GATE 2022 Signals and Systems Questions and Solutions

A discrete-time periodic signal \(x[n]=\sum_{k=0}^{N-1}a_k e^{j(2\pi kn/N)}\) of period \(N=3\) has non-zero Fourier-series coefficients \(a_{-3}=2\) and \(a_{4}=1\). The signal is:

1. \(2+2e^{-j(2\pi/3)n}\cos\!\big(\tfrac{2\pi}{6}n\big)\)
2. \(1+2e^{\,j(2\pi/3)n}\cos\!\big(\tfrac{2\pi}{6}n\big)\)
3. \(1+e^{\,j(2\pi/3)n}\)
4. \(2+2e^{\,j(2\pi/3)n}\cos\!\big(\tfrac{2\pi}{6}n\big)\)

Solution

The coefficients are periodic with \(N=3\), so \(a_{-3}=a_0=2\) and \(a_{4}=a_1=1\), with \(a_2=0\). Therefore

Checking sample values: \(x[0]=2+1=3\) and \(x[1]=2+e^{j2\pi/3}=1.5+j\,0.866\). The official key is (B).

Note: the printed options were heavily corrupted by OCR (the exponent denominators show "\(\delta\)" instead of "3", and two options were duplicated). The verified signal is \(x[n]=2+e^{j2\pi n/3}\); please confirm the option wording against the original paper.

An input \(x(t)=2\sin(10\pi t)+5\cos(15\pi t)+7\sin(42\pi t)+4\cos(45\pi t)\) is passed through an LTI system with impulse response \(h(t)=2\left(\dfrac{\sin 10\pi t}{\pi t}\right)\cos 40\pi t\). The output is:

1. \(2\sin(10\pi t)+5\cos(15\pi t)\)
2. \(7\sin(42\pi t)+5\cos(15\pi t)\)
3. \(7\sin(42\pi t)+4\cos(45\pi t)\)
4. \(2\sin(10\pi t)+4\cos(45\pi t)\)

Solution

The factor \(\dfrac{\sin10\pi t}{\pi t}\) is an ideal low-pass response of bandwidth \(10\pi\). Multiplying by \(2\cos40\pi t\) shifts this band up to be centred at \(\omega=40\pi\), giving a unity-gain band-pass filter:

The input tones sit at \(5,\,7.5,\,21,\,22.5\) Hz; only \(21\) Hz (\(42\pi\)) and \(22.5\) Hz (\(45\pi\)) fall in the passband.

A system is described by \(y(t)=x(e^{t})\). The system is:

1. linear and causal
2. linear and non-causal
3. non-linear and causal
4. non-linear and non-causal

Solution

The mapping acts only on the time argument, so scaling and summing inputs carry through unchanged — the system is linear. For causality, \(y(0)=x(e^{0})=x(1)\) depends on the input one second in the future, so the system is non-causal.

A causal LTI system obeys \(y(t)+\dfrac14\dfrac{dy}{dt}=2x(t)\). Its impulse response is:

1. \(2e^{-t/4}u(t)\)
2. \(2e^{-4t}u(t)\)
3. \(8e^{-t/4}u(t)\)
4. \(8e^{-4t}u(t)\)

Solution

Taking the Laplace transform, \(Y\big(1+\tfrac{s}{4}\big)=2X\), so