Phase margin calculation in paper !!!

Hi all,

I am studying the concept of phase margin. I understand the meaning but I have a question related to the calculation of the phase margin in paper.

For example if I have a open loop system a(s) = 10 / ( (s+1) * (s+2) ), the phase margin in Matlab is approximately 55 degrees. So closed loop system is stable but I want to verify this value with hand calculations. So the equation is:

angle = 180° - arctg (W/|1|) - arctg (W/|2|)

the value of W correspond to the frequency where the modulus of the open loop system is equal to 1, but what make feel confuse is that I don´t understand why I am doing the arctg of the frequency W where the gain is 0 dB and the poles of the open loop system.

I will be very grateful if you can help me, please.

Best regards

Joaquin

Instability occurs when the loop gain is > 0dB at 180deg (-ve feedback system adds the extra 180deg phase shift).

Loop gain is the open loop gain. Therefore your analysis will be on the open loop system response.

In your system, each pole adds a phase shift. Therefore you want to find the phase shift of each pole at the
frequency where loop gain = 0dB (i.e at the crossover frequency wx). To do this, map your poles into the frequency domain (set dampning = 0)
and then get the phase shift by arctan(jwx/1) and artcan(jwx/2). The sum of these being the total phase shift
contribution of both poles, when subtracted from 180 then gives you the PM.

diarmuid said:

Instability occurs when the loop gain is > 0dB at 180deg (-ve feedback system adds the extra 180deg phase shift).

Loop gain is the open loop gain. Therefore your analysis will be on the open loop system response.

In your system, each pole adds a phase shift. Therefore you want to find the phase shift of each pole at the
frequency where loop gain = 0dB (i.e at the crossover frequency wx). To do this, map your poles into the frequency domain (set dampning = 0)
and then get the phase shift by arctan(jwx/1) and artcan(jwx/2). The sum of these being the total phase shift
contribution of both poles, when subtracted from 180 then gives you the PM.

Click to expand...

Thank you for your response !. I understood what you told me but for example if one pole is after the crossover frequency why I have to considerate that pole in the calculation of the phase margin ?

Best regards !

Joaquín

Hi Joaquin,

Consider a one pole system .... the phase starts from 0deg and falls all the way down to -90deg at infinity .....
The phase at the corner frequency (i.e at the frequency at which the pole is placed) is -45deg ... This means even before the occurrence of the pole the phase plot has already started falling.
So in a two pole system, even if the second pole is out side my UGB or gain cross over frequency ... its impact starts much earlier in the phase plot.
That means when we are expecting a 90deg phase margin in a two pole system with one pole within UGB and the other outside UGB .... the Phase margin is actually much lesser than 90deg.

Now if the second pole is just outside UGB then its impact in the phase is greater than is its 1 or 2 decade outside .... So you have to take a call depending on the second pole position whether to include it or not in your hand calculations ...

Try to visualize the phase margin values keeping the 1st pole fixed and moving the second pole from within UGB to outside it ...

Hope it will help ....

If a pole occurs at > 10x the crossover frequency it wont degrade phase margin.

If it is < 10x the crossover frequency it will and so needs to be included in the PM calculation.

This is because a pole starts to affect the phase response of a system at ~ 1/10th the pole frequency.