(New page: = 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 <math>\omeg...) |
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| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
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| + | Topic: Filter Design | ||
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| + | </center> | ||
| + | ([[:Category:Problem_solving|Practice Question]] 5, [[ECE438]] Fall 2010, [[User:Mboutin|Prof. Boutin]] ) | ||
---- | ---- | ||
| + | ==Question== | ||
Define a two-pole band-pass filter such that | Define a two-pole band-pass filter such that | ||
#The center of its band-pass is at <math>\omega=\pi/2 </math>. | #The center of its band-pass is at <math>\omega=\pi/2 </math>. | ||
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Post Your answer/questions below. | Post Your answer/questions below. | ||
| − | * | + | |
| + | * Transfer function | ||
| + | |||
| + | <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: | ||
| + | *<math>p_1 = re^{j\theta}</math> | ||
| + | *<math>p_2 = re^{-j\theta}</math> | ||
| + | |||
| + | So | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | H(z) &= \frac{1}{(1-p_1z^{-1})(1-p_2z^{-1})} \\ | ||
| + | &= \frac{1}{(1-re^{j\theta}z^{-1})(1-re^{-j\theta}z^{-1})} \\ | ||
| + | &= \frac{1}{1-2rcos(\theta)z^{-1}+r^2z^{-2}} | ||
| + | \end{align}</math> | ||
| + | |||
| + | Then the frequency response of the filter is | ||
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| + | <math>H(e^{j\omega}) = \frac{1}{1-2rcos(\theta)e^{-j\omega}+r^2e^{-j2\omega}}</math> | ||
| + | |||
| + | Constant input gain is zero. | ||
| + | |||
| + | <math>H(e^{j\omega})|_{\omega=\frac{\pi}{2}} = \frac{1}{1-2rcos(\theta)+r^2} = 1</math>(*) | ||
| + | |||
| + | Filter has zero frequency response at <math>\omega = 0,\pi</math> | ||
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| + | <math>H(e^{j\omega})|_{\omega=0} = \frac{1}{1-2rcos(\theta)+r^2} = 0</math> | ||
| + | |||
| + | <math>H(e^{j\omega})|_{\omega=\pi} = \frac{1}{1+2rcos(\theta)-r^2} = 0</math> | ||
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| + | 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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| + | ---- | ||
*Answer/question | *Answer/question | ||
*Answer/question | *Answer/question | ||
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[[Practice_Question_4_ECE438F10|Previous practice problem]] | [[Practice_Question_4_ECE438F10|Previous practice problem]] | ||
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[[2010 Fall ECE 438 Boutin|Back to 2010 Fall ECE 438 Boutin]] | [[2010 Fall ECE 438 Boutin|Back to 2010 Fall ECE 438 Boutin]] | ||
[[Category:2010_Fall_ECE_438_Boutin]] | [[Category:2010_Fall_ECE_438_Boutin]] | ||
Latest revision as of 12:00, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Filter Design
(Practice Question 5, ECE438 Fall 2010, Prof. Boutin )
Question
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_1z^{-1})(1-p_2z^{-1})} \\ &= \frac{1}{(1-re^{j\theta}z^{-1})(1-re^{-j\theta}z^{-1})} \\ &= \frac{1}{1-2rcos(\theta)z^{-1}+r^2z^{-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=\frac{\pi}{2}} = \frac{1}{1-2rcos(\theta)+r^2} = 1 $(*)
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!
- Answer/question
- Answer/question
- Answer/question
