Concepts of Stability

Question 1
The characteristic equation of a linear time-invariant (LTI) system is given by
\Delta (s)=s^4+3s^3+3s^2+s+k=0
The system is BIBO stable if
A
0 \lt k \lt \frac{12}{9}
B
k \gt 3
C
0 \lt k \lt \frac{8}{9}
D
k \gt 6
GATE EE 2019   Control Systems
Question 1 Explanation: 
Routh array is

For BIBO stability,
\frac{\left ( \frac{8}{3}-3k \right )}{\left ( \frac{8}{3} \right )} \gt 0
\Rightarrow k \lt \frac{8}{9}
and k \gt 0
\therefore \;\; 0 \lt k \lt \frac{8}{9}
Question 2
The number of roots of the polynomial,
s^{7}+s^{6}+7s^{5}+14s^{4}+31s^{3}+73s^{2}+25s+200,
in the open left half of the complex plane is
A
3
B
4
C
5
D
6
GATE EE 2018   Control Systems
Question 2 Explanation: 
Characteristic equation, s^7+s^6+7s^5+14s^4+31s^3+73s^2+25s+200=0

Auxillary equation, A(s)=8s^4+48s^2+200
\frac{d}{ds}A(s)=32s^3+96s
Total number of poles =7
Two sign change above auxillary equation=2 poles in RHS.
Two sign changes below auxillary equation implies out of 4 symmetric roots about origin, two poles are in LHS and two poles are in RHS.
Therefore 3 poles in LHS and 4 poles in RHS.
Question 3
The range of K for which all the roots of the equation s^{3}+3s^{2}+2s+K=0 are in the left half of the complex s-plane is
A
0 \lt K \lt 6
B
0 \lt K \lt 16
C
6 \lt K \lt 36
D
6 \lt K \lt 16
GATE EE 2017-SET-2   Control Systems
Question 3 Explanation: 
From the given equation, s^3+3s^2+2s+K=0
Using Routh's criterion, we get
K \lt 6 and K \gt 0 or 0 \lt K \lt 6
Question 4
A closed loop system has the characteristic equation given by s^{3}+Ks^{2}+(K+2)s+3=0. For this system to be stable, which one of the following conditions should be satisfied?
A
0 \lt K \lt 0.5
B
0\lt K \lt 1
C
0 \lt K \lt 1
D
K \gt 1
GATE EE 2017-SET-1   Control Systems
Question 4 Explanation: 
Characteristic equation is,
s^3+Ks^2+(K+2)s+3=0
For this system to be stable, using Routh's criterion, we can write,
3 \lt K(K+2)
or, K^2+2K-3 \gt 0
or, (K+3)(K-1) \gt 0
Here, the valid answer will be out of all the options given.
i.e K \gt 1.
Question 5
The open loop transfer function of a unity feedback control system is given by
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}, K\gt 0,T \gt 0
The closed loop system will be stable if,
A
0\lt T \lt\frac{4(K+1)}{K-1}
B
0 \lt K \lt \frac{4(T+2)}{T-2}
C
0 \lt K \lt \frac{T+2}{T-2}
D
0 \lt T \lt \frac{8(K+1)}{K-1}
GATE EE 2016-SET-2   Control Systems
Question 5 Explanation: 
Open loop transfer function:
G(s)=\frac{K(s+1)}{s(1+Ts)(1+2s)}; \; K \gt 0 \; and \; t \gt 0
For closed loop system stability, characteristic equation is,
1+G(s)H(s)=0
1+\frac{K(s+1)}{s(1+Ts)(1+2s)}\cdot 1=0
s(1+Ts)(1+2s)+K(s+1)=0
2Ts^3+(2+T)s^2+(1+K)s+K=0
Using Routh's criteria,

For stability, K \gt 0
and (2+T)(1+K)-2TK \gt 0
K(2+T-2T)+(2+T) \gt 0
or
-(T-2)K+2(2+T) \gt 0
-K \gt -\frac{(2+T)}{(T-2)} or K \lt \frac{T+2}{(T-2)}
Hence for stability,
0 \lt K \lt \frac{T+2}{T-2}
Question 6
Given the following polynomial equation s^{3}+5.5s^{2}+8.5s+3=0,
the number of roots of the polynomial, which have real parts strictly less than -1, is ________ .
A
1
B
2
C
3
D
4
GATE EE 2016-SET-1   Control Systems
Question 6 Explanation: 
s^3+5.5s^2+8.5s+3=0
Putting, s=s_1-1
(s_1-1)^3+5.5(s_1-1)^2+(8.5)(s_1-1)+3=0
s_1^3+2.5s_1^2+0.5s_1-1=1

As there is one sign change, hence, two roots of given polynomial will lie to the left of s=-1.
Question 7
The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at (2-j3). List all the poles and zeroes.
A
Poles at (2\pmj3), no zeroes.
B
Poles at (\pm2-j3), one zero at origin.
C
Poles at (2-j3),(-2+j3), zeroes at (-2-j3),(2+j3).
D
Poles at (2\pmj3), zeroes at (-2\pmj3).
GATE EE 2015-SET-1   Control Systems
Question 7 Explanation: 
Response of transfer function is unit for all \omega.
M=1; P_1=2-j3
Second order system, hence number of poles =2
Therefore, second pole P_2=2+j3
Now for M=1, and due to x-axis symmetry of root locus of transfer function, position of zeroes must be
Z_1=-2-j3 and Z_2=-2+j3
Question 8
A single-input single output feedback system has forward transfer function G(s) and feedback transfer function H(s). It is given that |G(s)H(s)| \lt1. Which of the following is true about the stability of the system ?
A
The system is always stable
B
The system is stable if all zeros of G(s)H(s) are in left half of the s-plane
C
The system is stable if all poles of G(s)H(s) are in left half of the s-plane
D
It is not possible to say whether or not the system is stable from the information given
GATE EE 2014-SET-3   Control Systems
Question 9
A system with the open loop transfer function
G(s)=\frac{K}{s(s+2)(s^{2}+2s+2)}
is connected in a negative feedback configuration with a feedback gain of unity. For the closed loop system to be marginally stable, the value of K is ______
A
4
B
5
C
6
D
7
GATE EE 2014-SET-2   Control Systems
Question 9 Explanation: 
Given, G(s)=\frac{K}{s(s+2)(s^2+2s+2)}

The characteristic equation of given unity-negative feedback control system is given by
1+G(s)H(s)=0
or, 1+\frac{K}{s(s+2)(s^3+2s+2)}=0
or, s(s+2)(s^3+2s+2)+K=0
or, s^4+4s^3+6s^2+4s+K=0
Forming Routh' array as shown below

For stability of the system, K \gt 0 and \frac{20-4K}{5} \gt 0 or K\gt 5
\therefore For stability, 0 \lt K \lt 5
For given system to be marginally stable

K=5
Question 10
For the given system, it is desired that the system be stable. The minimum value of \alpha for this condition is ______.
A
0.22
B
0.46
C
0.86
D
0.62
GATE EE 2014-SET-1   Control Systems
Question 10 Explanation: 
Given, G(s)=\frac{(s+\alpha )}{s^3+(1+\alpha )s^2+(\alpha -1)s+(1-\alpha )}
and H(s)=1
Characteristic equation of given control system is given by
1+G(s)H(s)=0
or, 1+\left [ \frac{s+\alpha }{s^3+(1+\alpha )s^2+(\alpha -1)s+(1-\alpha )} \right ]=0
or, s^3+(1+\alpha )s^2+(\alpha -1)s+(1-\alpha )+s+\alpha =0
or, s^3+(1+\alpha )s^2+\alpha s+1=0
Routh's array is

For the given system to be stable, there should not be any sign change in the first column of Routh's array.
Therefore, 1+\alpha \gt 0 \;or, \; \alpha \gt -1\;\; ...(i)
Also, \frac{\alpha ^2 +\alpha -1}{1+\alpha } \gt 0
or, \alpha ^2 +\alpha -1 \gt 0
or, (\alpha +1.618)(\alpha -0.618) \gt 0
or, \alpha \lt -1.618
or, \alpha \gt 0.618
As \alpha \gt -1 (using equation (i))
therefore, \alpha \gt 0.618 \;\;...(ii)
Combining condition (i) and (ii),
-1 \lt \alpha \lt 0.618
Thus, for the system to be stable minimum value of \alpha =0.618.
There are 10 questions to complete.
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