Solution to Q1 of Week 12 Quiz Pool
a) DFT is a frequency sampling of DTFT and both are related such that $ H_9[k]=H(w)|_{w=\frac{2\pi}{9}k} $.
Thus, we need to find the DTFT of $ h[n] $.
$ \begin{align}h[n]=\delta[n-1]+\delta[n-2] \;\; \Leftrightarrow \;\; H(w)&=e^{-jw}+e^{-j2w}=e^{-jw}(1+e^{-jw}) \\ &=e^{-jw}e^{-j\frac{w}{2}}\left(e^{j\frac{w}{2}}+e^{-j\frac{w}{2}}\right) \\ &=2\text{cos}\left(\frac{w}{2}\right)e^{-j\frac{3w}{2}} \end{align} $
Therefore, $ H_9[k]=2\text{cos}\left(\frac{\pi}{9}k\right)e^{-j\frac{\pi}{3}k} \;\;\; \text{for} \;\; k=0,1,...,7,8. $.
b) Note that $ y_9[n] $ is the 9-pt inverse DFT of $ Y_9[k]=X_9[k]H_9[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 9-pt inverse DFT of $ Y_9[k] $, $ y_9[n] $, is related such that $ y_9[n]=H_9[k]e^{j\frac{2\pi}{9}kn}(u[n]-u[n-9]) $.
- for any length-9 input with frequency corresponding to $ w=\frac{2\pi}{9}k $, i.e. $ e^{j\frac{2\pi}{9}kn}(u[n]-u[n-9]) $.
Note that,
$ \begin{align} x[n]&=\text{cos}\left(\frac{2\pi}{3}n\right)(u[n]-u[n-9]) \\ &=\left(\frac{1}{2}e^{j\frac{2\pi}{3}n}+\frac{1}{2}e^{-j\frac{2\pi}{3}n}\right)(u[n]-u[n-9]) \\ &=\left(\frac{1}{2}e^{j\frac{2\pi}{9}3n}+\frac{1}{2}e^{-j\frac{2\pi}{9}3n}\left(e^{j2\pi n}\right)\right)(u[n]-u[n-9]) \;\; (\text{since }e^{j2\pi n}=1) \\ &=\left(\frac{1}{2}e^{j\frac{2\pi}{9}3n}+\frac{1}{2}e^{j\frac{2\pi}{9}6n}\right)(u[n]-u[n-9]) \end{align} $
Hence, $ y_9[n]=\left(\frac{1}{2}H_9[3]e^{j\frac{2\pi}{9}3n} + \frac{1}{2}H_9[6]e^{j\frac{2\pi}{9}6n}\right)(u[n]-u[n-9]) $.
$ H_9[3]=2\text{cos}\left(\frac{3\pi}{9}\right)e^{-j\frac{3\pi}{3}}=2\frac{\sqrt{3}}{2}(-1)=-\sqrt{3} $
$ H_9[6]=2\text{cos}\left(\frac{6\pi}{9}\right)e^{-j\frac{6\pi}{3}}=2\frac{1}{2}(1)=1 $
Therefore, $ y_9[n]=\left(-\frac{\sqrt{3}}{2}e^{j\frac{2\pi}{9}3n} + \frac{1}{2}e^{j\frac{2\pi}{9}6n}\right)(u[n]-u[n-9]) $.
Credit: Prof. Mike Zoltowski
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