Why the stability margin is typically computed by the open loop response or transfer function rather than the closed loop response or transfer function, how could I measure/compute the stability margin thru a closed loop response or transfer function?
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$\begingroup$ possibly related $\endgroup$– AJNCommented Dec 31, 2021 at 6:42
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Well the closed loop transfer function is considered stable if no physically feasible frequency can excite the system in such a way the open loop response becomes -1. This can be simply reasoned from the closed loop equation: $$\frac{GC}{1+GC}$$ if $GC\rightarrow-1$ the closed-loop gain will explode, hence destabilize the plant.
As such, we evaluate closed-loop stability through the open-loop response.
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$\begingroup$ Thanks, but I have to verify the margin by checking the closed loop response on the real system, that is doing the frequency sweeping on the command and compare it with the feedback, how could I tell the margin from the closed loop results? $\endgroup$– LHXCommented Dec 14, 2021 at 1:00
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$\begingroup$ Are you allowed to introduce or vary the gain in the closed loop system? Or are you only allowed to sweep the input over a frequency range? $\endgroup$– AJNCommented Dec 14, 2021 at 12:00
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1$\begingroup$ @AJN I can only do the input sweeping. $\endgroup$– LHXCommented Dec 31, 2021 at 5:37
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Closing the loop inevitably fixes the gain to $1$ or $-1$, depending how you define it, because if the output is connected to the input, the disturbance at the output must be the same as the disturbance at the input. The boundary between stability and instability is a curve in complex gain space (the curve along which the imaginary part of the excitation frequency is zero), most or (more likely) all of which is at gain values other than $\pm 1$. Hence, in order to investigate that boundary, and thereby determine the stability margin, one has to take measurements at gain values that cannot be achieved if the loop is closed; i.e., one has to take measurements with the loop open.