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F O U R
Time Response
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: Open-Loop Response
The forward transfer function for angular velocity is,
G(s) =
ω0(s) V (^) P(s) =^
(s+150)(s+1.32)
a. ω0(t) = A + Be-150t^ + Ce-1.32t
b. G(s) =
s 2 +151.32s+
. Therefore, 2ζωn =151.32, ωn = 14.07, and ζ = 5.38.
c. ω0(s) =
s(s 2 +151.32s+198)
Therefore, ω0(t) = 0.12121 + .0010761 e -150t^ - 0.12229e -1.32t^.
d. Using G(s),
ω 0
••
- 198 ω 0 = 24 v (^) p ( t ) Defining, x 1 = ω 0
x 2 =ω 0
Thus, the state equations are,
x 1
= x 2
x 2
= − 198 x 1 − 151.32 x 2 + 24 v p ( t )
y = x 1
In vector-matrix form,
x
⎦⎥^
x +
⎦⎥^
v p ( t ); y = [ 1 0 ] x
Solutions to Case Studies Challenges 75
e. Program: 'Case Study 1 Challenge (e)' num=24; den=poly([-150 -1.32]); G=tf(num,den) step(G)
Computer response: ans =
Case Study 1 Challenge (e)
Transfer function: 24
s^2 + 151.3 s + 198
Ship at Sea: Open-Loop Response
a. Assuming a second-order approximation: ωn^2 = 2.25, 2ζωn = 0.5. Therefore ζ = 0.167, ωn = 1.5.
Ts =
ζωn = 16; TP^ =^
ωn 1-ζ^2
%OS = e-ζ^ π^ /^ 1 -^ ζ
2 x 100 = 58.8%; ωnTr = 1.169 therefore, Tr = 0.77.
b. θ s 2.
s s^2 + 0.5 s +2.
= =^1
s
s +0. s^2 + 0.5 s +2.
=^1
s
s + 0.25 0.
s + 0.25^2 +2.
Solutions to Case Studies Challenges 77
Tp =
pos =
Tr =
ans =
(b)
K =
-0.3431 + 0.1058i -0.3431 - 0.1058i
p =
-3. -1.0000 + 1.1180i -1.0000 - 1.1180i 0
k =
[]
ans =
(c)
78 Chapter 4: Time Response
ANSWERS TO REVIEW QUESTIONS
1. Time constant 2. The time for the step response to reach 67% of its final value 3. The input pole 4. The system poles 5. The radian frequency of a sinusoidal response 6. The time constant of an exponential response 7. Natural frequency is the frequency of the system with all damping removed; the damped frequency of oscillation is the frequency of oscillation with damping in the system. 8. Their damped frequency of oscillation will be the same. 9. They will all exist under the same exponential decay envelop. 10. They will all have the same percent overshoot and the same shape although differently scaled in time. 11. ζ, ωn, TP, %OS, Ts 12. Only two since a second-order system is completely defined by two component parameters 13. (1) Complex, (2) Real, (3) Multiple real 14. Pole's real part is large compared to the dominant poles, (2) Pole is near a zero 15. If the residue at that pole is much smaller than the residues at other poles 16. No; one must then use the output equation 17. The Laplace transform of the state transition matrix is (sI - A )- 18. Computer simulation 19. Pole-zero concepts give one an intuitive feel for the problem. 20. State equations, output equations, and initial value for the state-vector 21. Det(s I - A ) = 0
SOLUTIONS TO PROBLEMS
1. a. Overdamped Case: C(s) =
s(s^2 + 9s + 9)
Expanding into partial fractions,
Taking the inverse Laplace transform,
c(t) = 1 + 0.171 e-7.854t - 1.171 e-1.146t
80 Chapter 4: Time Response
The evaluation of the constants in the numerator are found the same way as they were for the
underdamped case. The results are K 2 = -1 and K 3 = 0. Hence,
Therefore,
c(t) = 1 - cos 3t d. Critically Damped
The constants are then evaluated as
Now, the transform of the response is
c(t) = 1 - 3t e-3t - e-3t 2. a. C(s) =
s(s+5) =
s -^
s+5. Therefore, c(t) = 1 - e
-5t (^).
Also, T =
5 , Tr^ =
a =
5 = 0.44, Ts^ =
a =
b. C(s) =
s(s+20) =
s -^
s+20. Therefore, c(t) = 1 - e
-20t (^). Also, T =^1 20 ,
Tr =
a =
20 = 0.11, Ts^ =
a =
Solutions to Problems 81
Program: '(a)' num=5; den=[1 5]; Ga=tf(num,den) subplot(1,2,1) step(Ga) title('(a)') '(b)' num=20; den=[1 20]; Gb=tf(num,den) subplot(1,2,2) step(Gb) title('(b)')
Computer response: ans =
(a)
Transfer function: 5
s + 5
ans =
(b)
Transfer function: 20
s + 20
Solutions to Problems 83
X ( s )
F ( s )
Ms
2
+ 8 s
Differentiating to yield the transfer function in terms of velocity,
sX ( s )
F ( s )
Ms + 8
1 / M
s +
M
Thus, the settling time, Ts , and the rise time, Tr, are given by
Ts =
8/ M
M ; Tr =
8 / M
= 0.275 M
Program: Clf M= num=1/M; den=[1 8/M]; G=tf(num,den) step(G) pause M= num=1/M; den=[1 8/M]; G=tf(num,den) step(G)
Computer response: Transfer function: M =
1
Transfer function: 1
s + 8
M =
2
Transfer function:
s + 4
84 Chapter 4: Time Response
From plot, time constant = 0.125 s.
From plot, time constant = 0.25 s.
8. a. Pole: -2; c(t) = A + Be-2t^ ; first-order response.
b. Poles: -3, -6; c(t) = A + Be-3t^ + Ce-6t^ ; overdamped response.
c. Poles: -10, -20; Zero: -7; c(t) = A + Be -10t^ + Ce-20t^ ; overdamped response.
86 Chapter 4: Time Response
3 7 C = 2 8 - D = 0
Transfer function: 5 s^2 + 136 s - 1777
s^3 - s^2 - 91 s + 67
poles =
-9.
Writing the node equation at the capacitor, VC(s) (
R 2 +
Ls + Cs) +
V (^) C(s) - V(s) R 1 = 0.
Hence,
VC(s) V(s) =
R 1
R 1 +
R 2 +
Ls + Cs
10s s^2 +20s+
. The step response is
s^2 +20s+
.The poles
are at
-10 ± j20. Therefore, vC(t) = Ae-10t^ cos (20t + φ). 13. Program: num=[10 0]; den=[1 20 500]; G=tf(num,den) step(G)
Computer response:
Transfer function: 10 s
s^2 + 20 s + 500
Solutions to Problems 87
The equation of motion is: (Ms^2 +f (^) vs+K (^) s)X(s) = F(s). Hence,
X(s) F(s) =^
Ms^2 +f (^) vs+K (^) s
s^2 +s+
The step response is now evaluated: X(s) =
s(s^2 +s+5)
s -
5 s +
(s+
2 )^
2 +^19
5 (s+
2 ) +^
(s+
2 )^
2 +^19
Taking the inverse Laplace transform, x(t) =
5 e
-0.5t (^) ( cos 19 2 t +^
sin
2 t )
=
19 e
-0.5t (^) cos ( 19 2 t - 12.^
o).
C(s) =
ωn^2 s(s^2 +2ζωns+ωn^2 )
s -^
s + 2ζωn s^2 +2ζωns+ωn^2
s -^
s + 2ζωn (s+ζωn) 2 + ωn^2 - ζ 2 ωn^2
s -^
(s + ζωn) + ζωn (s+ζωn) 2 + (ωn 1 - ζ^2 ) 2
s -
(s+ζωn) +
ζωn ωn 1 - ζ^2
ωn 1 - ζ^2
(s+ζωn) 2 + (ωn 1 - ζ^2 ) 2
= 1 - e
cos ωn 1 - ζ
2
t +
ζ
sin ω (^) n 1 - ζ 2
t
Hence, c(t) ⎟
Solutions to Problems 89
e.
f.
a. N/A
b. s^2 +9s+18, ωn^2 = 18, 2ζωn = 9, Therefore ζ = 1.06, ωn = 4.24, overdamped.
c. s^2 +30s+200, ωn^2 = 200, 2ζωn = 30, Therefore ζ = 1.06, ωn = 14.14, overdamped.
d. s^2 +6s+144, ωn^2 = 144, 2ζωn = 6, Therefore ζ = 0.25, ωn = 12, underdamped.
e. s^2 +9, ωn^2 = 9, 2ζωn = 0, Therefore ζ = 0, ωn = 3, undamped.
f. s^2 +20s+100, ωn^2 = 100, 2ζωn = 20, Therefore ζ = 1, ωn = 10, critically damped.
X(s) =
s(s^2 +100s+100^2 )
s -^
s+ (s+50)^2 +
s -^
(s+50) + 50 (s+50)^2 +
s -
(s+50) +
(s+50)^2 +
Therefore, x(t) = 1 - e-50t^ (cos 7500 t +
sin 7500 t)
e-50t^ cos (50 3 t - tan-^
a. ωn^2 = 16 r/s, 2ζωn = 3. Therefore ζ = 0.375, ωn = 4. Ts =
ζωn = 2.667 s; TP^ =^
ωn 1-ζ^2
0.8472 s; %OS = e-ζ^ π^ /^ 1 -^ ζ
2 x 100 = 28.06 %; ωnTr = (1.76ζ^3 - 0.417ζ^2 + 1.039ζ + 1) = 1.4238;
therefore, Tr = 0.356 s.
90 Chapter 4: Time Response
b. ωn^2 = 0.04 r/s, 2ζωn = 0.02. Therefore ζ = 0.05, ωn = 0.2. Ts =
ζωn = 400 s; TP^ =^
ωn 1-ζ^2
15.73 s; %OS = e-ζ^ π^ /^ 1 -^ ζ
2 x 100 = 85.45 %; ωnTr = (1.76ζ^3 - 0.417ζ^2 + 1.039ζ + 1); therefore,
Tr = 5.26 s.
c. ωn^2 = 1.05 x 10^7 r/s, 2ζωn = 1.6 x 10^3. Therefore ζ = 0.247, ωn = 3240. Ts =
ζωn = 0.005 s; TP =
ωn 1-ζ^2
= 0.001 s; %OS = e-ζ^ π^ /^ 1 -^ ζ
2 x 100 = 44.92 %; ωnTr = (1.76ζ^3 - 0.417ζ^2 + 1.039ζ +
1); therefore, Tr = 3.88x10 -4^ s. 21. Program: '(a)' clf numa=16; dena=[1 3 16]; Ta=tf(numa,dena) omegana=sqrt(dena(3)) zetaa=dena(2)/(2omegana) Tsa=4/(zetaaomegana) Tpa=pi/(omeganasqrt(1-zetaa^2)) Tra=(1.76zetaa^3 - 0.417zetaa^2 + 1.039zetaa + 1)/omegana percenta=exp(-zetaapi/sqrt(1-zetaa^2)) subplot(221) step(Ta) title('(a)') '(b)' numb=0.04; denb=[1 0.02 0.04]; Tb=tf(numb,denb) omeganb=sqrt(denb(3)) zetab=denb(2)/(2omeganb) Tsb=4/(zetabomeganb) Tpb=pi/(omeganbsqrt(1-zetab^2)) Trb=(1.76zetab^3 - 0.417zetab^2 + 1.039zetab + 1)/omeganb percentb=exp(-zetabpi/sqrt(1-zetab^2)) subplot(222) step(Tb) title('(b)') '(c)' numc=1.05E7; denc=[1 1.6E3 1.05E7]; Tc=tf(numc,denc) omeganc=sqrt(denc(3)) zetac=denc(2)/(2omeganc) Tsc=4/(zetacomeganc) Tpc=pi/(omegancsqrt(1-zetac^2)) Trc=(1.76zetac^3 - 0.417zetac^2 + 1.039zetac + 1)/omeganc percentc=exp(-zetacpi/sqrt(1-zetac^2)) subplot(223) step(Tc) title('(c)')
Computer response: ans =
(a)
92 Chapter 4: Time Response
Trb =
percentb =
ans =
(c)
Transfer function: 1.05e
s^2 + 1600 s + 1.05e
omeganc =
3.2404e+
zetac =
Tsc =
Tpc =
Trc =
3.8810e-
percentc =
Solutions to Problems 93
Program: T1=tf(16,[1 3 16]) T2=tf(0.04,[1 0.02 0.04]) T3=tf(1.05e7,[1 1.6e3 1.05e7]) ltiview
Computer response: Transfer function: 16
s^2 + 3 s + 16
Transfer function:
s^2 + 0.02 s + 0.
Transfer function: 1.05e
s^2 + 1600 s + 1.05e
