How to prove final value theorem?

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Final value theorem

Hello all
the final value theorem is defined by



the final value theorem is valid only if F(S) has poles only in the left half-plane and ,at most ,one pole at the origin. If F(S) has poles in the right half-plane or poles on the imaginary axis other than at the origin , the final value theorem is invalid~~~~~ how to prove this and why

thanks a lot
 

Re: Final value theorem

I am not sure but as far as i know, if F(S) has poles in the right half-plane or poles on the imaginary axis then the system described by F(s) is unstable(oscillating) thus there is no final(steady state value) and therefore the final value theorem is invalid.
 

Final value theorem

a Systen having poles in the right side is simply unstable. it does not have a settling value as u say in time domain or in Laplace it has no steady state value.or simply put

F(t)=(limt-->inf)f(t)
 

Re: Final value theorem

Perhaps you have forgotten the mathermatical equation of s=jw.
We know s->0 in s-domain is the same as saying t->∞ in time domain.
We know the function of a system is stable to infinite time.
Therefore if s is not zero, then jw is a non-zero.
To examine this, any pole given by (s-x) is represented in Z-Transform of Z=exp(jw). Now, if jw is a non-zero, then Z is not a one. Given a unit circle, for a stable system, Z must lie within the unit circle. If Z=1, it is likewise saying pole is placed along the Y-axis or more correctly called the Imaginary Axis of the Argand Diagram of a Root Locus Plot. Poles in the right-hand plane of the Argand Diagram is the same as having Z>1 outside the unit circle.
 

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