 ### Network # 2nd order circuit's bode plot

#### BlackMelon

Aug 7, 2012
188
Hello,

The pictures show the circuit and the gain and phase plot. In the plot, I don't know about the letter phi and the phrase "link the close loop with open loop". Could you give me some advice or link to study further?

Thank you
BlackMelon

#### LvW

Apr 12, 2014
604
The plot shows the loop gain (magnitude and phase).
The stability criterion requires to determine the phase difference between the actual loop phase at the "cross-over frequency" (magnitude 0dB) and zero phase. This difference must be positive (for a stable closed-loop) and, hence , is called "phase margin (phi)m".
The open-loop response - and, in particular, the phase margin determine the closed-loop response (magnitude peaking) in the "critical" area (around the cross.over frequency). This magnitude peaking is related to the Q factor (pole quality). This Q factor is defined in the denominator of the last equation .
The angular frequencies wo and w2 can also be derived from the plot.

EDIT: The whole plot is misleading. The same plot was used on page 9 of the contribution for explaining phase and gain margin. That`s OK. And now - the same plot was used in conjunction with given functions for open and closed-loop gains. This is misleading because the shown plot does NOT belong to a simple 2nd order function with real poles only (as given with the function).
So - what is the problem to be solved?

Last edited:

#### BlackMelon

Aug 7, 2012
188
The plot shows the loop gain (magnitude and phase).
The stability criterion requires to determine the phase difference between the actual loop phase at the "cross-over frequency" (magnitude 0dB) and zero phase. This difference must be positive (for a stable closed-loop) and, hence , is called "phase margin (phi)m".

According to this:http://www.mathworks.com/help/control/ref/margin.html, the phase margin is measured at the gain cross-over frequency as you said. But it's done by subtracting the phase at that point by -180 degree, not 0 degree. So, did you state it in another point of view? (I mean what you said might be correct but in a different explanation from www.mathworks.com)

#### BlackMelon

Aug 7, 2012
188
So - what is the problem to be solved?

I just want to study the app note, DC-DC Converters Feedback and Control, to design a feedback compensation for my forward converter. I have to take phase lag/lead between an AC variation of my output and duty cycle variation into account, because I don't want the converter to act like positive feedback.

Replies
1
Views
1K
Replies
7
Views
967
Replies
3
Views
1K
Replies
7
Views
5K
U
Replies
8
Views
4K
MooseFET
M