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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.
The Z-Transform of a sequence {fn} is the discrete analogue of the Laplace Transform, and
| 𝒵{fn} = F(z) = ∑ n=0∞f nz−n. |
This series converges in the region outside the circle |z| = |z0| = limsup n→∞.
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| > ρ then ∃ a sequence {fn} with
𝒵{fn} = F(z) given by
fn =  ∮
F(z)zn−1dz
|
where k,λ ∈ N−{0} and A,B are fractions or variables (B > 0) and α,β, & ϕ are angles in radians.
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
 | (16.99) |
with initial conditions f0 = h0, f1 = h1, …, fk−1 = hk−1 (where hj are given), it
is possible to solve it in the following way. If the coefficients a1,…,ak
are constants, then the Z-Transform of (16.99) can be calculated using
the shift equation, and results in a solvable linear equation for 𝒵{fn}.
Application of the Inverse Z-Transform then results in the solution of
(16.99).
If the coefficients a1,…,ak are polynomials in n then the Z-Transform of (16.99)
constitutes a differential equation for 𝒵{fn}. If this differential equation can be
solved then the Inverse Z-Transform once again yields the solution of
(16.99). Some examples of these methods of solution can be found in
§16.84.6.
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 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)
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
fn+5 − 2fn+3 + 2fn+2 − 3fn+1 + 2fn = 0
with initial conditions f0 = 0, f1 = 0, f2 = 9, f3 = − 2, f4 = 23. The
Z-Transform of the left hand side can be written as F (z) = P(z)∕Q (z)
3 2 5 3 2
where P2(z) = 9z2 − 2z + 5z and Q (z ) = z − 2z + 2z − 3z + 2 =
(z − 1)(z + 2)(z + 1), which can be inverted to give
fn = 2n + (− 2)n − cos π2n.
The following REDUCE session shows how the present package can
be used to solve the above problem.](manual792x.png)
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
![II (See [BS81 ], p. 651, Example 2).
Consider the inhomogeneous difference equation:
f − 4f + 3f = 1
n+2 n+1 n
with initial conditions f0 = 0, f1 = 1. Giving
( )
F (z ) = 𝒵 {1} -2-1---+ -2-z---
z −4z+3 z− 4z+3
-z- (---1--- ---z---)
= z−1 z2−4z+3 + z2− 4z+3 .
The Inverse Z -Transform results in the solution
1 (3n+1−1 )
fn = 2 2 − (n + 1) .
The following REDUCE session shows how the present package can
be used to solve the above problem.](manual793x.png)
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
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)
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