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I was trying to compare a Bode plot and a Nyquist plot of the same system and realized that I was probably comparing apples to oranges

Bode plot is plotting the entire closed loop transfer function, whereas the Nyquist plot is only plotting the loop gain

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Given the above system, the loop gain is defined as \$C(s) \times P(s)\$, whereas the closed loop transfer function is defined as $$\frac{1}{1+C(s) \times P(s)}$$

Why does Nyquist only care about the loop gain (but not the entire closed loop tf) and does it make sense to compare Bode with Nyquist?

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    \$\begingroup\$ Actually, the closed-loop transfer function is $$\frac{C(s)P(s)}{1+C(s)P(s)}$$ \$\endgroup\$
    – Zulu
    Commented Apr 13, 2015 at 18:48
  • \$\begingroup\$ the loop gain is $$-C(s)\cdot P(s)$$ because of the subtraction \$\endgroup\$
    – endolith
    Commented Mar 1, 2016 at 16:25

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When evaluating the stability of a control system, it's most insightful to plot the loop gain, either as a Bode plot or as a Nyquist plot. They do not differ in this respect.

The Nyquist plot is useful for employing the Nyquist stability criterion. In summary, loop gain encirclements of the point \$(-1, 0)\$ on the Nyquist plot indicate instability. Unless the system was already unstable (i.e., has RHP poles), in which case a counter-clockwise encirclement must be made for each RHP pole.

On a Bode plot, the usual technique for evaluating stability is to investigate the gain margin and phase margin of the loop gain. If both of these values are greater than zero, then the system is stable (as long as it doesn't have RHP poles). This technique isn't quite as general as the Nyquist criterion, but for the vast majority of control systems it's good enough. It's possible to evaluate the Nyquist criterion by looking at a Bode plot, but it's more difficult.

So, why would you evaluate stability with a Bode plot when the Nyquist criterion is more general? Because the Bode plot gives you a lot of insight that the Nyquist plot doesn't. The Bode plot shows gain and phase versus frequency, helping you identify what frequencies to place compensating poles and zeroes, as well as lending insights into closed-loop response that are impossible to see on a Nyquist plot (such as closed-loop bandwidth).

Finally, once you've determined your system is stable, you can re-use the Bode plot for a meaningful demonstration of closed-loop response as well. Plotting closed-loop gain on a Nyquist plot isn't nearly as meaningful.

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  • \$\begingroup\$ could you explain why loop gain is important for stability and not the entire closed loop TF? It is because you assume unity feedback and therefore want to avoid the points on bode (gain and phase margins) where C(s)P(s) = -1. \$\endgroup\$
    – user1084113
    Commented Feb 20, 2016 at 1:46
  • \$\begingroup\$ A system is unstable if it has any right-half plane poles. The poles of the fraction \$\frac{N(s)}{D(s)}\$ occur at the zeros of the denominator \$D(s)\$, so for the feedback system considered above whose closed-loop TF is \$T=\frac{C(s)P(s)}{1+C(s)P(s)}\$, the poles occur when \$1+C(s)P(s)=0\$. If all of these zeros occur on the left-half plane of \$s\$, the closed-loop TF will be stable; if any occur on the right-half plane, it will be unstable. Thus, the open-loop TF gives you all the information needed to know whether the closed-loop TF will be stable. This is very useful indeed. \$\endgroup\$
    – Zulu
    Commented Feb 21, 2016 at 23:41

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