puzzling concept that I'm unable to find a clear
with poles on the jw axis or in the right half plane
the proof. They mealy state poles in RHP cause
simply based on the fact all poles must be in the
Without contradicting Mr. Popelish's reply, I will add or clarify: The RHP poles do not *cause* instability. They represent a way of describing an unstable system. The relationship is tautological.
-- --Larry Brasfield email: donotspam_larry_brasfield@hotmail.com Above views may belong only to me.
The two dimensional S plane maps all combinations of sinusoidal and exponential frequency response. The entire left half plane maps all the decaying responses, purely exponential on the axis, and decaying sinusoids above and below it (the two quarters are symmetrical, because a sine wave going forward in time looks exactly like a sine wave going backward in time.
The right half plane maps all the responses that grow exponentially with time. A pole is a point in that map that has infinite response (it happens with no external forcing signal, but is a natural response of the system, if it is just left alone) at that combination of exponentially increasing sine wave (or pure exponential on the horizontal axis).
-- John Popelish
I'm trying to grasp some concepts in control loop design, and there is one puzzling concept that I'm unable to find a clear definitive explanation for. Why does a systems closed loop transfer function with poles on the jw axis or in the right half plane (RHP) make it unstable (i.e its transient response will never settle) ?
The two text books I'm reading and my web searching haven't actually given me the proof. They mealy state poles in RHP cause instability then go on to explaining the Nyquist stability criteria, which is simply based on the fact all poles must be in the left half plane for stability.
Adam
Take the inverse Laplace transform. If there is a pole in the right half plane, the result of the inverse transform (i.e. what you get in the time domain) will have e^t as a factor. This grows without limit, i.e. it is unstable.
This comes from the transform pair:
Ke^(-at).u(t) K/(s+a)
If there's a complex conjugate pair of poles on the jw axis, the result of the inverse Laplace transform will have a term which is a sinusoid that never dies away. A complex conjugate pair in the left half plane will cause a response that dies away (a damped sinusoid) and a pair in the right half plane will increase in amplitude.
Perhaps you should review your Laplace transform basics. You should learn a table of commonly encounted transform - inverse transform pairs.
Regards, Allan
Thanks for the replies. After getting a better understanding of the inverse Laplace transforms it did show me how the exponent term in the time domain will represent the fate of stability. If the real part of any pole location is equal to or above zero then the exponent term in the time domain function will never converge. I guess the texts I am reading assumed this was intuitive.
Adam
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