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Why Bode plots are on jw axis when determining gain and phase margin? Usually it is explained as 1+G(s)H(s) should not be zero and margins represent how far away from this we are. But for showing this ("distance" to G(s)H(s) = -1, i.e. 0dB, -180 degree phase combination) we use bode plot which are GH values on jw axis. I would rather expect it to be on the whole s plane.

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  • $\begingroup$ Do you mean why is the Bode plot constructed using signals with frequencies along the $j \omega$ axis? $\endgroup$
    – TimWescott
    Jul 22 '19 at 19:40
  • $\begingroup$ Usually it is explained as 1+G(s)H(s) should not be zero and margins represent how far away from this we are. But for showing this ("distance" to G(s)H(s) = -1, i.e. 0dB, -180 degree phase combination) we use bode plot which are GH values on jw axis. I would rather expect it to be on the whole s plane. $\endgroup$
    – Pasha
    Jul 22 '19 at 23:23
  • $\begingroup$ Please edit your question with that comment. $\endgroup$
    – TimWescott
    Jul 22 '19 at 23:25
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The idea behind Bode plot analysis is that you want to know how close the poles of the closed-loop transfer function are to the stability boundary. If a pole is on the stability boundary, then the closed-loop gain would calculate as infinite.

In the case of Laplace-domain analysis, the stability boundary is the $s = j \omega$ line -- any pole that has a positive real part is outside of the stability region, and the system is unstable.

When you do an open-loop Bode plot, you are concerned with how close the open-loop gain gets to -1, because if it were exactly -1 then there would be one or more poles sitting right on the stability boundary. The degree to which the open-loop gain "misses" -1 is a measure of the degree to which the resulting closed-loop poles are far away from the stability boundary.

Note that a drawback of the Bode plot is that you only know how far the closed-loop poles are from the stability boundary -- not whether they are actually stable. For that you need to apply some intuition, or you need to make a Nyquist plot, and know how many unstable zeros the system has.

Note, too, that for a sampled-time system expressed in the z domain you can do a Bode plot by tracing $z = e^{j \omega \theta}$ for $0 \le \theta \le \pi$, because that's the stability boundary in the z domain.

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  • $\begingroup$ Thank you. I was thinking this way too, but then it is very simple approach. Do you think that in real practical situations, if, for example, GH value somewhere in right half s plane gets very near to -1 value the system can go unstable? $\endgroup$
    – Pasha
    Jul 23 '19 at 0:19
  • $\begingroup$ Are you asking what happens if the value of GH that is *computed for a value of s in the right half plane? That's an immaterial question. First, because if you're doing that you may as well figure out if it's exactly -1, and at what value of s. And second, because you want to design with plenty of gain and phase margin, because when you're short on one or the other then your system misbehaves; with lots of time-domain ringing or overshoot, and peaking in the frequency domain. $\endgroup$
    – TimWescott
    Jul 23 '19 at 3:45
  • $\begingroup$ In real engineering I don't think that "exactly" is most of the time relevant. Here I assume that we know the locations of all the poles of the system, and say there is no in the right half plane. I am just saying that if GH for some s value in the right half plane is very close to -1, small errors in the modeling can cause it be a pole. Shouldn't we also take this closeness to -1 as a margin against unstability? $\endgroup$
    – Pasha
    Jul 23 '19 at 11:11

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