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GATE 2023 EC – Question 48

Networks, Signals and Systems · Discrete-time Signals · 2 marks · Multiple choice

Let an input $x[n]$ having discrete time Fourier transform $X(e^{j\Omega})=1-e^{-j\Omega}+2e^{-3j\Omega}$ be passed through an LTI system. The frequency response of the LTI system is $H(e^{j\Omega})=1-\dfrac12e^{-j2\Omega}$. The output $y[n]$ of the system is

  1. $\delta[n]+\delta[n-1]-\frac12\delta[n-2]-\frac52\delta[n-3]+\delta[n-5]$
  2. $\delta[n]-\delta[n-1]-\frac12\delta[n-2]-\frac52\delta[n-3]+\delta[n-5]$
  3. $\delta[n]-\delta[n-1]-\frac12\delta[n-2]+\frac52\delta[n-3]-\delta[n-5]$
  4. $\delta[n]+\delta[n-1]+\frac12\delta[n-2]+\frac52\delta[n-3]+\delta[n-5]$

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Correct answer: (C) $\delta[n]-\delta[n-1]-\frac12\delta[n-2]+\frac52\delta[n-3]-\delta[n-5]$

Explanation

$Y=XH=\left(1-e^{-j\Omega}+2e^{-3j\Omega}\right)\left(1-\tfrac12e^{-2j\Omega}\right)=1-e^{-j\Omega}-\tfrac12e^{-2j\Omega}+\left(2+\tfrac12\right)e^{-3j\Omega}-e^{-5j\Omega}$. So $y[n]=\delta[n]-\delta[n-1]-\tfrac12\delta[n-2]+\tfrac52\delta[n-3]-\delta[n-5]$.