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* Transfer function | * Transfer function | ||
− | <math>H(z) = \frac{1}{(1- | + | <math>H(z) = \frac{1}{(1-p_1z^{-1})(1-p_2z^{-1})}, \text{where } p_1 \text{and } p_2 \text{ are poles of the filter.}</math> |
In order for the filter's impulse response to be real-valued, the two poles must be complex conjugates. So we assume that: | In order for the filter's impulse response to be real-valued, the two poles must be complex conjugates. So we assume that: |
Revision as of 19:31, 29 November 2010
Practice Question 5, ECE438 Fall 2010, Prof. Boutin
Filter Design
Define a two-pole band-pass filter such that
- The center of its band-pass is at $ \omega=\pi/2 $.
- There is no gain at the center of its band-pass
- The filter has a zero frequency response at $ \omega=0 $ and $ \omega=\pi $.
Express the system using a constant coefficient difference equation.
Post Your answer/questions below.
- Transfer function
$ H(z) = \frac{1}{(1-p_1z^{-1})(1-p_2z^{-1})}, \text{where } p_1 \text{and } p_2 \text{ are poles of the filter.} $
In order for the filter's impulse response to be real-valued, the two poles must be complex conjugates. So we assume that:
- $ p_1 = re^{j\theta} $
- $ p_2 = re^{-j\theta} $
So
$ \begin{align} H(z) &= \frac{1}{(1-p_1p^{-1})(1-p_2p^{-1})} \\ &= \frac{1}{(1-re^{j\theta}p^{-1})(1-re^{-j\theta}p^{-1})} \\ &= \frac{1}{1-2rcos(\theta)p^{-1}+r^2p^{-2}} \end{align} $
Then the frequency response of the filter is
$ H(e^{j\omega}) = \frac{1}{1-2rcos(\theta)e^{-j\omega}+r^2e^{-j2\omega}} $
Constant input gain is zero.
$ H(e^{j\omega})|_{\omega=0} = \frac{1}{1-2rcos(\theta)+r^2} = 0 $(*)
Filter has zero frequency response at $ \omega = 0,\pi $
$ H(e^{j\omega})|_{\omega=0} = \frac{1}{1-2rcos(\theta)+r^2} = 0 $
$ H(e^{j\omega})|_{\omega=\pi} = \frac{1}{1+2rcos(\theta)-r^2} = 0 $
I am unsure if this is correct way to tackle this problem. I don't wish to continue until the posted steps have been verified. Thanks!
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