Revision as of 05:25, 29 December 2010

Practice Question 5, ECE438 Fall 2010, Prof. Boutin

Filter Design

Define a two-pole band-pass filter such that

1. The center of its band-pass is at $ \omega=\pi/2 $.
2. There is no gain at the center of its band-pass
3. 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

Previous practice problem

Next practice problem

Back to 2010 Fall ECE 438 Boutin