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: F = (s-1)/(s+1)(s+2) Step Response

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
    \$\begingroup\$ +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. \$\endgroup\$ – zebonaut Apr 17 '12 at 9:01

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

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