Solution to Q2 of Week 11 Quiz Pool
a) DFT is a frequency sampling of DTFT and both are related such that $ H_8[k]=H(w)|_{w=\frac{2\pi}{8}k} $.
Thus, we need to find the DTFT of $ h[n] $.
$ \begin{align}h[n]=\delta[n]-\delta[n-1] \;\; \Leftrightarrow \;\; H(w)&=1+e^{-jw}=e^{-j\frac{w}{2}}\left(e^{j\frac{w}{2}}-e^{-j\frac{w}{2}}\right) \\ &=2j\text{sin}\left(\frac{w}{2}\right)e^{-j\frac{w}{2}} \end{align} $
Therefore, $ H_8[k]=2j\text{sin}\left(\frac{\pi k}{8}\right)e^{-j\frac{\pi k}{8}} \;\;\; \text{for} \;\; k=0,1,...,6,7. $.
b) Note that $ y_8[n] $ is the 8-pt inverse DFT of $ Y_8[k]=X_8[k]H_8[k] $.
But we know that when the input $ x[n] $ is in the form of $ e^{jw_0 n} $, then the output is $ y[n]=H(w_0)e^{jw_0 n} $.
Thus, the 8-pt inverse DFT of $ Y_8[k] $, $ y_8[n] $, is related such that $ y_8[n]=H_8[k]e^{j\frac{2\pi}{8}kn}(u[n]-u[n-8]) $.
- for the length-8 input of frequency $ w=\frac{2\pi}{8}k $, i.e. $ e^{j\frac{2\pi}{8}kn}(u[n]-u[n-8]) $.
Note that,
$ \begin{align} x[n]&=\text{cos}\left(\pi n\right)(u[n]-u[n-8]) \\ &=\left(\frac{1}{2}e^{j\pi n}+\frac{1}{2}e^{-j\pi n}\right)(u[n]-u[n-8]) \\ &=\left(\frac{1}{2}e^{j\frac{2\pi}{8}4n}+\frac{1}{2}e^{-j\frac{2\pi}{8}4n}\left(e^{j2\pi n}\right)\right)(u[n]-u[n-8]) \;\; (\text{since }e^{j2\pi n}=1) \\ &=\left(\frac{1}{2}e^{j\frac{2\pi}{8}4n}+\frac{1}{2}e^{j\frac{2\pi}{8}4n}\right)(u[n]-u[n-8]) \\ &=e^{j\frac{2\pi}{8}4n}(u[n]-u[n-8]) \end{align} $
Hence, $ y_8[n]=H_8[4]e^{j\frac{2\pi}{8}4n}(u[n]-u[n-8]) $.
$ H_8[4]=2j\text{sin}\left(\frac{4\pi}{8}\right)e^{-j\frac{4\pi}{8}}=2j(-j)=2 $
Therefore, $ y_8[n]=2e^{j\frac{2\pi}{8}4n}(u[n]-u[n-8])=2e^{j\pi n}(u[n]-u[n-8]) $
Credit: Prof. Mike Zoltowski
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