Q2GATE 2020MCQ

Which of the following statements is true about the two sided Laplace transform?

Q3GATE 2019MCQ

The output response of a system is denoted as y(t), and its Laplace transform is given by $Y(s)=\frac{10}{s(s^{2+}s+100\sqrt{2})}$ The steady state value of y(t) is

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📖 Explanation

Steady state value of y(t)

\begin{aligned} &=\lim_{s \to 0}sY(s)\\ &=\lim_{s \to 0}\frac{10s}{s(s^{2+}s+100\sqrt{2})}\\ &=\frac{10}{100\sqrt{2}}=\frac{1}{10\sqrt{2}} \end{aligned}

Q4GATE 2019MCQ

A system transfer function is $H(s)=\frac{a_1s^{2+}b_1s+c_1}{a_2s^{2+}b_2s+c_2}$ . If $a_1=b_1=0$ , and all other coefficients are positive, the transfer function represents a

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📖 Explanation

\begin{aligned} H(s)&=\frac{c_1}{a_2s^{2+}b_2s+c_2}\\ & as \;\; a_1=b_1=0\\ &=\frac{c_1}{(1+s\tau _1)(1+s\tau _2)}\\ \text{Put }s&=0,\; H(0)=\frac{c_1}{c_2}\\ \text{Put }s&=\infty ,\; H(\infty )=0 \end{aligned}

which represents second order low pass filter.

Q5GATE 2019MCQ

The inverse Laplace transform of $H(s)=\frac{s+3}{s^{2+}2s+1} \; for \; t\geq 0$ is

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📖 Explanation

\begin{aligned} L^{-1}\left ( \frac{s+3}{s^{2+}2s+1} \right )&=L^{-1}\left ( \frac{s+1+2}{(s+1)^2} \right )\\ &=L^{-1}\left ( \frac{1}{s+1}+\frac{2}{(s+1)^2} \right )\\ &=e^{-t}+2te^{-t} \end{aligned}

Q6GATE 2016MCQ

The Laplace Transform of $f(t)=e^{2t}sin(5t)u(t)$ is

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📖 Explanation

Laplace transform of $\sin 5t u(t)\rightarrow \frac{5}{s^{2+}25}$ $e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^{2+}25}=\frac{5}{s^{2-}4s+29}$

Q7GATE 2016MCQ

The transfer function of a system is $\frac{Y(s)}{R(s)}=\frac{s}{s+2}$ . The steady state output y(t) is $A \cos (2t+\varphi)$ for the input cos(2t). The values of A and $\varphi$ , respectively are

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📖 Explanation

\begin{aligned} \frac{Y(s)}{R(s)}&=\frac{s}{s+2} \\ y(t)&=A \cos (2t+\phi ), \\ r(t)&=\cos 2t \\ \because \;H(s) &=\frac{s}{(s+2)} \\ H(j\omega )&=\frac{j\omega }{j\omega +2} \\ |H(j\omega )| &=\frac{\omega }{\sqrt{\omega ^{2+}4 }} \\ \angle H(j\omega ) &=90^{\circ} -\tan ^{-1}\left ( \frac{\omega }{2} \right ) \\ \because \; \omega &= 2 \text{ (as given)}\\ |H(j\omega )| &=\frac{2}{\sqrt{4+4}}=\frac{1}{\sqrt{2}} \\ |H(j\omega )| &=90^{\circ} -\tan ^{-1}(1)=45^{\circ} \\ \because \; \text{hence, }A &=1 \times |H(j\omega )|_{\omega =2}\\ &=1 \times \frac{1}{\sqrt{2}}=0.707\\ \phi &= 45^{\circ} \end{aligned}

Q8GATE 2016NAT

Consider a linear time-invariant system with transfer function $H(s)=\frac{1}{(s+1)}$ If the input is cos(t) and the steady state output is $A \cos (t+\alpha )$ , then the value of A is _________.

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📖 Explanation

\begin{aligned} H(s)&=\frac{1}{(s+1)}\\ \text{Put, }s&=j\omega \\ H(j\omega)&=\frac{1}{j\omega+1}\\ |H(j\omega)|&=\frac{1}{\sqrt{\omega ^2 +1}}\\ \because \; \text{ input }x(t)&=\cos (t)\\ \text{Here, } \omega &=1 \text{rad/sec}\\ \text{and }|x(t)|&=1\\ \text{hence, }&\text{steady state output}\\ y(t)&=|H(j\omega)|_{\omega=1} \cos (t+\angle H(j\omega)|_{\omega =1})\\ A&=|H(j\omega)|_{\omega =1}=\frac{1}{\sqrt{2}}=0.707 \end{aligned}

Q9GATE 2015MCQ

The Laplace transform of $f(t)=2\sqrt{t/\pi }$ is $s^{-3/2}$ . The Laplace transform of $g(t)=\sqrt{1/\pi t}$ is

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📖 Explanation

Given that, $f(t)=2\sqrt{\frac{t}{\pi}}\rightleftharpoons F(s)=s^{-3/2}$ By using property of differentiation In time,

\begin{aligned} \frac{df(t)}{dt}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{2}{\sqrt{\pi}}\cdot \frac{1}{2}t^{-1/2}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s\cdot s^{-3/2} \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s^{-1/2} \end{aligned}

Q10GATE 2014MCQ

Consider an LTI system with transfer function $H(s)=\frac{1}{s(s+4)}$ If the input to the system is cos(3t) and the steady state output is Asin(3t+ $\alpha$ ) , then the value of A is

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📖 Explanation

Given,

\begin{aligned} H(s)&=\frac{1}{s(s+4)}\\ r(t)&=\text{input}= \cos (3t)=\cos \omega t\\ \therefore \; \omega &=3 \text{ rad/s}\\ H(j\omega)&=\frac{1}{(j\omega)(j\omega+4)}\\ \text{Now, }|H(j\omega)|&=\frac{1}{\omega \sqrt{\omega^{4+}4^2}}\\ &=\frac{1}{3\sqrt{25}}=\frac{1}{15} \;\;\;(at\; \omega=3)\\ \angle H(j\omega)&=-90^{\circ}-\tan ^{-1}\frac{\omega}{4}\\ &=-90^{\circ}-\tan ^{-1}\frac{3}{4}=-126.86^{\circ}\\ \therefore \; c(t)&=\frac{1}{15} \cos (3t-126.86^{\circ})\\ &=\frac{1}{15} \sin (3t-36.86^{\circ})\;\;...(i)\\ c(t)&=A \sin (3t+\alpha )\;\;...(ii)\\ \text{Comparing }&\text{ eq. (i) and (ii), we have,}\\ A&=\frac{1}{15} \end{aligned}

Q11GATE 2014MCQ

Let $X(s)=\frac{3s+5}{s^{2}+10s+21}$ be the Laplace Transform of a signal x(t). Then, x( $0^+$ ) is

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📖 Explanation

Given, $X(s)=\left[ \frac{3s+5}{s^{2+}10s+21} \right]$ Using initial value theorem,

\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left[ \frac{s(3s+5)}{s^{2+}10+21} \right]\$ \\ &= \lim_{s \to \infty }\left[ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right]\$=3 \end{aligned}

Q12GATE 2013MCQ

Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is

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📖 Explanation

\begin{aligned} Y(s)&=\frac{1}{s}U(s)=\frac{1}{s^2}\\ y(t)&=tu(t) \end{aligned}

Q13GATE 2013MCQ

Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system?

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Q14GATE 2012MCQ

Consider the differential equation $\frac{d^{2}y(t)}{dt^{2}}+2\frac{dy(t)}{dt}+y(t)=\delta (t)$ with $y(t)|_{t=0^{-}}=-2$ and $\frac{dy}{dt}|_{t=0^{-}}=0$ The numerical value of $\frac{dy}{dt}|_{t=0^{+}}$ , is

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📖 Explanation

\begin{aligned} &\frac{d^2y(t)}{dt^2}+2\frac{dy(t)}{dt}+y(t)=\delta (t)\\ &\Rightarrow \; s^2y(s)-sy(0)-y'(0)+2[sy(s)-y(0)]\\&+y(s)=1\\ &\Rightarrow \; y(s)\$[s^22s+1]\$-(s \times -2)-y(0)\\&-(2 \times -2)=1\\ &\Rightarrow \; y(s)\$[s^{2+}2a+1]\$=-2s-3\\ &\Rightarrow \;y(s)=\frac{-2s-3}{s^{2+}2s+1}=\frac{-2}{s+1}-\frac{1}{(s+1)^2}\\ &\Rightarrow \; y(t)=-2e^{-t}-te^{-t}\\ &\frac{dy(t)}{dt}=2e^{-t}-(-te^{-t}+e^{-t})\\ &\frac{dy(t)}{dt}=e^{-t}+te^{-t}\\ &\frac{dy(t)}{dt}|_{t=0^+}=1+0=1 \end{aligned}

Q15GATE 2012MCQ

The unilateral Laplace transform of $f(t)$ is $\frac{1}{s^{2}+s+1}$ . The unilateral Laplace transform of $tf(t)$ is

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📖 Explanation

\begin{aligned} \text{If }L[f(t)]&=\bar{f}(s)\\ \text{then, }L\$[t^nf(t)]\$&=(-1)^n\frac{d^n}{ds^n}\bar{f}(s)\\ &\text{In this problem,}\\ \text{Given, }L[f(t)]&=\frac{1}{s^{2+}s+1}=\bar{f}(s)\\ &\text{We need,}\\ L[tf(t)]&=(-1)^n\frac{d^1}{ds^1}\bar{f}(s)\\ &=-\frac{d}{ds}\bar{f}(s)\\ &=-\frac{d}{ds}\left[ \frac{1}{s^{2+}s+1} \right]\$\\ &=-\left[ \frac{-1}{(s^{2+}s+1)^2} \right]\$ \times (2s+1)\\ &=\frac{2s+1}{(s^{2+}s+1)^2} \end{aligned}

Q16GATE 2011MCQ

Let the Laplace transform of a function F(t) which exists for $t \gt 0 \; be \; F_1(s)$ and the Laplace transform of its delayed version $f(t-\tau ) \; be \; F_{2}(s)$ . Let $F_1^*(s)$ be the complex conjugate of $F_1(s)$ with the Laplace variable set $s=\sigma +i \omega$ . If $G(s)=\frac{F_{2}(s)F_{1}^*(s)}{|F_{1}(s)|^{2}}$ , then the inverse Laplace transform of G(s) is

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📖 Explanation

\begin{aligned} F_2(t)&=L(f(t-\tau ))=e^{-s\tau }F_1(s)\\ G(s)&=\frac{e^{-s\tau }F_1(s)F_1^*(s)}{|F_1(s)|^2}=e^{-s\tau }\\ G(t)&=\delta (t-\tau ) \end{aligned}

Q17GATE 2010MCQ

Given f(t) and g(t)as shown below: The Laplace transform of g(t) is

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📖 Explanation

\begin{aligned} g(t) &=\left\{\begin{matrix} 1 &3 \lt t \lt 5 \\ 0& t \lt 3, t \gt 5 \end{matrix}\right. \\ \Rightarrow \; g(t) &=u(t-3)-u(t-5) \\ L(g(t))&=L(u(t-3))-L(u(t-5)) \\ &= \frac{1}{s}e^{-3s}-\frac{1}{s}e^{-5s}\\ &= \frac{1}{s}(e^{-3s}-e^{-5s})\\ &= \frac{e^{-3s}}{s}(1-e^{-2s}) \end{aligned}

Q18GATE 2010MCQ

Given f(t) and g(t)as shown below: g(t) can be expressed as

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📖 Explanation

Since, g(t) has width of 2-unit and f(t) has 1 unit, therefore we have to first expand f(t) by 2 unit and for this we have to scale f(t) by 1/2. i.e. $f_1(t)=f\left ( \frac{1}{2}t \right )$ Now shift it by three unit to get g(t) $g(t)=f_1(t-3)=f\left ( \frac{t-3}{2} \right )=f\left ( \frac{t}{2}-\frac{3}{2} \right )$

Q19GATE 2006MCQ

The running integration, given by $y(t)=\int_{-\infty }^{t}x(t)dt$

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Q20GATE 2005MCQ

The Laplace transform of a function f(t) is $F(s)=\frac{5s^{2}+23s+6}{s(s^{2}+2s+2)}. \; As \; t\rightarrow \infty$ , f(t) approaches

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📖 Explanation

$F(s)=\frac{5s^{2+}23s+6}{s(s^{2+}2s+2)}$ As F(s) has only left hand poles. By final value theorem, $\lim_{t \to \infty }F(t)=\lim_{s \to 0}sF(s)=\frac{6}{2}=3$

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