1 1 2 0
3 2 0 1
2 3 2 1
4 1 2 1
− 
 
 =
 −
 
 
A 
Solution 
 
1 1 2 0 1 1 2 0 1 1 2 0
3 2 0 1 0 5 6 1 0 5 6 1
REF( )
2 3 2 1 0 5 6 1 0 0 0 0
4 1 2 1 0 5 6 1 0 0 0 0
− − −     
     − −     = → → =
     − −
     −     
A A 
 
Therefore, rank( ) 2=A . 
 
 
13. Determine the value of a such that rank( ) 2=A . 
 
 
2 1 0 2
1 3 1 4
1 2 1 6
5 1 0a
− 
 − =
 − −
 − 
A 
Solution 
2 1 0 2 1 3 1 4 1 3 1 4 1 3 1 4
1 3 1 4 2 1 0 2 0 5 2 10 0 5 2 10
1 2 1 6 1 2 1 6 0 5 2 10 0 0 0 0
5 1 0 5 1 0 0 15 4 20 0 5 0 0a a a a
− − − −       
      − − − − − −       = → → →
       − − − − −
       − − − − −       
A 
 
Therefore, 5a = . 
 
 
55 
 
 
14. Given 
 
1
2
0 1 0 0
0 0 1 , 0
1 2 1
  
  = =   
  − − −   
A B 
 find the rank of 
 2 
 B AB A B 
Solution 
 
1 1
2 2
2 21 1 1 1
2 2 2 2
1 1 1 1 1
2 2 2 2 2
0 0 0
 , 0
     
      = = − ⇒ = −      
     − − − −     
AB A B B AB A B 
 
The rank of this matrix is readily seen to be 3. 
 
 
 
In Problems 15 through 18, evaluate the determinant. 
 
15. 
1 2 3 1
6 0 1 3
0 1 5 1
1 3 0 1
−
−
−
−
 
Solution 
using the 4th row
1 2 3 1
2 3 1 1 3 1 1 2 3
6 0 1 3
0 1 3 3 6 1 3 6 0 1 (18) 3(34) ( 41) 161
0 1 5 1
1 5 1 0 5 1 0 1 5
1 3 0 1
−
− −
−
= − − − − + = − − + − = −
−
− −
−
 
 
16. 
0 0 3 2
0 0 1
 , , parameters
1 2
0 0 0 2
b
a b
a a
a
− −
=
−
 
Solution 
2
0 0 3 2
0 0 3
0 0 1
2 0 0 6
1 2
1
0 0 0 2
b
a b a b
a a
a a
a
− −
−
= =
−
−
 
 
 
56 
 
 
 
 
17. 
1 0 0
0 0 1
 , parameter
1 2 1 0
0 1 3 2
s
s
s
s
s
−
−
=
+
+
 
Solution 
using the first row
3 2 4 3 2
1 0 0
0 1 0 0 1
0 0 1
2 1 0 1 1 0
1 2 1 0
1 3 2 0 3 2
0 1 3 2
 ( 3 3 5) ( 3) 3 3 5 3
s
s
s
s s s
s
s s
s
s s s s s s s s
−
− −
−
= + + +
+
+ +
+
= + + − + − = + + − −
 
 
 
18. 
5 2 0 0 0
2 4 0 0 10
0 0 3 0 0
0 0 0 1 2
0 0 1 3 5
−
−
−
 
Solution 
block upper-triangular
2 2 and 3 3
5 2 0 0 0
2 4 0 0 10
(24)(33) 7920 0 3 0 0
0 0 0 1 2
0 0 1 3 5
× ×
−
= =−
−
 
 
 
In Problems 19 through 24, find the inverse of the matrix. 
 
19. 
cos sin
 , parameter
sin cos
a a
a
a a
 
= = − 
A 
Solution 
2 2cos sin 1a a= + =A . The adjoint matrix is formed as 
cos sin
adj( )
sin cos
a a
a a
− 
=  
 
A . Finally, 
 1 cos sin
sin cos
a a
a a
− − 
=  
 
A 
 
 
 
 
57 
 
20. 
1 0 2
0 2 1
0 0 3
 
 = − 
  
A 
Solution 
Upper-triangular matrix; 6=A . Then, 
 
 
2
3
1 1 1
2 6
1
3
6 0 4 6 0 4 1 0
1adj( ) 0 3 1 , 0 3 1 0
6
0 0 2 0 0 2 0 0
−
 − − −   
    = = =     
         
A A 
 
 
21. 
1 1 0
3 0 0
2 0 2
− 
 =  
 − 
A 
Solution 
6= −A . Then, 
 
1
3
1 1
3
1 1
3 2
0 2 0 0 2 0 0 0
1adj( ) 6 2 0 , 6 2 0 1 0
6
0 2 3 0 2 3 0
−
 − −   
    = − = − = −    −     − − −     
A A 
 
 
22. 
1 0 1
0 2 , parameter
1 0 2
a
a a
a
+ 
 = = 
 + 
A 
Solution 
3 2 23 ( 3 1)a a a a a a= + + = + +A . Next, 
 
 2
( 2) 0
adj( ) 2 3 1 2( 1)
0 ( 1)
a a a
a a a
a a a
+ − 
 
= + + − + 
 − + 
A 
Finally, 
 1 2
2
( 2) 0
1 2 3 1 2( 1)
( 3 1) 0 ( 1)
a a a
a a a
a a a a a a
−
+ − 
 
= + + − + + +  − + 
A 
 
 
 
 
58 
 
23. 
0 0
0 3 0 , parameter
1 2 3( 1)
a
a a
a a a
 
 = + = 
 + + + 
A 
Solution 
Lower-triangular matrix; 3 ( 1)( 3)a a a= + +A . Then, 
 
 
3( 1)( 3) 0 0
adj( ) 0 3 ( 1) 0
( 1)( 3) ( 2) ( 3)
a a
a a
a a a a a a
+ + 
 = + 
 − + + − + + 
A 
Finally, 
 
 1
3( 1)( 3) 0 0
1 0 3 ( 1) 0
3 ( 1)( 3)
( 1)( 3) ( 2) ( 3)
a a
a a
a a a
a a a a a a
−
+ + 
 = + + +
 − + + − + + 
A 
 
 
 
24. ( ) ( )
( ) ( )
1 1 2 1 2 2 1 2
1 2 1 2
1 1 2 1 2 2 1 2
sin sin sin
 , , , , parameters
cos cos cos
L L L
L L
L L L
θ θ θ θ θ
θ θ
θ θ θ θ θ
− − + − + 
= = + + + 
A 
 
Solution 
 ( ) ( )
( ) ( )
2 1 2 2 1 21
1 1 2 1 2 1 1 2 1 21 2 2
cos sin1
cos cos sin sinsin
L L
L L L LL L
θ θ θ θ
θ θ θ θ θ θθ
− + + 
=  − − + − − + 
A 
 
 
25. Prove that the inverse of a (non-singular) symmetric matrix is symmetric. 
Solution 
Suppose A is symmetric so that T =A A . Then 
 
 1 1 1( ) ( )T T− − −= =A A A 
 
This proves that 1−A is symmetric, as asserted. 
 
26. Given n n×A and scalar 0k ≠ , show that 1 11( )k
k
− −=A A . 
Solution 
By definition, we have 1 1( ) adj( )k k
k
− =A A
A
. But by one of the determinant properties listed 
earlier, nk k=A A . It can also be shown that 1adj( ) adj( )nk k −=A A . As a result, 
 
 
59 
 
 1 1 11 1 1 1 1( ) adj( ) adj( ) adj( )n
nk k k
k k k k
− − −= = = =A A A A A
A A A
 
 
 
 
Problem Set 3.2 
In Problems 1 through 6, solve the linear system Ax = b using Gauss elimination. 
 
1. 
2 2 1 1
3 1 4 , 2
1 2 5 8
−   
  = = −  
  − −   
A b 
Solution 
Swap 1st and More EROs Back substitution
3rd rows 51 51
7 7
2 2 1 1 1 2 5 8 1 2 5 8 1
3 1 4 2 3 1 4 2 0 7 19 26 1
1 2 5 8 2 2 1 1 0 0 1
  −  − −  − −  
      − → − → − ⇒ = −     
     − − − − −      
x
 
 
2. 
1 0 4.2 10.4
2.3 3 1.5 , 4.6
5 3.2 1 4.8
−   
  = − = −  
   −   
A b 
Solution 
EROs Back substitution
1 0 4.2 10.4 1 0 4.2 10.4 2
2.3 3 1.5 4.6 0 3 11.16 19.32 1
5 3.2 1 4.8 0 0 33.9040 67.9040 2
 −  −  − 
    − − → − ⇒ =     
    −     
x 
 
 
3. 
3 1.4 2.1 4.10
1.5 3.2 1 , 11.9
0.75 2.5 2 4.75
−   
  = − =   
  − −   
A b 
Solution 
EROs Back substitution
3 1.4 2.1 4.10 3 1.4 2.1 4.10 3
1.5 3.2 1 11.9 0 3.9 2.05 9.85 2
0.75 2.5 2 4.75 0 0 0.3449 0.3449 1
 −   −   
    − → − ⇒ = −    
    − − − −     
x 
 
 
 
60 
 
4. 
3 0 2 0
1 4 3 , 1
2 3 2 4
−   
  = − − =   
     
A b 
Solution 
Swap 1st two rows Back substitution
and more EROs
3 0 2 0 1 4 3 1 2
1 4 3 1 0 12 7 3 2
2 3 2 4 0 0 1.5833 4.75 3
 −   − −   
    − − → − ⇒ = −    
         
x 
 
 
5. 
2 1 0 3 10
1 3 1 4 14
 , 
2 0 3 5 9
1 2 4 6 11
−   
   − −  = =        − −   
A b 
Solution 
EROs Back substitution
2 1 0 3 10 2 1 0 3 10 1
1 3 1 4 14 0 3.5 1 5.5 19 2
 
2 0 3 5 9 0 0 4.7143 3.5714 2.4286 1
1 2 4 6 11 0 0 0 1.5152 3.0303 2
 −   −   
     − − − − −    → ⇒ =      −      − −        
x 
 
 
 
6. 
2 4 3 1 7
3 2 6 1 6
 , 
1 2 3 2 1
4 8 9 3 3
−   
   − − −  = =   − −     − −   
A b 
Solution 
1EROs Back substitution
2
1
3
2 4 3 1 7 1 2 3 2 1 1
3 2 6 1 6 0 8 3 7 9
 
1 2 3 2 1 0 0 3 5 9
4 8 9 3 3 0 0 0 28 56 2
−   − −  − 
     − − − − −     → ⇒ =     − − −       − −        
x 
 
 
 
In Problems 7 through 10, solve the linear system Ax = b using 
(a) The inverse of the coefficient matrix. 
(b) The "\" operator in MATLAB. 
 
 
61 
 
7. 
2 1 2 4
3 0 2 , , parameter
1 2 1 0
a
a a
−   
  = = =  
  − −   
A b 
Solution 
(a) 1
4 3 2 4
1 1 4 10 0
19
6 5 3 0
a a
a
a
−
− −     
    = = − − =    −     − − −     
x A b 
(b) 
>> syms a 
>> A = [2 1 -2;3 0 2;-1 2 -1]; b = [4*a;a;0]; x = A\b 
 
x = 
 
 a 
 0 
 -a 
 
 
8. 
1 2 1 1 5
1 2 1 , 4 , parameter
1 2 1 3 1 7 2
a a a
a a a
a a a
− + −   
  = − − = − =  
  − + + +   
A b 
Solution 
(a) 
2 2
1 2
3
2
1 6 6 7 2 4 1 5 1
1 3 3 2 2 2 4 2
6 2 2
1 (2 1) 2 1 7 2 1
a a a a a
a a a a a
a a
a a a a
−
 − − − − − −    
−      = = − − − − − =    + +     − + − − +    
x A b 
(b) 
>> syms a 
>> A = [a-1 2*a+1 -1;1 -2*a -1;-1 2*a+1 3*a+1]; b = [5*a;-4*a;7*a+2]; x = A\b 
 
x = 
 
 1 
 2 
 1 
 
 
 
 
 
9. 
31
2 2
31
2 2
2 2 0 0 0
0 0 1
 , 
0 0 1
0 0 1 1 2
   
   
  = =   − −    −   
A b 
 
 
62 
 
Solution 
(a) 
3
4
1
1 4
3
2
0 11 0 0
1 11 0 01
11 0 0 2 2
2 10 0 2 1
−
−−     
     −     = = =     −− − −         −− −     
x A b 
(b) 
>> A = [2 2 0 0;1/2 3/2 0 0;0 0 -1/2 1;0 0 1 -1]; b = [0;1;-3/2;2]; x = A\b 
 
x = 
 
 -1 
 1 
 1 
 -1 
 
 
 
10. 
1 4 1
2 3 6
3 1
2 2
3 2 0 0 1
1 0
 , 
1 0 1
0 01 1 2
−   
   − −  = =    −    −   
A b 
Solution 
(a) 
10
3
9 15 1
4 21 6
4 1
3 2
54
3 2
5 2 2 1 1
3 3 12
2 3 3 111
2 3 2 1
−
− − −     
     − − − −−     = = =    − − − −        − − −    
x A b 
(b) 
>> A = [3 2 0 0;1/2 4/3 -1 0;1 0 3/2 1;0 0 1 -1]; b = [-1;-1/6;-1/2;2]; 
>> x = A\b 
 
x = 
 
 -1.0000 
 1.0000 
 1.0000 
 -1.0000 
 
 
In Problems 11 through 14, 
(a) Solve the linear system Ax = b using Cramer’s rule. 
(b) Repeat (a) in MATLAB. 
11. 
1 5 0 13
1 2 1 , 0
2 5 3 1
   
  = − =   
  −   
A b 
 
 
63 
 
Solution 
(a) The determinant of the coefficient matrix is 
 
1 5 0
1 2 1 6
2 5 3
D = − =
−
 
Since 0D ≠ , we proceed by evaluating 
 1 2 3
13 5 0 1 13 0 1 5 13
0 2 1 18 , 1 0 1 12 , 1 2 0 6
1 5 3 2 1 3 2 5 1
D D D= = = − = = − = −
− −
 
Subsequently, 
 1 2 3
18 12 63 , 2 , 1
6 6 6
x x x −
= = = = = = − 
(b) 
>> A = [1 5 0;-1 2 1;-2 5 3]; b = [13;0;1]; d = det(A); 
>> A1=A; A1(:,1)=b; d1=det(A1); 
>> A2=A; A2(:,2)=b; d2=det(A2); 
>> A3=A; A3(:,3)=b; d3=det(A3); 
>> x1=d1/d; x2=d2/d; x3=d3/d; x=[x1;x2;x3] 
 
x = 
 
 3 
 2 
 -1 
 
 
 
12. 
1
2
1 1
2 2
2 1 19
1 1 2 , 26
1 8
 −  
  = − =   
  − −  
A b 
Solution 
(a) The determinant of the coefficient matrix is 
 
1
2
1 1
2 2
2 1
91 1 2
4
1
D
−
= − = −
−
 
Since 0D ≠ , we find 
 
1 1
2 2
1 2 3
1 1 1 1
2 2 2 2
19 1 2 19 1 2 19
2726 1 2 9 , 1 26 2 , 1 1 26 18
2
8 1 8 1 8
D D D
− −
= − = − = − = − = =
− − − − −
 
so that 
 31 2
1 2 34 , 6 , 8DD Dx x x
D D D
= = = = = = − 
 
 
64 
 
(b) 
>> A = [2 1/2 -1;1 1 -2;1/2 -1 1/2]; b = [19;26;-8]; d = det(A); 
>> A1=A; A1(:,1)=b; d1=det(A1); 
>> A2=A; A2(:,2)=b; d2=det(A2); 
>> A3=A; A3(:,3)=b; d3=det(A3); 
>> x1=d1/d; x2=d2/d; x3=d3/d; x=[x1;x2;x3] 
x = 
 4.0000 
 6.0000 
 -8.0000 
 
 
13. 
2 1 1
3 3
1 1
3 3
 , , , parameters
0
Fs s
s F
s
 + + −  
= = =   
− +    
A b 
Solution 
(a) The determinant of the coefficient matrix is 
 
 
2 1 1
3 23 3 4 2
3 31 1
3 3
s s
D s s s
s
+ + −
= = + +
− +
 
 
We proceed by evaluating 
 
 
21 1
3 31 1
1 23 31 1
3 3
( ) , 
0 0
F s s F
D s F D F
s
− + +
= = + = =
+ −
 
so that 
 
1 1
3 3
1 23 2 3 24 2 4 2
3 3 3 3
( ) , s F Fx x
s s s s s s
+
= =
+ + + +
 
(b) 
>> syms s F 
>> A = [s^2+s+1/3 -1/3;-1/3 s+1/3]; b = [F;0]; d = det(A); 
>> A1=A; A1(:,1)=b; d1=det(A1); 
>> A2=A; A2(:,2)=b; d2=det(A2); 
>> x1=d1/d; x2=d2/d; x=[x1;x2] 
 
x = 
 
(F*(3*s + 1))/(s*(3*s^2 + 4*s + 2)) 
 F/(s*(3*s^2 + 4*s + 2)) 
 
 
 
14. 
81 4
32 3
33
42
1
2
73 2 0 0
1 0
 , 
1 0 1
0 0 1 1
  
   −−   = =  − −        
A b 
 
 
65 
 
Solution 
(a) The determinant of the coefficient matrix is 
 
 
1 4
2 3
3
2
3 2 0 0
1 0 9
1 0 1 2
0 0 1 1
D
−
= = −
−
 
Since 0D ≠ , we proceed by evaluating 
 
 
8 84 1
3 3 2 3
1 23 3 3 3
4 2 4 2
1 1
2 2
7 2 0 0 3 7 0 0
1 0 1 09 , 9
0 1 1 12
0 1 1 0 1 1
D D
− −−
− −
= = − = = −
−
 
 
 
8 81 4 1 4
2 3 3 2 3 3
3 43 3 3
4 2 4
1 1
2 2
3 2 7 0 3 2 0 7
0 19 9 , 
1 0 1 1 04 2
0 0 1 0 0 1
D D
− −− −
− −
= = = = −
− −
 
Finally, 
 1
1 2 3 421 , 2 , , 1x x x x= = = − = 
 
(b) 
>> A = [3 2 0 0;1/2 -4/3 1 0;-1 0 3/2 1;0 0 1 1]; b = [7;-8/3;-3/4;1/2]; 
>> d = det(A); 
>> A1=A; A1(:,1)=b; d1=det(A1); 
>> A2=A; A2(:,2)=b; d2=det(A2); 
>> A3=A; A3(:,3)=b; d3=det(A3); 
>> A4=A; A4(:,4)=b; d4=det(A4); 
>> x1=d1/d; x2=d2/d; x3=d3/d; x4=d4/d; x=[x1;x2;x3;x4] 
 
x = 
 
 1.0000 
 2.0000 
 -0.5000 
 1.0000 
 
 
In Problems 15 through 17, solve the homogeneous linear system Ax = 0 , where the 
components of x are 1 2, , ... , nx x x . 
15. 
1 3 4
2 2 1
1 1 5
− 
 = − 
  
A 
 
 
66 
 
Solution 
Since the coefficient matrix is singular, the system has infinitely many solutions. The row-
echelon from of the system is obtained as 
 
2 2 1 0
0 4 9 0
0 0 0 0
 − 
 
 
  
 
 
The last row being entirely zero suggests that there is one free variable. The second row yields 
2 34 9 0x x+ = . Choosing 3x as the free variable, this yields 9
2 34x x= − . Using this information 
in the first row, we find 11
1 34x x= − . Therefore, 
 
 
11
1 4
9
2 34
3 1
x
x x
x
 − 
   = = −   
   
   
x 
For example, letting 3 4x = generates one of the infinitely many solutions, as 
11
9
4
− 
 − 
 
 
. 
16. 
2 0 1
1 1 3
2 2 3
 
 = − 
 − 
A 
Solution 
Since the coefficient matrix is non-singular, 18=A , the system only has a trivial solution. 
 
17. 
5 3 2
0 2 4
3 2 6
− 
 =  
 − 
A 
Solution 
Since the coefficient matrix is non-singular, 68=A , the system only has a trivial solution. 
 
18. Determine a so that the following system has a non-trivial solution: 
 
1
2
3
1
42
2 1 0 0 0
1 0 0 0
0 0 2 1 0
0 0 1 0
x
a x
a x
x
−     
     
     =             −     
 
Solution 
Since the determinant of the coefficient matrix is (2 1)( 1)a a+ + , the system has a non-trivial 
solution for both cases of 1a = − or 1
2a = − . 
 
 
67 
 
 
Problem Set 3.3 
In Problems 1 through 8, 
(a) Find the eigenvalues and eigenvectors of the matrix. 
(b) Confirm the results of (a) in MATLAB. 
 
1. 
2 4
0 3
 
=  − 
A 
Solution 
(a) Upper-triangular matrix; ( ) 2, 3λ = −A . For 1 2λ = we solve 
 
 [ ]
EROs
1 1 1 1
0 4 0 0 1 0 1
2 
0 5 0 0 0 0 0
         
− = ⇒ = ⇒ = ⇒ =        −         
A I v 0 v v v 
 
For 2 3λ = − we solve 
 
 [ ] 2 2 2
5 4 0 4
3 
0 0 0 5
     
+ = ⇒ = ⇒ =     −     
A I v 0 v v 
(b) 
>> A = [2 4;0 -3]; [V,D] = eig(A) 
 
V = 
 
 1.0000 -0.6247 
 0 0.7809 
 
D = 
 
 2 0 
 0 -3 
 
 
2. 
5 2
2 0
− 
=  
 
A 
Solution 
(a) ( ) 1, 4λ =A . For 1 1λ = we solve 
 
 [ ]
EROs
1 1 1 1
4 2 0 2 1 0 1
 
2 1 0 0 0 0 2
− −         
− = ⇒ = ⇒ = ⇒ =        −         
A I v 0 v v v 
 
For 2 4λ = we solve 
 
 
 
68 
 
 [ ]
EROs
2 2 2 2
1 2 0 1 2 0 2
4 
2 4 0 0 0 0 1
− −         
− = ⇒ = ⇒ = ⇒ =        −         
A I v 0 v v v 
(b) 
>> A = [5 -2;2 0]; [V,D] = eig(A) 
 
V = 
 
 0.8944 0.4472 
 0.4472 0.8944 
 
D = 
 
 4 0 
 0 1 
 
 
3. 
0 3
3 0
 
=  
 
A 
Solution 
(a) ( ) 3λ = ±A . For 1 3λ = we solve 
 
 [ ]
EROs
1 1 1 1
3 3 0 1 1 0 1
3 
3 3 0 0 0 0 1
− −         
− = ⇒ = ⇒ = ⇒ =        −         
A I v 0 v v v 
 
For 2 3λ = − we solve 
 
 [ ]
EROs
2 2 2 2
3 3 0 1 1 0 1
3 
3 3 0 0 0 0 1
         
+ = ⇒ = ⇒ = ⇒ =         −         
A I v 0 v v v 
(b) 
>> A = [0 3;3 0]; [V,D] = eig(A) 
 
V = 
 
 -0.7071 0.7071 
 0.7071 0.7071 
 
 
D = 
 
 -3 0 
 0 3 
 
 
4. 
1 1 0
3 3 0
0 0 2
 
 =  
 − 
A 
 
 
69 
 
Solution 
(a) Block diagonal matrix; ( ) 0, 4, 2λ = −A . For 1 0λ = we solve 
 
 [ ]
EROs
1 1 1 1
1 1 0 0 1 1 0 0 1
 3 3 0 0 0 0 0 0 1
0 0 2 0 0 0 1 0 0
         
        = ⇒ = ⇒ = ⇒ = −        
        −         
A v 0 v v v 
 
For 2 4λ = we solve 
 
 [ ]
EROs
2 2 2 2
3 1 0 0 3 1 0 0 1
4 3 1 0 0 0 0 0 0 3
0 0 6 0 0 0 1 0 0
− −         
        − = ⇒ − = ⇒ = ⇒ =        
        −         
A I v 0 v v v 
 
For 3 2λ = − we solve 
 
 [ ]
EROs
3 3 3 3
3 1 0 0 1 0 0 0 0
2 3 5 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 1
         
        + = ⇒ = ⇒ = ⇒ =        
             