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The graph below shows a linear phase response of FIR Hilbert transformer with a constant diff of -0.06135923. I am not sure if I understand the concept of the phase response correctly, but as the quadrature is the input signal shifted by PI/2, why is the phase response not a straight horizontal line at PI/2? enter image description here

The code to generate the figure was copied directly from Scipy's documentation (the example at the bottom of the page) except that I did not unwrap the phase angle and I used the following filter coefficients:

N = 10 b = [2*np.sin(np.pi*n/2)**2/(np.pi*n) for n in range(1, N+1)] b = np.append(-1*np.flip(b, 0), np.append([0], b))*np.hamming(2*N+1)

I tried the same with different values of N and each time the behavior of the phase response line was the same, just the steepness was varying.

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The slope represents a delay of $N$ samples. You can center the impulse response to time zero by multiplying the transfer function by $e^{i\omega N}$ (transfer function of a negative delay of $N$ samples) or np.exp(1j*w*N) in Python:

from scipy import signal import numpy as np N = 10 b = [2*np.sin(np.pi*n/2)**2/(np.pi*n) for n in range(1, N+1)] b = np.append(-1*np.flip(b, 0), np.append([0], b))*np.hamming(2*N+1) w, h = signal.freqz(b) h = h*np.exp(1j*w*N) import matplotlib.pyplot as plt fig = plt.figure() plt.title('Digital filter frequency response') ax1 = fig.add_subplot(111) plt.plot(w, 20 * np.log10(abs(h)), 'b') plt.ylabel('Amplitude [dB]', color='b') plt.xlabel('Frequency [rad/sample]') ax2 = ax1.twinx() angles = np.angle(h) plt.plot(w, angles, 'g') plt.ylabel('Angle (radians)', color='g') plt.grid() plt.axis('tight') plt.show()

Most of the above is from Scipy documentation which you link to. This gets plotted:


Figure 1. Frequency response of the Hilbert transformer centered at time zero.

The phase goes to zero at frequency zero, because the phase frequency response transitions from $-\pi/2$ for positive frequencies to $\pi/2$ for negative frequencies.

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why is the phase response not a straight horizontal line at PI/2?

It can't be! Because group delay is the derivative of phase over frequency, that'd mean that you get zero group delay (because a constant $\frac\pi2$ doesn't depend on $f$).

Since this is a causal filter, you can't have zero delay. That'd be awesome, but impossible for causal systems.

So, think of this as your constant-phase thing, just shifted in time by half the filter length to become causal. In frequency domain, a time shift becomes a multiplication with a complex oscillation, and that has linear phase!

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You have a linear phase, which is expected for a causal filter: the filter is delayed in the time domain, not centered about $t=0$.

Recall the shift theorem of the Fourier Transform. A shift in the time domain corresponds to modulation by a complex exponential in the frequency domain.

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