Stability Analysis

Question 1
Consider a unity feedback system, as in the figure shown, with an integral compensator \frac{K}{s} and open-loop transfer function

G(s)=\frac{1}{s^2+3s+2}

where k \gt 0. The positive value of K for which there are exactly two poles of the unity feedback system on the j\omega axis is equal to ______ (rounded off to two decimal places).

A
2.45
B
4.28
C
6.0
D
6.25
GATE EC 2019   Control Systems
Question 1 Explanation: 
\frac{Y(s)}{X(s)}=\frac{K}{s^{3}+3 s^{2}+2 s+K}
Two poles of this system lie the system is moro:
System is marginally stable.
\qquad k_{\text{mar}}=3 \times 2=6
Question 2
A unity feedback control system is characterized by the open-loop transfer function
G(s)=\frac{2(s+1)}{s^{3}+ks^{2}+2s+1}
The value of k for which the system oscillates at 2 rad/s is ________.
A
0.6
B
0.69
C
0.89
D
0.75
GATE EC 2017-SET-2   Control Systems
Question 2 Explanation: 
The given open loop transfer function is,
G(s)=\frac{2(s+1)}{s^{3}+K s^{2}+2 s+1}
The closed loop transfer function is,
T(s)=\frac{G(s)}{1+G(s)}=\frac{2(s+1)}{s^{3}+K s^{2}+4 s+3}
When system oscillates, i.e., when system is marginally stable,
(1)(3)=K(4)
\mathrm{So}, \quad K=0.75
Question 3
Which one of the following options correctly describes the locations of the roots of the equation s^{4}+s^{2}+1=0 on the complex plane?
A
Four left half plane (LHP) roots
B
One right half plane (RHP) root, one LHP root and two roots on the imaginary axis
C
Two RHP roots and two LHP roots
D
All four roots are on the imaginary axis
GATE EC 2017-SET-1   Control Systems
Question 3 Explanation: 
q(s)=s^{4}+s^{2}+1=0
\begin{array}{r|rrr} s^{4} & 1 & 1 & 1\\ s^{3} & 4 & 2 & 0\\ s^{2} & 0.5 & 1 & 0 \\ s^{1} & -6 & 0 & 0 \\ s^{0} & 1 & 0 & 0 \end{array} \frac{d A(s)}{d s}=4 s^{3}+2 s^{1}
There are two sign Changes in the first column of the R-H table and the order of auxiliary equation is 4. So, four poles are symmetric about origin.
\therefore 2 RHP roots and 2 LHP roots.

Question 4
The first two rows in the Routh table for the characteristic equation of a certain closed-loop control system are given as

The range of K for which the system is stable is
A
-2.0 \lt K \lt 0.5
B
0 \lt K \lt 0.5
C
0\lt K\lt \infty
D
0.5\lt K\lt \infty
GATE EC 2016-SET-3   Control Systems
Question 4 Explanation: 
\begin{array}{c|cc} s^{3}& 1 & (2 k+3) \\ s^{2}&2 k & 4 \\ s&\frac{4 k^{2}+6 k-4}{2 k} & 0\\ s^{0}&4 \end{array}
For stability, K \gt 0
\begin{aligned} 4 K^{2}+6 K-4& \gt 0 \\ 2 K^{2}+3 K-2& \gt 0 \\ (2 K-1)(K+2)& \gt 0 \\ \Rightarrow \quad K& \gt \frac{1}{2} or, k \lt -2 \\ \text{Hence, }\quad 0.5& \lt K \lt \infty \end{aligned}
Question 5
The transfer function of a linear time invariant system is given by
H(s)=2s^{4}-5s^{3}+5s-2
The number of zeros in the right half of the s-plane is ________
A
1
B
2
C
3
D
4
GATE EC 2016-SET-1   Control Systems
Question 5 Explanation: 
2 s^{4}-5 s^{3}+5 s-2=0
By Routh Array,
\begin{array}{c|ccc} s^{4} & 2 & 0 & -2 \\ s^{3} & -5 & 5 & \\ s^{2} & 2 & -2 & \\ s^{1} & 0(2) & & \\ s^{0} & -2 & & \end{array}
Number at sign changes = number at roots (zeros) in right half of s-plane =3
Question 6
Match the inferences X, Y, and Z, about a system, to the corresponding properties of the elements of first column in Routh?s Table of the system characteristic equation.
A
X\rightarrowP, Y\rightarrowQ, Z\rightarrowR
B
X\rightarrowQ, Y\rightarrowP, Z\rightarrowR
C
X\rightarrowR, Y\rightarrowQ, Z\rightarrowP
D
X\rightarrowP, Y\rightarrowR, Z\rightarrowQ
GATE EC 2016-SET-1   Control Systems
Question 6 Explanation: 
When all elements are positive, the system is stable. When any element is zero, the test breaks down. When there is change in sign of coefficients, the system is unstable.
Question 7
The characteristic equation of an LTI system is given by F(s)=s^{5}+2s^{4}+3s^{3}+6s^{2}-4s-8=0 The number of roots that lie strictly in the left half s-plane is _________.
A
0
B
1
C
2
D
3
GATE EC 2015-SET-3   Control Systems
Question 7 Explanation: 
Using Routh's tabular or:
\begin{array}{c|lll} s^{5} & 1 & 3 & -4 \\ s^{4} & 2 & 6 & -8 \\ s^{3} & 0 8 & 0(12) & 0 \\ s^{2} & 3 & -8 & 0 \\ s^{1} & \frac{100}{3} & 0 & 0 \\ s^{0} & -8 & & \end{array}
Auxiliary equation:
\begin{aligned} A &=2 s^{4}+6 s^{2}-8=0 \\ \frac{d A}{d s} &=8 s^{3}+12 s=0 \end{aligned}
As one time row of zero occur in Routh's table therefore a pair of imaginary pole exists, also number of sign change in Routh's table is one. Hence, two poles lie strictly in the left half of s-plane.
Question 8
A plant transfer function is given as G(s)=(K_{P}+\frac{K_{I}}{s})\frac{1}{s(s+2)}. When the plant operates in a unity feedback configuration, the condition for the stability of the closed loop system is
A
K_{P}\gt \frac{K_{I}}{2} \gt  0
B
2K_{I}\gt K_{P} \gt 0
C
2K_{I} \lt K_{P}
D
2K_{I} \gt K_{P}
GATE EC 2015-SET-1   Control Systems
Question 8 Explanation: 
G(s)=\left(K_{p}+\frac{K_{I}}{s}\right)\left(\frac{1}{s(s+2)}\right)
The closed loop transfer function for unity feedback
\begin{aligned} \frac{G(s)}{1+G(s)} &=\frac{\left(K_{p} s+K_{I}\right)}{s^{2}(s+2)+\left(K_{p} s+K_{I}\right)} \\ &=\frac{K_{p} s+K_{I}}{s^{3}+2 s^{2}+K_{p} s+K_{I}} \end{aligned}
Using Routh's tabular form:
\begin{array}{c|cc} s^{3} & 1 & K_{p} \\ s^{2} & 2 & K_{I} \\ s^{1} & \frac{2 K_{p}-K_{I}}{2} & 0 \\ s^{0} & K_{p} & \end{array}
For system to be stable;
\begin{aligned} K_{p}& \gt 0\\ \text{and }\frac{2 K_{p}-K_{I}}{2}& \gt 0 \\ \text{or} \quad 2 K_{p}-K_{I}& \gt 0 \\ \text{or} \quad K_{p}& \gt \frac{K_{I}}{2} \gt 0 \end{aligned}
Question 9
Consider a transfer function G_{p}(s)=\frac{ps^{2}+3ps-2}{s^{2}+(3+p)s+(2-p)} with p a positive real parameter. The maximum value of p until which G_{p} remains stable is ____.
A
1
B
2
C
3
D
4
GATE EC 2014-SET-4   Control Systems
Question 9 Explanation: 
Given transfer function
G_{p}(s)=\frac{p s^{2}+3 p s-2}{s^{2}+(3+p) s+(2-p)} \qquad\ldots(i)
The characteristic equation is
=s^{2}+(3+p) s+(2-p)\qquad\ldots(ii)
Using Routh Hurtwiz criterion,
\begin{array}{l|ll} s^{2} & 1 & (2-P) \\ s^{1} & (3+p) & 0 \\ s^{0} & (2-p) & 0 \end{array}
For system to be remain stable
\begin{aligned} 2-p & \geq 0 \\ p_{\max } &=1.99 \end{aligned}
Question 10
The forward path transfer function of a unity negative feedback system is given by
G(s)=\frac{K}{(s+2)(s-1)}
The value of K which will place both the poles of the closed-loop system at the same location, is ______.
A
4.5
B
2.25
C
1.25
D
0.5
GATE EC 2014-SET-1   Control Systems
Question 10 Explanation: 
Given that, \quad G(s)=\frac{K}{(s+2)(s-1)}
Using root locus method, the break point can be obtain as


\begin{aligned} \Rightarrow \quad 1+G(s)&=0 \\ 1+\frac{K}{(s+2)(s-1)} &=0 \\ \text{or }\quad K&=-(s+2)(s-1)\\ \frac{d K}{d s} &=-2 s-1=0 \\ \text{or }\quad s &=-0.5 \end{aligned}
To have, both the poles at the same directions
\begin{aligned} |G(s)|_{s=0.5} &=1 \\ K &=2.25 \end{aligned}
There are 10 questions to complete.
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