Topic: Filter Design

(Practice Question 5, ECE438 Fall 2010, Prof. Boutin )

## Question

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.

• 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! 