The Zeta transform: definition, theorems, examples

The zeta transform plays the same role for discrete-time systems and signals as the Laplace transform for continuous systems.

Zeta Transform

Definition of Z transform and first examples

The Z transform of a sequence f(n) with n=0,1,2.. is defined by:

Z[f(n)]=F(z)=\sum_{0}^{\infty }f(n)z^{-n}

If the series converges, the convergence region, which is the set of values of z in the complex plane for which the series converges, is different for each f(n). The transform is defined by n≥0 so as to allow a two-way correspondence between the function in time and its transform. Consequently, the time function will also have to be defined for n≥0, otherwise, its transform would ignore all the function for n<0.

Example 1. Consider f(n)=1, constant for any n≥0 (the so-called unit step):

Z(1)=\sum_{0}^{\infty }(1)z^{-n}=1+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+...

As you can see, this is the geometric series of reason 1/z that converges to:

Z(1)=\frac{1}{1-(\frac{1}{z})}=\frac{z}{z-1}   per \left|z \right|> 1

The convergence region is therefore made up of all complex z-s outside the circle of unit radius.

Example 2. Consider the Z transform of f(n)=d^{n} :

Z[d^{n}]=\sum_{0}^{\infty }d^{n}z^{-n}=1+(\frac{d}{z})+(\frac{d}{z})^{2}+(\frac{d}{z})^{3}+\cdots

The series of reason d/z converges to:

Z[d^{n}]=\frac{1}{1-(\frac{d}{z})}=\frac{z}{z-d}   per \left|\frac{d}{z} \right|< 1

and the convergence region consists of all complex z values outside the circle of radius |d|.

Zeta transform of common functions
Zeta transform of elementary functions. Clic to enlarge.

Fundamental theorems related to the zeta transform

Linearity. Assumed Z[f1(n)]=F1(z) and Z[f2(n)]=F2(z) with c1, c2 constants:

Z[c_{1}f_{1}(n)+c_{2}f_{2}(n)]=c_{1}F_{1}(z)+c_{2}F_{2}(z)

Translation over time. Assumed Z[f(n)]=F(z) , then:

Z[f(n+1)]=zF(z)-zf(0)

In particular, if the initial conditions are null, you will have:

Z[f(n+1)]=zF(z)

and, similarly:

Z[f(n-1)]=z^{-1}F(z)

Translation in z. Assumed Z[f(n)]=F(z), then placing a=constant:

Z[e^{anT}f(n)]=F(ze^{-aT})

Final value. If Z[f(n)]=F(z) converges by |z|>1 and all poles of (1-z)F(z) are within the circle of unit radius, then:

\displaystyle \lim_{n \to \infty }=\displaystyle \lim_{z \to 1}[(1-z^{-1})F(z)]

Initial value. If Z[f(n)]=F(z) and there exists the \displaystyle \lim_{z \to \infty }F(z) , then:

\displaystyle \lim_{n \to \o}f(n)=\displaystyle \lim_{z \to \infty }F(z)

Partial derivative. If Z[f(n,a)]=F(z,a), then:

Z[\frac{\partial f(n,a)}{\partial a}]=\frac{\partial F(z,a)}{\partial a}

Convolution over time. If Z[f_{1}(n)]F_{1}(z) and Z[f_{2}(n)]F_{2}(z), then:

Z[f_{1}(n)\ast f_{2}(n)]=F_{1}(z) \times F_{2}(z)

Applying the Z transform in solving equations to finite differences

The most common application of the Z transform is the one in which it is used to analytically solve finite difference equations. It allows to transform these equations into an algebraic equation that, once solved in z, will allow to obtain the solution in time y(n) by operating the inverse transform and calculating it for each n.

Example. Have the following equation at finite differences of the second order:

y(n+2)-y(n+1)+0,21y(n)=3x(n+1)+1,2x(n)

and suppose we want to find the answer to the unit step in the case of null initial conditions.

Applying the zeta transform to the equation, remembering that we are in the hypothesis of null initial conditions:

(z^{2}-z+0,21)Y(z)=(3z+1,2)X(z)

from here we can get the discrete transfer function (ratio of output to input in the variable z:)

G(z)=\frac{Y(z)}{X(z)}=\frac{3z+1,2}{z^{2}-z+0,21}

We can see that the function G(z) has a zero in 0.4 and two poles in p1=0.3 and p2=0.7.

To obtain the output in response to a unit step, we will now have to multiply the G(z) by X(z). The latter is nothing more than the zeta transform of the unit step which, as can be seen from the table, holds:

X(z)=\frac{z}{z-1}

Thus, the answer to the unit step in the z domain will be:

Y(z)=G(z)X(z)=\frac{z(3z+1,2)}{(z-1)(z^{2}-z+0,21)}

For the calculation of the y(n), we now proceed with the inverse transform, first applying the decomposition into partial fractions:

Y(z)=\frac{20}{z-1}+\frac{7,5z}{z-0,3}-\frac{27,5z}{z-0,7}

Using the table already seen, we can now calculate y(n) through the inverse transform:

y(n)=20+7,5(0,3)n-27,5(0,7)n

Cover photo by: Jorge Salvador

About Carlo Bazzo 9 Articles
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