Solution to Q6 of Week 6 Quiz Pool
$ \begin{align} \text{(a)} \quad & x[n]=\delta[n] \;\; \Rightarrow \;\; X(e^{jw})=1 \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & Y(e^{jw}) = X(e^{jw})H(e^{jw}) = (1)\text{rect}\Big(\frac{w}{\pi}\Big) \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & \Rightarrow \;\; y[n] = \frac{\text{sin}\big(\frac{\pi}{2}n\big)}{\pi n} = \frac{1}{2}\frac{\text{sin}\big(\frac{\pi}{2}n\big)}{\pi \frac{n}{2}} \\ & z[n] = y[2n] = \frac{1}{2}\frac{\text{sin}\big(\pi n\big)}{\pi n} \\ \end{align} \,\! $
$ \begin{align} \text{(b)} \quad & x[n] = \delta[n-1] \;\; \Rightarrow \;\; X(e^{jw}) = e^{-jw} \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & Y(e^{jw}) = X(e^{jw})H(e^{jw}) = e^{-jw}\text{rect}\Big(\frac{w}{\pi}\Big) \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & \Rightarrow \;\; y[n] = \frac{\text{sin}\big(\frac{\pi}{2}(n-1)\big)}{\pi (n-1)} = \frac{1}{2}\frac{\text{sin}\big(\frac{\pi}{2}(n-1)\big)}{\pi \frac{(n-1)}{2}} \\ & z[n] = y[2n] = \frac{1}{2}\frac{\text{sin}\big(\pi (n-\frac{1}{2})\big)}{\pi (n-\frac{1}{2})} \\ \end{align} \,\! $
Note that, since it is downsampled by a factor of 2, $ z[n] $ is shifted 0.5 to the right compared to $ z[n] $ of (a), even though the input $ x[n] $ is shifted 1 to the right compared to the input $ x[n] $ of (a).
$ \begin{align} \text{(c)} \quad & x[n] = 1 \;\; \Rightarrow \;\; X(e^{jw})=2\pi\delta(w) \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & Y(e^{jw}) = X(e^{jw})H(e^{jw}) = 2\pi\delta(w)\text{rect}\Big(\frac{w}{\pi}\Big) = 2\pi\delta(w) \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & \Rightarrow \;\; y[n]=1 \\ & z[n]=y[2n]=1 \\ \end{align} \,\! $
$ \begin{align} \text{(d)} \quad & x[n] = \text{cos}\big(\frac{\pi}{4}n\big) \;\; \Rightarrow \;\; X(e^{jw}) = \pi \Big[ \delta \big( w - \frac{\pi}{4} \big) + \delta \big( w + \frac{\pi}{4} \big) \Big] \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & Y(e^{jw}) = X(e^{jw})H(e^{jw}) = \pi \Big[ \delta \big( w - \frac{\pi}{4} \big) + \delta \big( w + \frac{\pi}{4} \big) \Big] \text{rect}\Big(\frac{w}{\pi}\Big) \;\; \text{for} \;\; w \in (-\pi,\pi) \\ & \Rightarrow \;\; y[n] = \text{cos}\big(\frac{\pi}{4}n\big) \\ & z[n] = y[2n] = \text{cos}\big(\frac{\pi}{4}(2n)\big) = \text{cos}\big(\frac{\pi}{2}n \big) \\ \end{align} \,\! $
Credit: Prof. Charles Bouman
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