Re: Low-pass Step Response

From: Gregory L. Hansen (glhansen_at_steel.ucs.indiana.edu)
Date: 03/11/05 

Date: Fri, 11 Mar 2005 01:22:50 +0000 (UTC)

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)

I shall do that, sir. I don't know whether it's the books I've been
reading or my own predilections, but I always think of the multiplicative
constants as sort of boring. It just multiplies straight through, so
focus on the part that actually transforms. 

-- 
"The result of this experiment was inconclusive, so we had to use 
statistics."  (Overheard at international physics conference)

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