Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been trying to figure this problem out for a while now. I've been given a transfer function $$H(s) = \frac{s(s+100)}{(s+2)(s+20)}.$$ I'm supposed to calculate the bode magnitude and frequency for it and then plot them. When I had done out the calculation I ended up getting $$|H(j\omega)| = \frac{\sqrt{\omega^4 - 10000\omega^2}}{\sqrt{\omega^4 - 484\omega^2 + 1600}}.$$

However this doesn't seem to be correct. I keep attempting my calculations but none of them seem to be correct. I'm also not sure how I can go about to plot it. Could anyone please help me out with this? Thanks!

share|cite|improve this question
up vote 2 down vote accepted

The magnitude of $H(j \omega)$ is

$$|H(j \omega)| = \frac{\sqrt{\omega^4 + 10000\, \omega^2}}{\sqrt{\omega^4+404\, \omega^2 + 1600}}$$

The reason for this is that $|a + i b| = \sqrt{a^2+b^2}$ for $a,b \in \mathbb{R}$.

The phase of $H(j \omega)$ may be computed by taking phases of individual terms:

$$\arg{H(j \omega)} = \frac{\pi}{2} + \arctan{\frac{\omega}{100}} - \arctan{\frac{\omega}{2}}- \arctan{\frac{\omega}{20}}$$

share|cite|improve this answer
Thanks, also for frequency, would it be then H(jw) = tan^-1(w/100) - tan^-1(w^2/(404w^2 + 1600)? – Somebody Apr 7 '13 at 23:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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