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It appears that your equalization for dispersion worked well for the positive frequencies if I read the plots correctly (at least on a error magnitude plot, to evaluate that visually better I recommend plotting in dB). If you only need frequencies below 6000 rad/s, consider decimating your signal to a lower rate as the first step in processing. Decimating to ...

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You're right that the freeverb "allpass" implementation actually isn't a perfect allpass filter. The implemented filter has a transfer function $$H(z)=\frac{(1+\alpha)z^{-N}-1}{1-\alpha z^{-N}}\tag{1}$$ The only case for which $(1)$ actually is a (scaled) allpass filter occurs if $\alpha$ satisfies $$\alpha(1+\alpha)=1\tag{2}$$ i.e., $\alpha=(\sqrt{... 0 A type I (odd length$N$, even symmetry) FIR highpass filter with transfer function $$H_{HP}(z)=\sum_{n=0}^{N-1}h_{HP}[n]z^{-n}\tag{1}$$ has a real-valued amplitude function $$A_{HP}(\omega)=H_{HP}\big(e^{j\omega}\big)e^{j\omega (N-1)/2}\tag{2}$$ If the highpass filter is normalized such that it approximates unity in its passband, it can be transformed into ... 3 Yes, of course. Any delay will look like this where the size of the delay determines the slope of the phase. If the the delay turns out to be an integer number of samples, than this very easy to implement. You can also do fractional delays but that's more work and can only be done approximately. 0 There's no such thing as "rejection of noise"; it's a filter, it doesn't care nor can't know whether what it filters is noise or signal. You can only compare these by making a statement on the spectrum of your noise, and how much of that spectrum makes it through the filter's frequency response. Also, it would very much makes sense to also compare ... 3 For an example analysis, I’ve picked up the low-shelf filter in Robert Bristow-Johnson’s Audio EQ Cookbook. In the book, the transfer function is given as; $$H(s) = A\frac{s^2 + \frac{\sqrt{A}}{Q}s + A}{As^2 + \frac{\sqrt{A}}{Q}s + 1}$$ Since the analysis is going to be done by hand, the asymptotic approximation method of Bode plot analysis can be followed. ... 3 The resonant frequency is related to the "significant frequency" (which is the shelf midpoint frequency) by a factor of$\frac{1}{\sqrt{A}}$for the lowShelf and the reciprocal of that for the highShelf. 0 However, the resonant peak of these filters doesn't seem to be at$f_0$It's not supposed to be. The center frequency you specify is the midpoint between the two parts of the shelf. So for a shelf with a gain of X dB, the center frequency is defined where the gain is X/2. If you increase the Q above$\sqrt{2}\$ you end up with TWO resonances: a peak and a ...

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Your signal has a massive DC bias so your output is dominated by the step response of the band pass filter. It will eventually get there but it's going to take a really long time. Initialize your state with zi = -26040*sosfilt_zi(). See my answer to your other question today. EDIT On second thought: while you can fix some of this in software, you probably ...

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But I don't understand what it does, and how it determines initial conditions from the sos argument and not from the actual signal to filter. From the documentation Compute an initial state zi for the sosfilt function that corresponds to the steady state of the step response. It assumes that the input signal is a unit step. That's useful if you input ...

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You can write your transfer function as the sum of three 2-nd order transfer functions (partial fractions), and then the impulse response will be the sum of the impulse response of each. To help you with the partial fractions you can use this calculator

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That's actually rather simple. A little bit of trial and error will do the trick. The standard specifies upper and lower limit for the transfer function of the filter. So you can just try a normal Butterworth bandpass and see what happens. The picture below shows a 3rd octave design at 1kHz and a sample rate of 48 kHz. A 2nd order bandpass isn't quite a good ...

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