# 0.5 Lab 5a - digital filter design (part 1)  (Page 2/5)

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$\begin{array}{ccc}\hfill X\left({e}^{j\omega }\right)& =& {\left(X,\left(,z,\right)|}_{z={e}^{j\omega }}\hfill \\ & =& \sum _{n=-\infty }^{\infty }x\left(n\right){e}^{-j\omega n}\hfill \end{array}$

From the definition of the Z-transform, a change of variable $m=n-K$ shows that a delay of $K$ samples in the time domain is equivalent to multiplication by ${z}^{-K}$ in the Z-transform domain.

$\begin{array}{ccc}\hfill x\left(n-K\right)& \stackrel{Z}{↔}& \sum _{n=-\infty }^{\infty }x\left(n-K\right){z}^{-n}\hfill \\ & =& \sum _{m=-\infty }^{\infty }x\left(m\right){z}^{-\left(m+K\right)}\hfill \\ & =& {z}^{-K}\sum _{m=-\infty }^{\infty }x\left(m\right){z}^{-m}\hfill \\ & =& {z}^{-K}X\left(z\right)\hfill \end{array}$

We may use this fact to re-write [link] in the Z-transform domain, by taking Z-transforms of both sides ofthe equation:

$\begin{array}{c}\hfill Y\left(z\right)=\sum _{i=0}^{N-1}{b}_{i}{z}^{-i}X\left(z\right)-\sum _{k=1}^{M}{a}_{k}{z}^{-k}Y\left(z\right)\\ \hfill Y\left(z\right)\left(1,+,\sum _{k=1}^{M},{a}_{k},{z}^{-k}\right)=X\left(z\right)\sum _{i=0}^{N-1}{b}_{i}{z}^{-i}\\ \hfill H\left(z\right)\stackrel{△}{=}\frac{Y\left(z\right)}{X\left(z\right)}=\frac{{\sum }_{i=0}^{N-1}{b}_{i}{z}^{-i}}{1+{\sum }_{k=1}^{M}{a}_{k}{z}^{-k}}\end{array}$

From this formula, we see that any filter which can be represented by a linear difference equation with constant coefficientshas a rational transfer function (i.e. a transfer function which is a ratio of polynomials).From this result, we may compute the frequency response of the filter by evaluating $H\left(z\right)$ on the unit circle:

$H\left({e}^{j\omega }\right)=\frac{{\sum }_{i=0}^{N-1}{b}_{i}{e}^{-j\omega i}}{1+{\sum }_{k=1}^{M}{a}_{k}{e}^{-j\omega k}}\phantom{\rule{4pt}{0ex}}.$

There are many different methods for implementing a general recursive difference equation of the form [link] . Depending on the application, some methods may be more robustto quantization error, require fewer multiplies or adds, or require less memory. [link] shows a system diagram known as the direct form implementation; it works for any discrete-time filterdescribed by the difference equation in [link] . Note that the boxes containing the symbol ${z}^{-1}$ represent unit delays, while a parameter written next to a signal path represents multiplication by that parameter. Direct form implementation for a discrete-time filter described by a linear recursive difference equation.

## Design of a simple fir filter

To illustrate the use of zeros in filter design, you will design a simple second order FIR filterwith the two zeros on the unit circle as shown in [link] . In order for the filter's impulse response to be real-valued,the two zeros must be complex conjugates of one another:

${z}_{1}={e}^{j\theta }$
${z}_{2}={e}^{-j\theta }$

where $\theta$ is the angle of ${z}_{1}$ relative to the positive real axis. We will see later that $\theta \in \left[0,\pi \right]$ may be interpreted as the location of the zeros in the frequency response.

The transfer function for this filter is given by

$\begin{array}{ccc}\hfill {H}_{f}\left(z\right)& =& \left(1-{z}_{1}{z}^{-1}\right)\left(1-{z}_{2}{z}^{-1}\right)\hfill \\ & =& \left(1-{e}^{j\theta }{z}^{-1}\right)\left(1-{e}^{-j\theta }{z}^{-1}\right)\hfill \\ & =& 1-2cos\theta {z}^{-1}+{z}^{-2}\phantom{\rule{4pt}{0ex}}.\hfill \end{array}$

Use this transfer function to determine the difference equation for this filter.Then draw the corresponding system diagram and compute the filter's impulse response $h\left(n\right)$ .

This filter is an FIR filter because it has impulse response $h\left(n\right)$ of finite duration. Any filter with only zeros and no poles other than thoseat 0 and $±\infty$ is an FIR filter. Zeros in the transfer function represent frequencies that arenot passed through the filter. This can be useful for removing unwanted frequencies in a signal.The fact that ${H}_{f}\left(z\right)$ has zeros at ${e}^{±j\theta }$ implies that ${H}_{f}\left({e}^{±j\theta }\right)=0$ . This means that the filter will not pass pure sine waves at a frequencyof $\omega =\theta$ .

Use Matlab to compute and plot the magnitude of the filter's frequency response $|{H}_{f}\left({e}^{j\omega }\right)|$ as a function of $\omega$ on the interval $-\pi <\omega <\pi$ , for the following three values of $\theta$ :

• $\theta =\pi /6$
• $\theta =\pi /3$
• $\theta =\pi /2$

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