4.2.1. Definition

The Z-transform of a sequence \(x[n]\) is given by

\[X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}\]

For the Z-transform to be meaningful the infinite summation has to converge to a finite value. Convergence of the summation over infinitely many terms is not guaranteed for all values of \(z\). The region of convergence (ROC) defines all values \(z\) for which the Z-transform converges (we say the Z-transform exists for those values of \(z\)). Determining the ROC is not a simple task. In these lecture notes we will most often only state the ROC without proof or use simple rules for systems/signals for which the ROC is known.

The Z-transform is such an important transform in practice because:

  • Discrete convolution in the time domain corresponds with multiplication in the \(z\)-domain. The Z-transform shares this property with the Fourier transform.

  • Any difference equation relating input and output of an LTI system is turned into an algebraic equation in \(z\) and the Z-transform of both input and output signal.

Especcially the last property is important in digital signal processing. Difference equations are the important characterizations of digital linear filters in practice.

The above definition of the Z-transform is the bilateral (or two-sided) Z-transform. It integrates from \(-\infty\) to \(+\infty\). The unilateral (or one-sided) Z-transform integrates from \(0\) to \(+\infty\). In signal processing this is a common definition for the Z-transform. In signal processing we are most often interested in causal systems and signals and for such systems both definitions of the Z-transform coincicde.

The inverse Z-transform is a mathematically complex transform. It is given as a line integral:

\[x[n] = \frac{1}{2\pi j}\oint_C X(z)z^{n-1}dz\]

where \(C\) is a closed path containing the origin and entirely in the ROC. Don’t be too much bothered in case you find it hard to see what is going on here. In (digital) signal processing we mostly use Z-transform pairs \(x[n]\longleftrightarrow X(z)\) of simple functions without too much use of the definition of the inverse transform.

Table 4.3 Z-Transform

Synthesis Equation

Analysis Equation

\(x[n] = \frac{1}{2\pi j}\oint_C X(z)z^{n-1}dz\)

\(X(z) = \sum_{n=\infty}^{\infty} x[n] z^{-n}\)

The proof of the synthesis equation given the analysis equation is remarkably simple and follows along the same line of thought as was used in the similar proof for the Fourier transform. This proof is not part of the obligatory theory for this course. The analysis equation is:

\[X(z) = \sum_{k=-\infty}^{\infty} x[k] z^{-k}\]

Multiplying both sides with \(z^{n-1}\) we get:

\[\begin{split}X(z) z^{n-1} &= \left( \sum_{k=-\infty}^{\infty} x[k] z^{-k} \right) z^{n-1}\\ &= \sum_{k=-\infty}^{\infty} x[k] z^{-k+n-1}\end{split}\]

Then integrating along a closed contour that is entirely in the region of convergence of \(X(z)\) and multiplying with \(1/(2\pi j)\) we get:

\[\begin{split}\frac{1}{2\pi j} \oint_C X(z) z^{n-1} dz &= \frac{1}{2\pi j} \oint_C \left(\sum_{k=-\infty}^{\infty} x[k] z^{-k+n-1}\right)dz\\ &= \frac{1}{2\pi j} \sum_{k=-\infty}^{\infty} x[k] \left(\oint_C z^{-(k-n+1)}dz \right)\end{split}\]

Consider the integral in the above expression:

\[\oint_C z^{-m}dz\]

The path C should be entirely in the ROC of \(X(z)\). As a path we select

\[z = r e^{j\phi}\]

the circle around the origin with radius \(r\). The radius \(r\) should be chosen such that the path is within the ROC. The path integral then becomes:

\[\begin{split}\oint_C z^{-m}dz &= \int_{0}^{2\pi} (r e^{j\phi})^{-m} \frac{dz}{d\phi} d\phi\\ &= \int_{0}^{2\pi} r^{-m} e^{-j\phi m} j r e^{j\phi}d\phi\\ &= j r^{1-m} \int_{0}^{2\pi} e^{j\phi (1-m)} d\phi\end{split}\]

For \(m=1\) we get \(2\pi j\) and for \(m\not=1\) the integration is always over one or more periods of the complex exponential and thus the integral is zero. Note that the value of \(r\) is irrelevant.

This leads to

\[\begin{split}\frac{1}{2\pi j} \oint_C X(z) z^{n-1} dz &= \frac{1}{2\pi j} \sum_{k=-\infty}^{\infty} x[k] 2\pi j\delta[k-n]\\ &= \sum_{k=-\infty}^{\infty} x[k] \delta[k-n]\\ &= x[n]\end{split}\]

So we have derived the synthesis equation starting from the analysis equation.