Stability from laplace transform analysis
in [DSP]
|
Prev: Map controller output to PWM duty-cycle
Next: Auto-correlation of a circularly shifted sequence in frequency domain
From: dhaval_shah on 13 Jan 2010 07:52 All, Can any one pls explain me how "sigma" in s in laplace transform > 0 means un-stable system and < 0 means stable system? My understanding is completely reverse to that. BR, Dhaval
From: Rune Allnor on 13 Jan 2010 08:03
On 13 Jan, 13:52, "dhaval_shah" <shah.dhavals...(a)gmail.com> wrote: > All, > > Can any one pls explain me how "sigma" in s in laplace transform > 0 means > un-stable system and < 0 means stable system? > > My understanding is completely reverse to that. The Laplace Transform contains an exponential term exp(+/-s) where s = sigma + jw. Depending on the choise of sign in that exponent, the sigma term corresponds to a time domain term that either grows or decays exponentially with increasing time. The *conventional* *choise* of signs is such that sigma < 0 corresponds to an exponentially decaying term in time domain. Rune |