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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?

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1 Answer 1

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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.

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  • \$\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

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