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I'm relatively new to control systems.

I got to know how critically damped system looks like in Nyquist plot (same gain and phase cross over frequencies. PS. correct me if I'm wrong) and Bode plot thanks to the Octave online simulator, but I'm really curious to know what a Nyquist plot for a undamped system looks like if it exists. If it doesn't exist, why not?

I tried to plotting plots for

num = [1];
den = [1,0,5];
s = tf(num,den);
bode(s)  # exists
nyquist(s)  # throws error
polar(s) # throws error

PS. I know the system would be unstable, just curious.

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  • \$\begingroup\$ Your title mentions "undamped damped" systems which seems like a contradiction, whilst your post body mentions that you know how a Nyquist plot for a critically damped system looks, but are curious to know what a nyquist plot for a critically damped system looks like. This is a bit confusing and (to the best of my understanding) self-contradictory. Can you double-check to make sure you specified the right types of systems in each case? \$\endgroup\$
    – nanofarad
    Sep 21 '20 at 20:23
  • \$\begingroup\$ @nanofarad edited the question \$\endgroup\$
    – yogesh
    Sep 21 '20 at 20:36
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The Nyquist plot of a stable 2nd order TF looks like an 8-shaped balloon.

As the TF approaches "undamped", the balloon grows and grows. There is no development of a sharp spike / pole as in a Bode plot.

The Nyquist plot of an undamped system has an infinite size balloon.

Here's an example of 3 systems with decreasing damping. (Ignore the H(s) expression at the top, it's the Matlab example. The TF vectors are what I entered based on your question)

enter image description here

In terms of errors, I think it depends how the engine handles the infinite magnitude and/or infinite re/im numbers. Matlab handles it quietly, Octave spits out errors. Both plot something resembling a horizontal line with the balloon curving back somewhere infinitely far away and out of screen.

Have a look here: https://www.mathworks.com/help/control/ref/nyquist.html

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