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The first statement is not true, it is true for convolution1, since the hilbert transform itself is a convolution $H(x) = h(t) * x(t)$ from the convolution properties $H(x*y) = x*H(y) = y*H(x)$ The second is correct 2, (the imaginary case is follows by the fact that the product of an analytic function and a constant is analytic) The meaning of the third is ...


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It looks like artifacts due to the derivative. I used this code in Octave: fs=300; t=[0:1/fs:2]; c=chirp(t, 20, 2, 100); s=c.*(1+0.5*cos(2*pi*10*t)); h=hilbert(s); m=abs(h); a=diff(unwrap(arg(h)))/2/pi*fs; plot(m,"",a) and this is what comes out: I also tested the equivalent of this in LTspice and, without any form of unwrapping, this is the ...


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I computed and compared the minimum phase HRIR and the original one. This is my final code: def min_phase_coversion(HRIR): ''' :param HRIR: the desired HRIR impulse response to convert into minimum phase :return: the minimum phase version of the original HRIR ''' HRIR_fft = fft(HRIR,44100) #computing magnitude, tested with sinusoid,...


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