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The frequency response of the corresponding filters must satisfy $H^2(\omega)=H(\omega)\tag{1}$ This was already pointed out in Hilmar's answer, but the conclusion is not that the only option is $H(\omega)=1$ for all $\omega$. The correct conclusion from $(1)$ is that the frequency response $H(\omega)$ must equal either $0$ or $1$ for any frequency. One ...


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In what ring of convolutions the given kernel is idempotent? Certainly not for the regular convolution kernel = np.array([[-1, -1, -1], [-1, 8, -1], [-1, -1, -1]]) scipy.signal.fftconvolve(kernel, kernel, mode='full') array([[ 1., 2., 3., 2., 1.], [ 2., -14., -12., -14., 2.], [ 3., -12., 72.,...


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Of course it should be $n\ge 2$, that's a typo in your edition. The sign of the formula in your question is wrong. It should be $$h[0]h_I[n]=-\sum_{k=1}^nh[k]h_I[n-k],\qquad n>0\tag{1}$$ With $h[0]=1$, $h[1]=-\alpha$, and $h[n]=0$ for $n>1$, Eq. $(1)$ simplifies to $$h_I[n]=-h[1]h_I[n-1]=\alpha h_I[n-k],\qquad n>0\tag{2}$$ And since $h_I[0]=1/h[0]=1$...


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(Poster and I have iterated on this in a chat here: https://chat.stackexchange.com/rooms/131792/isi ) The OP mentions that the filter should not impact the signal since the signal is within the passband- this would only be true if this was a linear phase filter, otherwise we must also give consideration to the phase response and distortion that can cause. ...


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In general, you can't: time delay would be calculated through group delay, and in general, that's different for different frequencies. Your input doesn't contain only a single frequency, so it will come out dispersed. There's an exception to that rule: Linear-phase filters have constant group delay over all frequencies. That's because the group delay is the ...


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They Don't. To address this question, linear interpolation is NOT a linear phase filter except for when we are interpolating to half a sample. The necessity for a linear phase filter is that the filter coefficients be complex conjugate symmetric or anti-symmetric. Linear interpolation as an FIR filter is done with two coefficients $a$ and $b$ such that: $$y[...


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Sometimes Stackexchange coughs up these ancient unanswered questions. Since I've started, I'll answer in spite of the age: iirdesign expects you to give it frequencies in the range 0..1, normalized to the Nyquist frequency (the latest version looks like it'll let you tell it the sampling rate for your own convenience). You can expect that any filter design ...


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