Norman Nise Control Systems Engineering 6th Solultions.PDF
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Norman Nise Control Systems Engineering 6th Solultions.PDF


DisciplinaControle Linear38 materiais291 seguidores
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%Solve for X2(t). 
x3=ilaplace(X(3)) %Solve for X3(t). 
y=C*[x1;x2;x3] %Solve for output, y(t). 
y=simplify(y) %Simplify y(t). 
'y(t)' %Display label. 
pretty(y) %Pretty print y(t). 
 
Computer response: 
ans = 
 
a 
 
A = 
 -3 1 0 
 0 -6 1 
 0 0 -5 
 
X0 = 
 1 
 1 
 0 
x1 = 
 
7/6*exp(-3*t)-1/3*exp(-6*t)+1/15+1/10*exp(-5*t) 
 
x2 = 
 
exp(-6*t)+1/5-1/5*exp(-5*t) 
 
x3 = 
 
1/5-1/5*exp(-5*t) 
 
y = 
 
2/5+exp(-6*t)-2/5*exp(-5*t) 
 
y = 
 
2/5+exp(-6*t)-2/5*exp(-5*t) 
 
ans = 
 
y(t) 
 2/5 + exp(-6 t) - 2/5 exp(-5 t) 
 
43. 
 |\u3bbI - A | = \u3bb2 + 5\u3bb +1 
 |\u3bbI - A | = (\u3bb + 0.20871) (\u3bb + 4.7913) 
 
 Solutions to Problems 4-39 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
 
 
 
 
 
 
Therefore, 
 
 
Solving for Ai's two at a time, and substituting into the state-transition matrix 
 
 
To find x(t), 
 
 
 
 
 
To find the output, 
 
 
 
44. 
 
 |\u3bbI - A | = \u3bb2 + 1 
 
4-40 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
 
 
 
Solving for the Ai's and substituting into the state-transition matrix, 
 
 
To find the state vector, 
 
 
 
 
 
 
(3, 4)
1 cos[ ]
(3,4)
sin[ ]
( 3cos[ ] 4sin[ ] 3)
y x
t
y
t
y t t
=
\u2212\u239b \u239e\u394 = \u239c \u239f\u239d \u23a0
= \u2212 + +
 
 
45. 
 |\u3bbI - A | = (\u3bb + 2) (\u3bb + 0.5 - 2.3979i) (\u3bb + 0.5 + 2.3979i) 
 Let the state-transition matrix be 
 
 
 Solutions to Problems 4-41 
Copyright © 2011 by John Wiley & Sons, Inc. 
 Since \u3c6(0) = I, \u3a6. (0) = A, and \u3c6..(0) = A2, we can evaluate the coefficients, Ai's. Thus, 
 
 
 
 
 
 
 Solving for the Ai's taking three equations at a time, 
 
 
 
U s i n g x (t ) = \u3c6 (t )x (0 ) + \u222b
0
t
\u3c6 (t -\u3c4 )B u (\u3c4 )d\u3c4 , an d y = 1 0 0 x (t ), 
 
 
= 
1
2 - 
1
2 e
-2t 
46. 
Program: 
syms s t tau %Construct symbolic object for 
 %frequency variable 's', 't', and 'tau. 
'a' %Display label. 
A=[-2 1 0;0 0 1;0 -6 -1] %Create matrix A. 
B=[1;0;0] %Create vector B. 
C=[1 0 0] %Create vector C. 
X0=[1;1;0] %Create initial condition vector,X(0). 
I=[1 0 0;0 1 0;0 0 1]; %Create identity matrix. 
'E=(s*I-A)^-1' %Display label. 
E=((s*I-A)^-1) %Find Laplace transform of state 
 %transition matrix, (sI-A)^-1. 
Fi11=ilaplace(E(1,1)); %Take inverse Laplace transform 
Fi12=ilaplace(E(1,2)); %of each element 
Fi13=ilaplace(E(1,3)); 
Fi21=ilaplace(E(2,1)); 
Fi22=ilaplace(E(2,2)); 
Fi23=ilaplace(E(2,3)); 
4-42 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
Fi31=ilaplace(E(3,1)); 
Fi32=ilaplace(E(3,2)); %to find state transition matrix. 
Fi33=ilaplace(E(3,3)); %of (sI-A)^-1. 
'Fi(t)' %Display label. 
Fi=[Fi11 Fi12 Fi13 %Form Fi(t). 
Fi21 Fi22 Fi23 
Fi31 Fi32 Fi33]; 
pretty(Fi) %Pretty print state transition matrix, Fi. 
Fitmtau=subs(Fi,t,t-tau); %Form Fi(t-tau). 
'Fi(t-tau)' %Display label. 
pretty(Fitmtau) %Pretty print Fi(t-tau). 
x=Fi*X0+int(Fitmtau*B*1,tau,0,t); 
 %Solve for x(t). 
x=simple(x); %Collect terms. 
x=simplify(x); %Simplify x(t). 
x=vpa(x,3); 
'x(t)' %Display label. 
pretty(x) %Pretty print x(t). 
y=C*x; %Find y(t) 
y=simplify(y); 
y=vpa(simple(y),3); 
y=collect(y); 
'y(t)' 
pretty(y) %Pretty print y(t). 
 
Computer response: 
ans = 
 
a 
 
 
 
 
A = 
 
 -2 1 0 
 0 0 1 
 0 -6 -1 
 
B = 
 
 1 
 0 
 0 
 
C = 
 
 1 0 0 
 
X0 = 
 
 1 
 1 
 0 
 
ans = 
 
E=(s*I-A)^-1 
 
E = 
 
[ 1/(s+2), (s+1)/(s+2)/(s^2+s+6), 1/(s+2)/(s^2+s+6)] 
[ 0, (s+1)/(s^2+s+6), 1/(s^2+s+6)] 
[ 0, -6/(s^2+s+6), s/(s^2+s+6)] 
 
 
 
 Solutions to Problems 4-43 
Copyright © 2011 by John Wiley & Sons, Inc. 
ans = 
 
Fi(t) 
 
 [ 13 
 [exp(-2 t) , - 1/8 exp(-2 t) + 1/8 %1 + --- %2 , 
 [ 184 
 
 ] 
 1/8 exp(-2 t) - 1/8 %1 + 3/184 %2] 
 ] 
 
 [ 
 [0 , 1/23 %2 + %1 , - 1/23 
 
 1/2 1/2 1/2 
 (-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t)) 
 
 ] 
 ] 
 
 [ 
 [0 , 6/23 
 
 1/2 1/2 1/2 
 (-23) (exp((-1/2 + 1/2 (-23) ) t) - exp((-1/2 - 1/2 (-23) ) t)) 
 
 ] 
 , - 1/23 %2 + %1] 
 
 1/2 
 %1 := exp(- 1/2 t) cos(1/2 23 t) 
 
 1/2 1/2 
 %2 := exp(- 1/2 t) 23 sin(1/2 23 t) 
 
ans = 
 
Fi(t-tau) 
 
 
 [ 
 [exp(-2 t + 2 tau) , 
 [ 
 
 13 1/2 
 - 1/8 exp(-2 t + 2 tau) + 1/8 %2 cos(%1) + --- %2 23 sin(%1) , 
 184 
 
 1/2 ] 
 1/8 exp(-2 t + 2 tau) - 1/8 %2 cos(%1) + 3/184 %2 23 sin(%1)] 
 ] 
 
 [ 1/2 1/2 
 [0 , 1/23 %2 23 sin(%1) + %2 cos(%1) , - 1/23 (-23) ( 
 
 1/2 
 exp((-1/2 + 1/2 (-23) ) (t - tau)) 
 
 1/2 ] 
 - exp((-1/2 - 1/2 (-23) ) (t - tau)))] 
 
 [ 1/2 1/2 
 [0 , 6/23 (-23) (exp((-1/2 + 1/2 (-23) ) (t - tau)) 
 
 1/2 
 - exp((-1/2 - 1/2 (-23) ) (t - tau))) , 
4-44 Chapter 4: Time Response 
 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 1/2 ] 
 - 1/23 %2 23 sin(%1) + %2 cos(%1)] 
 
 1/2 
 %1 := 1/2 23 (t - tau) 
 
 %2 := exp(- 1/2 t + 1/2 tau) 
 
ans = 
 
x(t) 
 
 [.375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t) 
 
 + .339 exp(-.500 t) sin(2.40 t) + .500] 
 
 [.209 exp(-.500 t) sin(2.40 t) + exp(-.500 t) cos(2.40 t)] 
 
 [1.25 i (exp((-.500 + 2.40 i) t) - 1. exp((-.500 - 2.40 i) t))] 
 
ans = 
 
y(t) 
 
 .375 exp(-2. t) + .125 exp(-.500 t) cos(2.40 t) 
 
 + .339 exp(-.500 t) sin(2.40 t) + .500 
47. 
 The state-space representation used to obtain the plot is, 
 
 
x. = 0 1
-1 -0.8
 x + 
0
1
 u(t); y(t) = 1 0 x
 
 Using the Step Response software, 
 
 
 
 Solutions to Problems 4-45 
Copyright © 2011 by John Wiley & Sons, Inc. 
 
 
Calculating % overshoot, settling time, and peak time, 
 
2\u3b6\u3c9n = 0.8, \u3c9n = 1, \u3b6 = 0.4. Therefore, %OS = e\u2212\u3b6\u3c0 / 1\u2212\u3b6 2 x100 = 25.38%, Ts = 4\u3b6\u3c9n = 10 sec, 
Tp = 
\u3c0
\u3c9n 1-\u3b62
 = 3.43 sec. 
 
 
 
 
 
 
 
 
 
48. 
 
 
4-46 Chapter 4: Time Response 
 
Copyright © 2011 by John