Q2GATE 2023MCQ

Let $m(t)$ be a strictly band-limited signal with bandwidth $B$ and energy $E$ . Assuming $\omega_{0}=10 B$ , the energy in the signal $m(t) \cos \omega_{0} t$ is

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📖 Explanation

\begin{aligned} \text { Energy }(E)&=\frac{1}{2 \pi} \int_{-B}^{B}(1)^{2} \cdot d \omega \\ E&=\frac{B}{\pi}\\ \text {Now, let } y(t)&=m(t) \cos \omega_{0} t \\ Y(\omega)&=\frac{1}{2}\left[M\left(\omega-\omega_{0}\right)+M\left(\omega+\omega_{0}\right)\right] \\ \text{Here; }\omega_{0}&=10 B \end{aligned}

Now,

\begin{aligned} Energy \left(E^{\prime}\right)&=\frac{1}{2 \pi} \int_{-\infty }^{\infty }|Y(\omega)|^{2} \cdot d \omega\\&=\frac{1}{2 \pi}\left[\int_{-11 B}^{-9 B}\left(\frac{1}{2}\right)^{2} \cdot d \omega+\int_{9 B}^{11 B}\left(\frac{1}{2}\right)^{2} \cdot d \omega\right]\$\\ & =\frac{1}{2 \pi}\left[\frac{1}{4} \times 2 B+\frac{1}{4} \times 2 B\right] \\ E^{\prime} & =\frac{B}{2 \pi}=\frac{1}{2}\left(\frac{B}{\pi}\right)=\frac{E}{2} \end{aligned}

Q3GATE 2023MCQ

In the table shown below, match the signal type with its spectral characteristics. [latex]

\begin{array}{ll} \textbf{Signal type} &\textbf{Spectral Characteristics}\\ \text{(i) Continuous, aperiodic}& \text{(i) Continuous, aperiodic}\\ \text{(ii) Continuous, periodic}& \text{(ii) Continuous, periodic}\\ \text{(iii) Discrete, aperiodic}& \text{(iii) Discrete, aperiodic}}\\ \text{(iv) Discrete, periodic} & \text{(iv) Discrete, periodic}} \end{array}

[/latex]

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Q4GATE 2019NAT

Consider the signal $f(t)=1+2cos(\pi t)+3sin\left ( \frac{2 \pi}{3}t \right )+4cos\left (\frac{\pi}{2}t+\frac{\pi}{4} \right )$ , where t is in seconds. Its fundamental time period, in seconds, is ____________

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📖 Explanation

\begin{aligned} f(t)=1+2 \cos (\pi t)&+3 \sin \left(\frac{2 \pi}{3} t\right)+4 \cos \left(\frac{\pi}{2} t+\frac{\pi}{4}\right) \\ \omega_{1}&=\pi \\ \omega_{2}&=\frac{2 \pi}{3} \\ \omega_{3}&=\frac{\pi}{2} \\ \omega_{0}&=G C D\left(\pi, \frac{2 \pi}{3}, \frac{\pi}{2}\right)=\frac{\pi}{6} \end{aligned}

Fundamental period, $N=\frac{2 \pi}{\omega_{0}}=\frac{2 \pi}{(\pi / 6)}=12$

Q5GATE 2018MCQ

Let the input be u and the output be y of a system, and the other parameters are real constants. Identify which among the following systems is not a linear system:

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📖 Explanation

$y=a u+b, b \neq 0$ is a non-linear system.

Q6GATE 2017MCQ

The input $x(t)$ and the output $y (t)$ of a continuous-time system are related as $y(t)=\int_{t-T}^{t}x(u)du$ . The system is

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📖 Explanation

Given that, $y(t)=\int_{t-T}^{t} x(u) d u$ since the given system satisfies both homogeneity and additivity properties, the system is linear. Check for time invariance: $y\left(t-t_{0}\right)=\int_{t-t_{0}-T}^{t-t_{0}} x(u) d u$ When the applied input is $x\left(t-t_{0}\right)$

\begin{aligned} y_{1}(t) &=\\ \int_{t-T}^{t} x\left(u-t_{0}\right) d u &=\int_{t-t_{0}-T}^{t-t_{0}} x(\tau) d \tau \\ &=y\left(t-t_{0}\right) \end{aligned}

$\Rightarrow$ System is time invariant

Q7GATE 2017MCQ

Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as

y[n]=\left\{\begin{matrix} n|x[n]| & for 0\leq n\leq 10\\ x[n]-x[n-1] & othrwise \end{matrix}\right.

Which one of the following statements is true about the system?

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Q8GATE 2016MCQ

A continuous-time function x(t) is periodic with period T. The function is sampled uniformly with a sampling period $T_{s}$ . In which one of the following cases is the sampled signal periodic?

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📖 Explanation

A signal is said to be periodic if $\frac{T}{T_{s}}$ is a rational number. Here, $T=1.2 T_{s}$ $\Rightarrow \frac{T}{T_{s}}=\frac{6}{5} \quad$ Which is a rational number

Q9GATE 2015NAT

Two sequences $x_{1}[n] \; and \; x_{2}[n]$ have the same energy. Suppose $x_{1}[n] = \alpha 0.5^{n} u[n]$ , where $\alpha$ is a positive real number and u[n] is the unit step sequence. Assume

x_{2}[n]=\left\{\begin{matrix} \sqrt{1.5} &for n=0,1 \\ 0 & otherwise. \end{matrix}\right.

Then the value of $\alpha$ is _________.

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📖 Explanation

\begin{array}{l} \text { Energy of } x_{1}[n]=\sum_{n=-\infty }^{\infty }\left|x_{1}[n]\right|^{2} \\ \begin{aligned} =\sum_{n=0}^{\infty } \alpha^{2} &\left(\frac{1}{4}\right)^{n}=\alpha^{2} \cdot \frac{1}{1-\frac{1}{4}}=\alpha^{2} \cdot \frac{4}{3} \\ \text { Energy of } x_{2}[n] &=1.5+1.5=3 \\ \Rightarrow \quad \alpha^{2} \frac{4}{3} &=3 \\ \alpha^{2}=\frac{9}{4} \\ \quad \alpha=1.5 \end{aligned} \end{array}

Q10GATE 2015MCQ

The waveform of a periodic signal x(t) is shown in the figure. A signal g(t) is defined by $g(t)=x(\frac{t-1}{2})$ . The average power of g(t) is _______.

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📖 Explanation

\begin{array}{l} x(t)=-3 t, \quad-1\lt t \lt 1 \\ x\left(\frac{t-1}{2}\right)=-\frac{3}{2}(t-1) \qquad-1 \lt t \lt 3 \\ \text { and } \quad T=6 \\ \text { Average power }=\frac{1}{6} \int_{-1}^{3}\left(-\frac{3}{2}(t-1)\right)^{2} d t=2 \end{array}

Q11GATE 2014MCQ

A discrete time signal $x[n]=sin(\pi ^{2 }n)$ , $n$ being an integer, is

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📖 Explanation

\begin{aligned} x[n] &=\sin \left(\pi^{2} n\right) \\ \omega_{0} &=\pi^{2} \\ \therefore \quad N &=\frac{2 \pi}{\omega_{0}} \cdot m \end{aligned}

where m is the smallest integer that converts $\frac{2 \pi}{\omega_{0}}$ into a integer value. $\therefore \quad N=\frac{2 \pi}{\pi^{2}} \cdot m=\frac{2}{\pi} \cdot m$ So, there exists no such integer value of m which could make the N integer, so the system is not periodic.

Q12GATE 2013MCQ

For a periodic signal $v\left ( t \right )=30\sin 100t+10\cos 300t+6\sin \left ( 500t+\frac{\pi }{4} \right )$ , the fundamental frequency in rad/s is

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📖 Explanation

\begin{array}{l} \omega_{1}=100 \\ \omega_{2}=300 \\ \omega_{3}=500 \end{array}

H.C.F. of $\left(\omega_{1}, \omega_{2} \text { and } \omega_{3}\right)=$ H.C.F. (100,300,500) $\omega=100 \mathrm{rad} / \mathrm{sec}$

Q13GATE 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

Step response = Integration of impulse response $\int_{-\infty }^{t} \delta(t-1) d t=u(t-1)$ $\int_{-\infty }^{t} \delta(t-3) d t=u(t-3)$

\begin{aligned} \text{at}\quad t&=2\\ y(t)&=1 \end{aligned}

Q14GATE 2012MCQ

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

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📖 Explanation

\begin{aligned} y&=\int_{-\infty }^{t} x(\tau) \cos (3 \tau) d \tau \\ y\left(t-t_{0}\right)&=\int_{-\infty }^{t-t_{0}} x(\tau) \cos (3 \tau) d \tau \\ y^{\prime}(t) \text { for input } &x\left(t-t_{0}\right) \text { , is }\\ y^{\prime}(t)&=\int_{-\infty }^{t} x\left(\tau-t_{0}\right) \cos 3 \tau d \tau \\ y^{\prime}(t)&=\int_{-\infty }^{\left(t-t_{0}\right)} x(\tau) \cos 3\left(\tau+t_{0}\right) d \tau \end{aligned}

$y^{\prime}(t) \neq y\left(t-t_{0}\right)$ so system is not time invariant for input $x(\tau)=\cos (3 \tau)$ (bounded input) $y(t)=\int_{-\infty }^{t} \cos ^{2}(3 \tau) d \tau \rightarrow \infty \text { as } t \rightarrow \infty$ So, for bounded i/p, o/p is not bounded therefore system is not stable.

Q15GATE 2008MCQ

Let x(t) be the input and y(t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3,R4

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Q16GATE 2008MCQ

The input and output of a continuous time system are respectively denoted by x(t) and y(t). Which of the following descriptions corresponds to a causal system ?

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Q17GATE 2007MCQ

A Hilbert transformer is a

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📖 Explanation

The Hilbert transformer is characterised by the impulse response