## Problem Consider [[discrete-time LTI system]] with 2 poles in $z_{1}=\frac{1}{4}$ and $z_{2}=-\frac{1}{2}$, 2 zeros at the origin, and unitary static gain. Knowing that it is a [[causal system]], find the [[transfer function]] of the system, and the [[difference equation]] that defines it. ![[ztran-o01s-pole-zero.png|300]] > [!Solution]- > > The system as a [[rational transfer function]] in the form: > $ > H(z)=\frac{P(z)}{Q(z)} > $ > Poles are the zeros of $Q(z)$: > $ > Q(z) = \left( z-\frac{1}{4} \right)\left( z+\frac{1}{2} \right) > $ > If the system has 2 zeros at the origin, the numerator is: > $ > P(z) = K z^{2} > $ > The transfer function is: > $ > H(z) = \frac{K z^{2}}{\left( z-\frac{1}{4} \right)\left( z+\frac{1}{2} \right)} > $ > Unitary static gain means that $H(e^{j0t})=1$, that is: > $ > K = \left(1 -\frac{1}{4} \right)\left(1 +\frac{1}{2} \right) = \frac{9}{8} > $ > Reshaping $H(z)$: > $ > \begin{align} > H(z) &= \frac{9}{8} \frac{1}{\left( 1-\frac{1}{4}z^{-1} \right)\left( 1+\frac{1}{2}z^{-1} \right)} \\ > \frac{Y(z)}{X(z)}&=\frac{\frac{9}{8}}{1+\frac{1}{4}z^{-1}-\frac{1}{8}z^{-2}} > \end{align} > $ > that is > $ > Y(z)+\frac{1}{4}Y(z)z^{-1}-\frac{1}{8}z^{-2}=\frac{9}{8}X(z) > $ > using the [[time shifting property of the z-transform]]: > $ > y(n) + \frac{1}{4}y(n-1)-\frac{1}{8}y(n-2)=\frac{9}{8} x(n) > $ > resulting in the [[difference equation]]: > $ > y(n) = \frac{9}{8}x(n)-\frac{1}{4}y(n-1)+\frac{1}{8}y(n-2) > $ > > > The [[impulse response]] of [[discrete-time LTI system]] can be computed from the [[transfer function]]. > > It can be computed using the `signal.lfilter()` function of the `scipy` Pyhthon package. > > > ```run-python > import numpy as np > from scipy import signal > import matplotlib.pyplot as plt > > def delta(n: np.array) -> np.array: > """Discrete-time unit impulse signal""" > return np.where(n ==0, 1, 0) > > # time axis > n = np.arange(-7,10) > > y = signal.lfilter([9/8], [1, 1/4, -1/8], delta(n)) > > plt.stem(n, y) > plt.grid() > plt.savefig('ztran-o01s.svg') > plt.show() > ``` > > That results in: > ![[ztran-o01s.svg]] > > The [[impulse response]] of the [[discrete-time LTI system]] is the inverse [[z-transform]] of the [[transfer function]]. > > To compute it we can expand $H(z)$ in partial fractions > $ > \begin{align} > H(z) &= \frac{\frac{9}{8}}{1+\frac{1}{4}z^{-1}-\frac{1}{8}z^{-2}} \\ > &= \frac{\frac{3}{8}}{1-\frac{1}{4}z^{-1}} + \frac{\frac{3}{4}}{1+\frac{1}{2}z^{-1}} > \end{align} > $ > > The region of convergence of a causal system is the outside of the circle that includes all the poles, that is, $|z| > \frac{1}{2}$. > > From the list of [[z-transform pairs]] we can select this pair: > $ > a^{n}u(n) \xrightarrow[\cal ZT]{} \frac{1}{1-az^{-1}}, \, |z|>|a| > $ > > Using the linearity of the z-transform, we can compute the impulse response: > $ > h(n) = \frac{3}{8} \left( \frac{1}{4} \right)^{n}u(n) + \frac{3}{4}\left( -\frac{1}{2} \right)^{n} u(n) > $ >