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Z - transform

The Z-transform of an arbitrary, time-discretized function xt is defined by  
X(Z) \eq \cdots x_{-2} \,Z^{-2} \ +\ x_{-1} \,Z^{-1} \ +\ x_0 
 \ +\ x_1 \,Z \ +\ x_2 \,Z^2 \ +\ \cdots\end{displaymath} (17)
Give Z the physical interpretation of time delay by one time unit. Then Z2 delays two time units. Expressions like $X(Z)\,U(Z)$ and $X(Z)\,U(1/Z)$ are useful because they imply convolution and cross-correlation of the time-domain coefficients. (See FGDP).

Going on to consider numerical values for the delay operator Z, we discover that it is useful to ask whether X(Z) is finite or infinite. Numerical values of Z that are of particular interest are Z = +1, Z = -1, and all those complex values of Z which are unit magnitude, say, |Z| = 1 or  
Z \eq e^{{i} \, \omega \, \Delta t}\end{displaymath} (18)
where $\omega$ is the real Fourier transform variable. Taking $\omega$ to be real means that Z is on the unit circle. Then the Z-transform is a discrete Fourier transform. Our attention can be restricted to time functions with a finite amount of energy by demanding that U(Z) be finite for all values of Z on the unit circle |Z| = 1. Filter functions are always restricted to have finite energy.

The most straightforward way to say that a filter is causal is to say that its time domain coefficients vanish before zero lag, that is ut = 0 for t<0. Another way to say it is to say that U(Z) is finite for Z=0. At Z=0 the Z-transform would be infinite if the coefficients $u_{-1} ,\ u_{-2} ,$ etc. were not zero. For a causal function, each term in |U(Z)| will be smaller if Z is taken inside the disk |Z|<1 rather than on it. Thus convergence at Z=0 and on the circle |Z|=1 implies convergence everywhere inside the unit disk. So boundedness combined with causality means convergence in the unit disk. Convergence at Z = 0 but not on the circle |Z| = 1 would refer to a causal function with infinite energy, a case of no practical interest. What kind of function converges on the circle, at $Z = \infty$,but not at Z = 0? What function converges at all three places, Z = 0, $Z = \infty$, and |Z| = 1?

The filter $ 1/(1 \ -\ 2Z) $ can be expanded into powers of Z in (at least) two different ways. These are
{1 \over 1 \ -\ 2Z} &=& 1 \ +\ 2Z \ +\ 4Z^2 \ +\ 8Z^3 \ +\ \cdo...
 ...+\ {1 \over 2Z }\ +\ {1 \over 4Z^2 } \ +\ \cdots \right]
\nonumber\end{eqnarray} (19)
Which of these two infinite series converges depends on the numerical value of Z. For |Z| = 1 the first series diverges, but the second converges. So the only acceptable filter is anticausal. Is a series expansion unique? It is if it converges. Complex-variable theory proves this.

Let bt denote a filter. Then at is its inverse filter if the convolution of at with bt is a delta function. In the Fourier domain, we would say that filters are inverse to one another if their Fourier transforms are inverse to one another. Z-transforms can be used to define the inverse filter, say, A(Z) = 1/B(Z). Whether the filter A(Z) is causal depends on whether it is finite everywhere inside the unit disk, or really on whether B(Z) vanishes anywhere inside the disk. For example, B(Z) = 1 - 2Z vanishes at Z = 1/2. There A(Z) = 1/B(Z) must be infinite, that is to say, the series A(Z) must be nonconvergent at Z = 1/2. Thus--as we have just seen--at is noncausal. A most interesting case, called minimum phase, occurs when both a filter B(Z) and its inverse are causal. In summary:

causal $\mid B(Z)\mid\ <\ \infty\ $ for $\ \mid Z\mid\ \leq 1$
causal inverse $\mid 1/B(Z)\mid\ <\ \infty\ $ for $\ \mid Z\mid\ \leq 1$
minimum phase both above conditions

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Next: Review of impedance filters Up: IMPEDANCE Previous: Beware of infinity!
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