ZTRANS: Z-Transform Package

REDUCE

20.66 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.

20.66.1 Z-Transform

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

∞ 𝒵 {f } = F(z) = ∑ f z− n . n n=0 n

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

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

20.66.2 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 {f_n} with 𝒵{f_n} = F(z) given by

∮ f = -1-- F (z)zn−1dz n 2πi

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

20.66.3 Input for the Z-Transform

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

1	e^αn

	sin(αn + ϕ)	e^αn sin(βn)
	cos(αn + ϕ)	e^αn cos(βn)
	sinh(αn + ϕ)
cosh(αn + ϕ)

Other Combinations

Linearity: 𝒵{af_n + bg_n} = a𝒵{f_n} + b𝒵{g_n}
Multiplication by n: 𝒵{n^k ⋅ f_n} = −z
Multiplication by λⁿ: 𝒵{λⁿ ⋅ f_n} = F
Shift Equation: 𝒵{f_n+k} = z^k
Symbolic Sums: 𝒵 = ⋅𝒵{f_n}
𝒵 combination of the above

where k,λ ∈N∖{0}; and a,b are variables or fractions; and p,q ∈Z or are functions of n; and α,β and ϕ are angles in radians.

20.66.4 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)) cosz(β) (sin(β)) cos(zβ) sin z e cos z e ∘ -z (∘ -z) (∘ -z) A-sin A- cos A- ∘ -- (∘ -) (∘ --) zA-sinh zA- cosh zA- ( ) (√ -------) z log √---z----- zlog --z2+Az+B- z2−Az+B z (-sin(β)-) arctan z+cos(β)

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

20.66.5 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

(20.89)

with initial conditions f₀ = h₀, f₁ = h₁, …, f_k−1 = h_k−1 (where h_j are given), it is possible to solve it in the following way. If the coeﬃcients a₁,…,a_k are constants, then the Z-Transform of (20.89) can be calculated using the shift equation, and results in a solvable linear equation for 𝒵{f_n}. Application of the Inverse Z-Transform then results in the solution of (20.89).
If the coeﬃcients a₁,…,a_k are polynomials in n then the Z-Transform of (20.89) constitutes a diﬀerential equation for 𝒵{f_n}. If this diﬀerential equation can be solved then the Inverse Z-Transform once again yields the solution of (20.89). Some examples of these methods of solution can be found in §20.66.6.

20.66.6 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) where P(z) = 9z³ − 2z² + 5z and Q(z) = z⁵ − 2z³ + 2z² − 3z + 2 = (z − 1)²(z + 2)(z² + 1), which can be inverted to give

f = 2n + (− 2)n − cos πn . n 2

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:

fn+2 − 4fn+1 + 3fn = 1

with initial conditions f₀ = 0, f₁ = 1. Giving

F(z)	= 𝒵{1}
	= .

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{f_n}. $(n+ 1) ⋅fn+1 − fn = 0$
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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