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\begin{document}
\noindent{\Large ECE 602}\hfill{\Large Fall 2010}
\begin{center}
{\large \bf The ${\cal Z}$-transform}
\end{center}

%The ${\cal Z}$-transform can be viewed formally without concern for conditions of convergence of the series defining the ${\cal Z}$-transform. In this approach we regard the variable $z$ as a symbol and $F(z)$ would be regarded as a series that is never actually being summed.
%This is the same approach that one sometimes takes in connection with polynomials, where the polynomials at hand are manipulated algebraically.
\subsection{Definition and Simple Examples}
\begin{figure}[h]
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\begin{picture}(60,20)
\thicklines
\put(14,9){\vector(1,0){8}}
\put(16,10){$f(kT)$}
\put(22,5){\framebox(10,8){${\cal Z}(\cdot)$}}
\put(32,9){\vector(1,0){9}}
\put(34,10){$F(z)$}
\end{picture}
\end{center}
\caption{The ${\cal Z}$-transform as an operator acting on a sequence of numbers resulting in a function of the variable $z$.}
\label{fig:z-transform}
\end{figure}

\begin{example}\rm\label{ex:unit_s_z_tr}
We find the ${\cal Z}$-transform of the sampled unit step, $f(kT)=1$. The ${\cal Z}$-transform of this sequence is
\begin{eqnarray*}
F(z)&=& \sum_{k=0}^{\infty} f(kT)z^{-k}\\
&=& 1 + \frac{1}{z} + \frac{1}{z^2} + \cdots
\end{eqnarray*}
If $z$ is a complex variable such that $|z| > 1$, then the above sequence converges and we have
\begin{eqnarray*}
F(z)&=& \frac{1}{1-1/z}\\
&=& \frac{z}{z-1}.
\end{eqnarray*}
\end{example}
\markex

\begin{example}\rm\label{ex:ramp_z_tr}
We find the ${\cal Z}$-transform of the sampled unit ramp, $f(kT)=kT$. The ${\cal Z}$-transform of this sequence is
\begin{eqnarray}
F(z)&=& \sum_{k=0}^{\infty} kT z^{-k}\nonumber\\
&=& T\left( z^{-1} + 2 z^{-2} + 3 z^{-3} + \cdots \right)
\label{eq:defn_ramp_z}
\end{eqnarray}
Multiplying both sides of the above by $z$ gives
\begin{equation}
z F(z)=T\left( 1 + 2 z^{-1} + 3 z^{-2} + \cdots \right)
\label{eq:ztimesF}
\end{equation}
Subtracting~\eqref{eq:defn_ramp_z} from~\eqref{eq:ztimesF} yields
\begin{eqnarray}
z F(z) - F(z) &=& (z-1) F(z)\nonumber\\
&=& T\left(1 + z^{-1} + z^{-2} + \cdots \right)\nonumber\\
&=& \frac{1}{1-z^{-1}}\nonumber\\
&=& \frac{z}{z-1}.
\label{eq:aftersubing}
\end{eqnarray}
Hence, the ${\cal Z}$-transform of $f(k T) = kT$ is
\begin{equation}
\fbox{$\displaystyle {\cal Z}(kT) = T \frac{z}{(z-1)^2} $}
\label{eq:z_tr_u_ram}
\end{equation}
\end{example}
\markex

\begin{example}\rm\label{ex:geo_s_z_tr}
We now find the ${\cal Z}$-transform of the geometric sequence
\begin{equation}
f(k) = a^{k T}.
\label{eq:geo_seq_w_T}
\end{equation}
We obtain
\begin{equation}
\fbox{$\displaystyle {\cal Z}(a^{k T}) = \sum_{k=0}^{\infty} \frac{a^{k T}}{z^k} = \frac{z}{z-a^T} $}
\label{eq:z_tr_geo_seq}
\end{equation}
for $|z| > \left|a^T  \right|$.
\end{example}
\markex

\subsection{Some Properties of the ${\cal Z}$-Transform }
We analyze now three basic properties that are very useful when computing transforms of more complicated sequences.

{\bf Linearity property:} The ${\cal Z}$-transform is a linear operator, that is,
\begin{itemize}
\item[(i)] if $F_1(z)={\cal Z}(f_1(kT))$ and $F_2(z)={\cal Z}(f_2(kT))$ then
    \[
    {\cal Z}(f_1(kT)+f_2(kT))=F_1(z) + F_2(z); \mb{ and}
    \]
\item[(ii)] if $F(z)={\cal Z}(f(kT))$ then
\[
{\cal Z}(cf(kT)) = cF(z),
\]
where $c$ is a real or complex number.
\end{itemize}
We can represent the above two properties as: if $F_1(z)={\cal Z}(f_1(kT))$ and $F_2(z)={\cal Z}(f_2(kT))$, and $c$ is a real or complex number, then
\[
{\cal Z}(f_1(kT)+cf_2(kT))=F_1(z) + cF_2(z).
\]

{\bf ${\cal Z}$-transform of a delayed sequence:} If the sequence $f(kT)$ has the ${\cal Z}$-transform $F(z)$, then the unit delayed sequence
\[
g(kT)=\left\{\begin{array}{lll}
f((k-1)T) &{}& k\ge 1\\
0 &{}& k=0
\end{array}  \right.
\]
has the ${\cal Z}$-transform
\begin{equation}
G(z)=z^{-1} F(z).
\label{eq:del_seq_z_tr}
\end{equation}
We use the definition of the ${\cal Z}$-transform to prove the above property;
\begin{eqnarray*}
G(z) &=& \sum_{k=0}^{\infty} z^{-k} g(k T)\\
&=& \sum_{k=1}^{\infty} z^{-k} f((k-1)T)\\
&=& z^{-1}\sum_{k=1}^{\infty} z^{-(k-1)} f((k-1)T)\\
&=& z^{-1}\sum_{j=0}^{\infty} z^{-j} f(j T)\\
&=& z^{-1} F(z).
\end{eqnarray*}
{\bf ${\cal Z}$-transform of an advanced sequence:} If the sequence $f(kT)$ has the ${\cal Z}$-transform $F(z)$, then the unit advanced sequence $h(kT)=f((k+1)T)$, $k\ge 0$, has the transform
\begin{equation}
{\cal Z}(f((k+1)T)) = z F(z) - z f(0).
\label{eq:z_tr_adva_seq}
\end{equation}
To prove the above, we apply the definition of the ${\cal Z}$-transform to obtain
\begin{eqnarray*}
H(z) &=& \sum_{k=0}^{\infty} z^{-k} h(k T) \\
&=& \sum_{k=0}^{\infty} z^{-k} f((k + 1) T) \\
&=& z\sum_{k=0}^{\infty} z^{- k - 1} f((k + 1) T)\\
&=& z \sum_{j=1}^{\infty} z^{-j} f(j T)\\
&=& z \sum_{j=0}^{\infty} z^{-j} f(j T) - z f(0)\\
&=& z F(z) - z f(0).
\end{eqnarray*}

\begin{example}\rm\label{ex:del_geo_ser}
We find the ${\cal Z}$-transform of the delayed geometric sequence,
\begin{equation}
f(kT) = \left\{\begin{array}{lll}
0 &\mb{for}& k=0\\
a^{(k-1)T} &\mb{for}& k>0 .
\end{array}  \right.
\label{eq:del_geo_ser}
\end{equation}
Applying~\eqref{eq:del_seq_z_tr} to the geometric sequence~\eqref{eq:geo_seq_w_T} yields
\begin{equation}
\fbox{$\displaystyle {\cal Z}(f(kT)=\frac{1}{z-a^T} $}
\label{eq:z_tr_dely_geo_s}
\end{equation}
\end{example}
\markex

\subsection{Solving Difference Equations Using the ${\cal Z}$-Transform }

The ${\cal Z}$-transform is a very effective tool for solving linear, constant-coefficients difference equations. The basis for the method is the ${\cal Z}$-transform of an advanced sequence given by~\eqref{eq:z_tr_adva_seq}. Repeated application of this property gives
\begin{equation}
{\cal Z}(f((k+2)T)) = z^2 F(z) - z^2 f(0) - z f(T).
\label{eq:advanc_twice}
\end{equation}
Agaian, applying~\eqref{eq:z_tr_adva_seq} to~\eqref{eq:advanc_twice} gives
\begin{equation}
{\cal Z}(f((k+3)T)) = z^3 F(z) - z^3 f(0) - z^2 f(T) - z f(2T)
\label{eq:advanc_thrice}
\end{equation}
and so on.
\begin{example}\rm\label{ex:simp_differe_eq}
We solve the difference equation
\[
x[k+1] + 3 x[k]= 2^k,\quad k=0, 1, \ldots ,
\]
where $x(0)=0$. Applying the ${\cal Z}$-transform to both sides of the above difference equation gives
\[
z X(z) - z x(0) + 3 X(z)=\frac{z}{z-2}.
\]
Hence,
\[
X(z)=\frac{z}{(z+3)(z-2)},
\]
which is a strictly proposer rational function of $z$. We represent $X(z)$ in partial fraction form,
\[
X(z)=\frac{A}{z+3} + \frac{B}{z-2},
\]
where
\[
A=\left. (z+3)X(z) \right\arrowvert_{z=-3}=\frac{3}{5}\quad\mb{and}\quad B=\left. (z-2)X(z) \right\arrowvert_{z=2}=\frac{2}{5}.
\]
Therefore,
\[
X(z)=\frac{3/5}{z+3} + \frac{2/5}{z-2}.
\]
We use~\eqref{eq:z_tr_dely_geo_s} when applying now the inverse ${\cal Z}$-transform to the above,
\[
x[k]=\left\{\begin{array}{lll}
0 &\mb{for}& k=0\\
\frac{3}{5}(-3)^{k-1}   + \frac{2}{5} 2^{k-1} &\mb{for}& k \ge 1
\end{array}
\right.
\]
\end{example}
\markex

\begin{example}\rm\label{ex:nonz_ic_differe_eq}
We solve the difference equation
\[
x[k+1] + 3 x[k]= 2^k,\quad k=0, 1, \ldots ,
\]
where $x(0)=5$. Applying the ${\cal Z}$-transform to both sides of the above difference equation gives
\[
z X(z) - z x(0) + 3 X(z)=\frac{z}{z-2}.
\]
Hence,
\[
X(z)=\frac{5z^2 - 9z}{(z+3)(z-2)}=\frac{z(5z-9)}{(z+3)(z-2)}.
\]
The above rational function is not strictly proper so we use a different approach than in Example~\ref{ex:simp_differe_eq}. In this example $z$ is a factor of the numerator so we can represent that above as
\[
\frac{X(z)}{z}=\frac{5z-9}{(z+3)(z-2)}
\]
and express that right hand side in partial fraction format
\[
\frac{X(z)}{z}=\frac{24/5}{z+3} + \frac{1/5}{z-2}.
\]
Thus
\[
X(z)=\frac{24}{5}\frac{z}{z+3} + \frac{1}{5}\frac{z}{z-2} .
\]
Taking into account~\eqref{eq:z_tr_geo_seq} when applying the inverse ${\cal Z}$-transform yields
\[
x[k]=\frac{24}{5}(-3)^{k} + \frac{1}{5} 2^{k},\quad k \ge 0
\]
\end{example}

\markex

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