Phase margin and stability

Question

"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:

Answer 1

"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: -

• 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.

Answer 2

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:

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\$.

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