why does a filter change the amplitude of a signal ?

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pavan sunder
pavan sunder 2016-2-2
编辑： Star Strider 2023-12-31
when i filtered a signal with a cutoff equal to the desired range. i got the signal of desired frequency but the amplitude was altered. can anybody explain what is the reason ?
i assume since we are giving the signal ( represented as combination of complex exponential) as input to the filter the output is scaled as we know that complex exponentials are eigen functions.. Is this justification correct ?
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Joshua Diamond
Joshua Diamond 2023-12-27
编辑：Joshua Diamond 2023-12-27
For me, filtering is actually increasing the amplitude. Can anyone help explain this, and whether/why it's expected behavior?
Here, is 88.
Paul
Paul 2023-12-27
Ran in:
ts = zeros(1,10000); ts(5000) = 1;
w = gausswin(250,2.5);
figure
plot(filtfilt(w,1,ts))
ts is a (shifted) unit pulse. Therefore, the output of filtfilt will essntially (not exaclty) be the convolution of w with itself shifted to the time of the pulse.
hold on
plot((0:498)+4750,conv(w,w),'o')
xlim([4500 5500])

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回答（1 个）

Star Strider
Star Strider 2016-2-2
编辑：Star Strider 2023-12-31
It depends on the filter design you choose. A filter with a more gradual rolloff (such as a Butterworth design) could attenuate your signal if the rolloff was too gradual. Increasing the filter order could decrease or eliminate the attenuation, however this would also lengthen the filter. A Chebyshev design with a sharper cutoff could also minimise the passband attenuation.
EDIT — (31 Dec 2023 at 11:38)
Filtering a signal removes energy from it, so the amplitude of a filtered signal in the time-domain would be expected to be less than the amplitude of the unfiltered (original) signal. The frequency domain magnitude in the passband should remain the same (except in the transition region).
Filters are generally designed to have unity gain in the passband, however this is not an absolute requirement. A filter can amplify or attenuate in the passband, depending on how it is designed.

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