Z Transforms
Why Are Z Transforms Used?
The Z Transform
Signals With Multiple Poles
Important Facts About Z Transforms
Problems
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Why Are Z Transforms Used?

        You should know that Laplace transform methods are widely used for analysis in linear systems.  Laplace transform methods are used when a system is described by a linear differential equation, with constant coefficients.  However:

        At this point, there are an incredible number of systems we use every day that have digital components which satisfy difference equations.

        In continuous systems Laplace transforms play a unique role.  They allow system and circuit designers to analyze systems and predict performance, and to think in different terms - like frequency responses - to help understand linear continuous systems.  They are a very powerful tool that shapes how engineers think about those systems.  Z-transforms play the role in sampled systems that Laplace transforms play in continuous systems.

        There are numerous sampled systems that look like the one shown below.

        The processing can take many forms.
Goals

        In sampled systems you will deal with sequences of samples, and you will need to learn Z-transform techniques to deal with those signals.  In this lesson many of your goals relate to basic understanding and use of Z-transform techniques.  In particular, work toward these goals.

Later you will need to learn about transfer functions in the realm of sampled systems.  As you move through this lesson, there are other things you should learn.
What Is A Z Transform?

        You will be dealing with sequences of sampled signals.  Let us assume that we have a sequence, yk.  The subscript "k" indicates a sampled time interval and that yk is the value of y(t) at the kth sample instant.

        It's easy to get a sequence of this sort if a computer is running an A/D board, and measuring some physical variable like temperature or pressure at some prescribed interval, T seconds.  A sampled sequence like this plays the same role that a continuous signal plays in a continuous system.  It carries information just like a continuous signal.

        The Z transform, Y[z],  of a sequence, yk is defined as:

We will use the following notation.  A large "z" denotes the operation of taking a Z-transform (i.e., performing the sum above) and the result is usually denoted with an upper-case version of the variable used for the sampled time function, yk.

        The definition is simple.  Take the sequence, and multiply each term in the sequence by a negative power of z.  Then sum all of the terms to infinity.  That's it.

        Let's look at the transform of some simple functions to show how this definition works.


Example

E1   We are going to calculate the Z-transform of a simple sequence.  So that you can see the sequence in all its glory, we have a sequence calculator for you.  The expression for the sequence is:

To use the calculator, input a in the text box and click the Start button.  If you want to see the sequence for a different value of a, click the Clear button to clear the plot, enter a new value for a and click the Start button to replot.  We have used a value of 1.0 for the starting value.

The simulator lets you see the sequence for different values of a.  We want to get the Z-transform of the sequence for a general value of a.  To do that we calculate the sum above.

But, we know that yk = yoak.  Put that expression into the sum to get.

If you do the last sum, you should find.Then we get:


        Things work differently in the z-plane.  It's not the same as the s-plane where a pole must be in the left half of the s-plane to represent a function that decays to zero.  Here, for a function to decay to zero, the pole must be inside the unit circle - shown in red on the plot.  Here is an example.
Example

E2   You have a decaying sampled signal.  The signal is 2.0*(.9)k.  The Z-transform of the signal is:

We can plot the pole and zero for this function, and that plot is shown below.


        Let's think about this signal a little bit more.         Now, let's look at another signal.  We'll just change things by making a negative.  That won't change the algebra that we do, but it will change how the function looks.
Example

E3   We are going to calculate the Z-transform of another sequence.  The only difference from the last situation is that we are going to consider negative values for a.  We didn't look at negative values before, but we did ignore the possibility.  It's time to rectify that.

We still have the same expression for the sequence.

In the calculator, you can input negative values for both the starting value.  You should notice and think about the following points.  Try both of those possibilities, and then ponder the following.
Observations & Comments

    When we do the algebra for the sequence in the example above, we have

That's what we have been working with.        Again, it pays to compare our results to continuous signal results and to sum up.         There is one other interesting correlation we can make with analog signals.  In analog signals, decaying exponentials are important.  Note the following. This is a pretty interesting correspondence between sampled and analog signals.  Clearly, if you sample a decaying exponential you get the kind of sequence we have been discussing earlier.  Conversely, any time you have a decaying sequence you might want to think of the decaying sequence as a sampled decaying exponential - and there may well be times when that is advantageous.

Other Sampled Signals

       As with Laplace transforms there are a number of simple signals that are important.  Besides decaying signals, two important signals are the unit impulse and the unit step.  Before we go much further we will look at the Z-transforms of those two signals because they are important.

        We will first examine the unit impulse in the sampled world.  We'll call that impulse Dk.

Here is a picture of the sampled impulse, Dk.

        It's pretty easy to compute the Z-transform of the unit impulse.         We can see that the sequence, Dk, is going to play the role that the unit impulse (Dirac Impulse) plays in continuous signals and systems.  Just like the unit impulse, the transform of Dk is 1.

        Another important signal is the unit step.  Here is a unit step in the sampled signal domain.

        To get the expression, U[z] = z/(z - 1), the series can be summed using standard techniques from calculus.  Or, you can divide out the result - z/(z - 1) - to generate the series.  Either way, you should convince yourself that the series is, in fact, correct.

        To this point we have considered some simple functions in the sampled time domain.  They include the following:

        There are other interesting signals.  The ones considered to this point are among the simplest and most fundamental signals.  There are more complex signals.
Signals With Multiple Poles

        Clearly there are lots of interesting situations with multiple poles, and we need to examine some situations there.  Let's look at a case with two real poles.


Example/Simulation

E4  Here is a simulator that will compute the inverse transform of:

Do the following with this simulator.



        Now, at this point you have seen several signals. These signals have some interesting properties, and we can make a few observations.         We know that there are other kinds of signals with oscillations.  We especially know that there are probably signals that take many sample periods to complete an oscillation.  Think of measuring temperature every hour during the day.  If you have two identical days in a row, you would have 24 samples in a period.  In the next section we will examine signals with those properties.

Sampled Decaying Sinusoids

        A signal with two real poles is a simple case of a more general situation.  In continuous signals we often encounter decaying sinusoids.  Those signals have a time representation given below.

f(t) = e-akTsin(bt)

Note, this signal starts at zero for t = 0.  A plot of a signal of this sort is shown below.


Example

E5   Imagine that we have a decaying sinusoid - as above - and that we sample the sinusoid at intervals of T seconds.  We would have a sampled signal:

fk = f(kT) = e-akTsin(bkT)


        The decaying sinusoid is similar to the alternating decaying signal, but it has significant differences:         Now, let us consider the Z-transform of our decaying sinusoid signal.

fk = f(kT) = e-akTsin(bkT)

        We can use the expansion for the sine to give us

We can take each term in this expansion separately.  Let's start with the first part of this expansion.  Define a new function for this first part.  Call that function f1k.

f1k = [e-akT+jbkT]

We can rewrite the new sequence as: We know how to take the z transform of the sequence, fk.  That sequence is just the sum of two of the decaying signal sequences - even though we now have complex values for "a".  So, let's take the Z-transform of the sequence. We can combine these two terms, if that is desired.  The result is:

 These poles are interesting.


Example/Simulation

E6  Let's look at the numbers we used above.  Here they are repeated.

These are the values in the expression for the sequence, fk = f(kT) = e-akTsin(bkT), used above.  With these values we can compute the pole location and the real and imaginary part of the pole location.  Here is the computation. The plot above, repeated here, shows the pole locations.  The plot is consistent with our calculations.


E7  Here is a simulator in which you can enter the real and imaginary parts for a pair of complex poles in the z-plane.  In this simulator, do the following.


An Observation About Decay Rate

        In a sampled system, decay rate is also important, just as it is in analog systems.  In a sampled system we will need to discuss things in terms of decay to a certain percentage after a number of sample periods, and then relate number of sample periods to time using the sample period, T.

        To get a handle on decay rate remember that the poles of a sampled system with two complex poles are:

The critical observation to be made is that the response has terms like the expression below, which is repeated from the material above. Then, we should realize that the critical term is the envelope of the response, and that is determined by: We can note the following critical observation about these poles.         It is possible to get even more insight into how pole position is related to response.         We can relate these features of the response to the pole position.  Let's look at the example sequence we looked at earlier.  Here's the sequence and the pole positions are shown in the figure at the right below. Now, note the following for this example. The magnitude of the pole - the distance of the pole from the origin - determines the decay rate.  That distance is shown on the plot, and it is equal to  e-aT.  That's the amount the envelope of the response decays each sample period.  Remember, the poles are at:         The magnitude of both poles is |e-aT|.  The factor, ejbT, only changes the angle of the first pole - and the factor, e-jbT,  changes the angle of the second pole - but in the opposite direction.

        The same angle, bT, appears in both poles - once positively, and once negatively.  That angle determines how much the angle of the sinusoidal signal (which is also decaying!) changes each sample period since the response is given by:

fk = f(kT) = e-akTsin(bkT)

We will call bT the angular rate.


Example

E8  Now, consider the example we have been using.

  • a = 0.05,
  • b = .3
  • T = 1.0
  •         With b = .3 and T = 1.0, we have bT = .3 radians or about 17.2 degrees.  That means that the sinusoidal part of the response will move through a complete cycle in a little over 21 sample periods.  Check that using the simulator.


    Example

    E9   Here is a simulator which allows you to input the decay rate and the anglular change between samples.


    Problem

    P1   Say that you want the response to decay to within 5% of the starting value in 20 sample periods.  What should the decay factor be?

    Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

    Your grade is:


            The pole position determines the significant features of the response.         Let's look at some particular cases.  In the process we shoul come to a better understanding of how pole position affects response.

            One interesting particular case is when the poles are on the imaginary axis.  Here is a copy of the simulator we used earlier.


    Example

    E10   Here is the simulator.  Do the following.


            The various examples show behavior that is much like the behavior you would get in a continuous system by changing the damping ratio.

            What is interesting about the response for a ninety degree angular rate is that there are a lot of points that are zero.  To explain that consider the following:

            We can sum up what you should have obtained from this part of the lesson. And these were part of the set of goals enumerated early in this lesson.
    Some Important Facts & Z-Transform Theorems

            When dealing with sampled signals, there are some relationships you need to know.  In continuous systems, multiplication by s comes about by taking the transform of a derivative.  That's important in continuous systems because that's what eventually lets you apply Laplace transforms to differential equations and develop concepts like the transfer function.

            In sampled systems, multiplication by z is what helps you solve difference equations, and eventually that will let us develop equivalent transfer function concepts for sampled systems.  Some simple facts are the results for multiplication by a constant and the linearity theorem.  In what follows, we assume that we have a signal sequence, yk, and the transform of that signal sequence is Y[z].


    Linearity Theorem

           These two theorems are fairly easy to show, and the first is really a special case of the second - the linearity theorem - so we will just show how the second one comes about.  Here is a statement of the linearity theorem again:

            Now, we can follow the following steps, starting with the definition



    and that's the theorem.


    Delayed Signals - Shifting Theorem

           The most important theorem for Z-transforms is the real translation theorem - also known as the shifting theorem.  The shifting theorem says:

    You need to understand what yk-1 is.  If you think about it, when k is 3, for example, the value of the function is  y2.  In other words, the signal  yk-1 is the same as the signal  yk except that it takes on specific values one sample period later than  yk.

            This one takes a little more effort to demonstrate.  First, let's look at the shifted function.

     A function is shown to the right in red.  The same function - delayed by one sample period - is shown in blue on the plot.

            Now, let us look at the Z-transform of the shifted function.  Here is the summation we want to perform.


    Final Value Theorem

            There are other important results you will need to know for control systems.  One of those is the final value theorem.  Here is a statement of the result of the theorem.

            Paraphrasing the result, we say that the limit of the sequence as time (k) becomes large is the limit in the z-domain of (z - 1)/z times the transform of the function, Y[z].  (Note that (1 -  z-1) is the same as (z - 1)/z.)

            Consider the transform of a sequence, yk.

    Now, also consider the transform of the same sequence delayed by one sample period.

    Then, consider taking the difference between these two transforms.

    Now, take the limit of this difference as z approaches 1.  The sums on the right hand side of this equation can be written as:

    [yo - y-1] + [y1 - y0] + [y2 - y1] + . . .

    Notice how y-1 = 0, and how every term gets cancelled except the very last.  In what is shown y2 is left.  In the limit, the "final value" is left.  So, we have the final value theorem as a result.


    Problems
    Links To Related Lessons

    Other Lessons On Sampled Systems

    Moving Along - More Advanced Material
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