How to determine gain and phase margins for a transfer function?

1. Mar 19, 2014

annas425

How would I determine gain and phase margins for the following transfer function:

G(s) = (0.38(s^2 + 0.1s +0.55)) / (s(s+1)(s^2 + 0.6s + 0.5))

We are learning about Bode plots but I am really struggling. I know that Bode plots display the relationship between magnitude vs. frequency and phase angle vs. frequency, but that's about it. I also know that s = iw (where w is \omega). Thank you so much, in advance! I am REALLY lost.

2. Mar 19, 2014

maajdl

You simply need to calculate this function as a function of w and determine the amplitude and the phase.
The phase is defined up to an indeterminacy of 2Pi.
Therefore, you might see 2Pi phase jumps in the graph.
Eventually you may find it nicer to smooth out these jumps.

3. Mar 19, 2014

annas425

Thanks for the response! Honestly I don't understand what you mean…is there any way you could give me the steps to get started?

Many thanks! :)

4. Mar 19, 2014

maajdl

For example, for w=1, you get:

G(i w) = -0.02024590163934427 - 0.1572950819672131 i = 0.15859268376650384 Exp(-1.6988053854680936 i)

the amplitude is 0.15859268376650384
the phase is -1.6988053854680936 = -0.54074654889676 Pi

And you need a definition (I don't know this terminoogy):

http://en.wikipedia.org/wiki/Phase_margin

Last edited: Mar 19, 2014
5. Mar 19, 2014

maajdl

Nicer in this way:

for w=1

G(iw) = -247/12200 - 1919/12200 i

and therefore

amplitude(G(iw)) = sqrt((247/12200)² + (1919/12200)²) = 19/20 Sqrt(17/610)
phase(G(iw)) = arctan(1919/247) = arctan(101/13)

6. Mar 19, 2014

maajdl

Make a plot as a function of w, for amplitude and phase.
I checked, there is no phase jumps.

7. Mar 19, 2014

donpacino

the phase margin is your phase + 180 degrees at the point when your gain is 0 db or 1 magnitude.

so look on your bode plot where the gain crosses over 0 db. then look at your phase at the same frequency and add 180. Phase margin is an extremely important concept to learn.