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I have the transfer function of the system, which is:

$$H(z) = \frac{1-z^{-1}}{5(1+2z^{-1})}$$

How do I sketch the magnitude and phase response?

I'm sorry for the bad formatting, it's my first time post a question.

Thank you very much in advance for the help!

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  • $\begingroup$ i think this question also asks this question sorta. $\endgroup$
    – robert bristow-johnson
    May 1, 2016 at 2:24
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    $\begingroup$ unlike analog (or more accurately, continuous-time) transfer functions, $H(s)$, i don't think it's as easy or natural to sketch the frequency response of a discrete-time transfer function $H(z)$ $\endgroup$
    – robert bristow-johnson
    May 1, 2016 at 2:29

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Use the transformation $z = e^{j\omega}$, you will get

$$H(e^{j\omega}) = \frac{1-e^{-j\omega}}{5\left(1+2e^{-j\omega}\right)}$$

Solving this (using $e^{j\omega} = \cos \omega + j \sin \omega$), you should get something like (please double check):

$$H(\omega) = \frac{\left(\cos\omega-1 \right) + j4\sin\omega}{5\left(5+4\cos\omega\right)}$$

Here,

$$\text{Re}\left(H(\omega)\right) = A = \frac{\left(\cos\omega-1 \right)}{5\left(5+4\cos\omega\right)}$$

and, $$\text{Im}\left(H(\omega)\right) = B = \frac{4\sin\omega}{5\left(5+4\cos\omega\right)}$$

Now, the magnitude response will be

$$\left|H(\omega)\right| = \sqrt{A^2+B^2}$$

and the phase response will be

$$\angle{H(\omega)} = \tan^{-1}\frac{B}{A}$$

You will get both responses as a function of $\omega$, just vary $\omega$, calculate the values and plot the response.

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  • $\begingroup$ First you need to be sure that it's a "stable" system to be able to substitude $z = e^{j\omega}$ into $H(z)$ to get $H(e^{j\omega})$ which is only then known to exist (converge) $\endgroup$
    – Fat32
    May 1, 2016 at 9:14
  • $\begingroup$ there's nothing stopping anyone from substituting $z \leftarrow e^{j\omega}$ to get a concept of frequency response. still doesn't make the filter stable, but frequency response and stability are not identical concepts. $\endgroup$
    – robert bristow-johnson
    May 2, 2016 at 0:45
  • $\begingroup$ Can anyone explain how we got $H(\omega)$ from $H(e^{j\omega})$ $\endgroup$
    – Hossam Houssien
    Jan 8, 2017 at 22:53
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For a rough sketch, you can eyeball or measure the distance of the poles and zeros to a point on the unit circle, multiply/divide to get a magnitude, and sum/difference the angles from the poles and zeros to that point to get a phase. A protractor and ruler might be useful.

The angles and distances change more rapidly when a pole or zero is near the unit circle, so you may have to plot more points around that portion of the unit circle before interpolating a spline or something in your sketch.

(Random historical note: This was how it was actually done in the age of slide-rules, mechanical adding machines, and drafting tables.)

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  • $\begingroup$ This would be a great answer if it provided a link or more details on how it is done. $\endgroup$
    – jpa
    Aug 10, 2017 at 14:37
  • $\begingroup$ mosaic.cnfolio.com/FileManagerBVM518ay2010/… explains it pretty well $\endgroup$
    – jpa
    Aug 10, 2017 at 16:19

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