How to determine a system is stable using pole zero analysis?

To my knowledge, as long as the poles of the transfer function are in the left half plane, then the system is stable. It is because the time response can be written as "a*exp(-b*t)" where 'a' and 'b' are positive. Therefore, the system is stable.

However, I saw people stated on websites that "Also no zero is allow in the right half plane". Why?

For a LTI system to be stable, it is sufficient that its transfer function has no poles on the right semi-plane.

Take this example, for instance: F = (s-1)/(s+1)(s+2). It has a zero at s=1, on the right half-plane. Its step response is: As you can see, it is perfectly stable.

The characteristic function of a closed-looped system, on the other hand, cannot have zeros on the right half-plane. The characteristic function of a closed loop system is the denominator of the overall transfer function, and therefore its zeros are the poles of the system. That's why you are mixing things up.

A very important concept, worth mentioning, is closely related with the existence of zeros on the right half-plane, though: minimum and maximum phase systems. I suggest you take a look at the wikipedia article about it.

For open-loop stability, all the poles of the open-loop transfer function G(s)H(s) have to be in the left half-plane.

For closed-loop stability (the one that matters), all the zeros of the transfer function F(s) = 1 + G(s)H(s) have to be in the left half-plane. These zeros are the same as the poles of the transfer function of the closed-loop system (G(s) / (1+G(s)H(s)).

So if you draw the poles and zeros of G(s)H(s) in a graph, the poles have to be in the left half-plane for open-loop stability.

But if you draw the poles and zeros of the closed-loop transfer function (G(s) / (1+G(s)H(S)) then if all the poles are in the left half-plane, the closed-loop system is stable.

But how do you then figure out the closed-loop stability from a G(s)H(s) function? You can either: 1) Find the roots of 1+G(s)H(s)=0 (simple) 2) Use the Routh stability criterion (moderate) 3) Use the Nyquist stability criterion or draw the Nyquist diagram (hard)

In summary, if you have the closed-loop transfer function of a system, only the poles matter for closed-loop stability. But if you have the open-loop transfer function you should find the zeros of the 1+G(s)H(s) transfer function and if they are in the left half-plane, the closed-loop system is stable.

• +1 Great! There are countless application notes about switching converters out there telling you that the RHP zero is bad, without even mentioning that it's bad for a closed-loop system. I wish all of these app'notes had this exact answer as their first paragraph, before diving into the RHP zero stuff over and over again, with no context info. – zebonaut Apr 17 '12 at 9:01

protected by Dave Tweed♦Dec 17 '14 at 2:49

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