0.1 Discrete-time signals  (Page 8/10)

solving for $X\left(z\right)$ gives

and inversion of this transform gives the solution $x\left(n\right)$ . Notice that two initial values were required to give a unique solution just as theclassical method needed two values.

These are very general methods. To solve an $n$ th order DE requires only factoring an $n$ th order polynomial and performing a partial fraction expansion, jobs that computers are well suited to. There are problemsthat crop up if the denominator polynomial has repeated roots or if the transform of $y\left(n\right)$ has a root that is the same as the homogeneous equation, but those can be handled with slight modifications givingsolutions with terms of the from $n{\lambda }^n$ just as similar problems gave solutions for differential equations of the form $t{e}^{st}$ .

The original DE could be rewritten in a different form by shifting the index to give

which can be solved using the second form of the unilateral z-transform shift property.

Region of convergence for the z-transform

Since the inversion integral must be taken in the ROC of the transform, it is necessary to understand how this region is determined and what it meanseven if the inversion is done by partial fraction expansion or long division. Since all signals created by linear constant coefficientdifference equations are sums of geometric sequences (or samples of exponentials), an analysis of these cases will cover most practicalsituations. Consider a geometric sequence starting at zero.

with a z-transform

Multiplying by $a{z}^{-1}$ gives

and subtracting from [link] gives

Solving for $F\left(z\right)$ results in

The limit of this sum as $M\to \infty$ is

for $\left|z\right|>\left|a\right|$ . This not only establishes the z-transform of $f\left(n\right)$ but gives the region in the $z$ plane where the sum converges.

If a similar set of operations is performed on the sequence that exists for negative $n$

the result is

for $\left|z\right|<\left|a\right|$ . Here we have exactly the same z-transform for a different sequence $f\left(n\right)$ but with a different ROC. The pole in $F\left(z\right)$ divides the z-plane into two regions that give two different $f\left(n\right)$ . This is a general result that can be applied to a general rational $F\left(z\right)$ with several poles and zeros. The z-plane will be divided into concentricannular regions separated by the poles. The contour integral is evaluated in one of these regions and the poles inside the contour give the part ofthe solution existing for negative $n$ with the poles outside the contour giving the part of the solution existing for positive $n$ .

Notice that any finite length signal has a z-transform that converges for all $z$ . The ROC is the entire z-plane except perhaps zero and/or infinity.

Relation of the z-transform to the dtft and the dft

The FS coefficients are weights on the delta functions in a FT of the periodically extended signal. The FT is the LT evaluated on the imaginaryaxis: $s=j\omega$ .