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All the example of bode plots I see in textbooks are of stable systems without delay. I wanted to know if bode plots can be drawn for unstable systems with time delay. Using the usual method to draw the plots would yield same result for the magnitude plot if the pole or zero were on either side of the y-axes and only the phase plot would probably change. Is this approach correct or is there something that I am missing?

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You are absolutely right. Since the magnitude is the absolute value of the complex function $F(j\omega)$ the plot does not change with the sign of the pole or zero. The Phase on the other hand does change with the sign of the pole making the system unstable. You can see this very good at this simple example (made with WolfamAlpha)

Stable system with pole at s = -1

This is the stable system. The Phase changes from 0° to -90° as expected since the pole is negative.

Unstable system with pole at s = 1

This is the unstable system. The magnitude stays the same, just as you predicted. But the phase on the other hand starts at -180° and rises to -90°.

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The magnitudes are the same but the phase changes:

tfms = {1/(s + 1), 1/(s - 1), E^-s/(s + 1), E^-s/(s - 1)};

BodePlot[tfms, PlotLegends -> "Expressions"]

enter image description here

Here's another perspective obtained by plotting them in the complex plane:

p = NyquistPlot[tfms, {0, 15}, PlotLegends -> "Expressions", AspectRatio -> 1]

enter image description here

And here's an animation showing how the vectors have the same magnitude but varying phases.

enter image description here

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