I read in a particular book that the fact that gain margin and phase margin are positive implies stability is only true for minimum phase systems. But, for non-minimum phase systems, system can be stable even with negative gain and phase margins

Can you please given an example of a non-minimum phase system, which is stable but has a negative gain margin or phase margin

  • \$\begingroup\$ The first sentence seems to imply that positive GM/PM for NMP systems can happen when the system is unstable, ie. the reverse of what you're asking. Please include the title and author of the "particular book". \$\endgroup\$ – Sven B Oct 30 '18 at 8:14
  • \$\begingroup\$ books.google.co.in/… \$\endgroup\$ – ShiS Oct 30 '18 at 15:09
  • \$\begingroup\$ Under the heading Gain margin of non minimum phase systems \$\endgroup\$ – ShiS Oct 30 '18 at 15:09

The author uses a different definition than I'm used to. According to wikipedia:

Systems that are causal and stable whose inverses are causal and unstable are known as non-minimum-phase systems.

While your book describes it using:

When a transfer function has either a pole or a zero in the right-half s-plane, it is called a nonminimum-phase transfer function.

So your book includes RHP poles, while Wikipedia's definition requires a stable system, so no RHP poles.

If RHP poles are allowed then it is relatively simple to find an example:

$$ F(s) = -\frac{2s\cdot (2s+3)}{(s-1)(s+5)} $$

The bode plot looks like this:


This shows a GM of about -3.67dB. So it technically violates the Bode criterion, but it is stable when closing the loop.

The closed-loop transfer function is:

$$ F_{cl}(s) = \frac{2s\cdot (2s + 3)}{3s^2 + 2s + 5} $$

which has only poles in the LHP. You can kind of see how I constructed it from the Nyquist plot:


In order to have a GM < 0, you want the transfer function to cross (not end/start at) the x-axis (real axis) left of \$s=-1\$. Given that I had a pole in the RHP, I want the curve to loop around \$-1\$ CCW once to make sure it is stable (\$Z = N + P, P = 1 \Rightarrow N = -1\$). So I just shifted and scaled the graph until \$s=-1\$ was in the right part of the \$\infty\$-figure.

This part has been added after the question in the comments

Is it possible to find such an example for stable systems?

Yes it is possible. Take for example the following stable system:

$$ F(s) = \frac{1}{150}\frac{(s + 10)(s^2 + 6s + 4)(s^2 - 0.1s + 2)}{(s+1)^2} $$

It has 2 zero's in the RHP so it is a nonminimum phase system. It does not contain poles in the RHP so it is also stable.

The bode plot looks like this:


But the closed loop response is stable. The Nyquist diagram shows you how that is possible.


The two low-frequency LHP poles and LHP zero will make the curve start downwards. Then we can use multiple zero's in left and right half planes to make the curve go move up and down. I made sure to use two RHP zero's and not one to avoid that the curve would start on the left instead of on the right (I wanted to start at \$0^\circ\$ phase for this example).

  • \$\begingroup\$ Is it possible to find such an example for stable systems? \$\endgroup\$ – ShiS Oct 31 '18 at 7:23
  • 1
    \$\begingroup\$ Check my edit for one that is stable. \$\endgroup\$ – Sven B Oct 31 '18 at 15:06

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