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"In control systems, an open loop transfer function having poles in left half plane, should have positive phase margin and positive gain margin to ensure closed loop stability."

I was trying to interpret this statement. I was able to figure out why is the statement correct for gsin margin. I did so by relating root locus and Nyquist plots as follows:

For root locus with open loop poles on left half plane, we get a stable closed loop system untill the locus crosses jw axis. This may happen for larger values of gain, K. So the system is stable for smaller values of gain K. This reflects that the Nyquist plot must have smaller K, and hence would pass the negative real axis at some place greater than -1 (analogous to jw axis.) This would ensure a positive phase margin.

I made many failed attempts to prove similarly for the positive phase margin, too. But I could not figure out why the phase margin is positive for a stable closed loop system (for a open loop function with left hand poles.)

EDIT:

enter image description here

enter image description here

enter image description here

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  • \$\begingroup\$ Where did the starting quote come from? \$\endgroup\$ – Andy aka Jan 14 '20 at 11:23
  • \$\begingroup\$ There were many sources stating similar fact, many pdf online, a video lecture by IIT Madras Professor. I am editing the question to add those \$\endgroup\$ – Bhuvnesh Jan 14 '20 at 14:32
  • \$\begingroup\$ The key words at the top of your page are "having poles in Right half plane". I didn't see that mentioned in the adds you put in. \$\endgroup\$ – Andy aka Jan 14 '20 at 14:47
  • \$\begingroup\$ My bad! I meant "poles in left half plane". I have edited the question. Second picture shows this fact written there \$\endgroup\$ – Bhuvnesh Jan 14 '20 at 15:57
  • \$\begingroup\$ 'Phase margin equal to the gain margin', etc. makes no sense. They are different measures, and can't be compared. \$\endgroup\$ – Chu Jan 14 '20 at 23:12
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"In Control Systems, an open loop Transfer Function having poles in Left half plane, should have positive phase margin and positive gain margin to ensure closed loop stability."

I would point you to this site where I took the picture below: -

enter image description here

  • A positive gain margin means that the gain magnitude has dropped below unity as the phase angle of the output reaches the point where it is fully inverted (the oscillatory point). This means it is stable because there cannot be enough gain to produce oscillation when used in a negative feedback control system.

  • A positive phase margin means the the phase angle of the output has not yet reached full inversion (the oscillatory point) as the gain magnitude drops to unity. This also means it is stable when applied within a negative feedback control system.

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  • \$\begingroup\$ Only true for minimum phase systems. \$\endgroup\$ – user110971 Jan 14 '20 at 19:45
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Your sources are wrong. This line of argument only work for minimum phase systems. A minimum phase system is a system such that both said system and its inverse are stable and causal.

Let me give you a counter example. Consider the open loop transfer function.

$$L = \frac{10 (s + 1)^2}{s^3}.$$

Here is the bode plot:

bode

Clearly the system should be unstable when the loop is closed, right? Well, not quite. The closed loop transfer function is

$$H = \frac{L}{L + 1} = \frac{10(s+1)^2}{s^3 + 10(s+1)^2}.$$

It can be seen that all the poles are stable from the plot of the denominator of the closed loop transfer function \$H\$.

poles

The system is stable, as can be seen from the step response of \$H\$.

step

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  • \$\begingroup\$ What is a 'minimum phase system' and what is a system that isn't called? On the graph showing 'poles' what do the x and y axes represent? In the transfer functions, what are 'H', 'L', and 's'? \$\endgroup\$ – Bruce Abbott Jan 15 '20 at 8:28
  • \$\begingroup\$ A minimum phase system is a system such that both said system and its inverse are causal and stable. Here we are talking about the open loop transfer function. I don’t think a system that is not a minimum phase has a particular name. The plot of the poles is simply the plot of the denominator of the closed loop transfer function H. The x-axis is s. L is the open loop transfer function and H is the closed loop transfer function, which you obtain by applying negative feedback with unity gain. s is the Laplace variable. \$\endgroup\$ – user110971 Jan 15 '20 at 9:14
  • \$\begingroup\$ @BruceAbbott Sorry, forgot to tag you. For more information on minimum phase systems, have a look at the wiki page. \$\endgroup\$ – user110971 Jan 15 '20 at 9:25
  • \$\begingroup\$ Well, as far as I know, a minimum phase system is indeed the system for which poles and zeros are on Left half comPlex plane. So I think the statement I made should be correct! \$\endgroup\$ – Bhuvnesh Jan 15 '20 at 13:18
  • \$\begingroup\$ @Bhuvnesh That’s only the stability half of the definition. If the denominator has a higher order than the numerator, then the inverse is not causal. Hence the statement is incorrect. Besides your original quote does not say anything about the zeros. \$\endgroup\$ – user110971 Jan 15 '20 at 13:27

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