Q2GATE 2025MCQ

The continuous-time unit impulse signal is applied as an input to a continuous-time linear timeinvariant system $\mathbf{S}$ . The output is observed to be the continuous-time unit step signal $u(t)$ . Which one of the following statements is true?

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📖 Explanation

Step 1: Given that the output of system $\mathcal{S}$ for an input $\delta(t)$ (unit impulse) is $u(t)$ (unit step), So, the impulse response of the system is: $h(t) = u(t)$ Step 2: A system is Bounded Input Bounded Output (BIBO) stable if its impulse response is absolutely integrable: $\int_{-\infty }^{\infty } |h(t)| dt < \infty$ Here, $h(t) = u(t)$ , so: $\int_{-\infty }^{\infty } |u(t)| dt = \int_{0}^{\infty } 1 , dt = \infty$ Hence, the system is not BIBO stable. Step 3: If a system is not BIBO stable, then there exists at least one bounded input that will produce an unbounded output. Final Answer: (B) It is possible to find a bounded input signal which when applied to $\mathbf{S}$ results in an unbounded output signal.

Q3GATE 2024MCQ

The input $x(t)$ and the output $y(t)$ of a system are related as $y(t)=e^{-t} \int_{-\infty }^{t} e^{\tau} x(\tau) d \tau, \quad-\infty \lt t \lt \infty$ The system is

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📖 Explanation

System will follow law of superposition. Therefore, system will be linear. Time-invariance check: $y\left(t-t_{0}\right)=e^{-\left(t-t_{0}\right)} \int_{-\infty }^{t-t_{0}} e^{\tau} \cdot x(\tau) d \tau$ For input $x\left(t-t_{0}\right)$ , output will be $y(t)=e^{-t} \int_{-\infty }^{t} e^{\tau} \cdot x\left(\tau-t_{0}\right) d \tau$ Put $\tau-t_{o}=k$ , then $d \tau=d k$ and for $\tau=t, k=t-t_{0}$

\begin{aligned} & =e^{-t} \int_{-\infty }^{t-t_{0}} e^{k+t_{0}} \cdot x(k) d k \\ & =e^{-t} \times e^{t o} \times \int_{-\infty }^{t-t_{0}} e^{k} \cdot x(k) d k=e^{-\left(t-t_{0}\right)} \times \int_{-\infty }^{t-t_{0}} e^{k} \cdot x(k) d k \\ & =e^{-\left(t-t_{0}\right)} \times \int_{-\infty }^{t-t_{0}} e^{\tau} \cdot x(\tau) d \tau \end{aligned}

Since, $y(t)=y\left(t-t_{0}\right)$ So, the system is time-invariant.

Q4GATE 2023MCQ

Which of the following statement(s) is/are true?

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📖 Explanation

For causal system, impuse response $h(n)=0 ; \quad n \lt 0$ Therefore, for step input also $h(n)=0 ; n \lt 0$

Q5GATE 2022MCQ

Consider the system as shown below where $y(t)=x(e^t)$ . The system is

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📖 Explanation

We know, a linear system follows the law of superposition. It is a combination of two laws: (i) Law of additivity: Both results are same, hence, it follows law of additivity. (ii) Law of Homogeneity: Here also both results are same, hence it follows law of Homogeneity. Therefore, System is linear. We know, a causal system is independent of future values of input at each & every instant of time them system will be causal. Given : $y(t)=x(e^t)$ Put $t=0$ $y(0)=x(e^0)=x(1)$ Because its depends on future value. Therefore, system is non-causal.

Q6GATE 2022MCQ

Let a causal LTI system be governed by the following differential equation $y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)$ , where $x(t)$ and $x(t)$ are the input and output respectively. Its impulse response is

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📖 Explanation

Given: $y(t)+\frac{1}{4}\frac{dy}{dt}=2x(t)$ Taking Laplace transform, $Y(s)+\frac{1}{4}(sY(s))=2X(s)$ Now, $H(s)=\frac{Y(s)}{X(s)}=\frac{2}{\frac{s}{4}+1}=\frac{8}{s+4}$ Taking inverse Laplace, transform, $h(t)=8e^{-4t}u(t)$

Q7GATE 2021MCQ

If the input x(t) and output y(t) of a system are related as $y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right )$ , then the system is

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📖 Explanation

\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}

Linearity check: at input $x_{1}(t)=-2$ , output $y_{1}(t)=0$ at input $x_{2}(t)=1$ , output $y_{2}(t)=1$ $\therefore$ system is non-linear because it violates law of additivity. Check for time-invariance : Delayed $\mathrm{O} / \mathrm{P}:$

y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.

$\mathrm{O} / \mathrm{P}$ of system when input is $x\left(t-t_{0}\right)=f(t)$

y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.

Therefore, system is time-invariant.

Q8GATE 2020MCQ

Which of the following options is true for a linear time-invariant discrete time system that obeys the difference equation: $y[n]-ay[n-1]=b_0x[n]-b_1x[n-1]$

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📖 Explanation

\begin{aligned} y(n)-ay(n-1)&=b_{0}x(n)-b_{1}x(n-2) \\ &\text{By applying ZT,} \\ Y(z)-az^{-1}Y(z)&=b_{0}X(z)-b_{1}z^{-1}X(z)\\ \Rightarrow \, \, H(z)&=\frac{Y(z)}{X(z)}=\frac{b_{0}-b_{1}z^{-1}}{1-az^{-1}} \end{aligned}

By taking right-sided inverse ZT, $h(n)=b_{0}a^{n}u(n)-b_{1}a^{n-1}u(n-1)$ By taking left-sided inverse ZT, $h(n)=-b_{0}a^{n}u(-n-1)+b_{1}a^{n-1}u(-n)$ Thus system is not necessarily causal. The impulse response is non-zero at infinitely many instants.

Q9GATE 2018MCQ

A continuous-time input signal x(t) is an eigenfunction of an LTI system, if the output is

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📖 Explanation

Eigen function is a type of input for which output is constant times of input. i.e. Where, $x(t)=$ System input = eigen function $H(s)=$ transfer function of system $y(t)=$ system output Here, $y(t)=H(s)|_{s=a}\; e^{at}=k \cdot x(t)$ where, k= eigen-value = $H(s)|_{s=a}$ $x(t)=$ eigen-function input

Q10GATE 2017MCQ

Let z(t)=x(t) * y(t) , where "*" denotes convolution. Let c be a positive real-valued constant. Choose the correct expression for z(ct).

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📖 Explanation

Time scaling property of convolution. If, $x(t)*y(t)=z(t)$ Then, $x(ct)*y(ct)=\frac{1}{c} z(ct)$ $z(ct)=c \times x(ct) * y(ct)$

Q11GATE 2016MCQ

Consider a causal LTI system characterized by differential equation $\frac{dy(t)}{dt}+\frac{1}{6}y(t)=3x(t)$ . The response of the system to the input $x(t)=3e^{-\frac{t}{3}}u(t)$ . where u(t) denotes the unit step function, is

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📖 Explanation

The differential equation

\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}

Q12GATE 2016MCQ

Consider a continuous-time system with input x(t) and output y(t) given by y(t) = x(t) cos(t) This system is

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📖 Explanation

\begin{aligned} y(t)&=x(t)\cos (t)\\ &\text{To check linearity,}\\ y_1(t)&=x_1(t)\cos (t)\\ &[y_1(t) \text{ is output for }x_1(t)]\\ y_2(t)&=x_2(t) \cos (t)\\ &[y_2(t) \text{ is output for }x_2(t)]\\ \text{so, the}& \text{ output for }(x_1(t)+ x_2(t)) \text{ will be}\\ y(t)&=[x_1(t)+ x_2(t)]\cos (t)\\ &=y_1(t)+y_2(t) \end{aligned}

So, the system is linear, to check time invariance. The delayed output, $y(t-t_0)=x(t-t_0)\cos (t-t_0)$ The output for delayed input, $y(t, t_0)=x(t-t_0)\cos (t)$ Since, $y(t-t_0)\neq y(t,t_0)$ System is time varying.

Q13GATE 2016MCQ

The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigen-signal of the system T, when $T\{z(t)\} = \gamma z(t)$ , where $\gamma$ is a complex number, in general, and is called an eigenvalue of T. Suppose the impulse response of the system T is real and even. Which of the following statements is TRUE?

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📖 Explanation

Given that impulse response is real and even, Thus $H(j\omega )$ will also be real and even. Since, $H(j\omega )$ is real and even thus, $H(j\omega )=H(-j\omega )$ Now, $\cos (t)$ is input i.e. $\frac{e^{jt}+e^{-jt}}{2}$ is input Output will be $\frac{H(j1)e^{jt}+H(-j1)e^{-jt}}{2}=H(j1)\left ( \frac{e^{jt}+e^{-jt}}{2} \right )= H(j1) \cos (t)$ If, $\sin (t)$ is input i.e. $\frac{e^{jt}+e^{-jt}}{2}$ is input Output will be $\frac{H(j1)e^{jt}+H(-j1)e^{-jt}}{2}=H(j1)\left ( \frac{e^{jt}-e^{-jt}}{2j} \right )= H(j1) \sin (t)$ So, $\sin (t)$ and $\cos (t)$ are eigen signal with same eigen values.

Q14GATE 2016NAT

Consider the following state-space representation of a linear time-invariant system.

\dot{x}(t)=\begin{bmatrix} 1 & 0\\ 0&2 \end{bmatrix}x(t),

$y(t)=c^{T}x(t),$

c=\begin{bmatrix} 1\\ 1 \end{bmatrix}

and

x(0)=\begin{bmatrix} 1\\ 1 \end{bmatrix}

The value of y(t) for $t= log_{e}2$ is______.

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Q15GATE 2015MCQ

For linear time invariant systems, that are Bounded Input Bounded Output stable, which one of the following statements is TRUE?

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Q16GATE 2015MCQ

The following discrete-time equations result from the numerical integration of the differential equations of an un-damped simple harmonic oscillator with state variables x and y. The integration time step is h. $\frac{x_{k+1}-x_{k}}{h}=y_{k}$ $\frac{y_{k+1}-y_{k}}{h}=-x_{k}$ For this discrete-time system, which one of the following statements is TRUE?

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Q17GATE 2015NAT

A moving average function is given by $y(t)=\frac{1}{T}\int_{t-T}^{t}u(\tau )d\tau$ . If the input u is a sinusoidal signal of frequency $\frac{1}{2\tau }$ Hz, then in steady state, the output y will lag u (in degree) by ______ .

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📖 Explanation

System input: $\sin \omega _0 t$

\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left[ \because \; \omega _0=\frac{t}{T} \right]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}

Q18GATE 2015MCQ

The impulse response g(t) of a system, G , is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure (b)?

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📖 Explanation

\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1

Q19GATE 2014MCQ

Consider an LTI system with impulse response $h(t)=e^{-5t}u(t)$ . If the output of the system is $y(t)=e^{-3t}u(t)-e^{-5t}u(t)$ then the input, x(t), is given by

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📖 Explanation

Impulse response of an LTI system = transfer function $=\frac{Y(s)}{X(s)}=H(s)$

\begin{aligned} \text{where, } y(t)&=e^{-3t}u(t)-e^{-5t}u(t) \\ \therefore \; Y(s) &=\frac{1}{(s+3)}-\frac{1}{(s+5)} \\ &=\frac{2}{(s+3)(s+5)} \\ \text{Also, } h(t)&=e^{-5t}u(t) \\ \therefore \; H(s) &=\frac{1}{(s+5)} \\ \text{Therefore, } X(s)&=\frac{Y(s)}{H(s)} \\ &=\frac{2}{(s+3)(s+5)} \times (s+5) \\ &= \frac{2}{(s+3)}\\ \therefore \; \text{Input}&=x(t)=2e^{-3t}u(t) \end{aligned}

Q20GATE 2014MCQ

x(t) is nonzero only for $T_x \lt t \lt T'_x$ , and similarly, y(t) is nonzero only for $T_y \lt t \lt T'_y$ . Let z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE ?

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📖 Explanation

$x(t)$ is non-zero for $T_x \lt t \lt T'_x$ and $y(t)$ is non-zero for $T_y \lt t \lt T'_y$ $\because \;\; z(t)=x(t)\otimes y(t)$ then the limits of the resultant signal is $T_x+T_y \lt t \lt T'_x+T'_y$ i.e. $z(t)$ is non-zero for $T_x+T_y \lt t \lt T'_x+T'_y$

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