Indian Institute of Technology Bombay

Department of Aerospace Engineering

Control Theory - AE308
Autumn 2023-2024

Practice Questions - 2 Solutions Autumn Semester 2023

1. 1. Estimate the values of ζ and ωn by comparing the closed loop transfer function
with the second order system and then compute the other properties.
2. Given, percentage overshoot of 5%,
( )
−ζπ
OS = exp p = 0.05 (1)
1 − ζ2

−ζπ
p = ln 0.05
1 − ζ2
ζ 2π2
2
= (ln 0.05)2
1−ζ
(ln 0.05)2
ζ2 = 2 =⇒ ζ = 0.69 (2)
π + (ln 0.05)2
Given, settling time of 2 sec,
4 2
Ts = = 2 =⇒ ωn = = 2.9 rad/s (3)
ζωn ζ

3. Referring to the system shown in Figure

Figure 1: Closed-loop system.

For the given system we have,

C(s) K ωn2
= 2 = 2 (4)
R(s) s + (2 + Kk)s + K s + 2ζωn s + ωn2

The system has a damping ratio ζ of 0.7 and an undamped natural frequency
ωn of 4 rad/sec.. Comparing with the general second order system,

K = ωn2 = 16 =⇒ K = 16 (5)
(2 + Kk) = 2ζωn = 2 × 0.7 × 4 =⇒ k = 0.225 (6)
2. System equation when mass m is set into motion by an unit impulse,

mẍ + kx = δ(t) (7)

Consider another impulse function with magnitude A and acts after T time to stop
the system. Thus, the equation for the system when the two impulse forces are given
is,

mẍ + kx = δ(t) + Aδ(t − T ); x(0−) = 0, ẋ(0−) = 0 (8)

Laplace transform of equation 8 gives,

(ms2 + k)X(s) = 1 + A exp{−sT } (9)

solving for X(s),

1 A exp{−sT }
X(s) = + (10)
ms2
+ kp ms2 + k p
1 k/m A k/m exp{−sT }
=√ 2
+√ (11)
km s + k/m km s2 + k/m

Taking inverse Laplace transform of X(s),

r ! r !
1 k A k
x(t) = √ sin t +√ sin (t − T ) u(t − T ) (12)
km m km m

If the motion of the mass m is to be stopped at t = T , then x(t) must be identically

zero for t ≥ 0.
π 3π 5π
x(t) = 0 ∀ t ≥ T ; if, A = 1, T = p ,p ,p ,... (13)
k/m k/m k/m

Thus motion of the mass m can be stopped by another impulse force, such as,
π 3π 5π
Aδ(t − T ) where, A = 1, T = p ,p ,p ,... (14)
k/m k/m k/m

3. G(s) is the open loop transfer function, Then the closed loop transfer function for a
unity feedback system is,
G(s) Ks + b Ks + b
= 2 =⇒ G(s) = (15)
1 + G(s) s + as + b s(s + a − K)

Consider ramp input to unity feedback system (R(s) = 1/s2 ), then error is obtained
is
R(s) 1
E(s) = = 2 (16)
1 + G(s) s (1 + G(s))
The steady state error due to ramp input is,
1
ess = lim sE(s) = lim (17)
s→0 s→0 s(1 + G(s))

1 s(s + a − K) a−K
= lim = lim = (18)
s→0 sG(s) s→0 s(Ks + b) b

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4. Given, open-loop transfer function is
K
G(s) = (19)
s(Js + B)

The closed loop transfer function (unity-feedback control system) is,

G(s) K ωn2
H(s) = = 2 = 2 (20)
1 + G(s) Js + Bs + K s + 2ζωn s + ωn2

Consider ramp input to unity feedback system (R(s) = 1/s2 ), then error is obtained
is
R(s) 1
E(s) = = 2 (21)
1 + G(s) s (1 + G(s))

The steady state error due to ramp input is,

1
ess = lim sE(s) = lim (22)
s→0 s→0 s(1 + G(s))

1 s(Js + B) B
= lim = lim = (23)
s→0 sG(s) s→0 sK K

The steady state error ess can be reduced by increasing gain K or decreasing the
viscous-friction coefficient B. Increasing gain (K = ωn ) or decreasing the viscous-
friction coefficient (B = 2ζωn ), however causes the damping ratio (ζ) to decrease,
with the result that the transient response of the system will become more oscillatory.

5. Reducing the block diagram shown in Figure P5.6 to a single block

Figure 2: Block diagram of system

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Page 4
 Reduced the block diagram is,

Let, Heq = G2 /G5 + G3 (G4 + G6 G7 ) + G8 and Geq = G1 G5 , then the closed loop
transfer function is,

C(s) Geq (s)

T (s) = = (24)
R(s) 1 + Geq (s)Heq (s)
G5 G1
= (25)
1 + G1 (G8 G5 + G7 G6 G5 G3 + G5 G4 G3 + G2 )

6. Finding the unity feedback system that is equivalent to the given system,

Figure 3: Block diagram of system

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The unity feedback system that is equivalent to the system is,

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 5s2 + 2s
T (s) = (26)
6s2 + 9s + 6

7. 1. X2 = G12 X1 + G32 X3 + G42 X4 + G52 X5

2. X3 = G23 X2

3. X4 = G34 X3 + G44 X4

4. X5 = G35 X3 + G45 X4

8. Obtaining C(s)/R(s) using Mason’s gain rule

Figure 4: Signal flow graph of a system

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 • Forward Path Gain:
P1 = 1 · G2 · G4 · G6 · 1;
P2 = 1 · G3 · G5 · G7 · 1;
P3 = 1 · G2 · G1 · G7 · 1;
(27)
P4 = 1 · G3 · G8 · G6 · 1;
P5 = 1 · G2 · G1 − H2 · G8 · G6 · 1;
P6 = 1 · G3 · G8 − H1 · G1 · G7 · 1;

• Loop Gains:
L1 = −H2 · G5 ;
L2 = −H1 · G4 ; (28)
L3 = G1 − H2 · G8 · −H1 ;

• ∆= 1- loop gains+ non touching loops taken two at a time- non touching
P P P
loops taken three at a time+ . . .
∆k = Value of ∆ for the part of graph not touching the k th forward path.
∆ = 1 − (L1 + L2 + L3 ) + L1 · L2 ;
∆1 = 1 − L1 ;
(29)
∆2 = 1 − L2 ;
∆3 = 1; ∆4 = 1; ∆5 = 1; ∆6 = 1;

C(s) 1
= [P1 ∆1 + P2 ∆2 + P3 ∆3 + P4 ∆4 + P5 ∆5 + P6 ∆6 ] (30)
R(s) ∆

9. For the characteristic eqaution,

s4 + 2s2 + (4 − k)s2 + 9s + 25 = 0 (31)

the Routh array of coefficients are,

s4 1 (4 + k) 25
s3 2 9 0
2k−1
s2 2
25
18k−109
s1 2k−1
0
s0 25

For stability we require,

2k − 1 18k − 109
> 0, >0
2 2k − 1
or, k > 0.5, 18k > 109

Thus, For stability we require k > 109/18 = 6.056

10. From the block diagram,

C(s) 20K
= 3 2
(32)
r(s) s + 5s + (4 + 20KKh )s + 20K

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 Figure 5: Servo system with tachometer feedback.

The stability of this system is determined by the characteristic polynomial. The

Routh array of the characteristic eqauation,

s3 + 5s2 + (4 + 20KKh )s + 20K = 0 (33)

the Routh array of coefficients are,

s3 1 (4 + 20KKh )
s2 5 20K
s1 20KKh − 4K + 4 0
s0 20K

For stability, we require,

20KKh − 4K + 4 > 0 10K > 0

5KKh > K − 1 K>0 (34)

The stable region in the K −Kh plane is the region that satisfies the two inequalities,
5KKh > K − 1 and K > 0.

Figure 6: stable-region

If a point lies in the shaded region the system is stable.

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