Q2GATE 2023MCQ

A closed loop system is shown in the figure where $k \gt 0$ and $\alpha \gt 0$ . The steady state error due to a ramp input $\left(R(s)=\alpha / s^{2}\right)$ is given by

🚀 Solve in Practice Mode

📖 Explanation

Given, $\quad$ input is $r(t)=\alpha t u(t)$ $R(s)=\frac{\alpha}{s^{2}}$ From the figure, $G(s) H(s)=\frac{K}{s(s+2)}$ Now steady state error for Ramp input is $e_{s s} =\frac{\alpha}{K_{v}}, \text { where } \alpha \text { is the magnitude of Ramp input }$

\begin{aligned}K_{v} & =\lim _{s \rightarrow 0}[s G(s) H(s)] \\ K_{v} & =\lim _{s \rightarrow 0}\left[\frac{s \times K}{s(s+2)}\right]=\frac{K}{2}\\ \therefore \quad e_{s s} & =\frac{\alpha \times 2}{K} \\ e_{s s} & =\frac{2 \alpha}{K} \end{aligned}

Q3GATE 2022MCQ

Two linear time-invariant systems with transfer functions $G_1(s)=\frac{10}{s^{2+}s+1}, G_2(s)=\frac{10}{s^{2+}s\sqrt{10}+10}$ have unit step responses $y_1(t)$ and $y_2(t)$ , respectively. Which of the following statements is/are true?

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} G_1(s)&=\frac{10}{s^{2+}s+1}\\ \omega _{n1}^2&=1\\ 2\xi _1 \times 1&=1\\ \xi _1&=1/2\\ G_2(s)&=\frac{10}{s^{2+}s\sqrt{10}+10}\\ \omega _{n2}^2&=10\\ 2\xi _1 \times \sqrt{10}&=\sqrt{10}\\ \xi _2&=1/2\\ \end{aligned}

$\because M_P$ depends on $\xi$ only $y_1(t) \; and \; y_2(t)$ have same percentage peak overshoot. $\omega _{d_1}=\omega _{n_1}\sqrt{1-\xi ^2}$ $\Rightarrow \omega _{d_1}\neq \omega _{d_2} \;\; \because \omega _{n_1}\neq \omega _{n_2}$ damped frequency of oscillation is not same.

\begin{aligned} T_s &=\frac{4}{\xi \omega _n}(2%\; setting \; time) \\ T_{s_1} &\neq T_{s_2} \;\because \omega _{n_1}\neq \omega _{n_2}\\ C_1(s) &= \frac{10 \times 1/s}{s^{2+}s+1}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_1(s) \\ &= 10\\ C_2(s) &= \frac{10 \times 1/s}{s^{2+}\sqrt{10}s+10}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_2(s) \\ &= \frac{10}{10}1=1 \end{aligned}

Steady - state value is not same.

Q4GATE 2022NAT

The block diagram of a closed-loop control system is shown in the figure. R(s),Y(s) and D(s) are the Laplace transforms of the time-domain signals $r(t),y(t),$ and $d(t)$ , respectively. Let the error signal be defined as $e(t)=r(t)-y(t)$ . Assuming the reference input $r(t)=0$ for all $t$ , the steady-state error $e(\infty )$ , due to a unit step disturbance $d(t)$ , is _________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

$Y(s)=\frac{R(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}} +\frac{D(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}}$ When, $r(t)=0\; and \; d(t)=U(t)\underleftrightarrow{L.T.}\frac{1}{s}$ $e(\infty )=-\lim_{s \to 0}\frac{s\cdot \frac{1}{s} \times 1}{s(s+10)+10} =-1/10=-0.1\;\;\;(\because e(t)=r(t)-y(t))$

Q5GATE 2020NAT

Consider the following closed loop control system where $G(s)=\frac{1}{s(s+1)}$ and $C(s)=K\frac{s+1}{s+3}$ . If the steady state error for a unit ramp input is 0.1, then the value of K is _______.

🚀 Solve in Practice Mode

📖 Explanation

Open loop transfer function for the system $=C(s)\times G(s)=\frac{K(s+1)}{ (s+3)}\times \frac{1}{s(s+1)}$ Since the system is type-1, so far a given unit ramp input steady state $e_{ss}=\frac{1}{K_{V}}$ where, $K_{V}=\lim_{s\rightarrow 0}S\times \frac{K}{S(S+3)}=\frac{K}{3}$ so, $\; e_{ss}=\frac{1}{K/3}=\frac{3}{K}$ Given that, $\; e_{ss}=0.1$ So, $0.1=\frac{3}{K}\Rightarrow K=30$

Q6GATE 2020NAT

A system with transfer function $G(s)=\frac{1}{(s+1)(s+a)}, a \gt 0$ is subjected an input $5\cos 3t$ . The steady state output of the system is $\frac{1}{\sqrt{10}} \cos (3t-1.892)$ . The value of $a$ is ____.

🚀 Solve in Practice Mode

📖 Explanation

Given that,

\begin{aligned}G(j\omega )&=\frac{1}{(1+j\omega )(\alpha +j\omega )} \\ \left |G(j\omega ) \right |&=\frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}} \\ \text{According to }& \text{question,}\\ \left | G(j\omega ) \right |_{\omega =3}&=\frac{1}{5\sqrt{10}}\\ \Rightarrow \frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}}&=\frac{1}{5\sqrt{10}} \\ \Rightarrow \frac{1}{\sqrt{10(a^{2}+9)}}&=\frac{1}{5\sqrt{10}} \alpha ^{2}+9=25 \\ \alpha ^{2}&=16 \\ \alpha &=4\end{aligned}

Q7GATE 2020NAT

The loop transfer function of a negative feedback system is $G(s)H(s)=\frac{K(s+11)}{s(s+2)(s+8)}$ The value of K, for which the system is marginally stable, is ________.

🚀 Solve in Practice Mode

📖 Explanation

Characteristic equation q(s) for the given open loop system will be $q(s)=s^{3}+10s^{2}+16s+Ks+11K=0$ Using R-H criteria, For System to be Marginally Stable

\begin{aligned}\frac{10(16+K)-11K}{10}&=0 \\ 160+10K-11K&=0 \\ K&=160\end{aligned}

Q8GATE 2019MCQ

Consider a causal second-order system with the transfer function $G(s)=\frac{1}{1+2s+s^2}$ with a unit-step $R(s)=\frac{1}{s}$ as an input. Let C(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value $\lim_{t \to \infty }c(t)$ , rounded off to two decimal places, is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s+1}=\frac{1}{(s+1)^{2}} \\ R(s)&=\frac{1}{s} \\ \alpha (s)&=G(s) R(s)=\frac{1}{s(s+1)^{2}} \end{aligned}

Using partial fraction expansion, we get,

\begin{aligned} C(s) & =\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+1)^{2}} \\ A\left(s^{2}+2 s+1\right)&+B\left(s^{2}+s\right)+C s=1 \\ A&=1 \\ A+B & =0 \Rightarrow B=-1 \\ 2 A+B+C & =0 \Rightarrow C=-1 \\ \therefore \quad C(s) & =\frac{1}{s}-\frac{1}{(s+1)}-\frac{1}{(s+1)^{2}} \\ \text{and}\quad c(t) & =\left(1-e^{-t}-t e^{-t}\right) u(t) \\ \lim _{t \rightarrow \infty } c(t) & =1 \end{aligned}

In order to reach 94 % of its steady-state value, $\left(1-e^{-t}-t e^{-t}\right)=0.94$ By trial and error, we get, $t \approx 4.50 \mathrm{sec}$

Q9GATE 2017NAT

The open loop transfer function G(s)= $\frac{(s+1)}{s^{P}(s+2)(s+3)}$ Where p is an integer, is connected in unity feedback configuration as shown in figure. Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is _________.

🚀 Solve in Practice Mode

📖 Explanation

Steady state error of type 1 system, for step input is zero, for ramp input is $\frac{1}{K_{v}}$ and for parabolic input is infinity. So, the given system must be of type- 1 and p=1

Q10GATE 2016NAT

The open-loop transfer function of a unity-feedback control system is given by $G(s)=\frac{K}{s(s+2)}$ For the peak overshoot of the closed-loop system to a unit step input to be 10%, the value of K is____________

🚀 Solve in Practice Mode

📖 Explanation

$G(s)=\frac{K}{s(s+2)} ; H(s)=1$ Characteristic equation $=1+G(s) H(s)=0$

\begin{aligned} 1+\frac{K}{s(s+2)} &=0 \\ s^{2}+2 s+K &=0 \Rightarrow \omega_{n}=\sqrt{K} \\ 2 \xi \omega_{n} &=2 ; \xi=\frac{1}{\sqrt{K}} \\ M_{p}&=e^{-\pi\xi/\sqrt{1-\xi^{2}}}-0.1\\ -\frac{\pi \xi}{\sqrt{1-\xi^{2}}} &=\ln (0.1) \\ \Rightarrow \quad\frac{\pi \xi}{\sqrt{1-\xi^{2}}} &=2.3 \\ \pi^{2} \xi^{2} &=(2.3)^{2}\left(1-\xi^{2}\right) \\ 15.16 \xi^{2} &=(2.3)^{2} \\ \Rightarrow \qquad \xi&=0.59 \\ \Rightarrow \quad \text{Also}, \quad K&=\frac{1}{\xi^{2}}=2.8 \end{aligned}

Q11GATE 2016MCQ

For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as $G(s)=\frac{2}{s(s+1)}$ . The steady state error $e_{ss}$ due to a unit step input is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} E(s) &=\frac{R(s)}{1+G(s) H(s)} \\ R(s) &=\frac{1}{s} ; G(s)=\frac{2}{s(s+1)}; \\ H(s) &=1 \\ E(s) &=\frac{\frac{1}{s}}{1+\frac{s}{s(s+1)}}=\frac{s+1}{s^{2}+s+2} \\ e_{s s} &=\lim _{s \rightarrow 0} s E(s) \\ &=\lim _{s \rightarrow 0} \frac{s(s+1)}{s^{2}+s+2} \\ e_{s s} &=0 \end{aligned}

Q12GATE 2016NAT

In the feedback system shown below $G(s)=\frac{1}{(s^{2}+2s)}$ . The step response of the closed-loop system should have minimum settling time and have no overshoot. The required value of gain k to achieve this is ________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s} \\ \frac{Y(s)}{R(s)}&=\frac{\frac{K}{s^{2}+2 s}}{1+\frac{K}{s^{2}+2 s}}=\frac{K}{s^{2}+2 s+K} \end{aligned}

Minimum settling time and no overshoot implies critical damping i.e.

\begin{aligned} \xi &=1 \\ \omega_{n} &=\sqrt{K} \\ \therefore \quad 2 \times \xi \cdot \omega_{n} &=2 \\ \omega_{n} &=1 \\ \Rightarrow \quad \sqrt{K} &=1 \text { or } K=1 \end{aligned}

Q13GATE 2016NAT

The response of the system $G(s)=\frac{s-2}{(s+1)(s+3)}$ to the unit step input u(t) is y(t). The value of $\frac{dy}{dt}$ at t= $0^{+}$ is _________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} L\left(\frac{d y}{d t}\right) &=(s Y(s)-y(0)) \\ Y(s) &=G(s) \times \frac{1}{s}=\frac{s-2}{s(s+1)(s+3)} \\ y(0)&=\lim _{s \rightarrow \infty } s Y(s) \\ \text{(Applying } & \text{initial value theorem) } \\ &=\lim _{s \rightarrow \infty } \frac{s-2}{(s+1)(s+3)} \\ &=\frac{\left(1-\frac{2}{s}\right)}{s\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)} \\ y(0)&=0 \\ L\left(\frac{d y}{d t}\right) &=s Y(s)=\frac{s \times (s-2)}{s(s+1)(s+3)} \\ &=\frac{s-2}{(s+1)(s+3)} \\ \left.\frac{d y}{d t}\right|_{t=0} &=\lim _{s \rightarrow \infty } s L\left(\frac{d y}{d t}\right) \\ &=\lim _{s \rightarrow \infty } \frac{s \times (s-2)}{(s+1)(s+3)}\\ &=\frac{\left(1-\frac{2}{s}\right)}{\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)}=1 \end{aligned}

Q14GATE 2015MCQ

The output of a standard second-order system for a unit step input is given as $y(t)=1-\frac{2}{\sqrt{3}}e^{-t}cos(\sqrt{3}t-\frac{\pi }{6})$ . The transfer function of the system is

🚀 Solve in Practice Mode

📖 Explanation

The output of a second order system for unit step input $y(t)=1-\frac{e^{-\xi \omega_{n} t}}{\sqrt{1-\xi^{2}}} \sin \left(\omega_{d} t-\theta\right) \qquad\ldots(i)$ From given data

\begin{aligned} y(t)&=1-\frac{2}{\sqrt{3}} e^{-t} \cos \left(\sqrt{3} t-\frac{\pi}{6}\right)\\ \therefore \quad \alpha&=\xi \omega_{n}=1 \\ \text{and }\quad \omega_{d}&=\sqrt{3} \\ \omega_{n}&=\sqrt{\alpha^{2}+\omega_{d}^{2}}=2 \\ \xi&=\frac{1}{2} \end{aligned}

Therefore, the transfer function $T(s)=\frac{\omega_{n}^{2}}{s^{2}+2 \xi \omega_{n} s+\omega_{n}^{2}}=\frac{4}{s^{2}+2 s+4}$

Q15GATE 2015NAT

A unity negative feedback system has an open-loop transfer function $G(s)=\frac{K}{s(s+10)}$ . The gain for the system to have a damping ratio of 0.25 is ________.

🚀 Solve in Practice Mode

📖 Explanation

$T(s)=\frac{G(s)}{1+G(s) H(s)}=\frac{K}{s^{2}+10 s+K}$ Comparing with standard second order transfer function

\begin{aligned} T(s)&=\frac{\omega_{n}^{2}}{s^{2}+2 \xi \omega_{n} s+\omega_{n}^{2}}\\ \text{We have, }\omega_{n}^{2}&=K \\ \text{and }\quad 2 \xi \omega_{n}&=10 \\ \because \quad \xi&=0.25 \quad (given)\\ \therefore \quad \omega_{n}&=\frac{10}{2 \times 0.25}=20 \\ \text{and }\quad k&=\omega_{n}^{2}=(20)^{2}=400 \end{aligned}

Q16GATE 2014NAT

The steady state error of the system shown in the figure for a unit step input is _____.

🚀 Solve in Practice Mode

📖 Explanation

After converting the system into equivalent unity feedback system. $\therefore \quad OLTF =\frac{4(s+4)}{s(s+2)}$ Type =1 $e_{s s}$ for step input =0.5

Q17GATE 2014MCQ

For the second order closed-loop system shown in the figure, the natural frequency (in rad/s) is

🚀 Solve in Practice Mode

📖 Explanation

The closed loop transfer function of the given system is $\frac{C(s)}{R(s)}=\frac{4}{s(s+4)+4}=\frac{4}{s^{2}+4 s+4}$ Comparing with standard second order transfer

\begin{aligned} \text{function }\quad &\frac{K \omega_{n}^{2}}{S^{2}+2 \xi \omega_{n}+\omega_{n}^{2}} \\ \text{we get, }\quad \omega_{n}^{2}&=4 \\ \text{or }\quad\omega_{n}&=\sqrt{4}=2 \mathrm{rad} / \mathrm{sec} \end{aligned}

Q18GATE 2014NAT

The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is ______.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text { Given, } & \omega_{n}=40 \mathrm{rad} / \mathrm{sec} \\ & \xi=0.3 \\ \therefore \quad & \omega_{d}=\omega_{n} \sqrt{1-\xi^{2}}=38.16 \mathrm{rad} / \mathrm{sec} \end{aligned}

Q19GATE 2014NAT

The characteristic equation of a unity negative feedback system is $1 + KG(s) = 0$ . The open loop transfer function G(s) has one pole at 0 and two poles at -1. The root locus of the system for varying K is shown in the figure. The constant damping ratio line for $\xi= 0.5$ , intersects the root locus at point A. The distance from the origin to point A is given as 0.5. The value of K at point A is ____.

🚀 Solve in Practice Mode

📖 Explanation

According to the question $G(s) H(s)=\frac{K}{s(s+1)^{2}} \qquad \ldots(i)$ The damping ratio is given as $\xi=0.5$ $\therefore \quad \theta=\cos ^{-1} \xi=60^{\circ}$ Let the co-ordinates of point 'A' is (-a+jb) where, $\cos 60^{\circ}=\frac{O B}{O A} \qquad \ldots(ii)$ Also, $\quad O A=0.5$ (Given) $\therefore O B=0.25$ Similarly,

\begin{array}{l} \sin 60^{\circ}=\frac{B A}{O A} \qquad \ldots(iii)\\ \therefore \quad B A=j b=0.433 j \end{array}

Hence, co-ordinates of $A = -0.25 + 0.43j$ $\therefore$ The value of system gain K can be obtained as $|G(s) H(s)|=1\qquad \ldots(iv)$ From (i) and (iv) we get $s=-0.25+j 4.33$ $K=0.37$

Q20GATE 2013MCQ

A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y(t) for t > 0, when the forcing function is x(t) and the initial condition is y(0). If one wishes to modify the system so that the solution becomes -2y(t) for t > 0, we need to

🚀 Solve in Practice Mode

📖 Explanation

$\frac{d y(t)}{d t}+k y(t)=x(t)$ Taking Laplace transform of both sides, we have $s Y(s)-y(0)+k Y(s)=X(s)$ $Y(s)[s+k]=X(s)+y(0)$ $\Rightarrow Y(s)=\frac{X(s)}{s+k}+\frac{y(0)}{s+k}$ Taking inverse Laplace transform, we have $y(t)=e^{-k t} x(t)+y(0) e^{-k t}$ So if we want -2y(t) as a solution both x(t) and y(0) has to be multiplied by -2; hence change x(t) by -2x(t) and y(0) by -2y(0).

Time Response Analysis Practice Questions

Get the full Time Response Analysis experience — 100% free