Re: Low-pass Step Response

From: richard miller (richard_at_microscitech.freeserve.co.uk)
Date: 03/11/05 

Date: Fri, 11 Mar 2005 04:51:08 -0000

"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
news:d0qrta$3to$1@rainier.uits.indiana.edu...
> In article <d0qebq$7he$1@newsg2.svr.pol.co.uk>,
> richard miller <richard@microscitech.freeserve.co.uk> wrote:
> >
> >"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
> >news:d0qasv$u1i$1@rainier.uits.indiana.edu...
>
> >>
> >> Somebody help me with a reality check. I tried to find the step
response
> >> of a low pass filter with inverse time constant a. The transfer
function
> >> would be
> >>
> >> 1/(s+a)
> >>
> >> and the unit step response is
> >>
> >> 1/s(s+a)
> >>
> >> Expand in partial fractions giving
> >>
> >> (1/a)/s - (1/a)/(s+a)
> >>
> >> and invert the Laplace transform to find
> >>
> >> 1/a - (1/a) exp(-at) = (1/a)(1 - exp(-at))
> >>
> >> And that's just what I'd expect, except for the multiplying factor.
The
> >> behavior as t->inf should be independent of a.
> >>
> >> What's going on there?
> ...
> >
> >I think you should look at how you got your constant a.
> >
> >A Good example
> >
> >r.dq/dt + q/c = v(t)
>

etc.

> Q.(s + k ) = v / r
>
> where k=rc, time constant
>
> Q = 1/r . 1/(s + k )

etc

It would be a good example if I had put k=1/rc, not k=rc. The former is
correct, not what I wrote

Richard Miller