ZTRANS: Z-Transform Package

REDUCE

20.72 ZTRANS: Z-Transform Package

This package is an implementation of the \(Z\)-transform of a sequence. This is the discrete analogue of the Laplace Transform.

Authors: Wolfram Koepf and Lisa Temme

20.72.1 \(Z\)-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 \(|z|=|z_0|= \limsup \limits _{n \rightarrow \infty } \sqrt [n]{|f_n|}\;.\)

SYNTAX: `ztrans`(\(f_n\), n, z)	where \(f_n\) is an expression, and \(n\),\(z\)
	are identiﬁers.

20.72.2 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

\[f_n = \frac {1}{2 \pi i}\oint F(z) z^{n-1} dz\]

SYNTAX: invztrans(\(F(z)\), z, n)	where \(F(z)\) is an expression,
	and \(z\),\(n\) are identiﬁers.

20.72.3 Input for the \(Z\)-Transform

This package can compute the \(Z\)-Transforms of the following list of functions \(f_n\), and certain combinations thereof.

\(1\)	\(e^{\alpha n}\)	\(\frac {1}{(n+k)}\)
\(\frac {1}{n!}\)	\(\frac {1}{(2n)!}\)	\(\frac {1}{(2n+1)!}\)
\(\frac {\sin (\beta n)}{n!}\)	\(\sin (\alpha n+\phi )\)	\(e^{\alpha n} \sin (\beta n)\)
\(\frac {\cos (\beta n)}{n!}\)	\(\cos (\alpha n+\phi )\)	\(e^{\alpha n} \cos (\beta n)\)
\(\frac {\sin (\beta (n+1))}{n+1}\)	\(\sinh (\alpha n+\phi )\)	\(\frac {\cos (\beta (n+1))}{n+1}\)
\(\cosh (\alpha n+\phi )\)	\(\binom {n+k}{m}\)

Other Combinations

Linearity	\({\cal Z} \{a f_n+b g_n \} = a{\cal Z} \{f_n\}+b{\cal Z}\{g_n\}\)
Multiplication by \(n\)	\({\cal Z} \{n^k \cdot f_n\} = -z \frac {d}{dz} \left ({\cal Z}\{n^{k-1} \cdot f_n,n,z\} \right )\)
Multiplication by \(\lambda ^n\)	\({\cal Z} \{\lambda ^n \cdot f_n\}=F \left (\frac {z}{\lambda }\right )\)
Shift Equation	\({\cal Z} \{f_{n+k}\} = z^k \left (F(z) - \sum \limits ^{k-1}_{j=0} f_j z^{-j}\right )\)
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 \(k,\lambda \in \mathbf {N} \setminus \{0\}\); and \(a,b\) are variables or fractions; and \(p,q \in \mathbf {Z}\) or are functions of \(n\); and \(\alpha ,\beta \) and \(\phi \) are angles in radians.

20.72.4 Input for the Inverse \(Z\)-Transform

This package can compute the Inverse \(Z\)-Transforms of any rational function, whose denominator can be factored over \(\mathbf {Q}\), in addition to the following list of \(F(z)\).

\[ \renewcommand {\arraystretch }{2} \begin {array}{cc} \sin \left (\frac {\sin (\beta )}{z} \right ) e^{\left (\frac {\cos (\beta )}{z} \ \right )} & \cos \left (\frac {\sin (\beta )}{z} \right ) e^{\left (\frac {\cos (\beta )}{z} \ \right )} \\ \sqrt {\frac {z}{A}} \sin \left ( \sqrt {\frac {z}{A}} \right ) & \cos \left ( \sqrt {\frac {z}{A}} \right ) \\ \sqrt {\frac {z}{A}} \sinh \left ( \sqrt {\frac {z}{A}} \right ) & \cosh \left ( \sqrt {\frac {z}{A}} \right ) \\ z \log \left (\frac {z}{\sqrt {z^2-A z+B}} \right ) & z \log \left (\frac {\sqrt {z^2+A z+B}}{z} \right ) \\ \arctan \left (\frac {\sin (\beta )}{z+\cos (\beta )} \right ) \end {array} \]

where \(k,\lambda \in \mathbf {N} \setminus \{0\}\) and \(A,B\) are fractions or variables (\(B>0\)) and \(\alpha ,\beta \), and \(\phi \) are angles in radians.

20.72.5 Application of the \(Z\)-Transform

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 \(\S \)20.72.6.

20.72.6 EXAMPLES

Here are some examples for the \(Z\)-Transform

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
(e

        sin(b)                 sin(b)
 *(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)

Here are some examples for the Inverse \(Z\)-Transform

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)

Examples: Solutions of Diﬀerence Equations

I: (See [BS81], 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.

1: operator f;

2: f(0):=0$ f(1):=0$ f(2):=9$ f(3):=-2$ f(4):=23$


7: equation :=
   ztrans(f(n+5)-2*f(n+3)+2*f(n+2)-3*f(n+1)+2*f(n),n,z);

                              5
equation := ztrans(f(n),n,z)*z

                                   3
             - 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

8: ztransresult:=solve(equation,ztrans(f(n),n,z));

ztransresult :=

                             2
                       z*(9*z  - 2*z + 5)
{ztrans(f(n),n,z)=----------------------------}
                    5      3      2
                   z  - 2*z  + 2*z  - 3*z + 2

9: result:=invztrans(part(first(ztransresult),2),z,n);

result :=

     n       n           n  n    n
                                                                     

                                                                     
  - i *( - 1)  + 2*( - 1) *2  - i  + 4*n
-----------------------------------------
                    2

II

(See [BS81], p. 651, Example 2).
Consider the inhomogeneous diﬀerence equation:

\[ f_{n+2} - 4 f_{n+1} + 3 f_{n} = 1 \]

with initial conditions \(f_0=0\), \(f_1=1\). Giving

\begin{align*} 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 ). \end{align*}

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.

10: clear(f)$ operator f$ f(0):=0$ f(1):=1$


14: equation:=ztrans(f(n+2)-4*f(n+1)+3*f(n)-1,n,z);

                               3
equation := (ztrans(f(n),n,z)*z

                                    2
              - 5*ztrans(f(n),n,z)*z

              + 7*ztrans(f(n),n,z)*z

                                      2
              - 3*ztrans(f(n),n,z) - z )/(z - 1)

15: ztransresult:=solve(equation,ztrans(f(n),n,z));

ztransresult :=

                            2
                           z
{ztrans(f(n),n,z)=---------------------}
                    3      2
                   z  - 5*z  + 7*z - 3

16: result:=invztrans(part(first(ztransresult),2),z,n);

              n
           3*3  - 2*n - 3
result := ----------------
                 4

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.

17: clear(f)$ operator f$ f(0):=1$ f(1):=1$


21: equation:=ztrans((n+1)*f(n+1)-f(n),n,z);

equation :=

                            2
 - (df(ztrans(f(n),n,z),z)*z  + ztrans(f(n),n,z))

22: operator tmp;

23: equation:=sub(ztrans(f(n),n,z)=tmp(z),equation);

                              2
equation :=  - (df(tmp(z),z)*z  + tmp(z))

24: load_package(odesolve);

25: 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)

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