Editing Z-transform (section)

==Relationship to Laplace transform==
{{further|Laplace transform#Z-transform}}

===Bilinear transform===
{{Main|Bilinear transform}}
The '''bilinear transform''' can be used to convert continuous-time filters (represented in the Laplace domain) into discrete-time filters (represented in the Z-domain), and vice versa. The following substitution is used:
:<math>s =\frac{2}{T} \frac{(z-1)}{(z+1)}</math>
to convert some function <math>H(s)</math> in the Laplace domain to a function <math>H(z)</math> in the Z-domain ([[Bilinear transform|Tustin transformation]]), or
:<math>z =e^{sT}\approx \frac{1+sT/2}{1-sT/2}</math>
from the Z-domain to the Laplace domain. Through the bilinear transformation, the complex ''s''-plane (of the Laplace transform) is mapped to the complex z-plane (of the z-transform). While this mapping is (necessarily) nonlinear, it is useful in that it maps the entire <math>j\omega</math> axis of the ''s''-plane onto the [[unit circle]] in the z-plane. As such, the Fourier transform (which is the Laplace transform evaluated on the <math>j\omega</math> axis) becomes the discrete-time Fourier transform. This assumes that the Fourier transform exists; i.e., that the <math>j\omega</math> axis is in the region of convergence of the Laplace transform.

=== Starred transform ===
{{Main|Starred transform}}
Given a one-sided Z-transform <math>X(z)</math> of a time-sampled function, the corresponding '''starred transform''' produces a Laplace transform and restores the dependence on <math>T</math> (the sampling parameter):
:<math>\bigg. X^*(s) = X(z)\bigg|_{\displaystyle z = e^{sT}}</math>

The inverse Laplace transform is a mathematical abstraction known as an ''impulse-sampled'' function.