I am reading Foundations of Oscillator Design by Guillermo Gonzales. When an amplifier is connected in a positive feedback configuration, the transfer function is \$A_{vf}(s)=\frac{A_v(s)}{1-\beta (s)A_v(s)}\$

In that case, the book explains that the Nyquist test states that \$N=Z-P\$ where N is the number of encirclements of 1+j0, Z is the number of zeros in the RHP and P is the number of poles in the RHP. Of course, the usual Nyquist criterion deal with encirclements of -1+j0 instead, but my understanding is that we have 1+j0 here because the denominator of the closed-loop transfer function is \$1-\beta(s)A_v(s)\$ rather than \$1+\beta(s)A_v(s)\$

Up to there, this matches what I have learned in control theory courses. I am, however, confused as to how the book draws the Nyquist plots of \$\beta(s)A_v(s)\$.

For example, in Example 1.1, \$\beta(s)A_v(s)=\frac{3}{s(s+1)(s+2)}\$.

When plotting that using WolframAlpha, I obtain the following Nyquist plot:

Nyquist plot obtained using WolframAlpha

The book draws them mirrored across the imaginary axis:

enter image description here

Why is this? Is one of them wrong?


1 Answer 1


The answer is simple: For the Nyquist plot, we need the loop gain.

This is the gain araound the loop when it is open - and it includes the sign at the summing junction. For positive feedback (oscillator case) the loop gain is positive - and the "critical point" is at "+1".

In some books/contributions the loop gain is defined as the product of the blocks/stages within the loop only - without consideration of the sign at the summing junction (that`s what you call "the usual Nyquist criterion"). In case of negative feedback, this results in a critical point which is at "-1".

For my opinion, this last definition is not a good one because each measurement/simulation of the loop gain includes, of course, the sign at the summing junction - in particular, the minus sign within a negative feedback loop.

  • \$\begingroup\$ I understand why the critical point is a 1 for positive feedback and -1 for negative feedback. But I still don't understand why WolframAlpha draws a mirrored version of the book's Nyquist plot? \$\endgroup\$
    – John Arg
    May 13, 2022 at 23:19
  • \$\begingroup\$ To elaborate on my previous comment, if I set \$\beta A_v\$ to be real, the only value I obtain is -0.5 (when \$j\omega=\pm \sqrt2\$), which matches the WolframAlpha plot. Therefore I believe that the Wolfram Alpha plot is the correct plot of the given open-loop gain. So it then seems to me that either I plot this and look at the encirclements of +1. Or I multiply this loop gain by -1, plot the now mirrored version as in the textbook, and look at encirclements of -1. I don't understand why plot both the mirrored version AND still look at encirclements of +1? \$\endgroup\$
    – John Arg
    May 14, 2022 at 3:57
  • \$\begingroup\$ As I have written - for negative feedback, the critical point is "-1" only when the minus sign is NOT included in the analysis. When the correct sign is included in the loop gain definition ("-" for negative feedback and "+" for pos. feedback), the critical point in BOTH cases is at "+1" because the stability criterion requires that the loop gain never is real and identical to or larger than "1". \$\endgroup\$
    – LvW
    May 14, 2022 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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