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<math>H(e^{j\omega})=H_f(z)|_{z=e^{j\omega}}=1-2rcos\theta e^{-j\omega}+r^2e^{-2j\omega}</math> | <math>H(e^{j\omega})=H_f(z)|_{z=e^{j\omega}}=1-2rcos\theta e^{-j\omega}+r^2e^{-2j\omega}</math> | ||
| − | Since the constant input gain is | + | Since the constant input gain is 2, therefore |
| − | <math>H(e^{j\omega})|_{\omega =0}=1-2rcos\theta +r^2= | + | <math>H(e^{j\omega})|_{\omega =0}=1-2rcos\theta +r^2=2\text{ (*)}</math> |
| − | Since the filter has a zero frequency response at <math>\omega =\frac{\pi}{ | + | Since the filter has a zero frequency response at <math>\omega =\frac{\pi}{2}</math> |
| − | <math>H(e^{j\omega})|_{\omega =\frac{\pi}{2}}=1-2rcos\theta(-j)+r^2(-1)=0</math>(**) | + | <math>H(e^{j\omega})|_{\omega =\frac{\pi}{2}}=1-2rcos\theta(-j)+r^2(-1)=0\text{ (**)}</math> |
| + | |||
| + | Combine (*) and (**), we can compute <math>r=1,\theta=\frac{\pi}{2}</math> | ||
| + | |||
| + | Therefore, the transfer function is | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | &H_f(z)=1+z^{-2} \\ | ||
| + | \Leftrightarrow &Y(z)=X(z)+z^{-2}X(z) \\ | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | By conducting z inverse transfer on both sides, we obtain the constant coefficient difference equation of the filter: | ||
| + | |||
| + | <math>y[n]=x[n]+x[n-2]</math> | ||
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Latest revision as of 16:23, 3 November 2010
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 2, therefore
$ H(e^{j\omega})|_{\omega =0}=1-2rcos\theta +r^2=2\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] $
