The Z-transform does to discrete-time signals and systems what the Laplace transform does to a continuous-time signals and systems. <\/p>\n\n\n\n\n\n\n\n
The Z-transform of the discrete-time signal $u[n]$, $n \\in \\mathbb{N}$, is the function of the complex variable $z$ defined by<\/p>\n\n\n\n
$$U[z]=\\mathcal{Z} \\{ u[n] \\} = \\sum_{n=0}^{\\infty} u[n] \\, z^{-n}.$$ <\/p>\n\n\n\n
As with the Laplace transform, $u[n]$ is assumed to be zero for negative $n$, that is $u[n] = 0$, $n < 0$.<\/p>\n\n\n\n
Exponential discrete-time signals are of great importance and the calculation<\/p>\n\n\n\n
$$
\\mathcal{Z}\\{ a^{n} \\} = \\sum_{n=0}^{\\infty} a^n \\, z^{-n} = \\frac{1}{1 – a z^{-1}}, \\quad |z|>|a|.
$$<\/p>\n\n\n\n
reveals many of the inner workings of the Z-transform. The Z-transform of the exponential has a domain of convergence which is the exterior of the disc of radius $|a|$. As with the Laplace transform, the Z-transform has no trouble handling unbounded signals, e.g. $|a| > 1$, by adapting its region of convergence.<\/p>\n\n\n\n
A sufficient condition for a discrete-time signal to admit a Z-transform is for it to be of exponential order, that is if there exists $a > 0$, and $M > 0$ such that<\/p>\n\n\n\n
$$|u[n]| \\leq M a^n, \\quad n \\in \\mathbb{N}.$$<\/p>\n\n\n\n
Indeed, with $z = r e^{j \\theta}$<\/p>\n\n\n\n
$$|U[z]|\\leq \\sum_{n=0}^{\\infty} |u[n]| \\, |z^{-n}| \\leq M \\sum_{n=0}^{\\infty} a^n \\, r^{-n} = \\frac{1}{1 + a r^{-1}}$$<\/p>\n\n\n\n
is bounded for all $r > a$. The region of convergence of a typical Z-transform is the exterior of a disc centered at the origin, whereas the region of convergence of a Laplace transform were a right half-plane. <\/p>\n\n\n\n
The Z-transform has many properties that are analogous to the properties of the Laplace transform. For example, one useful property is the delay property<\/p>\n\n\n\n
$$\\mathcal{Z} \\{ u[n-k] \\} = z^{-k} U[z], \\quad k \\geq 0,$$<\/p>\n\n\n\n
obtained by assuming $u[n] = 0$, $n < 0$, and changing indices as in <\/p>\n\n\n\n
$$\\mathcal{Z} \\{ u[n-k] \\} = \\sum_{n=0}^{\\infty} u[n-k] \\, z^{-n} = \\sum_{m=-k}^{\\infty} u[m] \\, z^{-m-k} =z^{-k} U[z].$$<\/p>\n\n\n\n