Evaluating the Inverse \(z\) Transform

Given a square-summable sequence \(\{c_j\}\), the \(z\)-transform is

\[ g(z) = \sum_{j=-\infty}^{\infty} c_j\, z^j, \]

well-defined at least on the unit circle \(z = e^{i\omega}\), \(\omega \in [0, 2\pi]\). The sequence \(\{c_k\}\) can be recovered by the inversion formula

(175)\[c_k = \frac{1}{2\pi i} \int_\Gamma g(z)\, z^{-k-1}\, dz, \qquad k = 0, \pm 1, \pm 2, \ldots\]

where \(\Gamma\) denotes the unit circle and the integral is a contour integral.[1] In this section we give simple formulas for evaluating (175) using residues. Virtually no knowledge of complex analysis is required to use these formulas.

Poles and Residues

A pole of \(g(z)\) is a point \(z_0\) in the complex plane where \(g(z) \to \infty\) as \(z \to z_0\). In this book, \(g(z)\) is almost always a rational function—a ratio of finite-order polynomials in \(z\)—so the poles are simply the zeros of the denominator.

Test for poles of order \(m\): If for some positive integer \(m\) the function

(176)\[\phi(z) = (z - z_0)^m\, g(z)\]

can be defined[2] with \(\phi(z_0) \neq 0\), then \(g(z)\) has a pole of order \(m\) at \(z_0\).

For example, if \(g(z) = 1/(1-\lambda z)^r\), then \(g(z)\) has a pole of order \(r\) at \(z = \lambda^{-1}\).

Definition of residue: Suppose \(g(z)\) has a pole of order \(m\) at \(z = z_0\), and define \(\phi(z) = (z-z_0)^m g(z)\). The residue at \(z_0\) is

(177)\[\operatorname{res}(z_0) = \frac{\phi^{(m-1)}(z_0)}{(m-1)!},\]

where \(\phi^{(m-1)}\) denotes the \((m-1)\)th derivative. When \(m = 1\) (a simple pole), this reduces to

(178)\[\operatorname{res}(z_0) = \lim_{z \to z_0} (z - z_0)\, g(z).\]

The Residue Formulas

The inversion integral (175) can be evaluated by either of two equivalent formulas:[3]

(179)\[\begin{split}c_j = \frac{1}{2\pi i}\int_\Gamma g(z)\, z^{-j-1}\, dz = \begin{cases} \text{sum of residues of } g(z^{-1})z^{j-1} \text{ at poles inside unit circle} \\[4pt] \text{sum of residues of } g(z)z^{-j-1} \text{ at poles inside unit circle.} \end{cases}\end{split}\]

Choose whichever branch avoids poles of order greater than one at \(z = 0\).

Example 1: \(g(z) = 1/(1-\lambda z)\), \(|\lambda| < 1\)

Using the first branch of (179), the function \(g(z^{-1})z^{-j-1} = z^{j-1}/(1-\lambda z^{-1})\) has:

• For \(j > 0\): a single simple pole at \(z = \lambda\) (inside the unit circle), with residue \(\lambda^j\). So \(c_j = \lambda^j\).

• For \(j < 0\): using the second branch, \(z^{-j-1}/(1-\lambda z)\) has no poles inside the unit circle, so \(c_j = 0\).

• For \(j = 0\): using the second branch, \(z^{-1}/(1-\lambda z)\) has a simple pole at \(z = 0\) with residue \(1\), so \(c_0 = 1\).

Thus,

\[\begin{split} c_j = \begin{cases} \lambda^j & j \geq 0, \\ 0 & j < 0. \end{cases} \end{split}\]

(These results are more easily obtained by expanding \(1/(1-\lambda z)\) as a geometric series, but residue calculations are faster for more complex examples.)

Example 2: The \(n\)th-Order AR Covariance Generating Function

Consider the covariance generating function

\[ g_y(z) = \frac{1}{\prod_{j=1}^n(1-\lambda_j z)\,\prod_{j=1}^n(1-\lambda_j z^{-1})} = \frac{z^n}{\prod_{j=1}^n(1-\lambda_j z)\,\prod_{j=1}^n(z-\lambda_j)}, \]

where \(|\lambda_j| < 1\) for \(j = 1, \ldots, n\). Applying (179),

(180)\[\begin{split}c_y(\tau) = \begin{cases} \text{sum of residues of } \dfrac{z^{n-\tau-1}}{\prod_{j=1}^n(1-\lambda_j z)\,\prod_{j=1}^n(z-\lambda_j)} \text{ inside } \Gamma, & \tau \leq 0, \\[10pt] \text{sum of residues of } \dfrac{z^{n+\tau-1}}{\prod_{j=1}^n(1-\lambda_j z)\,\prod_{j=1}^n(z-\lambda_j)} \text{ inside } \Gamma, & \tau \geq 0. \end{cases}\end{split}\]

Use the first line for \(\tau \leq 0\) and the second for \(\tau \geq 0\) to avoid higher-order poles at \(z = 0\). In each case, the only poles inside the unit circle are the simple poles at \(\lambda_1, \ldots, \lambda_n\). For \(\tau \geq 0\) the residue at \(\lambda_j\) is

\[ \operatorname{res}(\lambda_j) = \frac{\lambda_j^{n+\tau-1}} {\prod_{k=1}^n(1-\lambda_k\lambda_j)\,\prod_{k=1,\, k\neq j}^n(\lambda_j-\lambda_k)}, \qquad \tau \geq 0. \]

Summing over \(j\) and using the symmetry \(c_y(\tau) = c_y(-\tau)\):

(181)\[c_y(\tau) = \sum_{j=1}^{n} \frac{\lambda_j^{n+|\tau|-1}} {\prod_{k=1}^{n}(1-\lambda_k\lambda_j)\,\prod_{k=1,\, k\neq j}^n(\lambda_j-\lambda_k)}.\]

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[1]

Churchill et al. (1974) is a good reference on the complex analysis used here. Gabel and Roberts (1973) is a good reference on operational methods for linear dynamic systems.

[2]

The function \(\phi(z)\) must also be “analytic” at \(z_0\), meaning its derivative exists at \(z_0\) and nearby. In our examples this requirement is routinely satisfied.

[3]

The second representation is standard (e.g., Churchill, Brown, and Verhey 1974). The first follows by noting that \(g(z^{-1}) = \sum_j c_{-j} z^j \equiv \sum_j d_j z^j\) with \(d_j = c_{-j}\); applying the second representation to compute \(d_j\) then gives \(c_j = \text{sum of residues of } g(z^{-1})z^{j-1}\) inside the unit circle.