What Is the z-Transform?
This Frequent Engineering Question gives a quick overview of an important mathematical technique used in digital signal processing, calculating the z-transform.
If you’ve studied the Laplace transform, you’re familiar with the concept of transforming a function of time into a function of frequency. The variable used in the Laplace transform is s, which represents complex frequency, i.e., it is frequency with a real and imaginary part:
\[s=\sigma+j\omega\]
You can think of the z-transform as a discrete-time version of the Laplace transform. We use the variable z, which is complex, instead of s, and by applying the z-transform to a sequence of data points, we create an expression that allows us to perform frequency-domain analysis of discrete-time signals.
With the z-transform, we can create transfer functions for digital filters, and we can plot poles and zeros on a complex plane for stability analysis. The inverse z-transform allows us to convert a z-domain transfer function into a difference equation that can be implemented in code written for a microcontroller or digital signal processor.
How to Calculate the z-Transform
The relationship between a discrete-time signal x[n] and its one-sided z-transform X(z) is expressed as follows:
\[X(z)=\sum_{n=0}^\infty x[n]z^{-n}\]
This summation begins as a sequence of individual values, and since we are summing from n = 0 to n = infinity, the sequence is of infinite length. What can we do with an infinite sequence of summed elements?
This is where convergence comes in.
Convergence with the z-Transform
Consider the unit step, which we define as follows:
\[u[n]=\begin{cases}0 & n < 0\\1 & n \geq 0\end{cases}\]
This results in the following summation:
\[X(z)=\sum_{n=0}^\infty u[n]z^{-n}=z^0+z^{-1}+z^{-2}+z^{-3}+\ …\]
An infinite sequence of summed numbers can converge to one number. For example:
\[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ …\ =2\]
If we continue the sequence according to the same pattern and sum all the elements, as the number of elements approaches infinity, the sum approaches the number 2. With the z-transform, the elements include a variable, but convergence can still occur—the sequence converges to a variable expression instead of a number.
The sequence shown above for the unit step converges as follows:
\[X(z)=\sum_{n=0}^\infty u[n]z^{-n}=z^0+z^{-1}+z^{-2}+z^{-3}+\ …=\frac{z}{z-1}\]
Not all z-transforms will converge. Here are examples of discrete-time signals that have “well-behaved” z-transforms; note that all of these x[n] functions are multiplied by the unit step, such that the z-transform operation is applied to a sequence that is zero for n < 0.
\[x[n]=nu[n]\ \ \ \ \ \ \ \ \ \ \ \ \ \ X(z)=\frac{z}{(z-1)^2}\]
\[x[n]=a^nu[n]\ \ \ \ \ \ \ \ \ \ \ \ \ \ X(z)=\frac{z}{z-a}\]
\[x[n]=\sin(\omega n)u[n]\ \ \ \ \ \ \ \ \ \ \ \ \ \ X(z)=\frac{z\sin(\omega)}{z^2-2z\cos(\omega)+1}\]
What additional questions do you have regarding the z-transform? Let us know in the comments below.
Robert - you said: “The inverse z-transform allows us to convert a z-domain transfer function into a difference equation.” Don’t you mean the “unit-sample” response? -Bernie
EXPLANATION of Comment Above
Suppose we have a transfer function:
H(z) =Y(z)/X(z) = 1 / [1 - 0.5z^(-1) - 0.3z^(-2)]
If we want the corresponding difference equation (DE) we DO NOT take the Inverse z-Transform (IZT) of H(z) (which would be the “impulse response” (IR), h(n) – or technically, the “unit-sample response”).
Since z^(-1) is a delay in the z-domain we can just WRITE DOWN the DE from H(z) by simple INSPECTION:
y(n) = x(n) + 0.5 y(n-1) + 0.3 y(n-2) {DE}
That’s it! Three terms. No inverse z-transform.
The IZT (thus the IR) is not an equation at all but rather a sequence (often infinite length) of numbers, which may or may not be represented by its own, assessable (by a variety of methods) closed-form expression.
In our example, the IZT/IR is conveniently found by numerical iteration of the DE with x(n)=1 for n=0, and x(n) = 0 for all other n: e.g., for n=0,1,2,3 …. ,
y = 0.50000, 0.55000, 0.42500, 0.37750, 0.31625, 0.27138, 0.23056, 0.19669, 0.16752, 0.14277, 0.12164, 0.10365…..
Thus h(n) - Lots of numbers!
-Bernie
How do you code the sampled value of an output variable using a digital filter using z^-1, z^-2, z^-3, et cetera ?