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!

  • $\begingroup$ i think this question also asks this question sorta. $\endgroup$ May 1, 2016 at 2:24
  • 1
    $\begingroup$ 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 $H(z)$ $\endgroup$ May 1, 2016 at 2:29

2 Answers 2


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):

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


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

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.

  • $\begingroup$ 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. $\endgroup$ May 2, 2016 at 0:45
  • $\begingroup$ Can anyone explain how we got $H(\omega)$ from $H(e^{j\omega})$ $\endgroup$ Jan 8, 2017 at 22:53

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.)

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.