How to prove final value theorem?

[v9260019]

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

 

[MSSN]

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.

 

[bangash]

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)

 

[SkyHigh]

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.