\documentstyle[11pt,reduce]{article}
\title{{\bf $Z$-Transform Package for {\tt REDUCE}}}
\author{Wolfram Koepf \\ Lisa Temme \\ email: {\tt Koepf@zib-berlin.de}}
\date{April 1995 : ZIB Berlin}
\begin{document}
\maketitle
\section{$Z$-Transform}
The $Z$-Transform of a sequence $\{f_n\}$ is the discrete analogue
of the Laplace Transform, and
\[{\cal Z}\{f_n\} = F(z) = \sum^\infty_{n=0} f_nz^{-n}\;.\] \\
This series converges in the region outside the circle
$|z|=|z_0|= \limsup\limits_{n \rightarrow \infty} \sqrt[n]{|f_n|}\;.$
\begin{tabbing}
{\bf SYNTAX:}\ \ {\tt ztrans($f_n$, n, z)}\ \ \ \ \ \ \ \
\=where $f_n$ is an expression, and $n$,$z$ \\
\> are identifiers.\\
\end{tabbing}
\section{Inverse $Z$-Transform}
The calculation of the Laurent coefficients of a regular function
results in the following inverse formula for the $Z$-Transform:
\\
If $F(z)$ is a regular function in the region $|z|> \rho$ then
$\exists$ a sequence \{$f_n$\} with ${\cal Z} \{f_n\}=F(z)$
given by
\[f_n = \frac{1}{2 \pi i}\oint F(z) z^{n-1} dz\]
\begin{tabbing}
{\bf SYNTAX:}\ \ {\tt invztrans($F(z)$, z, n)}\ \ \ \ \ \ \ \
\=where $F(z)$ is an expression, \\
\> and $z$,$n$ are identifiers.
\end{tabbing}
\section{Input for the $Z$-Transform}
\begin{tabbing}
This pack\=age can compute the \= $Z$-Transforms of the \=following
list of $f_n$, and \\ certain combinations thereof.\\ \\
\>$1$
\>$e^{\alpha n}$
\>$\frac{1}{(n+k)}$ \\ \\
\>$\frac{1}{n!}$
\>$\frac{1}{(2n)!}$
\>$\frac{1}{(2n+1)!}$ \\ \\
\>$\frac{\sin(\beta n)}{n!}$
\>$\sin(\alpha n+\phi)$
\>$e^{\alpha n} \sin(\beta n)$ \\ \\
\>$\frac{\cos(\beta n)}{n!}$
\>$\cos(\alpha n+\phi)$
\>$e^{\alpha n} \cos(\beta n)$ \\ \\
\>$\frac{\sin(\beta (n+1))}{n+1}$
\>$\sinh(\alpha n+\phi)$
\>$\frac{\cos(\beta (n+1))}{n+1}$ \\ \\
\>$\cosh(\alpha n+\phi)$
\>${n+k \choose m}$\\
\end{tabbing}
\begin{tabbing}
\underline {{\bf Other Combinations}}\= \\ \\
\underline {Linearity}
\>${\cal Z} \{a f_n+b g_n \} = a{\cal Z} \{f_n\}+b{\cal Z}\{g_n\}$
\\ \\
\underline {Multiplication by $n$}
\>${\cal Z} \{n^k \cdot f_n\} = -z \frac{d}{dz} \left({\cal Z}\{n^{k-1} \cdot f_n,n,z\} \right)$
\\ \\
\underline {Multiplication by $\lambda^n$}
\>${\cal Z} \{\lambda^n \cdot f_n\}=F \left(\frac{z}{\lambda}\right)$
\\ \\
\underline {Shift Equation}
\>${\cal Z} \{f_{n+k}\} =
z^k \left(F(z) - \sum\limits^{k-1}_{j=0} f_j z^{-j}\right)$
\\ \\
\underline {Symbolic Sums}
\> ${\cal Z} \left\{ \sum\limits_{k=0}^{n} f_k \right\} =
\frac{z}{z-1} \cdot {\cal Z} \{f_n\}$ \\ \\
\>${\cal Z} \left\{ \sum\limits_{k=p}^{n+q} f_k \right\}$
\ \ \ combination of the above \\ \\
where $k$,$\lambda \in$ {\bf N}$- \{0\}$; and $a$,$b$ are variables
or fractions; and $p$,$q \in$ {\bf Z} or \\
are functions of $n$; and $\alpha$, $\beta$ \& $\phi$ are angles
in radians.
\end{tabbing}
\section{Input for the Inverse $Z$-Transform}
\begin{tabbing}
This \= package can compute the Inverse \= Z-Transforms of any
rational function, \\ whose denominator can be factored over
${\bf Q}$, in addition to the following list \\ of $F(z)$.\\ \\
\> $\sin \left(\frac{\sin (\beta)}{z} \ \right)
e^{\left(\frac{\cos (\beta)}{z} \ \right)}$
\> $\cos \left(\frac{\sin (\beta)}{z} \ \right)
e^{\left(\frac{\cos (\beta)}{z} \ \right)}$ \\ \\
\> $\sqrt{\frac{z}{A}} \sin \left( \sqrt{\frac{z}{A}} \ \right)$
\> $\cos \left( \sqrt{\frac{z}{A}} \ \right)$ \\ \\
\> $\sqrt{\frac{z}{A}} \sinh \left( \sqrt{\frac{z}{A}} \ \right)$
\> $\cosh \left( \sqrt{\frac{z}{A}} \ \right)$ \\ \\
\> $z \ \log \left(\frac{z}{\sqrt{z^2-A z+B}} \ \right)$
\> $z \ \log \left(\frac{\sqrt{z^2+A z+B}}{z} \ \right)$ \\ \\
\> $\arctan \left(\frac{\sin (\beta)}{z+\cos (\beta)} \ \right)$
\\
\end{tabbing}
where $k$,$\lambda \in$ {\bf N}$ - \{0\}$ and $A$,$B$ are fractions
or variables ($B>0$) and $\alpha$,$\beta$, \& $\phi$ are angles
in radians.
\section{Application of the $Z$-Transform}
\underline {{\bf Solution of difference equations}}\\
In the same way that a Laplace Transform can be used to
solve differential equations, so $Z$-Transforms can be used
to solve difference equations.\\ \\
Given a linear difference equation of $k$-th order
\begin{equation}
f_{n+k} + a_1 f_{n+k-1}+ \ldots + a_k f_n = g_n
\label{eq:1}
\end{equation}
with initial conditions
$f_0 = h_0$, $f_1 = h_1$, $\ldots$, $f_{k-1} = h_{k-1}$ (where $h_j$
are given), it is possible to solve it in the following way.
If the coefficients $a_1, \ldots , a_k$ are constants, then the
$Z$-Transform of (\ref{eq:1}) can be calculated using the shift
equation, and results in a solvable linear equation for
${\cal Z} \{f_n\}$. Application of the Inverse $Z$-Transform
then results in the solution of \ (\ref{eq:1}).\\
If the coefficients $a_1, \ldots , a_k$ are polynomials in $n$ then
the $Z$-Transform of (\ref{eq:1}) constitutes a differential
equation for ${\cal Z} \{f_n\}$. If this differential equation can
be solved then the Inverse $Z$-Transform once again yields the
solution of (\ref{eq:1}).
Some examples of these methods of solution can be found in
$\S$\ref{sec:Examples}.
\section{EXAMPLES}
\label{sec:Examples}
\underline {{\bf Here are some examples for the $Z$-Transform}}\\
\begin{verbatim}
1: ztrans((-1)^n*n^2,n,z);
z*( - z + 1)
---------------------
3 2
z + 3*z + 3*z + 1
2: ztrans(cos(n*omega*t),n,z);
z*(cos(omega*t) - z)
---------------------------
2
2*cos(omega*t)*z - z - 1
3: ztrans(cos(b*(n+2))/(n+2),n,z);
z
z*( - cos(b) + log(------------------------------)*z)
2
sqrt( - 2*cos(b)*z + z + 1)
4: ztrans(n*cos(b*n)/factorial(n),n,z);
cos(b)/z sin(b) sin(b)
e *(cos(--------)*cos(b) - sin(--------)*sin(b))
z z
---------------------------------------------------------
z
5: ztrans(sum(1/factorial(k),k,0,n),n,z);
1/z
e *z
--------
z - 1
6: operator f$
7: ztrans((1+n)^2*f(n),n,z);
2
df(ztrans(f(n),n,z),z,2)*z - df(ztrans(f(n),n,z),z)*z
+ ztrans(f(n),n,z)
\end{verbatim}
\underline {{\bf Here are some examples for the Inverse $Z$-Transform}}
\begin{verbatim}
8: invztrans((z^2-2*z)/(z^2-4*z+1),z,n);
n n n
(sqrt(3) - 2) *( - 1) + (sqrt(3) + 2)
-----------------------------------------
2
9: invztrans(z/((z-a)*(z-b)),z,n);
n n
a - b
---------
a - b
10: invztrans(z/((z-a)*(z-b)*(z-c)),z,n);
n n n n n n
a *b - a *c - b *a + b *c + c *a - c *b
-----------------------------------------
2 2 2 2 2 2
a *b - a *c - a*b + a*c + b *c - b*c
11: invztrans(z*log(z/(z-a)),z,n);
n
a *a
-------
n + 1
12: invztrans(e^(1/(a*z)),z,n);
1
-----------------
n
a *factorial(n)
13: invztrans(z*(z-cosh(a))/(z^2-2*z*cosh(a)+1),z,n);
cosh(a*n)
\end{verbatim}
\underline {{\bf Examples: Solutions of Difference Equations}}\\ \\
\begin{tabbing}
{\bf I} \ \ \ \ \ \ \=
(See \cite{BS}, p.\ 651, Example 1).\\
\> Consider the \= homogeneous linear difference equation\\ \\
\>\> $f_{n+5} - 2 f_{n+3} + 2 f_{n+2} - 3 f_{n+1} + 2 f_{n}=0$\\ \\
\> with \ initial conditions \ $f_0=0$, $f_1=0$, $f_2=9$, $f_3=-2$,
$f_4=23$. \ The\\
\> $Z$-Transform of the left hand side can be written as
$F(z)=P(z)/Q(z)$ \\
\> where \ $P(z)=9z^3-2z^2+5z$ \
and \ $Q(z)=z^5-2z^3+2z^2-3z+2$ \ $=$\\
\> $(z-1)^2(z+2)(z^2+1)$, \ which can be inverted to give\\ \\
\>\> $f_n = 2n + (-2)^n - \cos \frac{\pi}{2}n\;.$ \\ \\
\> The following REDUCE session shows how the present package can
\\ \> be used to solve the above problem.
\end{tabbing}
\begin{verbatim}
14: operator f$ f(0):=0$ f(1):=0$ f(2):=9$ f(3):=-2$ f(4):=23$
20: equation:=ztrans(f(n+5)-2*f(n+3)+2*f(n+2)-3*f(n+1)+2*f(n),n,z);
5 3
equation := ztrans(f(n),n,z)*z - 2*ztrans(f(n),n,z)*z
2
+ 2*ztrans(f(n),n,z)*z - 3*ztrans(f(n),n,z)*z
3 2
+ 2*ztrans(f(n),n,z) - 9*z + 2*z - 5*z
21: ztransresult:=solve(equation,ztrans(f(n),n,z));
2
z*(9*z - 2*z + 5)
ztransresult := {ztrans(f(n),n,z)=----------------------------}
5 3 2
z - 2*z + 2*z - 3*z + 2
22: result:=invztrans(part(first(ztransresult),2),z,n);
n n n n
2*( - 2) - i *( - 1) - i + 4*n
result := -----------------------------------
2
\end{verbatim}
\begin{tabbing}
\\ \\
{\bf II} \ \ \ \ \ \ \=
(See \cite{BS}, p.\ 651, Example 2).\\
\> Consider the \= inhom\=ogeneous difference equation:\\ \\
\>\> $f_{n+2} - 4 f_{n+1} + 3 f_{n} = 1$\\ \\
\> with initial conditions $f_0=0$, $f_1=1$. Giving \\ \\
\>\> $F(z)$\>$ = {\cal Z}\{1\} \left( \frac{1}{z^2-4z+3} + \frac{z}{z^2-4z+3} \right)$\\ \\
\>\>\> $ = \frac{z}{z-1} \left( \frac{1}{z^2-4z+3} + \frac{z}{z^2-4z+3} \right)$.
\\ \\
\> The Inverse $Z$-Transform results in the solution\\ \\
\>\>
$f_n = \frac{1}{2} \left( \frac{3^{n+1}-1}{2}-(n+1) \right)$.\\ \\
\> The following REDUCE session shows how the present package can\\
\> be used to solve the above problem.
\end{tabbing}
\begin{verbatim}
23: clear(f)$ operator f$ f(0):=0$ f(1):=1$
27: equation:=ztrans(f(n+2)-4*f(n+1)+3*f(n)-1,n,z);
3 2
equation := (ztrans(f(n),n,z)*z - 5*ztrans(f(n),n,z)*z
2
+ 7*ztrans(f(n),n,z)*z - 3*ztrans(f(n),n,z) - z )/(z - 1)
28: ztransresult:=solve(equation,ztrans(f(n),n,z));
2
z
result := {ztrans(f(n),n,z)=---------------------}
3 2
z - 5*z + 7*z - 3
29: result:=invztrans(part(first(ztransresult),2),z,n);
n
3*3 - 2*n - 3
result := ----------------
4
\end{verbatim}
\begin{tabbing}
\\ \\
{\bf III} \ \ \ \ \ \ \=
Consider the \=following difference equation, which has a
differential\\
\> equation for ${\cal Z}\{f_n\}$.\\ \\
\>\> $(n+1) \cdot f_{n+1}-f_n=0$\\ \\
\> with initial conditions $f_0=1$, $f_1=1$. It can be solved in REDUCE\\
\> using the present package in the following way.\\
\end{tabbing}
\begin{verbatim}
30: clear(f)$ operator f$ f(0):=1$ f(1):=1$
34: equation:=ztrans((n+1)*f(n+1)-f(n),n,z);
2
equation := - (df(ztrans(f(n),n,z),z)*z + ztrans(f(n),n,z))
35: operator tmp;
36: equation:=sub(ztrans(f(n),n,z)=tmp(z),equation);
2
equation := - (df(tmp(z),z)*z + tmp(z))
37: load(odesolve);
38: ztransresult:=odesolve(equation,tmp(z),z);
1/z
ztransresult := {tmp(z)=e *arbconst(1)}
39: preresult:=invztrans(part(first(ztransresult),2),z,n);
arbconst(1)
preresult := --------------
factorial(n)
40: solve({sub(n=0,preresult)=f(0),sub(n=1,preresult)=f(1)},
arbconst(1));
{arbconst(1)=1}
41: result:=preresult where ws;
1
result := --------------
factorial(n)
\end{verbatim}
\begin{thebibliography}{9}
\bibitem{BS} Bronstein, I.N. and Semedjajew, K.A.,
{\it Taschenbuch der Mathematik},
Verlag Harri Deutsch, Thun und Frankfurt(Main),
1981.\\ISBN 3 87144 492 8.
\end{thebibliography}
\end{document}