GATE ECE · Control Systems

Stability Analysis – GATE ECE Control SystemsPractice Questions & PYQs

Generate GATE-level questions on Stability. Focus on: 1. Routh-Hurwitz criterion: Determining absolute stability and number of roots in RHP. 2. Stability of feedback systems: BIBO stability and characteristic equation analysis. 3. Relative stability and marginal stability cases.

25 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MCQ

Consider the polynomial $p(s)=s^{5}+7 s^{4}+3 s^{3}-33 s^{2}+2 s-40$ . Let $(L,I,R)$ be defined as follows. $L$ is the number of roots of $p(s)$ with negative real parts. $I$ is the number of roots of $\mathrm{p}(\mathrm{s})$ that are purely imaginary. $R$ is the number of roots of $p(s)$ with positive real parts. Which one of the following options is correct?

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📖 Explanation

Given $p(s)=s^{5}+7 s^{4}+3 s^{3}-33 s^{2}+2 s-40$

\begin{array}{l|lrl} s^{5} & 1 & 3 & 2 \\ s^{4} & 7 & -33 & -40 \\ s^{3} & \frac{54}{7} & \frac{54}{7} & 0 \\ s^{2} & -40 & -40 & 0 \\ s^{1} & 0(-80) & 0 & 0 \\ s^{0} & -40 & & \\ \end{array}

Auxiliary equation [A.E] $-40 \mathrm{~s}^{2}-40=0$ A.E $\Rightarrow \mathrm{s}^{2}+1=0 \Rightarrow \mathrm{~s}^{2}=-1 \Rightarrow \mathrm{~s}= \pm 1 \mathrm{j}$ So number of imaginary roots $[\mathrm{I}]=2$ In the RH tabular, number of sign changes in the first column $=1$ , hence number of poles in the right half of s-plane (or) roots with positive real parts $[R]=1$ , number of poles in the left half of s-plane $[\mathrm{L}]=5-2-1=2$

Q2GATE 2024MCQ

In the feedback control system shown in the figure below $G(s)=\frac{6}{s(s+1)(s+2)}$ . $R(s), Y(s)$ , and $E(s)$ are the Laplace transforms of $r(t), y(t)$ , and $e(t)$ , respectively. If the input $r(t)$ is a unit step function, then _____

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📖 Explanation

Given system is $G(s)=\frac{6}{s(s+1)(s+2)}$ The characteristic equation,

\begin{aligned} 1+G(s) & =0 \\ 1+\frac{6}{s(s+1)(s+2)} & =0 \\ s(s+1)(s+2)+6 & =0 \\ s^{3}+3 s^{2}+2 s+6 & =0 \end{aligned}

Clearly, here the internal coefficient product is equal to external coefficient product i.e., $3 \times 2=6 \times 1$ Hence, the given system is marginally stable system. $\therefore e(t)$ not exist because system is oscillatory system.

Explanation Figure 2

Q3GATE 2022MCQ

Consider an even polynomial $p(s)$ given by $p(s)=s^{4+}5s^{2+}4+K$ , where $K$ is an unknown real parameter. The complete range of $K$ for which $p(s)$ has all its roots on the imaginary axis is ________.

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📖 Explanation

$p(s)=s^{4+}5s^{2+}4+k$

\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 0 & 0 & \\ s^2& & & \\ s& & & \\ s^0& & & \end{matrix}

$s^{4+}5s^{2+}(4+k)=0$ $4s^{3+}10s=0$ $(4s^{2+}10)=0$ $s=\pm j\sqrt{5/2}$

\begin{matrix} s^4 & 1& 5 &(4+k) \\ s^3& 4 & 10 & \\ s^2& 5/2 & (4+k) & \\ s^1&10-\frac{4(4+k)}{5/2} & & \\ s^0& (4+k) & & \end{matrix}

\begin{aligned} (4+k) & \gt 0\\ k & \gt -4\\ 10 \times \frac{5}{2}& \gt 4(4+K)\\ (4+K)& \lt \frac{25}{4}\\ K& \lt 9/4 \end{aligned}

$\Rightarrow$ All roots be on imaginary axis $-4\leq K\leqslant 9/4$

Q4GATE 2019NAT

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).

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📖 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. $\qquad k_{\text{mar}}=3 \times 2=6$

Q5GATE 2017NAT

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 ________.

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📖 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, $(1)(3)=K(4)$ $\mathrm{So}, \quad K=0.75$

Q6GATE 2017MCQ

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?

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📖 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.

Explanation Figure 1

Q7GATE 2016MCQ

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

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📖 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}

Q8GATE 2016MCQ

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.

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Q9GATE 2016NAT

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 ________

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📖 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

Q10GATE 2015NAT

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 _________.

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📖 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.

Q11GATE 2015MCQ

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

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📖 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}

Q12GATE 2014NAT

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 ______.

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📖 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}

Explanation Figure 1

Q13GATE 2014NAT

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 ____.

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📖 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}

Q14GATE 2012MCQ

The feedback system shown below oscillates at 2 rad/s when

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📖 Explanation

Characteristic equation is $1+G(s)H(s)=0$ $1+\frac{K(s+1)}{s^{3}+a s^{2}+2 s+1}=0$ $s^{3}+a s^{2}+(2+K) s+(1+K)=0$ array for this is Routh

\begin{array}{l|lr} s^{3} & 1 & (2+K) \\ s^{2} & a & (1+K) \\ s^{1} & \frac{a(2+K)-(1+K)}{a} \\ s^{0} & (1+K) & \end{array}

For oscillation $\frac{a(2+K)-(1+K)}{a}=0$ $\Rightarrow \quad a=\left(\frac{1+K}{2+K}\right)$ Now

\begin{aligned} a s^{2}+(1+K)&=0 \\ -a \omega^{2}+(1+K)&=0 \\ \text{given }\quad \omega&=2 \mathrm{rad} / \mathrm{sec} \\ -4 a+(1+K)&=0 \\ -4 \frac{(1+K)}{(2+K)}+(1+K)&=0 \\ -4(1+K)+(2+K)(1+K)&=0 \\ (1+K)[(2+K)-4]&=0 \\ K=-1,2 \end{aligned}

but K=-1 is not possible as system will not oscillate for this as a=0 So, K=2 $a=\frac{1+K}{2+K}=\frac{3}{4}=0.75$

Q15GATE 2008MCQ

A certain system has transfer function $G(s)=\frac{s+8}{s^{2}+\alpha s-4}$ , where $\alpha$ is a parameter. Consider the standard negative unity feedback configuration as shown below Which of the following statements is true?

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📖 Explanation

Closed loop gain is $\frac{G(s)}{1+G(s)}=\frac{s+8}{s^{2}+\alpha s-4+s+8}$ Characteristic equation $q(s)=s^{2}+(\alpha+1) s+4$ Closed loop system is stable only for $\alpha \gt -1.$ . Therefore, for all positive of $\alpha$ , the closed loop system is stable.

Q16GATE 2008MCQ

The number of open right half plane of $G(s)=\frac{10}{s^{5}+2s^{4}+3s^{3}+6s^{2}+5s+3}$ is

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📖 Explanation

Characteristic equation $q(s)=s^{5}+2 s^{4}+3 s^{3}+6 s^{2}+5 s+3$ Putting $s=\frac{1}{z}$ $q^{\prime}(z)=3 z^{5}+5 z^{4}+6 z^{3}+3 z^{2}+2 z+1$ Routh array is

\begin{array}{l|lll} z^{5} & 3 & 6 & 2 \\ z^{4} & 5 & 3 & 1 \\ z^{3} & 21 / 5 & 7 / 5 & \\ z^{2} & \frac{63}{51} =\frac{4}{3} & 1 &\\ z^{1} & \frac{\frac{28}{15}-\frac{21}{5}}{4 / 3} =\frac{-35 / 15}{4 / 3}=\frac{-7}{4} & &\\ z^{0} & 1 & & \end{array}

Sinc_e sign changes twice in Routh-array, therefore, there are two poles on right half plane.

Q17GATE 2007MCQ

If the closed-loop transfer function of a control system is given as $T(s)=\frac{s-5}{(s+2)(s+3)}$ , then it is

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Q18GATE 2006MCQ

Consider a unity-gain feedback control system whose open-loop transfer function is: $G(s)=\frac{as+1}{s^{2}}$ With the value of "a" set for a phase - margin of $\frac{\pi }{4}$ , the value of unit-impulse response of the open-loop system at t=1 second is equal to

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📖 Explanation

\begin{aligned} G(s)&=\frac{0.84 s+1}{s^{2}}\\ H(s) &=1, R(s)=1 \\ c(s) &=G(s) \cdot R(s) \\ \therefore \quad C(s) &=\frac{0.84 s+1}{s^{2}} \\ c(t) &=L^{-1}\left[\frac{1+0.84 s}{s^{2}}\right]\$ \\ &=L^{-1}\left[\frac{1}{s^{2}}+\frac{0.84}{s}\right]\$ \\ c(t) &=[t+0.84] U(t) \\ \text{At}\quad t &=1, c(t)=1+0.84=1.84 \end{aligned}

Q19GATE 2006MCQ

The positive values of "K" and "a" so that the system shown in the figures below oscillates at a frequency of 2 rad/sec respectively are

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📖 Explanation

\begin{array}{l} \text { 1+ } G(s) H(s)=1+\frac{K(s+1)}{s^{3}+a s^{2}+2 s+1}=0 \\ \Rightarrow \frac{s^{3}+a s^{2}+2 s+1+K s+K}{s^{3}+a s^{2}+2 s+1}=0 \end{array}

\begin{array}{c|cc} s^{3}& 1 &2+K\\ s^{2}&a & K+1\\ s&\frac{a(2+K)-(K+1)}{a} \end{array}

\begin{array}{l} \text { For oscillation, } \frac{a(2+K)-(K+1)}{a}=0 \\ \qquad \begin{array}{l} a=\frac{K+1}{K+2} \\ a s^{2}+K+1=0 \\ s=j \omega, s^{2}=-\omega^{2}=-4 \\ \Rightarrow-4 a+K+1=0 \\ a=\frac{K+1}{4} \Rightarrow \frac{K+1}{4}=\frac{K+1}{K+2} \Rightarrow K=2 \\ a=0.75 \end{array} \end{array}

Q20GATE 2004MCQ

The gain margin for the system with open-loop transfer function $G(s)H(s)=\frac{2(1+s)}{s^{2}}$ , is

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📖 Explanation

\begin{aligned} \angle G(s) H(s) &=-180^{\circ}+\tan ^{-1} \omega \\ \text{For}\quad\omega_{\theta} &=-180^{\circ}+\tan ^{-1} \omega=-180^{\circ} \\ \omega &=0 \\ |G(s) H(s)| &=\frac{2 \sqrt{1+\omega^{2}}}{\omega^{2}}=\infty \\ G \cdot M \cdot &=\frac{1}{\infty }=0 \\ \ln \quad d B G \cdot M &=\infty \end{aligned}

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