Until now we have assumed that time is continuous. In other words, time can take on any value, and this is generally true of linear physical systems involving components as capacitors, masses, thermal resistances... However, computers are increasingly being integrated into systems. For the computer, time is not continuous, it passes in discrete intervals. So whenever a computer is being used, it is important to understand the ramifications of the inherently discrete nature of time. As with the Laplace Transform, we will assume that functions of interest are equal to zero for time less than zero.
Although some data associated with physical systems is inherently discrete-time (daily high temperatures, stock closings...) most data starts out as continuous time data and becomes discrete through a process called sampling. For physical systems of the type considered in this course, the sampling process is usually carried out in practice through the use of an analog-to-digital (A/D) convertor. The process is often depicted as shown in the diagram at the right. In this diagram x(t) represents a continuous-time signal that is sampled every T seconds, the resulting signal is called x*(t). This represents a continuous-time signal that is measured by a computer every T seconds that results in a sampled signal. A common example of this is music stored on a CD. The continous-time signal is generated by the human voice, pianos, guitars... and is sampled 44,100 times per second. The resulting sequence of numbers is stored on the disc for replay at a later time (Note).
The signal x(t) is also a continous-value signal; i.e., it can take on any one of an infinite number of values. After it is sampled and stored in a computer it is necessarily approximated by one of a finite number of values. This process is call quantization. We will ignore this aspect of the conversion process and assume that x*(t) can take on any value.
There are a number of ways to represent the sampling process mathematically. One way that is commonly used is to immediately represent the sampled signal by a series x[n]. In this scheme the first sample (i.e., the first time the switch closes, is x, the next sample is x, and so on. This technique has the advantage of being very simple to understand, but makes the connection of the sampled signal to the Laplace Transform easy to understand. We will use a different technique that is a bit more complicated mathematically but will, in the long run, yield some physical insights not afforded by the simpler technique. To understand the sampling process in this paradigm, first consider a signal that is a train of evenly spaced impulse functions. We start the sampling function with k=0 because all of the functions that we consider will be zero for time less than zero. (Note)
There is a double-sided Z Transform that takes the limit of the summation from negative to positive infinity much like the double-sided Laplace Transform. We will not consider the double sided transform here.
This function is shown below:
Now we can define
Since the product
is equal to zero except at t=kT, we can rewrite the last equation as
This sampling process can be clarified with an image. The graph below shows x(t),
If we multiply this by the impulse train, we get the sampled function, x*(t). For this figure, T=0.1, and the function x(t) is shown as a thin dotted red line for reference.
The key insight of this whole process is that the sampled function, x*(t), is simply a train of impulses whose magnitude is given by the height of the function x(t) (at the times x(kT)).We can rewrite the sampled signal as
where x[k] is the magnitude of the kth element in the sequence and x[k]=x(kT). Generally when we consider a sequence we will implicitly assume a sampling interval, T, and simply use x[k].
Since we now have a time domain signal, we wish to see what kind of analysis we can do in a transformed domain. Let's start by taking the Laplace Transform of the sampled signal:
Since x[k] is a constant, we can (because of Linearity) bring the Laplace Transform inside the summation.
This obviously looks different than the Laplace Transforms we have seen in the past. For one thing, there is an infinite sum. For another, we are used to seeing functions in the Laplace domain that are the ratios of polynomial in the variable 's,' not exponentials in 's.'
To simplify the expression a little bit, we will use the notation
We will call this the Z Transform and define it as
We say that X(z) is the Z Transform of x[k], and we can write this as
Likewise x[k] is the inverse Z Transform of X(z), or
As a simple example, consider the sequence of sampled numbers, x[k]
For this sequence, the Z Transform of x[k] is given by:
This may seem a complicated way to define sequences, but it turns out that many sequences of interest to us (impulse, unit step, ...) will have Z Transforms that can be expressed in a compact form.
This section uses a few infinite series.
In discrete time systems the unit impulse is defined somewhat differently than in continuous time systems.
The Z Transform is given by
From the definition of the impulse, every term of the summation is zero except when k=0. So
Note that this is the same as the Laplace Transform of a unit impulse in continuous time. The fact that the Z Transform of an impulse is unity will yield many of the same properties, and allow for many of the same analysis techniques (i.e., transfer functions...) to be used for discrete time systems that were used for continuous time systems.
The unit step is one when k is zero or positive (Note).
u[k] is more commonly used to represent the step function, but u[k] is also used to represent other things. We choose gamma (γ) to avoid confusion (and because in the Z domain (Γ(z)) it looks a little like a step input).
The Z Transform is given by
We will use the latter form, a ratio of polynomials of positive powers of z (Note).
As a historical note, practitioners of Digital Signal Processing (DSP) use the first form (a ratio of polynomial of negative powers of z). We will use the second form, with positive powers of z.
Consider the exponential function:
Note that this function is just the exponential function that we are used to seeing (f(t)=e-αt, with t>0) after it has been sampled at t=kT.
With the Z Transform it is more common to get solutions in the form of a power series:
This is the same as the exponential function with a=e-αT. Just as with continuous time systems, most of our systems will have behaviors that consist of a sum of such exponentials.
The Z Transform of some other functions is given in the Table of Transforms.
Find the Z Transform of
We can write
As we found with the Laplace Transform, it will often be easier to work with the Z Transform if we develop some properties of the transform itself.
As with the Laplace Transform, the Z Transform is linear.
and a and b are constants.
An important property of the Z Transform is the time shift. To see why this might be important consider that a discrete-time approximation to a derivative is given by:
Note that this approximation involves not only t[k] but a time-shifted version as well, f[k+1]. This is referred to as a difference equation. Let's examine what effect such a shift has upon the Z Transform. The difference equation will serve the same purpose with discrete time systems and the Z Transform that differential equations served with continuous time systems and the Laplace Transform.
Consider the function x[k] shown below, which represents an arbitrary sequence. In this figure, and subsequent figures, the samples are shown as a "stem" plot (small circles on the end of a line).
If we shift the function to the right by 3 intervals, we get the function x[k-3], shown below.
To find the Z Transform of this shifted function, start with the definition of the transform:
Since the first three elements (k=0, 1, 2) of the transform are zero, we can start the summation at k=3.
Apply a change of variables
In general, a time delay of n samples, results in multiplication by z-n in the z domain.
A shift to the left is a bit more difficult. Let's explore it in the same way as we did the shift to the right. Consider the same sequence, x[k], as before. This time we shift it to the left by three samples to get x[k+3].
Let's try to develop the Z Transform in the same way as we did previously.
Now we run into a problem because we can't easily make the lower bound on the summation equal to zero. However, we can add in and subtract off the first three points, without changing the result.
Now we can combine the first two sums into one longer summation, and finish.
What we have done, in effect, is to subtract off the points that were shifted to negative values of k (highlighted with a darker gray background below).
Once these points are subtracted off, we end up with a sequence that is zero for k less than zero (below).
Let's consider a couple examples.
A shift of one to the left gives us:
A shift of two to the left gives us:
A shift of three gives us:
This is reminiscent of the differentiation property of the Laplace Transform (except instead of initial conditions on the differentiated variable with the Laplace Transform, we have initial values of the shifted variable with the Z Transform). Each successive difference results in a polynomial in z that is of increasingly higher order.
In general, for a shift to the left:
The convolution property for the Z Transform can be proved in much the same way as it was for the Laplace Transform. It is stated here without proof. If y[k] is the convolution of x[k] and h[k],
These and other properties of the Z Transform are found on the Z Transform Properties table.
Go on to the Inverse Z Transform