GATE 2011 Signals and Systems Questions and Solutions

A zero-mean random signal is uniformly distributed between \(-a\) and \(+a\), and its mean-square value equals its variance. The r.m.s. value of the signal is:

1. \(\dfrac{a}{\sqrt{3}}\)
2. \(\dfrac{a}{\sqrt{2}}\)
3. \(a\sqrt{2}\)
4. \(a\sqrt{3}\)

Solution

For a zero-mean signal, mean-square value = variance. For a uniform distribution on \([-a,a]\),

The r.m.s. value is \(\sqrt{\overline{x^2}} = \dfrac{a}{\sqrt{3}}\).

Let \(F_1(s)=\mathcal{L}\{f(t)\}\) and \(F_2(s)=\mathcal{L}\{f(t-\tau)\}\), with \(F_1^*(s)\) the complex conjugate of \(F_1(s)\) (\(s=\sigma+j\omega\)). If \(G(s)=\dfrac{F_2(s)F_1^*(s)}{|F_1(s)|^2}\), then the inverse Laplace transform of \(G(s)\) is:

1. An ideal impulse \(\delta(t)\)
2. An ideal delayed impulse \(\delta(t-\tau)\)
3. An ideal step function \(u(t)\)
4. An ideal delayed step function \(u(t-\tau)\)

Solution

By the time-shift property, \(F_2(s)=e^{-s\tau}F_1(s)\). Then

since \(F_1(s)F_1^*(s)=|F_1(s)|^2\). The inverse transform of \(e^{-s\tau}\) is \(\delta(t-\tau)\).

Given \(x(t)=e^{-t}\) and \(y(t)=e^{-2t}\) for \(t>0\), the convolution \(z(t)=x(t)*y(t)\) is:

1. \(e^{-t}-e^{-2t}\)
2. \(e^{-3t}\)
3. \(e^{t}\)
4. \(e^{-t}+e^{-2t}\)

Solution

Multiply transforms and use partial fractions:

The inverse transform gives \(z(t) = e^{-t} - e^{-2t}\).

The Fourier series \(f(t)=a_0+\sum_{n=1}^{\infty}\big(a_n\cos n\omega t + b_n\sin n\omega t\big)\) of the periodic signal shown will contain which nonzero terms?

1. \(a_0\) and \(b_n,\ n=1,3,5,\dots\)
2. \(a_0\) and \(a_n,\ n=1,2,3,\dots\)
3. \(a_0,\ a_n\) and \(b_n,\ n=1,3,5,\dots\)
4. \(a_0\) and \(a_n,\ n=1,3,5,\dots\)

Solution

The waveform is an even function, so all sine coefficients vanish (\(b_n=0\)). Its symmetry suppresses the even cosine harmonics, leaving the DC term \(a_0\) together with cosine terms at odd harmonics only.