1.3. \(z\)-Transform

• The \(z\)-transform \(X(z)\) of a signal \(x[n]\) is the power series (or infinite-degree polynomial) with the values of \(x[n]\) as coefficients:

  (1.1)\[\begin{equation} X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}. \end{equation}\]

  Notation

  We use the notation \(x[n] \stackrel{z}{\longleftrightarrow} X(z)\) or \(X(z)=\mathcal{Z}(x[n])\) to say that \(X(z)\) is the \(z\)-transform of \(x[n]\).

• The region of convergence (ROC) of \(X(z)\) is the set of complex numbers at which the power series converges.

  Tip

  From the standard power series results [Rud76], there must be non-negative \(r_+\) and \(r_-\) (both could be \(\infty\)), specifying the regions \(\mathcal{R}_+ =\{z \in \mathbb{C}: |z| > r_+ \}\) and \(\mathcal{R}_- =\{z \in \mathbb{C}: |z| < r_- \}\) such that \(\sum_{n=-\infty}^{\infty} x[n] z^{-n}\) converges for \(z \in \mathcal{R}_+ \cap \mathcal{R}_-\) and diverges for \(z \notin \mathcal{\bar R}_+ \cap \mathcal{\bar R}_-\), where the bar denotes the closure of a subset in \(\mathbb{C}\).

  • If \(r_- < r_+\), then the power series does not converge at any \(z \in \mathbb{C}\) and \(x[n]\) does not have a \(z\)-transform.

  • If \(x[n]\) is causal (i.e., \(x[n]=0\) for all \(n<0\)), then \(r_- = \infty\) and hence the ROC includes \(\mathcal{R}_+\).

  • If \(x[n]\) is anti-causal (i.e., \(x[n]=0\) for all \(n \geq 0\)), then \(r_+ = 0\) (and the power series also trivially converges at \(z=0\)) and hence the ROC includes \(\mathcal{R}_-\).

  • Convergence on the boundaries of \(\mathcal{R}_+\) and \(\mathcal{R}_-\) needs to be worked out for each specific \(x[n]\).

1.3.1. Examples

1. Consider the causal signal \(x_1[n] = a^n u[n]\), where \(a\) is a complex number. From (1.1), we have

   \[\begin{align*} X_1(z) &= \sum_{n=-\infty}^{\infty} x_1[n] z^{-n} \\ &= \sum_{n=-\infty}^{\infty} a^n u[n] z^{-n} \\ &= \sum_{n=0}^{\infty} \left( a z^{-1} \right)^n \\ &= \begin{cases} \frac{1}{1-az^{-1}}, & |z|>|a| \\ \text{diverges}, & |z| \leq |a|. \end{cases} \end{align*}\]

   Thus, the \(z\)-transform of \(x_1[n]\) is \(X_1(z) = \frac{1}{1-az^{-1}}\) with ROC \(= \{|z|>|a|\}\).

   Notation

   Instead of writing \(\{ z \in \mathbb{C}: |z|>|a|\}\) all the time, we will henceforth use the shorthand notation \(\{|z|>|a|\}\).

2. Consider the anti-causal signal \(x_2[n] = -a^n u[-n-1]\), where \(a\) is a complex number. From (1.1), we have

   \[\begin{align*} X_2(z) &= \sum_{n=-\infty}^{\infty} x_2[n] z^{-n} \\ &= -\sum_{n=-\infty}^{\infty} a^n u[-n-1] z^{-n} \\ &= -\sum_{m=1}^{\infty} \left( a^{-1} z \right)^m \\ &= 1 - \sum_{m=0}^{\infty} \left( a^{-1} z \right)^m \\ &= \begin{cases} 1-\frac{1}{1-a^{-1}z}, & |z|<|a| \\ \text{diverges}, & |z| \geq |a|. \end{cases} \end{align*}\]

   Thus, the \(z\)-transform of \(x_2[n]\) is \(X_2(z) = \frac{1}{1-az^{-1}}\) with ROC \(= \{|z|<|a|\}\).

   Caution

   • Note that \(X_1(z)\) and \(X_2(z)\) have the same expression but different ROCs.

   • In general, the expression AND the ROC of \(X(z)\) together uniquely determine \(x[n]\).

3. Consider the signals \(u[n]\), \(x_3[n] = \frac{1}{n} u[n-1] \) and \(x_4[n] = \frac{1}{n^2}u[n-1] \).

   • From Example 1. above, the \(z\)-transform of \(u[n]\) is \(U(z) = \frac{1}{1-z^{-1}}\) with ROC \(= \{ |z|>1\}\).

   • Clearly, \(X_3(z) = \sum_{n=1}^{\infty} \frac{z^{-1}}{n}\) converges for \(|z|>1\) and diverges for \(|z|<1\). It turns out that \(X_3(z)\) also converges for every \(z\) on the unit circle (i.e., \(|z|=1\)) except at \(z=1\) where the power series diverges. Hence, the ROC of \(X_3(z)\) is \(\{ |z| \geq 1\} \setminus \{z=1\}\).

   • It can also be shown that \(X_4(z) = \sum_{n=1}^{\infty} \frac{z^{-n}}{n^2}\) converges for \(|z| \geq 1\) and diverges for \(|z| < 1\). Hence, the ROC of \(X_4(z)\) is \(\{ |z| \geq 1\}\).

1.3.2. Properties and tables

Typically, it is more convenient to find the \(z\)-transform of a signal by consulting a pair of tables listing common \(z\)-transform properties and \(z\)-transform pairs. For example, see the tables from Wikipedia.

Use-of-tables example:

4. Consider

   \[\begin{equation*} x_5[n] = n \cos(\hat\omega_0 n) u[n] = \frac{1}{2} n e^{j \hat\omega_0 n} u[n] + \frac{1}{2} n e^{-j \hat\omega_0 n} u[n]. \end{equation*}\]

   Hence, the \(z\)-transform of \(x_5[n]\) is

   \[\begin{align*} X_5(z) &= \frac{1}{2} \mathcal{Z}\left( ne^{j \hat\omega_0 n} u[n] \right) + \frac{1}{2} \mathcal{Z}\left( ne^{-j \hat\omega_0 n} u[n] \right) & \text{(linearity)} \\ & = \frac{e^{j \hat\omega_0} z^{-1}}{2(1 - e^{j \hat\omega_0} z^{-1})^2} + \frac{e^{-j \hat\omega_0} z^{-1}}{2(1 - e^{-j \hat\omega_0} z^{-1})^2} & \text{(directly from $z$-transfrom pair table)} \\ & = \frac{(\cos\hat\omega_0) z^{-1} -2 z^{-2} + (\cos\hat\omega_0) z^{-3}}{\left( 1 - (2 \cos\hat\omega_0) z^{-1} + z^{-2} \right)^2} \end{align*}\]

   with ROC \(=\{ |z|>1\}\).

1.3.3. Inverse \(z\)-transform

Let \(x[n] \stackrel{z}{\longleftrightarrow} X(z)\). The inverse \(z\)-transform formula is

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

where \(C\) is any closed counterclockwise contour in the ROC of \(X(z)\) enclosing the origin.

Tip

• Since \(X(z)\) is analytic (a power series), the contour integral may be calculated with the help of the Cauchy integral formula and/or the residue theorem [BC14].

• If \(X(z)\) is rational, inverse \(z\)-transform of \(X(z)\) may be performed by long division, or more commonly by the method of partial fraction expansion:

  1. Obtain the partial fraction expansion of \(X(z)\).

  2. Perform table lookup to obtain the inverse \(z\)-transform of each partial fraction.

  3. Use linearity to combine the inverse \(z\)-transforms of all component partial fractions.

• Watch this video by Prof. Oppenhiem from MIT Open Courseware for more discussions and examples.

1.3.4. Transfer function

• The transfer function \(H(z)\) of an LTI system is the \(z\)-transform of its impulse response \(h[n]\), i.e., \(h[n] \stackrel{z}{\longleftrightarrow} H(z)\).

• From the fact that an LTI system is stable if and only if \(\sum_{n=-\infty}^{\infty} |h[n]| < \infty\), the LTI system is stable if and only if the ROC of its transfer function \(H(z)\) contains the unit circle.

• For a causal FIR filter, \(H(z) = \sum_{k=0}^M b_k z^{-k}\).

  For a causal IIR filter, \(H(z) = \frac{\sum_{k=0}^M b_k z^{-k}}{\sum_{k=0}^N a_k z^{-k}}\)

  In both cases, \(H(z) = \frac{B(z)}{A(z)}\) is rational, where \(B(z)\) and \(A(z)\) are a numerator polynomial and a denominator polynomial in \(z^{-1}\).

1.3.5. Poles and zeros

• Pole: \(z_p \in \mathbb{C}\) is a pole of \(H(z)\) if \(\lim_{z \rightarrow z_p} |H(z)| = \infty\).

• Zero: \(z_0 \in \mathbb{C}\) is a zero of \(H(z)\) if \(H(z_0)=0\).

Tip

For a rational \(H(z) = \frac{B(z)}{A(z)}\):

• The roots of \(B(z)\) gives the zeros of \(H(z)\).

• The roots of \(A(z)\) gives the poles of \(H(z)\).

• If \(z_0\) is a root for both \(B(z)\) and \(A(z)\), then it does not give a pole or zero because of cancellation of the same factor in \(B(z)\) and \(A(z)\).

• For a causal LTI system with a rational transfer function \(H(z)\), the system is stable if and only if all its poles are strictly inside the unit circle.