GATE EE · Control Systems

Time Response Analysis Practice Questions

Generate GATE-level questions on Time Domain analysis. Focus on: 1. Standard test signals and Response of First and Second order systems. 2. Time domain specifications: Rise time, Peak time, Settling time, and Overshoot. 3. Steady-state error and Error constants (Kp, Kv, Ka).

27 questions · 7 PYQs · 0 AI practice · GATE EE 2027

Q21GATE 2008MCQ

The transfer function of a system is given as $\frac{100}{s^{2}+20s+100}$ . This system is

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📖 Explanation

$M(s)=\frac{100}{s^{2+}20s+100}$ Comparing with standard form, $M(s)=\frac{\omega _n^2}{s^{2+}2 \xi \omega _n s+\omega _n^2}$ $\therefore \;\;2 \xi \omega _n =20$ $\;\;\omega _n^2=100$ $\;\; \xi=1$ $\;\;\omega _n=10$ $\therefore$ The system is critically damped.

Q22GATE 2008MCQ

The transfer function of a linear time invariant system is given as $G(s)=\frac{1}{s^{2}+3s+2}$ The steady state value of the output of the system for a unit impulse input applied at time instant t=1 will be

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📖 Explanation

$r(t)=$ unit impulse applied at $t=1$ $=\delta (t-1)$ $R(s)=1[r(t)]=e^{-s}$ $G(s)=\frac{C(s)}{R(s)}=\frac{1}{s^{2+}3s+2}$ $C(s)=R(s)G(s)=\frac{e^{-s}}{s^{2+}3s+2}$ Steady state value of output, using final value theoram, $C_{ss}=\lim_{s \to 0}sC(s)$ $\;\;=\lim_{s \to 0}\frac{se^{-s}}{s^{2+}3s+2}=0$

Q23GATE 2007MCQ

Consider the R-L-C circuit shown in figure For a step-input $e_i$ , the overshoot in the output $e_o$ will be

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📖 Explanation

$e_i=R_i+L\frac{di}{dt}+\frac{1}{c}\int i \;dt$ Taking Laplace transform,

\begin{aligned}E_i(s)&=\left ( R+Ls+\frac{1}{Cs} \right )I(s) \\ I(s)&=\frac{E_i(s)}{R+Ls+\frac{1}{Cs} } \\ e_0&=\frac{1}{C}\int i \;dt \\ \Rightarrow \; E_0(s)&=\frac{1}{Cs}I(s) \\ &=\frac{1}{Cs}\left[ \frac{E_i(s)}{R+Ls+\frac{1}{Cs} } \right] \\ t_0(s)&=\frac{E_i(s)}{RCs+LCs^{2+}1} \\ \frac{E_0(s)}{E_i(s)}&=\frac{1}{LC\left ( s^{2+}\frac{R}{L}s+\frac{1}{LC} \right )} \\ &\text{Characteristic equation,} \\ \Rightarrow \; & s^{2+}\frac{R}{L}s+\frac{1}{LC}=0 \\ &\text{Comparing with,} \\ &s^{2+}2 \xi \omega _n s+\omega _n^2 \\ \omega _n&=\frac{1}{\sqrt{LC}} \\ 2 \xi \omega _n&=\frac{R}{L} \\ \xi&=\frac{R}{L} \times \frac{1}{2\omega _n} \\ &=\frac{R}{L} \times \frac{\sqrt{LC}}{2}=\frac{R}{2}\sqrt{\frac{C}{L}} \\ \xi &=\frac{10}{2}\sqrt{\frac{10 \times 10^{-6}}{1 \times 10^{-3}}}=0.5 \\ \text{Overshot }&=P_P =e^{-\pi \xi/\sqrt{1-\xi^2}} \\ &=e^{-\pi\times 0.5/\sqrt{1-0.5^2}} \\ & =0.163 or 16.3\%\end{aligned}