Question

Let an input $x[n]$ having discrete-time Fourier transform \[ X(e^{j\Omega}) = 1 - e^{-j\Omega} + 2e^{-j3\Omega} \] be passed through an LTI system whose frequency response is \[ H(e^{j\Omega}) = 1 - \tfrac12 e^{-j2\Omega}. \] The output $y[n]$ of the system is

• A. $\delta[n]+\delta[n-1]-\tfrac12\delta[n-2]-\tfrac52\delta[n-3]+\delta[n-5]$
• B. $\delta[n]-\delta[n-1]-\tfrac12\delta[n-2]-\tfrac52\delta[n-3]+\delta[n-5]$
• C. $\delta[n]-\delta[n-1]-\tfrac12\delta[n-2]+\tfrac52\delta[n-3]-\delta[n-5]$
• D. $\delta[n]+\delta[n-1]+\tfrac12\delta[n-2]+\tfrac52\delta[n-3]-\delta[n-5]$

Hint:

Multiplying spectra corresponds to convolving sequences: if $H(e^{j\Omega})=1-\tfrac12e^{-j2\Omega}$ then $h[n]=\delta[n]-\tfrac12\delta[n-2]$ and $y[n]=x[n]-\tfrac12x[n-2]$.

Answer 1

The Correct Option is C

Step 1: Find x[n] and h[n].
\(X(e^{j\Omega}) = 1 - e^{-j\Omega} + 2e^{-j3\Omega} \;\Rightarrow\; x[n] = \delta[n] - \delta[n-1] + 2\delta[n-3].\)
\(H(e^{j\Omega}) = 1 - \tfrac{1}{2}e^{-j2\Omega} \;\Rightarrow\; h[n] = \delta[n] - \tfrac{1}{2}\delta[n-2].\)

Step 2: Convolution for output.
\(y[n] = x[n] * h[n] = x[n] - \tfrac{1}{2}x[n-2].\)

Step 3: Compute term-by-term.
\(x[n] = \delta[n] - \delta[n-1] + 2\delta[n-3].\)
\(x[n-2] = \delta[n-2] - \delta[n-3] + 2\delta[n-5].\)
Therefore,
\[ \begin{aligned} y[n] &= \big(\delta[n] - \delta[n-1] + 2\delta[n-3]\big) - \tfrac{1}{2}\big(\delta[n-2] - \delta[n-3] + 2\delta[n-5]\big) \\ &= \delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]. \end{aligned} \]

\[ \boxed{y[n] = \delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]} \]