2

I was reading Razavi's Design of Analog CMOS ICs book and came across this in the Stability and Freq. Compensation chapter,

What the book says

The closed-loop transfer function of a system is given by the following: enter image description here

It then goes on to say that, enter image description here

Okay, that makes sense to me. As loop-gain (beta or Aol) increases, our pole moves further into the LHP, making any oscillation die out much quicker. Now, a few pages later, the book has this, enter image description here He says here that by reducing the feedback factor (hence loop gain) our system is more stable and this makes sense to me from the bode plot.

My Question

The two bolded statements above contradict each other. If I reduce my feedback factor (hence loop gain), my pole will move towards the RHP and thus from the s-plane perspective, the system is less stable but from the bode plot perspective, the system is more stable. Why does the s-plane and bode plot contradict each other?

AlfroJang80
  • 3,103
  • 2
  • 32
  • 59

1 Answers1

3

You are comparing a first-order system with a second-order system.

The first equation you show (10.4) describes a first-order system. The pole moves farther along the negative real axis as you increase the feedback, as the text describes.

The Bode plot in and discussion around Figure 10.7 is a second-order system. In a second order system with real open-loop poles, the poles first move toward one another along the real axis with increasing feedback, then after meeting along the real axis, move away from one another in the positive and negative imaginary directions. Thus the angle they make with the imaginary axis is decreasing, leading to a less stable system with increasing feedback. Thus the Bode analysis and the root-locus are consistent.

rpm2718
  • 2,812
  • 1
  • 6
  • 14
  • Thanks for the response. Even if the second case was a first-order system, reducing the gain from a bode plot view would cause the gain crossover to move towards the origin, thereby increasing stability? – AlfroJang80 Jul 22 '20 at 20:27
  • In the first order system with an open-loop pole along the negative real axis, the phase can't get beyond 90 degrees. So when the gain crossover happens, you never get close to the criteria for instability, which is 180 degree phase shift at the gain crossover. So stability is not really the issue. System response speed is more the issue there. The Bode analysis really centers around understanding how to keep the phase of the loop gain well under 180 degrees at the gain crossover point. – rpm2718 Jul 22 '20 at 20:43
  • Ah I see. One last question, what's the relationship between the angle of the pole and the stability? – AlfroJang80 Jul 22 '20 at 21:12
  • The settling time will be the same as the poles move in the imaginary direction, but the overshoot (to a step input) increases. You get a faster, higher climb to the first oscillatory peak as the poles move away from the real axis. So by this measure, you could say the system is less stable. Another way to look at it is the closer the angle of the poles is to the imaginary axis, the less margin there is for some unwanted additional phase shift in your system to push you toward an unstable system, where the poles actually cross the imaginary axis. – rpm2718 Jul 22 '20 at 21:26
  • Ah awesome. Thanks very much. May I ask, do you know any good resources on the s-plane (pole-zero diagrams). – AlfroJang80 Jul 22 '20 at 21:28
  • 1
    I'm sure there are some good websites, but I don't happen to know of one to recommend. Sorry. There are several great introductory books on Control Systems Engineering. The one written by Norman Nise is one of the standard ones that is well regarded. – rpm2718 Jul 22 '20 at 21:33
  • Thank you very much! Really appreciate the help. – AlfroJang80 Jul 22 '20 at 22:47