GATE 2025 Signals and Systems Questions and Solutions

Let \(x_1(t)=1\) on \([0,1]\), \(2-t\) on \([1,2]\), \(0\) otherwise; and \(x_2(t)=t\) on \([0,1]\), \(2-t\) on \([1,2]\), \(0\) otherwise. For \(y(t)=x_1(t)*x_2(t)\), the value of \(\int_{-\infty}^{\infty}y(t)\,dt\) is:

1. 1.5
2. 2.5
3. 3.5
4. 4

Solution

The area of a convolution equals the product of the areas of the two signals:

For \(x(t)=-t^{2}\{u(t+4)-u(t-4)\}\), the value of \(\int_{-\infty}^{\infty}x(t)\,\delta(t+3)\,dt\) is:

1. \(-9\)
2. \(9\)
3. \(3\)
4. \(-3\)

Solution

By the sifting property, \(\int x(t)\delta(t+3)\,dt=x(-3)\). At \(t=-3\), the window \(u(t+4)-u(t-4)=1\), so

A periodic signal \(x(t)=1+2\cos2\pi t+2\cos4\pi t+2\cos6\pi t\) has period \(T\). Then \(\dfrac{1}{T}\int_0^{T}|x(t)|^{2}\,dt=\) ____ (nearest integer).

Solution

This average is the signal power. By Parseval, the DC term contributes \(1^2\) and each cosine of amplitude \(2\) contributes \(2^2/2=2\):

An ideal low-pass filter has \(H(j\omega)=1\) for \(|\omega|\le200\pi\) and \(0\) otherwise. With \(h(t)\) its impulse response, \(h(0)=\) ____ (nearest integer).

Solution

Inverting the rectangular response,

Taking \(t\to0\), \(h(0)=\dfrac{200\pi}{\pi}=200\).

A unit impulse applied to a continuous-time LTI system \(\mathbf{S}\) produces the unit step \(u(t)\). Which statement is true?

1. Every bounded input to \(\mathbf{S}\) gives a bounded output.
2. It is possible to find a bounded input that produces an unbounded output.
3. Any input to \(\mathbf{S}\) gives a bounded output.
4. Any input to \(\mathbf{S}\) gives an unbounded output.

Solution

The impulse response is \(h(t)=u(t)\), which is not absolutely integrable (\(\int_0^{\infty}1\,dt=\infty\)). The system is therefore not BIBO stable, so some bounded input produces an unbounded output.

A discrete-time LTI system \(S\) has \(S\{\delta[n]\}=1\) for \(n\in\{0,1,2\}\) and \(0\) otherwise. For an input \(x[n]\), the output \(y[n]\) is:

1. \(x[n]+x[n-1]+x[n-2]\)
2. \(x[n-1]+x[n]+x[n+1]\)
3. \(x[n]+x[n+1]+x[n+2]\)
4. \(x[n+1]+x[n+2]+x[n+3]\)

Solution

The impulse response is \(h[n]=\delta[n]+\delta[n-1]+\delta[n-2]\). Convolving,