The purpose of this post is to answer some questions posed to me by ChatGPT regarding my understanding of the \(z\)-transform in digital signal processing and mathematics more broadly. My inquiry to it was simply: “Ask me some questions to test my understanding of the \(z\)-transform”. Here were ChatGPT’s questions, followed by my answers.
Question #1: What is the general definition of the \(z\)-transform for a discrete-time signal \(x_n\)?
My Answer: Given a sequence \((x_n)_{n=-\infty}^{\infty}\) of (possibly complex) numbers, the \(z\)-transform \(X(z)\) of the sequence \((x_n)_{n\in\textbf Z}\) is a map \(z\in\textbf C\to X(z)\in\textbf C\) on the complex plane \(\textbf C\) defined by the formal Laurent series about the origin:
\[X(z):=\sum_{n\in\textbf Z}x_n z^{-n}\]
It is also denoted \(X(z)=\mathcal Z\{x\}(z)\). More precisely, this is the bilateral \(z\)-transform and one can also consider sequences \((x_n)_{n=0}^{\infty}\) defined only for non-negative \(n\geq 0\) along with the unilateral \(z\)-transform \(X(z):=\sum_{n=0}^{\infty}x_n z^{-n}\). This then resembles the form of an (ordinary) generating function for the sequence \((x_n)_{n=0}^\infty\).
Question #2: For the discrete unit impulse signal \(x_n=\delta_{n,0}\), what is the z-transform?
My Answer: As an Einstein series over \(n\in\textbf Z\), we have \(X(z)=\delta_{n,0}z^{-n}\) which contracts to \(X(z)=z^{-0}=1\).
Question #3: What does the region of convergence (ROC) in the \(z\)-transform tell you about a system or signal?
My Answer: The region of convergence (ROC), perhaps more descriptively called the annulus of convergence, is simply the set of all \(z\in\textbf C\) for which the formal Laurent series defining the \(z\)-transform is actually a convergent infinite series. It provides information about both stability and causality.
Question #4: How do you determine if a discrete-time dynamical system is stable using the \(z\)-transform?
My Answer: Recalling that a discrete-time dynamical system \((x_n)_{n=-\infty}^{\infty}\) is said to be stable (or asymptotically stable) iff \(\lim_{n\to\infty}x_n=0\), the sufficient and necessary criterion for stability is that the poles of its \(z\)-transform \(X(z)\) must all lie in the interior of \(U(1)\) i.e. \(|z_{\text{poles}}|<1\).
Question #5: Compute the \(z\)-transform of \(x_n=\left(-\frac{1}{2}\right)^nu_n\) where \(u_n\) is the unit step function \(u_n:=[n\geq 0]\).
My Answer:
\[X(z)=\sum_{n=-\infty}^{\infty}\left(-\frac{1}{2}\right)^nu_nz^{-n}=\sum_{n=0}^{\infty}\left(-\frac{1}{2z}\right)^n=\frac{1}{1+1/2z}\]
where the annulus of convergence is \(|z|>1/2\).
Question #6: Explain how the \(z\)-transform simplifies the process of convolution between two discrete-time signals \(x_n,y_n\).
My Answer: The \(z\)-transform of the discrete convolution \((x*y)_n\) of two discrete-time signals \(x_n,y_n\) is the product of their individual \(z\)-transforms. This is one version of the convolution theorem:
\[\mathcal Z\{x*y\}(z)=\mathcal Z\{x\}(z)\mathcal Z\{y\}(z)\]
Put another way, to compute the convolution \((x*y)_n\) of two discrete-time signals, one can instead compute \((x*y)_n=\mathcal Z^{-1}(\mathcal Z\{x\}(z)\mathcal Z\{y\}(z))_n\).
Question #7: For the \(z\)-transform \(X(z)=\frac{z+1}{z^2-3z/4+1/8}\), what is the corresponding time-domain signal \(x_n\)?
My Answer: The \(z\)-transform can first be simplified using partial fractions:
\[X(z)=\frac{z+1}{z^2-3z/4+1/8}=\frac{z+1}{(z-1/2)(z-1/4)}=\frac{6}{z-1/2}-\frac{5}{z-1/4}\]
Using the standard complex analysis, one can check that the inverse \(z\)-transform is given by the contour integral:
\[x_n=\frac{1}{2\pi i}\oint_{\Gamma}X(z)z^{n-1}dz\]
where \(\Gamma\) is any counterclockwise-oriented curve that has winding number \(\text{Wind}(\Gamma,0)=1\) around the origin \(0\in\textbf C\) and lies within the annulus of convergence of the \(z\)-transform \(X(z)\) (which in this case is just \(\textbf C-\{1/2,1/4\}\)). Thus, the inverse \(z\)-transform yields:
\[x_n=6(1/2)^{n-1}-5(1/4)^{n-1}\]
Question #8: Explain how the \(z\)-transform is related to the discrete-time Fourier transform (DTFT).
The discrete time Fourier transform of a sequence \((x_n)_{n\in\textbf Z}\) is simply its \(z\)-transform \(X(z)\) but evaluated on \(U(1)\) i.e. \(z=e^{i\omega}\). Thus, the DTFT is:
\[X(e^{i\omega})=\sum_{n\in\textbf Z}x_ne^{-i\omega n}\]