Lesson 15

LTI System Applications

Transfer Functions

The Z-transform properties are particularly useful when you have an LTI system described by an LCCDE.

\begin{displaymath}\sum_{k=0}^{N}a_k y[n-k] = \sum_{k=0}^M b_k x[n-k]\end{displaymath}

\begin{displaymath}\sum_{k=0}^{N}a_k z^{-k}Y(z) = \sum_{k=0}^M b_k z^{-k}X(z)\end{displaymath}

\begin{displaymath}H(z) = \frac{Y(z)}{X(z)}
= \frac{\sum_{k=0}^M b_k z^{-k}}{\sum_{k=0}^{N}a_k z^{-k}}\end{displaymath}

We can use this to determine outputs of LTI systems by multiplying the Z-transform with the input with $H(z)$ to get the Z-transform of the output. Then we can recover the time domain output using the Inverse Z-transform.

\fbox{Ex.} Given a difference equation,

\begin{displaymath}y[n] - .3 y[n-1] = x[n] \end{displaymath}

find the Z-transform of the equation and then find the response $Y(z)$ of the system to an input $x[n] = (.6)^n u[n]$.

What if you wanted to find the response in the time domain?

$\Rightarrow$ We can use Partial Fraction Expansion to invert the Z-transform.

As we saw for Laplace Transforms,

\begin{displaymath}Y(z) = { N(z) \over D(z) } = \sum_{k=1}^N {r_k z \over z-p_k } \end{displaymath}

\begin{displaymath}p_k = {\rm pole} \quad r_k = \mbox{residue} \end{displaymath}

\begin{displaymath}r_k = [\frac{Y(z)}{z} (z - p_k)]\vert _{z=p_k} \end{displaymath}

Then use tables to invert the Z-transform, e.g.

\begin{displaymath}a^n u[n] \leftrightarrow {z \over z-a } \end{displaymath}

\fbox{Ex.} Find Inverse Z-Transform of

\begin{displaymath}X(z) = { 2z^2 - 5z \over (z-2) (z-3) } , \vert z\vert>3\end{displaymath}


\begin{displaymath}{X(z) \over z} = {2z - 5 \over (z-2) (z-3)}\end{displaymath}

\fbox{Ex.} Given $h[n] = a^n u[n]$ ($\vert a\vert < 1$) and $x[n] = u[n]$, find $y[n] = x[n] \ast h[n].$

What if $ x[n] = u[n-2]$?

\fbox{Ex.} Find the output $y[n]$ to an input $x[n] = u[n]$ and an LTI system with impulse response

\begin{displaymath}h[n]= - 3^n u[-n-1]. \end{displaymath}

Another method to invert Z-transforms is the Power Series Expansion. Using

\begin{displaymath}\delta[n-k] \longleftrightarrow z^{-k}\end{displaymath}

X(z) = & \sum_{k=0}^{\infty} x[k] z^{-k} & = x[0] +
...& = x[0]\delta[n] +
x[1]\delta[n-1] + x[2]\delta[n-2] + \cdots

So if you can expand $X(z)$ like this as a series in $z^{-1}$, you can pick off $x[n]$ as the coefficients of the series.

\fbox{Ex.} Find the Inverse Z-Transform of

\fbox{Ex.} Find the Inverse Z-Transform of

\begin{displaymath}X(z) = { z \over z - 1/2 },  \vert z\vert > 1/2 \end{displaymath}

Divide $z- \frac{1}{2}$ into $z$:

\fbox{Ex.} Find the inverse Z-transform of

\fbox{Ex.} Find the inverse Z-transform of

\begin{displaymath}H(z) = {{2 z^2 - \frac{5}{2} z} \over {z^2 - \frac{5}{2} z + 1}},\end{displaymath}

\begin{displaymath}\frac{1}{2} < \vert z\vert < 2.\end{displaymath}

Would a system having this Z-transform be BIBO stable?

\fbox{Ex.} Find the inverse Z-transform of

\begin{displaymath}W(z) = \frac{z^{-4}}{z^2 - 2z - 3}, \vert z\vert > 3\end{displaymath}


As we saw earlier, for BIBO stability of a causal LTI system, all roots of the system characteristic equation lie within the unit circle in the $z$-plane.

This is equivalent to stating that all poles of the transfer function $H(z)$ must lie within the unit circle on the z-plane. We point out that $H(z)$ does not converge at its poles.

Because causal systems have Regions of Convergence that lie outside the largest magnitude pole, an equivalent condition for BIBO stability is that the ROC must contain the unit circle.

\fbox{Ex.} Find the Z-Transform of the unit step $u[n]$. Would an LTI system with $u[n]$ as its system function be BIBO stable?

\fbox{Ex.} Find the Z-transform of $x[n] = (.9)^n u[n].$ Would an LTI system with $x[n]$ as its system function be BIBO stable?


\begin{displaymath}h[n] \ast h_i[n] = \delta [n] \Rightarrow H(z) H_i(z)=1 \end{displaymath}

Ex. Find the inverse system $h_i[n]$ of $h[n]=a^n u[n]. $

Check your results by taking the convolution of $h[n]$ with $h_i[n]$.

\fbox{Ex.} Find the inverse system of $h[n]$ where

\begin{displaymath}H(z) = { z-a \over z-b }. \end{displaymath}

For BIBO stability of both systems (assuming they are both causal), where must all poles and zeros of $H(z)$ lie?

Frequency Response

If we evaluate $H(z)$ at $z=e^{j\Omega}$, i.e., on the unit circle (as long as unit circle is in the ROC), then we get the DTFT $H(\Omega)$, which we call the frequency response. $H(\Omega)$ is periodic with period $2 \pi$. You will discuss the DTFT in Chapter 12.

Ex. Find the magnitude of $H(\Omega)$ for $h[n] = a^n u[n]$, $0 < a < 1. $

\begin{displaymath}\vert H(\Omega)\vert = \sqrt{ H(\Omega) H^\ast(\Omega) } \end{displaymath}