IIR filters¶
import numpy as np
import matplotlib.pyplot as plt
import sdr
%config InlineBackend.print_figure_kwargs = {"facecolor" : "w"}
%matplotlib inline
# %matplotlib widget
sdr.plot.use_style()
Create an IIR filter¶
The user creates an IIR filter with the sdr.IIR class by specifying the feedforward coefficients \(b_i\)
and feedback coefficients \(a_j\).
Alternatively, an IIR filter may be created by specifying the zeros and poles in sdr.IIR.ZerosPoles.
Below is an IIR filter with one real zero and two complex-conjugate poles.
zero = 0.6
pole = 0.8 * np.exp(1j * np.pi / 8)
iir = sdr.IIR.ZerosPoles([zero], [pole, pole.conj()])
print(iir)
sdr.IIR:
order: 2
b_taps: (2,) shape
[1.0, -0.6]
a_taps: (3,) shape
[1.0, -1.478207252018059, 0.6400000000000001]
zeros: (1,) shape
[0.6]
poles: (2,) shape
[(0.7391036260090295+0.3061467458920719j), (0.7391036260090295-0.3061467458920719j)]
streaming: False
print(f"Feedforward taps: {iir.b_taps}")
print(f"Feedback taps: {iir.a_taps}")
Feedforward taps: [ 1. -0.6]
Feedback taps: [ 1. -1.47820725 0.64 ]
Examine the impulse response, \(h[n]\)¶
The impulse response of the IIR filter is computed and returned from the sdr.IIR.impulse_response() method.
The impulse response \(h[n]\) is the output of the filter when the input is an impulse \(\delta[n]\).
h = iir.impulse_response(30)
print(h)
[ 1.00000000e+00 8.78207252e-01 6.58172329e-01 4.10862468e-01
1.86109590e-01 1.21565653e-02 -1.01140214e-01 -1.57286400e-01
-1.67772160e-01 -1.47338728e-01 -1.10422993e-01 -6.89312837e-02
-3.12240078e-02 -2.03953322e-03 1.69685122e-02 2.63882791e-02
2.81474977e-02 2.47193366e-02 1.85259041e-02 1.15647504e-02
5.23851924e-03 3.42176895e-04 -2.84684395e-03 -4.42721858e-03
-4.72236648e-03 -4.14721649e-03 -3.10813095e-03 -1.94024315e-03
-8.78877688e-04 -5.74077567e-05]
The impulse response is conveniently plotted using the sdr.plot.impulse_response() function.
plt.figure()
sdr.plot.impulse_response(iir, 30)
plt.show()
Examine the step response, \(s[n]\)¶
The step response of the IIR filter is computed and returned from the sdr.IIR.step_response() method.
The step response \(s[n]\) is the output of the filter when the input is a unit step \(u[n]\).
s = iir.step_response(30)
print(s)
[1. 1.87820725 2.53637958 2.94724205 3.13335164 3.1455082
3.04436799 2.88708159 2.71930943 2.5719707 2.46154771 2.39261642
2.36139242 2.35935288 2.3763214 2.40270968 2.43085717 2.45557651
2.47410241 2.48566716 2.49090568 2.49124786 2.48840102 2.4839738
2.47925143 2.47510421 2.47199608 2.47005584 2.46917696 2.46911955]
The step response is conveniently plotted using the sdr.plot.step_response() function.
plt.figure()
sdr.plot.step_response(iir, 30)
plt.show()
Examine the zeros and poles¶
Zeros are \(z\) values that set the numerator of \(H(z)\) to zero.
print(iir.zeros)
[0.6]
Poles are \(z\) values that set the denominator of \(H(z)\) to zero. The poles define the stability of the IIR filter.
print(iir.poles)
[0.73910363+0.30614675j 0.73910363-0.30614675j]
The zeros and poles are conveniently plotted in the complex plane using the sdr.plot.zeros_poles() function.
plt.figure()
sdr.plot.zeros_poles(iir)
plt.show()
Examine the frequency response, \(H(\omega)\)¶
The frequency response is the transfer function \(H(z)\) evaluated at the complex exponential \(e^{j \omega}\), where \(\omega = 2 \pi f / f_s\).
The two-sided frequency response is conveniently plotted using the sdr.plot.magnitude_response() function.
plt.figure()
sdr.plot.magnitude_response(iir)
plt.show()
The one-sided frequency response, with logarithmic scale, can be plotted using the x_axis="log" keyword argument.
plt.figure()
sdr.plot.magnitude_response(iir, x_axis="log")
plt.show()
Examine the group delay, \(\tau_g(\omega)\)¶
The group delay \(\tau_g(\omega)\) is the time shift of the envelope of a signal passed through the filter as a function of its frequency \(\omega\).
The group delay is conveniently plotted using the sdr.plot.group_delay() function.
plt.figure()
sdr.plot.group_delay(iir)
plt.show()
plt.figure()
sdr.plot.group_delay(iir, x_axis="log")
plt.show()
Fully analyze an IIR filter¶
The user can easily analyze the perform of a given IIR filter using the sdr.plot.filter() function.
Here is an IIR filter with one real zero and 8 complex poles.
zeros = np.array([0.8])
poles = 0.6 * np.exp(1j * np.linspace(np.pi / 8, np.pi / 4, 4, endpoint=False))
poles = np.concatenate((poles, poles.conj()))
iir = sdr.IIR.ZerosPoles(zeros, poles)
print(iir)
sdr.IIR:
order: 8
b_taps: (2,) shape
[1.0, -0.8]
a_taps: (9,) shape
[1.0, -4.092337035029908, 7.71104892744724, -8.684365018955985, 6.37868538208862, -3.1263714068241546, 0.9993519409971622, -0.1909320767063554, 0.016796159999999997]
zeros: (1,) shape
[0.8]
poles: (8,) shape
[(0.46380627201816593+0.3806359704987265j), (0.46380627201816593-0.3806359704987265j), (0.4988817673791569+0.3333421398098428j), (0.4988817673791569-0.3333421398098428j), (0.5543277195052645+0.22961005941814036j), (0.5543277195052645-0.22961005941814036j), (0.5291527586123693+0.28283804209787544j), (0.5291527586123693-0.28283804209787544j)]
streaming: False
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
Poles and digital filter stability¶
Reference:
R. Lyons, Understanding Digital Signal Processing 3rd Edition, Section 6.3.1.
When the pole is real and inside the unit circle, the impulse response \(h[n]\) is an exponential decay.
zeros = []
poles = [0.8]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
When the poles are complex conjugates and inside the unit circle, the impulse response \(h[n]\) is a decaying sinusoid.
zeros = []
pole = 0.8 * np.exp(1j * np.pi / 8)
poles = [pole, pole.conj()]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
When the pole is real and on the unit circle, the impulse response \(h[n]\) is constant. This filter is an integrator.
zeros = []
poles = [1]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
When the poles are complex conjugates and on the unit circle, the impulse response \(h[n]\) is a sinusoid.
zeros = []
pole = 1 * np.exp(1j * np.pi / 8)
poles = [pole, pole.conj()]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
When the pole is real and outside the unit circle, the impulse response \(h[n]\) is an exponential. This filter is unstable.
zeros = []
poles = [1.2]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
When the poles are complex conjugates and outside the unit circle, the impulse response \(h[n]\) is an exponentially-increasing sinusoid. This filter is unstable.
zeros = []
pole = 1.2 * np.exp(1j * np.pi / 8)
poles = [pole, pole.conj()]
iir = sdr.IIR.ZerosPoles(zeros, poles)
plt.figure(figsize=(8, 6))
sdr.plot.filter(iir, n_time=30)
plt.show()
