We can summarize the Z-Transform as follows:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) = \sum \limits_{n=-\infty}^{\infty} x(n) z^{-n} \\
RoC(x) &= \{ z \in \C \st \sum \limits_{n=-\infty}^{\infty} \abs{x(n) z^{-n}} \lt \infty \}
\end{align*}
\end{nicebox}

Since the convention used is $z = Re^{i\omega}$, we can also write the region of convergence as

\begin{nicebox}
\begin{align*}
RoC(x) &= \{ z \in \C \st \sum \limits_{n=-\infty}^{\infty} \abs{x(n) R^{-n}} \lt \infty \}
\end{align*}
\end{nicebox}

This is because only $R \bydef \abs{z}$ is important in defining the region of convergence. The math is also very clear:

\begin{align*}
\abs{x(n)z^{-n}} = \abs{x(n)R^{-n}e^{-i\omega n}} = \abs{x(n)R^{-n}}
\end{align*}

\begin{example}

$x(n) = \alpha^n u(n)$ is causal. Determine the Z-Transform.

\begin{align*}
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} x(n) z^{-n} \\
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) z^{-n} \\
\hat{X}(z) &= \sum \limits_{n= 0}^{\infty} \lr{\alpha z^{-1}}^n \\
\hat{X}(z) &= \frac{1}{1 - \alpha z^{-1} } \quad \mbox{ if } \abs{\alpha z^{-1}} \lt 1 \\
\end{align*}

We find the the Z-Transform is $\frac{1}{1 - \alpha z^{-1} }$ with a radius of convergence $\abs{\alpha} \lt \abs{z}$.
%\img{images/roc/roc/causal.ps}

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\end{example}

\begin{example}

Given $y(n) = -\alpha^n u(-n-1)$, determine the Z-Transform.

\begin{align*}
\hat{Y}(z) &= \sum \limits_{n=-\infty}^{\infty} y(n) z^{-n} \\
\hat{Y}(z) &= \sum \limits_{n=-\infty}^{\infty} -\alpha^n u(-n-1) z^{-n} \\
\hat{Y}(z) &= -\sum \limits_{n=-\infty}^{-1} \lr{\alpha z^{-1}}^n \\
\hat{Y}(z) &= -\sum \limits_{n= 1}^{\infty} \lr{\alpha^{-1} z}^n \quad \mbox{ let } n = -n \\
\hat{Y}(z) &= -\alpha^{-1} z \sum \limits_{n= 0}^{\infty} \lr{\alpha^{-1} z}^n  \\
\hat{Y}(z) &= -\frac{\alpha^{-1} z}{1 - \alpha^{-1} z} \quad \mbox{ if } \abs{\alpha^{-1} z} \lt 1 \\
\hat{Y}(z) &= \frac{1}{1 - \alpha z^{-1}} \\
\end{align*}

We find the the Z-Transform is $\frac{1}{1 - \alpha z^{-1} }$ with a radius of convergence $\abs{\alpha} \gt \abs{z}$.

%\img{images/roc/roc/anticausal.ps}

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\end{example}


\begin{claim}

If $x$ is causal ($x(n) = 0 \mbox{ } \forall n  \lt0$), and $R \in RoC(x)$, then $\tilde{R} \gt R \in Roc(x)$. In other words, causal signals have $RoC$'s that extend from outside a circle all of the way to infinity. 

\end{claim}

\begin{proof}

First determine a relationship between $\tilde{R}^{-n}$ and $R^{-n}$:
\begin{align*}
\tilde{R} &\gt R \\
\tilde{R}^{-1} &\lt R^{-1} \\
\tilde{R}^{-n} &\lt R^{-n} \\
\end{align*}

We need to consider the absolute summability of the tranfer function. $\sum \limits_{n=-\infty}^{\infty} \abs{x(n) z^{-n}} = \sum \limits_{n=0}^{\infty} \abs{x(n) R^{-n}}$

If sum converges absolutely for $R$, then certainly when $R^{-n}$ is replaced with a smaller term $\tilde{R}^{-n}$, the sum will still converge. Therefore, if $R \in RoC(x)$, then $\tilde{R} \in Roc(x)$.

\end{proof}

\begin{example}

Given our signal from a few examples ago, $x(n) = \alpha^n u(n)$, find the Z-transform of $g(n) = x(n+1) = \alpha^{n+1} u(n+1)$. This example brings out one of the properties of the Z-transform which you will use frequently, which is the shifting property. Lets solve this for an arbitrary shift.

\begin{align*}
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} g(n) z^{-n} \\
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} x(n-N) z^{-n} \\
\hat{G}(z) &= \sum \limits_{n=-\infty}^{\infty} x(m) z^{-\lr{m+N}} \quad \mbox{ let } m = n-N \\
\hat{G}(z) &= z^{-N}\sum \limits_{n=-\infty}^{\infty} x(m) z^-m = z^{-N}\hat{X}(z)\\
\hat{G}(z) &= z^{-N}\hat{X}(z)\\
\end{align*}

This defines the shifting property for the Z-transform:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) \\
x(n-N) &\ztp z^{-N}\hat{X}(z) \\
\end{align*}
\end{nicebox}

In our case we have $N = -1$, which means that $\hat{G}(z) = z\hat{X}(z) = \frac{z}{1-\alpha z^{-1}}$, where $\abs{\alpha z^{-1}} \lt 1$. What happens at infinity? Lets write out $\hat{G}(z)$ term by term:

$$
\hat{G}(z) = z + \alpha + \alpha^2 z^{-1} + \alpha^3 z^{-2} + \cdots 
$$

Notice that as $z \to \infty$, it blows up. This demonstrates how right-sided signals that are not causal have $RoC$s of the form $R \lt \abs{z} \lt \infty$. An analogous statement can be made about left-sided signals that are not anti-causal, they are of the form $0 \lt \abs{z} \lt R$. The first term that blows up these types of signals is $x(1)z^{-1}$.

\end{example}

Region of Convergence

We have discussed right- and left-sided signals, but what about two-sided signals? Consider the figure of an annular region of convergence of a Z-transform:

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The causal part of the signal is represented by the region outside of $R_1$, and the anti-causal part of the signal corresponds to the region inside $R_2$. The intersection is the region of convergence for the transform. If, for a given signal, the anti-causal and causal regions do not overlap, the Z-transform is not defined.

\begin{example}

Find the Z-transform for $f(n) = \lr{\frac{1}{2}}^n u(n) - \lr{\frac{3}{4}}^n u(-n-1)$.

\begin{align*}
\hat{F}(z) &= \frac{1}{1 - \frac{1}{2}z^{-1}} + \frac{1}{1-\frac{3}{4}z^{-1}} \quad \abs{z} \lt \frac{3}{4} \mbox{ and } \abs{z} \gt \frac{1}{2} \\
\hat{F}(z) &= \frac{z}{z - \frac{1}{2}} + \frac{z}{z-\frac{3}{4}} \\
\hat{F}(z) &= \frac{2z\lr{z - \frac{5}{8}}}{\lr{z-\frac{1}{2}}\lr{z-\frac{3}{4}}} \\ 
\end{align*}

%\img{images/roc/examples/twosided/twosided.ps}

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The poles are where the denominator is zero, $z = \frac{1}{2}$, and $z = \frac{3}{4}$. The zeros come from the numerator, $z = 0$, and $z=\frac{5}{8}$. Below is the pole-zero diagram for $\hat{F}(z)$.

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The region of convergence is bounded by the poles. The zeros play no role. However, there can be pole/zero cancellations. Recall that the Z-transform can only exist when there is a non-trivial overlap for the region of convergence. For example, the signal $q(n) = \lr{\frac{3}{4}}^n u(n) - \lr{\frac{1}{2}}^n u(-n-1)$ has no overlap for the region of convergence, hence its Z-transform doesn't exist.

\end{example}

Two-sided signals

Lets look at the poles and zeros of a finite duration signal $v(n) = \delta(n-N)$.

\begin{align*}
\hat{V}(z) &= \sum \limits_{n=-\infty}^{\infty} \delta(n-N) z^{-n} \\
\hat{V}(z) &= z^{-N}
\end{align*}

Lets consider the cases for different values of $N$:



I) $N \gt 0$

$$
\hat{V}(z) = z^{-N} = \frac{1}{z^N}
$$

There are $N$ poles at zero, hence, the region of convergence must exclude $0$.


II) $N = 0$

$$
\hat{V}(z) = 1
$$

The $RoC$ includes all $z$ including $0$ and $\infty$.


III) $N \lt 0$

$$
\hat{V}(z) = z^{-N} = z^M \mbox { where } M \gt 0
$$


We have $M$ poles at $\infty$ which is excluded from the $RoC$.

\begin{example}

Find the poles for the Z-transform of the signal $r(n) = \delta(n-1) + 2\delta(n-2) + 3\delta(n-3)$.

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\begin{align*}
\hat{R}(z) &= \sum \limits_{n = -\infty}^{\infty} r(n) z^{-n} \\
\hat{R}(z) &= \sum \limits_{n = -\infty}^{\infty}\lr{\delta(n-1) + 2\delta(n-2) + 3\delta(n-3)} z^{-n} \\
\hat{R}(z) &= z^{-1} + 2z^{-2} + 3z^{-3} \\
\hat{R}(z) &= \frac{z^2 + 2z + 3}{z^3} \\
\end{align*}

We have a triple pole at 0. $RoC(v): 0 \lt \abs{z}$.

\end{example}

\begin{example}

Find the poles for the anti-casual signal $g(n)$ represented in the figure below.

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\begin{align*}
\hat{G}(z) &= \sum \limits_{n = -\infty}^{\infty} g(n) z^{-n} \\
\hat{G}(z) &= 1 + 2z + 3z^2 \\
\end{align*}

The poles are at infinity, $RoC(g): \abs{z} \lt \infty$.
\end{example}

\begin{example}

Find the poles for the two-sided signal $h(n)$ represented in the figure below.

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\begin{align*}
\hat{H}(z) &= \sum \limits_{n = -\infty}^{\infty} h(n) z^{-n} \\
\hat{H}(z) &= z^{-1} + 2 + 3z \\
\hat{H}(z) &= \frac{1 + 2z + 3z^2}{z} \\
\end{align*}

Poles are at 0 and $\infty$, $RoC(h): 0 \lt \abs{z} \lt \infty$.

\end{example}

Poles and Zeros

We can write $\hat{R}(z)$ as a rational function of a polynomial $P(z)$ and $Q(z)$:

$$
\hat{R}(z) = \frac{P(z)}{Q(z)} = \frac{b_M z^M + b_{M-1}z^{M-1} + \cdots + b_0}{a_N z^N + a_{N-1} z^{N-1} + \cdots + a_0}
$$

For rational Z-transforms, we say that

\begin{enumerate}
\item $\hat{R}$ is strictly proper is $M \lt N$ (the power in the numerator is less than the power in the denominator).
\item $\hat{R}$ is proper if $M \leq N$.
\item $\hat{R}$ is improper if $M \gt  N$.
\item The roots of $Q(z)$ are the poles.
\item The roots of $P(z)$ are the zeros.
\item If $M \lt N \Rightarrow M - N $ poles at $\infty$.
\item If $N \lt M \Rightarrow N - M $ zeros at $\infty$.
\item The number of poles is equal to the number of zeros if we account for what happens at infinity.
\end{enumerate}

In general, the set of signals that produce rational Z-transforms are obtained from, and only from, linear combinations of $\alpha^n u(n)$, $\alpha^n u(-n)$, $n^k \alpha^n u(n)$, and $n^k \alpha^n u(-n)$, where $k \in \N$, and shifted versions thereof.

\begin{example}

\begin{align*}
\hat{R}(z) &= \frac{(z-1)(z-2)}{(z-3)} \\
\end{align*}
Here $M \gt  N$, so we have a pole at infinity. Poles are $3,\infty$. Zeros are $1,2$.

\end{example}

\begin{example}

\begin{align*}
\hat{R}(z) = \frac{(z-3)}{(z-1)(z-2)} \\
\end{align*}

Here $N\gt M$, so we have a zero at infinity. Poles are $1,2$. Zeros are $3,\infty$.

\end{example}

\begin{example}

\begin{align*}
\hat{R}(z) &= \frac{z^2 + 1}{z} \\
\end{align*}

Can $r$ be causal?

\begin{align*}
\hat{R}(z) &= \frac{z^2 + 1}{z} \\
\hat{R}(z) &= \frac{(z-i)(z+i)}{z} \\
\end{align*}

We have zeros at $\pm i$ and poles at $0,\infty$. If there is a pole at infinity, then $r$ cannot be causal. If there is a pole at 0, then $r$ cannot be anti-causal. We can also see this fact using an inverse transform $\hat{R}(z) = z + \frac{1}{z} \iztp \delta(n-1) + \delta(n+1)$.

\end{example}

Rational Z-Transforms

Lets look into the relation between the Z-transform and the Fourier Transform:

\begin{nicebox}
\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty} x(n) e^{-i\omega n} \\
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} x(n) z^{-n} \\
\end{align*}
\end{nicebox}

There is an affiliated $RoC(x)$ with the Z-Transform. If the $RoC$ included the unit circle, then the following statement is true:

\begin{nicebox}
$$
X(\omega) = \left. \hat{X}(z) \right|_{z = e^{i\omega}}
$$
\end{nicebox}

\begin{example}

Lets look at the function $x(n) = \alpha^n u(n)$ for two cases, the first where $\abs{\alpha} \lt 1$, and the other where $\abs{\alpha} \gt 1$.

\begin{align*}
\hat{X}(z) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) z^{-n} \\
\hat{X}(z) &= \sum \limits_{n = 0}^{\infty} \lr{\alpha z^{-1}}^n \\
\hat{X}(z) &= \frac{1}{1- \alpha z^{-1}} \quad \mbox{ if } \abs{\alpha z^{-1}} \lt 1 \\
\end{align*}

This means that $RoC(x): \abs{z} \gt \abs{\alpha}$.


I) $\abs{\alpha} \lt 1$ We have a fourier transform.

This works because the $RoC$ includes all values such that $\abs{z} \gt \abs{\alpha}$. Since $1 \gt \abs{\alpha}$, 1 is a value in the $RoC$.

\begin{align*}
X(\omega) &= \left. \hat{X}(z) \right|_{z = e^{i\omega}} \\
X(\omega) &= \frac{1}{1-\alpha e^{-i\omega }} \\
\end{align*}


II) $\abs{\alpha} \gt 1$ This function doesn't have a fourier transform.

Since $\abs{\alpha} \gt 1$, then $\abs{z} \gt \abs{\alpha} \gt 1$. This means that the unit circle is excluded from the region of convergence.

\end{example}

Relation to Fourier Transform

If a system is BIBO stable, then the unit circle is included in the $RoC$. First recall that a system is BIBO if and only if its impulse response is absolutely summable.

\begin{align*}
\sum \abs{h(n)} \lt \infty \iff BIBO
\end{align*}

We can see that the absolute summability relies on $R$, and that the $RoC$ must include the unit circle for stability:
\begin{align*}
\sum \limits_{n \in \Z} \abs{h(n) z^{-n}} = \sum \limits_{n \in \Z} \abs{h(n) R^{-n}} \lt \infty
\end{align*}

BIBO Stability and the RoC

When cascading to systems $g$ and $h$, what happens if we pass in a signal $x(n) = z^n$?

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\psline{->}(3.5,0)(4,0)

\end{pspicture}
\end{center}


Let $f = g * h$. If we follow the path of the signal piecewise, we can first solve the system for $q(n)$, and then pass $q(n)$ through $h$ to determine $y$.

\begin{align*}
q(n) &= x(n) * g(n) \\
q(n) &= \sum \limits_{k=-\infty}^{\infty} g(k) x(n-k) \\  
q(n) &= \sum \limits_{k=-\infty}^{\infty} g(k) z^{n-k} \\  
q(n) &= z^n \sum \limits_{k=-\infty}^{\infty} g(k) z^{-k} \\  
q(n) &= z^n \hat{G}(z) \\  
\end{align*}

Now the second part of the system:

\begin{align*}
y(n) &= q(n) * h(n) \\
y(n) &= z^n \hat{G}(z) * h(n) \\
y(n) &= \sum \limits_{k=-\infty}^{\infty} h(k) z^{n-k} \hat{G}(z) \\
y(n) &= z^n \hat{G}(z) \sum \limits_{k=-\infty}^{\infty} h(k) z^{-k} \\
y(n) &= z^n \hat{G}(z) \hat{H}(z) \\
\end{align*}

We find that convolution in the time domain is multiplication in the transform domain---a relationship similar to what you knew from the fourier transform:

\begin{nicebox}
$$
f(n) = (g * h)(n) \ztp \hat{F}(z) = \hat{G}(z) \hat{H}(z)
$$
\end{nicebox}

\begin{proof}

\begin{align*}
f(n) &= (g*h)(n) \\
\hat{F}(z) &= \sum \limits_{n\in\Z}(g*h)(n)z^{-n} \\
\hat{F}(z) &= \sum \limits_{n\in\Z} \sum \limits_{k\in\Z}g(k)h(n-k)z^{-n} \\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) \sum \limits_{n\in\Z}h(n-k)z^{-n} \quad \mbox{swap summations}\\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) \sum \limits_{m\in\Z}h(m)z^{-(m+k)} \quad \mbox{let } m = n-k\\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) z^{-k} \sum \limits_{m\in\Z}h(m)z^{-m} \\
\hat{F}(z) &= \hat{G}(z) \hat{H}(z) \\
\end{align*}

For the system $f = g * h$, the $RoC(g) \cap RoC(h) \subset RoC(f)$. If $\hat{G}(z)$ and $\hat{H}(z)$ are rational and there are no pole-zero cancellations, then $RoC(g) \cap RoC(h) = RoC(f)$.

\end{proof}

\begin{example}

Two systems are in cascade, $\hat{G}$ and $\hat{H}$. Find the overall $RoC$ of the system.

\begin{align*}
\hat{G}(z) = \frac{1}{1 - \frac{1}{2}z^{-1}} \quad \mbox{(causal)} \\
\hat{H}(z) = \frac{z - \frac{1}{2}}{z + \frac{1}{3}} \quad \mbox{(causal)} \\
\end{align*}

First find $\hat{F}$ and solve for the poles and zeros:

\begin{align*}
\hat{F}(z) &= \hat{G}(z)\hat{H}(z) \\
\hat{F}(z) &= \lr{\frac{z}{z - \frac{1}{2}}} \lr{\frac{z - \frac{1}{2}}{z + \frac{1}{3}}} \\
\hat{F}(z) &= \frac{z\lr{z - \frac{1}{2}}}{\lr{z - \frac{1}{2}}  \lr{z + \frac{1}{3}}} \\
\hat{F}(z) &= \frac{z}{\lr{z + \frac{1}{3}}} \\
\end{align*}

So we have a pole at $-\frac{1}{3}$ and a zero at 0.

\begin{align*}
\hat{F}(z) = \frac{z}{\lr{z + \frac{1}{3}}} \iztp -\frac{1}{3}^n u(n) \quad \mbox{ where } \abs{z} \gt \frac{1}{3}
\end{align*}

Looking at the regions of convergence we find that the pole-zero cancellation changed the region of convergence.

\begin{align*}
RoC(g): &\frac{1}{2} \lt \abs{z} \\
RoC(h): &\frac{1}{3} \lt \abs{z} \\
RoC(f): &\frac{1}{2} \lt \abs{z} \\
\end{align*}

In this case $RoC(f) \gt RoC(g) \cap Roc(h)$.



In general, when given two systems in cascade, there is a good method for solving for the system function without calculating it explicitly. However, you must be given the input/output pair. For example, given a system

\begin{align*}
x \to &\fbox{h} \to y \\
y(n) = (x*h)(n) &\ztp \hat{Y}(z) = \hat{X}(z)\hat{H}(z) \\ 
\end{align*}

We can use the relation $\hat{H}(z) = \frac{\hat{Y}(z)}{\hat{X}(z)}$. You should be able to determing the system, but $RoC(x) \cap RoC(h)$ must have non-trivial overlap. This is because $RoC(y) \supseteq RoC(x) \cap RoC(h)$.

\end{example}

Systems in Cascade

Lets look into the shifting property for the Z-transform:

\begin{nicebox}
\begin{align*}
x(n) &\ztp \hat{X}(z) \\
x(n-N) &\ztp z^{-N}\hat{X}(z) \\
\end{align*}
\end{nicebox}

There are situations that modify the $RoC$.


I) $N \gt 0$ (shift to the right)


\begin{align*}
x(n-N) &\ztp z^{-N}\hat{X}(z) \\
x(n-N) &\ztp \frac{\hat{X}(z)}{z^{N}} \\
\end{align*}

Potentially $z=0$ has up to $N$ poles, so 0 may be excluded from the $RoC$.


II) $N \lt 0$ (shift to the left)


Potentially add poles at $z=\infty$, and $N$ zeros at 0. For both cases of $N$, the $RoC$ may exclude 0, $\infty$, or neither if there are cancellations.


Shifting Property


\begin{align*}
y(n) &= \alpha y(n-1) + x(n) - x(n-1) \\
\end{align*}

Lets utilize the fact $\hat{H}(z) = \frac{\hat{Y}(z)}{\hat{X}(z)}$:

\begin{align*}
y(n) &= \alpha y(n-1) + x(n) - x(n-1) \\
\hat{Y}(z) &= \alpha z^{-1}\hat{Y}(z) + \hat{X}(z) - z^{-1}\hat{X}(z) \\
\hat{Y}(z)\lr{1 - \alpha z^{-1}} &= \hat{X}(z)\lr{1 - z^{-1}} \\
\hat{H}(z) &= \frac{\lr{1 - z^{-1}}}{\lr{1 - \alpha z^{-1}}} \\
\hat{H}(z) &= \frac{\lr{z - 1}}{\lr{z - \alpha}} \\
\end{align*}



10 Week 10