Q21GATE 2013MCQ

The impulse response of a continuous time system is given by $h(t)=\delta(t-1) \delta (t-3)$ . The value of the step response at t=2 is

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📖 Explanation

For step response, impulse response can be integrated.

\begin{aligned} y(t)&=u(t-1)+u(t-3)\\ y(2)&=u(1)+u(-1)\\ &=1+0=1 \end{aligned}

Q22GATE 2013MCQ

Two systems with impulse responses $h_{1}(t) \; and \; h_{2}(t)$ are connected in cascade. Then the overall impulse response of the cascaded system is given by

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📖 Explanation

In cascade connection,

\begin{aligned} H(s) &=H_1(s)\cdot H_2(s)\\ \Rightarrow \; h(t)&=h_1(t) * h_2(t) \end{aligned}

Q23GATE 2013MCQ

The impulse response of a system is h(t)=tu(t). For an input u(t-1), the output is

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📖 Explanation

\begin{aligned} h(t) &=tu(t) \\ H(s) &=\frac{1}{s^2} \\ \Rightarrow \; \frac{Y(s)}{U(s)} &=\frac{1}{s^2} \\ Y(s) &=\frac{1}{s^2} \frac{e^{-s}}{s}\\ \Rightarrow \; y(t) &=\frac{(t-1)^2}{2}u(t-1) \end{aligned}

Q24GATE 2012MCQ

The input x(t) and output y(t) of a system are related as $y(t)=\int^{t}_{-\infty }x(\tau ) cos (3\tau )d\tau$ . The system is

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📖 Explanation

\begin{aligned} y&=\int_{-\infty }^{t}x{\tau } \cos (3\tau )d\tau \\ y(t-t_0)&=\int_{-\infty }^{t-t_0}x{\tau } \cos (3\tau )d\tau\\ y(t) & \text{ for input }x(t-t_0) \text{ is}\\ y'(t)&=\int_{-\infty }^{t}x{\tau -t_0} \cos (3\tau )d\tau\\ y'(t)&=\int_{-\infty }^{t-t_0}x{\tau} \cos 3(\tau +t_0 )d\tau\\ y'(t) &\neq y(t-t_0) \end{aligned}

So, system is not time invariant for input $x{\tau } =\cos (3\tau )$ bounded input $y(t)=\int_{-\infty }^{t} \cos ^2 (3\tau )d\tau \rightarrow \infty \; as \; t \rightarrow \infty$ So, for bounded input, output is not bounded therefore system is not stable.

Q25GATE 2012MCQ

Let y[n] denote the convolution of h[n] and g[n], where h[n]= $(1/2)^{n}$ u[n] and g[n] is a causal sequence. If y[0]=1 and y[1]=1/2, then g[1] equals

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📖 Explanation

Given, g[n] is causal sequence Therefore, g[n] will be zero for $n \lt 0$

\begin{aligned} y[n]&=\sum_{k=-\infty }^{\infty }g[k]h[n-k]\\ h[n]&=\left ( \frac{1}{2} \right )^n u[n]\\ h[n]&=\left\{\begin{matrix} \left ( \frac{1}{2} \right )^n \; ;& n\geq 0\\ 0& n \lt 0 \end{matrix}\right.\\ h[0]&=0, h[1]=\frac{1}{2}, h[2]=\frac{1}{4}\\ y[0]&=g[0] h[0]+g[1]h[-1]+...\\ 1&=g[0] \times 1+0+0+...\\ g[0]&=1\\ y[1]&=g[0]h[1]+g[1]h[0]+g[2]h[-1]+...\\ \frac{1}{2}&=1\cdot \frac{1}{2}+g[1]\cdot 1+0\\ \Rightarrow \;g[1]&=0 \end{aligned}

Q26GATE 2011MCQ

Given two continuous time signals $x(t) =e^{-t}$ and $y(t) =e^{-2t}$ which exist for $t \gt 0$ , the convolution z(t)=x(t)*y(t) is

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📖 Explanation

\begin{aligned} x(s)&=\frac{1}{s+1} \\ y(s)&=\frac{1}{s+2} \\ z(s)&=x(s)\cdot y(s) \\ &=\frac{1}{s+1} \times \frac{1}{s+2} \\ &=\frac{1}{s+1} - \frac{1}{s+2} \\ z(t)&=e^{-t}-e^{-2t} \end{aligned}

Q27GATE 2010MCQ

Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is

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📖 Explanation

x[n] = { 1 , -1}, M=2 y[n] = { 1 ,0,0,0,-1}, $N_1=5$ Since, output hasnumber of elements $N_1=M+N-1$ where N is number of elements in impulse response h[n]. $\therefore \;\; 5=2+N-1$ $N=4$ Let it be $a_1,a_2,b_1,b_2$ $y[n]=[a_1,(a_2-a_1), (b_1-a_2), (b_2-b_1), -b_2]$ Comaparing with $y[n]=1, 0,0,0,-1]$ $a_1=1$ $a_2=a_1=1$ $b_1=a_1=1$ $b_2=b_1=1$ Therefore, $h[n]=\{1,1,1,1 \}$

Q28GATE 2010MCQ

The system represented by the input-output relationship $y(t)=\int_{-\infty }^{5t}x(\tau )d\tau , \; t\gt0$

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📖 Explanation

Integrator is always a linear system.

\begin{aligned} \text{Since, }y(t)&=\int_{-\infty }^{5t}x(\tau )d\tau \;\;t \gt 0 \\ &\text{For t=1,} \\ y(t)&= \int_{-\infty }^{5}x(\tau )d\tau \end{aligned}

here value at t=1 depends on future values like at t=2,3,... of input x(t). So, it is a noncausal systems.

Q29GATE 2009MCQ

A Linear Time Invariant system with an impulse response h(t) produces output y(t) when input x(t) is applied. When the input x(t- $\tau$ ) is applied to a system with impulse response h(t- $\tau$ ), the output will be

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📖 Explanation

Case-1: $Y(s)=H(s) \cdot X(s)$ Case-1: input $=x(t-\tau )\Rightarrow X(s)e^{-s\tau }$ impulse response $=h(t-\tau )\Rightarrow H(s)e^{-s\tau }$ Output: $Y'(s)=X(s)e^{-s\tau }H(s)e^{-s\tau }$ $Y'(s)=X(s)H(s)e^{-2s\tau }=Y(s)e^{-2s\tau}$ $y'(t)=y(t-2 \tau)$

Q30GATE 2009MCQ

A cascade of three Linear Time Invariant systems is causal and unstable. From this, we conclude that

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📖 Explanation

Since in cascade overall impulse response $h(t)=h_1(t)*h_2(t)*h_3(t)$ $h_1(t),h_2(t),h_3(t)$ are impulse response of individual systems. Since, initial point where h(t) is nonzero is $t \geq 0$ and since in convolution initial point $=t_1+t_2+t_3$ where, $t_1,t_2,t_3$ are initial point of $h_1(t), h_2(t),h_3(t)$ respectively. So for it to be greater than or equal to zero at least one of them $t_1, t_2\; or \; t_3$ must be +ve i.e. greater than zero so atleast one of them must be causal. Similarly if one (atleast) of the system become unstable then overall system will become unstable.

Q31GATE 2008MCQ

A signal $e^{-\alpha t}sin(\omega t)$ is the input to a real Linear Time Invariant system. Given K and $\phi$ are constants, the output of the system will be of the form $Ke^{-\beta t}sin(vt+\phi )$ where

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Q32GATE 2008MCQ

The impulse response of a causal linear time-invariant system is given as h(t). Now consider the following two statements : Statement (I): Principle of superposition holds Statement (II): h(t)=0 for $t \lt 0$ Which one of the following statements is correct ?

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📖 Explanation

Since, system is casual therefore its impulse response will be zero for $t \lt 0$ . So, statement-II is correct. Since, system is linear, it also satisfies the principle of superposition. So, both the statements are correct.

Q33GATE 2008MCQ

A system with x(t) and output y(t) is defined by the input-output relation : $y(t)=\int_{-\infty }^{-2t}x(\tau)d\tau$ The system will be

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📖 Explanation

First of all we will check for causality,

\begin{aligned} y(t)&=\int_{-\infty }^{-2t}x(\tau ) d\tau \\ \text{So for }t&=2 \\ y(-2)&=\int_{-\infty }^{4}x(\tau ) d\tau \end{aligned}

So, output depends on future values of input along with past and present values of input, so system is non-casual. Let us find output for shifted input $x(t-t_0)$

\begin{aligned} y(t)&=\int_{-\infty }^{-2t}x(\tau -t_0) d\tau \\ y'(t)&=\int_{-\infty }^{2(t-t_0)}x(\tau ) d\tau\;\;...(i) \end{aligned}

Now, shift the output by $t_0$ then, $y(t-t_0)=\int_{-\infty }^{2(t-t_0)}x(\tau ) d\tau \;\;...(ii)$ so from eq. (i) and (ii), $y'(t) \neq y(t-t_0)$ Therefore system is time varient. Non-causal and time variant is present only in option (D).

Q34GATE 2007MCQ

Let a signal $a_1 \sin(\omega _{1}t + \phi_1)$ be applied to a stable linear time invariant system. Let the corresponding steady state output be represented as $a_2 F(\omega _{2}t + \phi _{2})$ . Then which of the following statement is true?

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📖 Explanation

The effect of linear time invariant systems is best understtod by considering that, at any given input frequency $\omega _1$ , the output is sinusoidal at same frequency,as the input applied, with amplitude given by output = |H| x input and phase equal to $\phi _2=\phi _H+\phi _1$ , where |H| is the magnitude to the frequency response and $\phi _H$ is its phase angle. Both |H| and $\phi _H$ are functions of frequency.

Q35GATE 2007MCQ

If u(t), r(t) denote the unit step and unit ramp functions respectively and u(t)*r(t) their convolution, then the function u(t+1)*r(t-2) is given by

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📖 Explanation

\begin{aligned} u(t+1)*r(t-1)&=\int_{-\infty }^{\infty }r(\tau -2)u(t+1-\tau )d\tau \\ z(t)&=\int_{-\infty }^{(t+1)}r(\tau -2)d\tau \\ z(t)&=\int_{2}^{(t+1)}r(\tau -2)d\tau \\ \text{For, }(t+1)& \geq 2\\ z(t)&=\left ( \frac{\tau ^2}{2}-2\tau \right )_2^{(t+1)}\\ &=\frac{1}{2}\$[\tau ^{2-}4\tau ]_2^{(t+1)}\\ \text{For, }t \geq 1&=\frac{1}{2}\$[(t+1)^{2-}4(t+1)-4+8]\$\\ z(t)&=\frac{1}{2}\$[t^{2+}2t+1-4t-4+4]\$\\ \text{For, }t &\geq 1\\ z(t)&=\frac{1}{2}(t-1)^2\\ \text{For, }t &\geq 1\\ \Rightarrow \; z(t)&=\frac{1}{2}(t-1)^2u(t-1) \end{aligned}

Q36GATE 2006MCQ

y[n] denotes the output and x[n] denotes the input of a discrete-time system given by the difference equation y[n]-0.8y[n-1]=x[n]+1.25x[n+1]. Its right-sided impulse response is

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📖 Explanation

From the given difference equation $y[n] - 0.8 y[n-1] = x [n] + 1.25 x [n+1]$ $H(z)=Y(z)X(z)=1+1.25z11-0.8z-1=11-0.8z-1+z1.251-0.8z-1$ For right sided impulse response, $h(n)$ the ROC is $|Z| \gt 0.8$ $h[n] = (0.8)^n u[n] + 1.25 (0.8)^n+1 u[n+1].$ $h [n]$ has positive value for any n, $h [n]$ is non-negative. It can be verified that $h[n]$ is non-causal, bounded and not periodic As $h(-1) = 1.25, \; \; h[n]$ is non-causal $h [n] \leq M, \; M\lt \infty ; \; h[n]$ is bounded . $h [n]$ is not periodic as it is a monotonically decaying sequence.

Q37GATE 2006MCQ

$x[n]=0;n \lt -1, n \gt 0$ , $x[-1]=-1,x[0]=2$ is the input and $y[n]=0;n \lt -1, n \gt 2$ , $y[-1]=-1=y[1],y[0]=3,y[2]=-2$ is the output of a discrete-time LTI system. The system impulse response h[n] will be

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Q38GATE 2002MCQ

Let s(t) be the step response of a linear system with zero initial conditions; then the response of this system to an input u(t) is

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