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16.84 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.

16.84.1 Z-Transform

The Z-Transform of a sequence {fn} is the discrete analogue of the Laplace Transform, and

∞ ∑ −n 𝒵{fn } = F (z) = fnz . n=0

This series converges in the region outside the circle |z| = |z0| = limsup n→∞∘ ---- n |fn|.

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

16.84.2 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| > ρ then ∃ a sequence {fn} with 𝒵{fn} = F(z) given by

1 ∮ fn = ---- F (z)zn−1dz 2πi

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

16.84.3 Input for the Z-Transform

This package can compute the Z -Transforms of the following list of fn, and certain combinations thereof. 1 eαn (n1+k)- 1- --1- --1--- n! (2n)! (2n+1)! sin(βn) sin(αn + ϕ ) eαn sin (βn) n! cos(βn) αn --n!-- cos(αn + ϕ ) e cos(βn ) sin(β(n+1)) sinh(αn + ϕ ) cos(β(n+1)) n+1 n+1 (n+k) cosh(αn + ϕ ) m

Other-Combinations--- Linearity- 𝒵 {afn + bgn} = a𝒵 {fn } + b𝒵 {gn } Multiplication by n 𝒵 {nk ⋅ f } = − z-d (𝒵 {nk−1 ⋅ f ,n,z} ) ------------------ n dz n n n (z ) Multiplication by-λ- 𝒵 {λ ⋅ fn} = F λ ( ) k∑−1 Shift Equation 𝒵 {fn+k} = zk F(z) − fjz −j -------------- j=0 { ∑n } Symbolic-Sums-- 𝒵 fk = zz−1 ⋅ 𝒵 {fn} k=0 { } n∑+q 𝒵 fk combination of the above k=p where k,λ ∈ N− {0 }; and a,b are variables or fractions; and p,q ∈ Z or are functions of n; and α, β & ϕ are angles in radians.

16.84.4 Input for the Inverse Z-Transform

This package can compute the Inverse Z -Transforms of any rational function, whose denominator can be factored over Q, in addition to the following list of F (z). (sin(β)) (cos(zβ) ) (sin(β)) (cos(zβ) ) sin z e cos z e -- -- -- ∘ z-sin (∘ -z) cos (∘ z-) A A A ∘ z- (∘ -z-) (∘ -z) A sinh A cosh A ( ) (√ --------) z log √---z----- z log --z2+Az+B- z2− Az+B z ( ) arctan -sin(β)-- z+cos(β)

where k,λ ∈ N−{0} and A,B are fractions or variables (B > 0) and α,β, & ϕ are angles in radians.

16.84.5 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

fn+k + a1fn+k−1 + ...+ akfn = gn
(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.

16.84.6 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       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.

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.

 
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

III Consider the following difference equation, which has a differential equation for 𝒵 {fn}. (n + 1) ⋅ f − f = 0 n+1 n with initial conditions f0 = 1, f1 = 1. It can be solved in REDUCE using the present package in the following way.

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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