Unstable System: Pole in Right Half - Why?

[ranju]

I have read that if pole of a function or say , a system lies in right half of a s-plane , then the system is unstable..! But I did'nt get the logic behind it..! What's the reasn of system being unstable if pole is lying in right half..?? Please elaborate...!

 

[anorlunda]

Because it will have positive feedback. Any small oscillation will grow in size without limit.

 

[LvW]

To understand the meaning of the pole location in the s-plane one should know about the following relation between time and frequency domain:

The denominator D(s) of a systems transfer function H(s) always is identical to the characteristic polynominal P(s) of the differential equation in the time domain.
To find the time domain solution we have to calculate the roots of the characteristic equation P(s)=0.
Therefore, the roots (zeros) of the characteristic equation are identical to denominator`s zeros - equivalent to the poles of H(s).

Now - for a system to be stable we require that the real part σ of the time domain solution [exp(σT)] is negative (decaying amplitude).
That means: Also the poles of the transfer function H(s) must have a negative sign (must be in the left half of the s.plane).

 

[jim hardy]

Look at the denominator of your transfer function while remembering that right half of plane is positive frequency.
If denominator has a term that goes to zero at some positive frequency, well, that's division by zero at that frequency
and since transfer function is output/input
a denominator of zero means it can have an output with zero input
and that's an oscillator.

So any time you get a quadratic denominator, watch out. There's a potential for oscillation.

It's been fifty years now since my controls course. Doubtless some younger member can phrase it better for you , and in today's terminology.

EDIT: oops i see Lvw already did !

old jim