# 11.8 Continuous time filter design

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Describes how to design a general filter from the Laplace Transform and its pole/zero plots.

## Introduction

Analog (Continuous-Time) filters are useful for a wide variety of applications, and are especially useful in that they are very simple to build using standard, passive R,L,C components. Having a grounding in basic filter design theory can assist one in solving a wide variety of signal processing problems.

## Estimating frequency response from z-plane

One of the motivating factors for analyzing the pole/zero plots is due to their relationship to the frequency responseof the system. Based on the position of the poles and zeros, one can quickly determine the frequency response. This is aresult of the correspondence between the frequency response and the transfer function evaluated on the unit circle in thepole/zero plots. The frequency response, or DTFT, of the system is defined as:

$H(w)=(z, e^{iw}, H(z))=\frac{\sum_{k=0}^{M} {b}_{k}e^{-(iwk)}}{\sum_{k=0}^{N} {a}_{k}e^{-(iwk)}}$
Next, by factoring the transfer function into poles and zeros and multiplying the numerator and denominator by $e^{iw}$ we arrive at the following equations:
$H(w)=\left|\frac{{b}_{0}}{{a}_{0}}\right|\frac{\prod_{k=1}^{M} \left|e^{iw}-{c}_{k}()\right|}{\prod_{k=1}^{N} \left|e^{iw}-{d}_{k}()\right|}$
From [link] we have the frequency response in a form that can be used to interpretphysical characteristics about the filter's frequency response. The numerator and denominator contain a product ofterms of the form $\left|e^{iw}-h\right|$ , where $h$ is either a zero, denoted by ${c}_{k}$ or a pole, denoted by ${d}_{k}$ . Vectors are commonly used to represent the term and its parts on the complex plane. The pole or zero, $h$ , is a vector from the origin to its location anywhere on the complex plane and $e^{iw}$ is a vector from the origin to its location on the unit circle. The vector connecting these twopoints, $\left|e^{iw}-h\right|$ , connects the pole or zero location to a place on the unit circle dependent on the value of $w$ . From this, we can begin to understand how the magnitude of the frequency response is aratio of the distances to the poles and zero present in the z-plane as $w$ goes from zero to pi. These characteristics allow us to interpret $\left|H(w)\right|$ as follows:
$\left|H(w)\right|=\left|\frac{{b}_{0}}{{a}_{0}}\right|\frac{\prod \mathrm{"distances from zeros"}}{\prod \mathrm{"distances from poles"}}$
In conclusion, using the distances from the unit circle to the poles and zeros, we can plot the frequency response of thesystem. As $w$ goes from $0$ to $2\pi$ , the following two properties, taken from the above equations, specify how one should draw $\left|H(w)\right|$ .

## While moving around the unit circle...

1. if close to a zero, then the magnitude is small. If a zero is on the unit circle, then the frequency response iszero at that point.
2. if close to a pole, then the magnitude is large. If a pole is on the unit circle, then the frequency responsegoes to infinity at that point.

## Drawing frequency response from pole/zero plot

Let us now look at several examples of determining the magnitude of the frequency response from the pole/zero plot ofa z-transform. If you have forgotten or are unfamiliar with pole/zero plots, please refer back to the Pole/Zero Plots module.

In this first example we will take a look at the very simple z-transform shown below: $H(z)=z+1=1+z^{-1}$ $H(w)=1+e^{-(iw)}$ For this example, some of the vectors represented by $\left|e^{iw}-h\right|$ , for random values of $w$ , are explicitly drawn onto the complex plane shown in the figure below. These vectors show how the amplitude of the frequency response changes as $w$ goes from $0$ to $2\pi$ , and also show the physical meaning of the terms in [link] above. One can see that when $w=0$ , the vector is the longest and thus the frequency responsewill have its largest amplitude here. As $w$ approaches $\pi$ , the length of the vectors decrease as does the amplitude of $\left|H(w)\right|$ . Since there are no poles in the transform, there is only this onevector term rather than a ratio as seen in [link] .

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