# 7.1. Introduction¶

Computing the output $$y[k] = \mathcal{H} \{ x[k] \}$$ of a linear time-invariant (LTI) system is of central importance in digital signal processing. This is often referred to as *filtering* of the input signal $$x[k]$$. We already have discussed the realization of non-recursive filters. This section focuses on the realization of recursive filters.

## 7.1.1. Recursive Filters¶

Linear difference equations with constant coefficients represent linear-time invariant (LTI) systems

$\sum_{n=0}^{N} a_n \; y[k-n] = \sum_{m=0}^{M} b_m \; x[k-m]$

where $$y[k] = \mathcal{H} \{ x[k] \}$$ denotes the response of the system to the input signal $$x[k]$$, $$N$$ the order, $$a_n$$ and $$b_m$$ constant coefficients, respectively. Above equation can be rearranged with respect to the output signal $$y[k]$$ by extracting the first element ($$n=0$$) of the left hand sum

$y[k] = \frac{1}{a_0} \left( \sum_{m=0}^{M} b_m \; x[k-m] - \sum_{n=1}^{N} a_n \; y[k-n] \right)$

It is evident that the output signal $$y[k]$$ at time instant $$k$$ is given as a linear combination of past output samples $$y[k-n]$$ superimposed by a linear combination of the actual $$x[k]$$ and past $$x[k-m]$$ input samples. Hence, the actual output $$y[k]$$ is composed from the two contributions

1. a non-recursive part, and
2. a recursive part where a linear combination of past output samples is fed back.

The impulse response of the system is given as the response of the system to a Dirac impulse at the input $$h[k] = \mathcal{H} \{ \delta[k] \}$$. Using above result and the properties of the discrete Dirac impulse we get

$h[k] = \frac{1}{a_0} \left( b_k - \sum_{n=1}^{N} a_n \; h[k-n] \right)$

Due to the feedback, the impulse response will in general be of infinite length. The impulse response is termed as infinite impulse response (IIR) and the system as recursive system/filter.

## 7.1.2. Transfer Function¶

Applying a $$z$$-transform to the left and right hand side of the difference equation and rearranging terms yields the transfer function $$H(z)$$ of the system

$H(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{m=0}^{M} b_m \; z^{-m}}{\sum_{n=0}^{N} a_n \; z^{-n}}$

The transfer function is given as a rational function in $$z$$. The polynominals of the numerator and denominator can expressed alternatively by their roots as

$H(z) = \frac{b_M}{a_N} \cdot \frac{\prod_{\mu=1}^{P} (z - z_{0\mu})^{m_\mu}}{\prod_{\nu=1}^{Q} (z - z_{\infty\nu})^{n_\nu}}$

where $$z_{0\mu}$$ and $$z_{\infty\nu}$$ denote the $$\mu$$-th zero and $$\nu$$-th pole of degree $$m_\mu$$ and $$n_\nu$$ of $$H(z)$$, respectively. The total number of zeros and poles is denoted by $$P$$ and $$Q$$. Due to the symmetries of the $$z$$-transform, the transfer function of a real-valued system $$h[k] \in \mathbb{R}$$ exhibits complex conjugate symmetry

$H(z) = H^*(z^*)$

Poles and zeros are either real valued or conjugate complex pairs for real-valued systems ($$b_m\in\mathbb{R}$$, $$a_n\in\mathbb{R}$$). For the poles of a causal and stable system $$H(z)$$ the following condition has to hold

$\begin{split}\max_{\nu} | z_{\infty\nu} | < 1\end{split}$

Hence all poles have to be located inside the unit circle $$|z| = 1$$. Amongst others, this implies that $$M \leq N$$.

## 7.1.3. Example¶

The following example shows the pole/zero diagram, the magnitude and phase response, and impulse response of a recursive filter with so called Butterworth lowpass characteristic.

In :

%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.markers import MarkerStyle
from matplotlib.patches import Circle
import scipy.signal as sig

N = 5  # order of recursive filter

def zplane(z, p):

fig = plt.figure(figsize=(5,5))
ax = fig.gca()
plt.hold(True)

color='black', ls='solid', alpha=0.9)
ax.axvline(0, color='0.7')
ax.axhline(0, color='0.7')
plt.axis('equal')
plt.xlim((-2, 2))
plt.ylim((-2, 2))
plt.grid()
plt.title('Poles and Zeros')
plt.xlabel(r'Re{$z$}')
plt.ylabel(r'Im{$z$}')

ax.plot(np.real(z), np.imag(z), 'bo', fillstyle='none', ms = 10)
ax.plot(np.real(p), np.imag(p), 'rx', fillstyle='none', ms = 10)

plt.hold(False)

# coefficients of recursive filter
b, a = sig.butter(N, 0.2, 'low')
# compute transfer function of filter
Om, H = sig.freqz(b, a)
# compute impulse response
k = np.arange(128)
x = np.where(k==0, 1.0, 0)
h = sig.lfilter(b, a, x)

# plot pole/zero-diagram
zplane(np.roots(b), np.roots(a))
# plot magnitude response
plt.figure(figsize=(10, 3))
plt.plot(Om, 20 * np.log10(abs(H)))
plt.xlabel(r'$\Omega$')
plt.ylabel(r'$|H(e^{j \Omega})|$ in dB')
plt.grid()
plt.title('Magnitude response')
# plot phase response
plt.figure(figsize=(10, 3))
plt.plot(Om, np.unwrap(np.angle(H)))
plt.xlabel(r'$\Omega$')
plt.ylabel(r'$\varphi (\Omega)$ in rad')
plt.grid()
plt.title('Phase')
# plot impulse response
plt.figure(figsize=(10, 3))
plt.stem(20*np.log10(np.abs(np.squeeze(h))))
plt.xlabel(r'$k$')
plt.ylabel(r'$|h[k]|$ in dB')
plt.grid()
plt.title('Impulse response');    Exercise

• Does the system have an IIR?
• What happens if you increase the order N of the filter?