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Impulse Response, Fourier and Laplace Transforms, and Parseval's Relation

I need help resolving a contradiction relating the area under an impulse response curve of a system and the system's frequency response.

According to Parseval's Relation the area under the impulse response must equal the area under the impulse response's Fourier Transform. Also, this Fourier Transform is the frequency response of the system having that impulse response. This is a consequence of both having to have the same energy. I have thought about this and it makes perfect sense to me. But when looking at Bode plots of a single pole low pass filter I see the area under the frequency response changing with changes in the time constant of the filter. If two such filters having different bandwidths have their outputs normalized to the same DC gain, the area under their impulse responses must also be the same because when the impulse response of each is convolved with a step function the final values they settle on must be the same, the DC gain value.

So what I do not understand is how a Bode Plot of a filter's Laplace Transform can show an area under the frequency reponse curve that obviously changes with the time constant of the exponential decay of their impulse response, yet the area under the Fourier Transform does not change with change with time constants but at the same time the Fourier Transform must also be the frequency response.

Are these two different kinds of bandwidths? I got puzzled by this apparent contradiction when I started thinking about how the time constant of the exponential decay of an impulse response relates to the square root of the area under a frequency response curve and resulting system output noise in response to white input noise.

Reply to
Artist
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Parseval's relation says that the time integral of the _square_ of the impulse response (i.e. the total energy) is equal to the frequency integral of the _square_ of the frequency response* (i.e. the total energy).

You have added to your difficulties by trying to look at the area under the frequency response in a Bode plot, which is a log-log plot, not a linear plot.

  • times a factor of 2 pi that I always have to look up.
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Reply to
Tim Wescott

Beside signal parts being passed through a fourpole, there are also reflected signal parts.

Rene

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Reply to
Rene Tschaggelar

Laplace transform has an atteneuation constant but fourier shows the frequency domain represtation of a signal as such.Fourier transform of a signal should give the exact time domain representation when inverse fourier transform is taken.

Reply to
kkrish

I think the output of a filter fed by a true step function has infinite energy, so you can't really sum it up. You have to use an input, like an impulse, that has finite energy.

{snipped}

-- John

Reply to
John O'Flaherty

I am not trying to calculatet the area under a Bode Plot. I use it only to make the observation the area under it changes.

I had not considered the squaring. I understand it now. The filter with the higher bandwidth will have an impulse response with more of its area higher off the baseline than one with a lower bandwidth. The frequency response curve never gets higher off baseline than its DC gain. So with squaring the result will be higher energy for the filter with higher bandwidth and more area under its frequency response curve.

Thanks

Reply to
Artist

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