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# Norman Nise Control Systems Engineering 6th Solultions.PDF

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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

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

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
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

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
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

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

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