Difference between revisions of "ECE438 Week11 Quiz Q5sol" - Rhea

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Week11 Quiz Question 5 Solution

Suppose the transfer function of the filter has the form

$ H_f(z)=(1-z_1 z^{-1})(1-z_2 z^{-1}) $

Where $ z_1,z_2 $ are zeros of the filter.

In order for the filter's impulse response to be real-valued, the two zeros must be complex conjugates of one another:

Assume $ z_1=re^{j\theta},z_2=re^{-j\theta} $, where $ \theta $ is the angle of $ z_1 $ relative to the positive real axis. Without losing generality, assume $ \theta \in [0,\pi] $. Then

$ \begin{align} H_f(z)&=(1-re^{j\theta}z^{-1})(1-re^{-j\theta}z^{-1}) \\ &=1-2rcos\theta z^{-1}+r^2z^{-2} \end{align} $

Then the frequency response of the filter is

$ H(e^{j\omega})=H_f(z)|_{z=e^{j\omega}}=1-2rcos\theta e^{-j\omega}+r^2e^{-2j\omega} $

Since the constant input gain is 1, therefore

$ H(e^{j\omega})|_{\omega =0}=1-2rcos\theta +r^2=1\text{ (*)} $

Since the filter has a zero frequency response at $ \omega =\frac{\pi}{}2 $

$ H(e^{j\omega})|_{\omega =\frac{\pi}{2}}=1-2rcos\theta(-j)+r^2(-1)=0\text{ (**)} $

Combine (*) and (**), we can compute $ r=1,\theta=\frac{\pi}{2} $

Therefore, the transfer function is

$ \begin{align} &H_f(z)=1+z^{-2} \\ \Leftrightarrow &Y(z)=X(z)+z^{-2}X(z) \\ \end{align} $

By conducting z inverse transfer on both sides, we obtain the constant coefficient difference equation of the filter:

$ y[n]=x[n]+x[n-2] $

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