An Introduction to Z Transforms

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:

There are numerous systems that are described by difference equations - not differential equations - and those systems are common and different from those described by differential equations.
Systems that satisfy difference equations include things like:

Computer controlled systems - systems that take measurements with digital I/O boards or GPIB instruments, calculate an output voltage and output that voltage digitally. Frequently these systems run a program loop that executes in a fixed interval of time.

Other systems that satisfy difference equations are those systems with Digital Filters - which are found anywhere digital signal processing - digital filtering is done. That includes:

Digital signal transmission systems like the telephone system.
Systems that process audio signals. For example, a CD contains digital signal information, and when it is read off the CD, it is initially a digital signal that can be processed with a digital filter.

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.

In continuous systems, inputs and outputs are related by differential equations and Laplace transform techniques are used to solve those differential equations.
In sampled systems, inputs and outputs are related by difference equations and Z-transform techniques are used to solve those differential equations.

In continuous systems, Laplace transforms are used to represent systems with transfer functions, while in sampled systems, Z-transforms are used to represent systems with transfer functions.

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

An analog signal is converted to a digital form in an A/D.
The digital signal is processed somehow.
The processed digital signal is converted to an analog signal for use in the analog world.

The processing can take many forms.

In a voice transmission situation, the processing might be to band-limit the signal and filter noise from the signal.
In a control situation, a measurement might be processed to calculate a signal to control a system.
And there are many other situations.

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.

Given a sequence of samples in time,

Be able to calculate the Z-transform of the sequence for simple sequences.

Given a Z-transform,

Be able to determine the poles and zeroes of the Z-transform.
Be able to locate and plot the poles and zeroes in the z-plane.

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.

Given a Z-transform of a signal, and the pole locations,

Be able to relate distance from the origin to decay rate.
Be able to relate angle off the horizontal to the number of samples in a cycle of signal oscillation.

What Is A Z Transform?

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

y_k could be generated from a sample of a time function. For example:

y_k = y(kT), where y(t) is a continuous time function, and T is the sampling interval.
We will focus on the index variable, k, rather than the exact time, kT, in all that we do in this lesson.

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, y_k 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, y_k.

Z[y_k] = Y[z]

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:

y_k = y_oa^k

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 y_k = y_oa^k. Put that expression into the sum to get.

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

Y[z] = 1/[1 - a/z] = z/[z - a]
And we see that this function has a pole at z = a, and a zero at z = 0.
The pole is in the right half of the z-plane.
Despite that, this is the transform of a signal that decays to zero!

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:

2z/(z - 0.9)

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.

This signal decays to zero, just like a decaying exponential (like e^-t/^t)
This signal could, conceivably, have been generated by sampling a decaying exponential.
In the sampled world, this signal is probably going to play the same role as the decaying exponential plays in the continuous world.
In the sampled world, the transform of this signal has a pole at z = a. In the continuous world, the transform of e^-t/^t has a pole at s = -1/t.

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.

y_k = y_oa^k

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.

When a is negative, successive points in the plot alternate sign. In other words, there are oscillations, but they only take two sample periods. Remember that behavior. We'll revisit it later when we consider multiple real poles.
When you take the Z-transform, this function has a single pole at z = -a.

Can you sketch where the pole is when a = -0.5?

The simulator still works when a is larger than 1.0 or less than -1.0 (like a = -2.5). However, the response is not well behaved for those values.

Where is the pole for a larger than 1.0 or less than -1.0?

Observations & Comments

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

y_k = y_oa^k

That's what we have been working with.

The transform is given by: Y[z] = y_oz/(z - a),
We still have a pole at z = a.
If a is positive, that pole is in the right half of the z-plane, but that doesn't bother us in the z-plane like it does in the s-plane. If a is positive, as long as a < 1, the response settles out. If a > 1, the response grows without limit.
If a is negative, the pole is in the left half plane, and it is on the negative real axis. Interestingly that leads to oscillations that decay. You can't get oscillations in continuous systems unless you have at least two poles, so that's something you might not have expected.

Again, it pays to compare our results to continuous signal results and to sum up.

A decaying signal, y_k = y_oa^k, has a pole at z = a. However, unlike a decaying exponential, if a is negative, we can have oscillations in the decaying signal.
For the oscillations to decay, we must have |a| < 1. However, a can be either positive or negative, and that leads to the possibility of oscillations when a is negative.

There is one other interesting correlation we can make with analog signals. In analog signals, decaying exponentials are important. Note the following.

Say you have a decaying exponential. We can represent that with a time constant description:

Y(t) = y_oe^-t/^t

Now, consider sampling that decaying exponential. Assume that you sample every T seconds. Then the k^th sample (taken a t = kT) is given by:

Y(kT) = y_k = y_oe^-kT/^t = y_oe^(-T/^t)k = y_o[e^(-T/^t)k]

Now, you can think of this as y_k = y_oa^k with:

a = [e^(-T/^t)]

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 D_k.

D_kis one for n = 0.

In the continuous world the impulse is infinite for t = 0.
That's a big difference.

D_k is zero for all other k's - like the continuous impulse is zero for times that are not zero.

Here is a picture of the sampled impulse, D_k.

Remember that this is a sampled signal so it is not defined except for integer values of k.

It's pretty easy to compute the Z-transform of the unit impulse.

Earlier, we defined the Z-transform of a sequence, y_k as a sum of the sequence multipled by negative powers of z.
D_k is zero for k>0, so all those terms are zero.
D_k is one for k = 0, so that is the only term in the sum.
That means that we have:

Z[D_k] = D_oz^o = 1

We can see that the sequence, D_k, 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 D_k is 1.

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

u_k is one for all k.
We use the same expression to compute the Z-transform of the unit step. Since all samples are one, we get:

U[z] = 1 + z^-1 + z^-2 + z^-3 + z^-4 + . . .
U[z] = z/(z - 1)
Brush up on sums of infinite series if you're not with it for this.

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:

The unit impulse, D_k,
The unit step, u_k,
Exponentially decaying sequences, y_k = y_oa^k
Alternating decaying sequences, which are exponentially decaying sequences with a < 0.

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

We haven't considered signals with more than one pole. Next, we will consider a signal with two poles.
There are tools that you have available from work with Laplace transforms.

For example, with two real poles you should be able to divide the transform into two parts, each with one real pole, using partial fractions. Then you can analyze each part separately.
Conversely, a sequence with two decaying exponential sequences should give two poles. That should generalize to 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.

Here is the z-function:

And, the partial fraction expansion for the z-function is:

Taking the inverse Z-transform, we find the following sequence. Note D_k is a unit impulse at k = 0.

And, you should observe that we could, in fact, have performed these steps in the opposite order, i.e.

We could have started with the expression above, with two decaying terms (.7^k and.9^k), and added in a unit impulse, then
We could have taken the transform of both terms, including the D_k term, and then,
We could have combined terms to get the function we started with above

10/[(z - 0.7)(z - 0.9)]

Example/Simulation

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

Y[z] = 1/[(z - p₁)(z - p₂)]
Enter the poles in the text boxes indicated, and click the Start button.

Do the following with this simulator.

Input the values above, i.e.

p₁ = 0.7
p₂ = 0.9

Observe the result, and, in particular, note the following features.

The function starts at zero, reaches a peak and decays back to zero.
You should expect the response to die back to zero. Both poles here satisfy the criterion for stability as we noted above for single poles.
The function does not start immediately. There is no zero at z = 0 as we had earlier, and this delays the start of the signal. That will be discussed in more detail later.

Input one negative value for a pole and observe the result, including the following features.

There are now oscillations in the response. Those oscillations take only two sample periods. as noted above for a single negative pole.

Input two negative value for the poles and observe the result, including the following features.

The oscillations still take only two sample periods.
The oscillations are more pronounced (wilder?).

Now, at this point you have seen several signals.

The unit impulse - with a transform that is a constant.
The unit step - with a transform with a pole at the origin.
The decaying "exponential" - with a transform with a single real pole
Two exponentials - with two poles.

These signals have some interesting properties, and we can make a few observations.

The number of decaying terms (a^k terms) determines the number of poles.
In the cases we considered, the poles were real.

With real pole any oscilations we encountered were of the type where the cycle period was just two sample periods, i.e. the signal went up, then it went down, then back up, etc.

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:

f_k = f(kT) = e^-^akTsin(bkT)

The decaying sinusoid is similar to the alternating decaying signal, but it has significant differences:

The signal does not alternate from positive to negative.
The signal looks like samples from a decaying sinusoid.

Now, let us consider the Z-transform of our decaying sinusoid signal.

f_k = f(kT) = e^-^akTsin(bkT)

Now, we have to evalulate the summation. That doesn't look easy. There is another way.
We can recognize that sin(bkT) can be represented with a sum of two exponentials.

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 f1_k.

f1_k = [e^-^akT+jbkT]

Earlier in this lesson we considered a sequence:

y_k = y_oa^k
We found the transform of this sequence to be

Y[z] = y_o/(z - a),

Now we have a new sequence:

f1_k = [e^-^akT+jbkT]/2j = [(e^-^aT-jbT)^k]/2j

This new sequence is really a generalization of the simple sequence above, so we can apply our previous result - without having to reinvent the wheel.
There is a similar sequence for the other part of the sampled sine function.

f2_k = -[e^-^akT-jbkT]/2j = -[(e^-^aT-jbT)^k]/2j

We can rewrite the new sequence as:

f_k = f1_k + f2_k
f_k = [(e^-^aT+jbT)^k]/2j - [(e^-^aT-jbT)^k]/2j
f_k = f_o[(a)^k - (a^*)^k]

Note, a^* is the complex conjugate of a, and a = e^-^aT+jbT.
f_o = 1/2j

We know how to take the z transform of the sequence, f_k. 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.

Z[f_k] = f_o[z/(z - a) + z/(z - a^*)]

We can combine these two terms, if that is desired. The result is:

Z[f_k] = f_o[z/(z - a) - z/(z - a^*)]
Z[f_k] = f_o[2Im(a)z/(z - a)(z - a^*)]
We know:

f_o = 1/2j
a = e^-^aT-jbT = e^-^aT[cos(bT) + jsin(bT)]

So, we have:

Z[f_k] = f_o[2Im(a)z/(z - a)(z - a^*)]
Z[f_k] = (1/2j)[2e^-^aTsin(bT)z/(z - e^-^aT-jbT )(z - e^-^aT+jbT )]

And, that is the Z-transform of the sequence:

f_k = f(kT) = e^-^akTsin(bkT)

There are two poles for this signal. Those poles are at:

z₁ = e^-^aT-jbT
z₂ = e^-^aT+jbT

Here is an example plot for the two poles. Parameters are:

a = 0.05,
b = .3
T = 1.0

The two poles are shown in the plot below. The poles are marked with x's, and we have shown a unit circle. The two poles lie just slightly within the unit circle.

These poles are interesting.

The poles are complex conjugates - much like we find complex conjugate poles for continuous systems with decaying oscillations.
The poles are in the right half of the z-plane, but they still represent decaying oscillations - contrasting with poles in continuous systems in the left half of the s-plane.
The poles are inside the unit circle.

The unit circle is the stability boundary for sampled systems, like the imaginary axis is for continuous systems.

Just as in continuous systems, proximity to the stability boundary implies low relative stability. Poles in the z-plane that are close to the unit circle will produce slowly decaying oscillations just like poles in the s-plane do when they are close to the jw-axis.

Example/Simulation

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

a = 0.05,
b = .3
T = 1.0

These are the values in the expression for the sequence, f_k = 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 pole is given by:

z₁ = e^-^aT+jbT
z₁ = e^-^aT[cos(bT) + jsin(bT)]

So, the real and imaginary parts are:

Re(z₁) = e^-^aTcos(bT) = 0.909
Im(z₁) = e^-^aTsin(bT) = 0.281

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.

Check the values used above, i.e. Real Part = 0.909 and Imaginary Part = +/-0.281. Actually, the simulator should have these values preset.
Determine if the period is correct. You will need to figure out what the period should be, and remember that the sample period, T, is one second for this simulation.
Determine if the number of samples in a period is correct.

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:

z₁ = e^-^aT-jbT
z₂ = e^-^aT+jbT

The critical observation to be made is that the response has terms like the expression below, which is repeated from the material above.

f_k = f(kT) = e^-^akTsin(bkT)

Then, we should realize that the critical term is the envelope of the response, and that is determined by:

Envelope_k = e^-^akT= (e^-^aT)^k
Or, in other words, the magnitude of the poles (And since they are complex conjugates, they both have the same magnitude.) determines the decay rate per sample period. That decay rate/sample period id:

Decay Rate/Sample Period = e^-^aT

We can note the following critical observation about these poles.

If the magnitude of the pole(s) is less than one, the response will eventually settle out to a constant value (possibly zero) because the transient part of the response will eventually die out.
If the magnitude of the pole(s) is greater than one, the response will grow indefinitely.
That's why the unit circle is the stability boundary for sampled systems. Poles outside the unit circle represent signals that grow in time, while poles inside the unit circle represent signals that eventually decay to zero.

It is possible to get even more insight into how pole position is related to response.

If we start with a sequence,f_k = f(kT) = e^-^akTsin(bkT)

In one sample period, the bounding envelope of the sinusoid, e^-^akT always becomes smaller by a factor e^-^aT.
In one sample period, the angle in the argument of the sinusoid always increases by bT radians.

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.

a = 0.05,
b = .3
T = 1.0

Now, note the following for this example.

The decay factor, e^-^aT = e^-0.05, so each sample interval, the bound on the sinusoid will decrease to e^-0.05 of the value the preceding sample period. That's approximately a 5% decrease to 0.951 times the preceding value.

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:

z₁ = e^-^aT-jbT
z₂ = e^-^aT+jbT

The magnitude of both poles is |e^-^aT|. The factor, e^j^bT, only changes the angle of the first pole - and the factor, e^-j^bT, 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:

f_k = 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?

The pole position determines the significant features of the response.

The distance from the origin determines the rate of decay. The closer to the origin the quicker the decay - as measured in sample periods.
The angle off the horizontal - measured from the origin - determines the number of sample periods in a period of the sinusoid.

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.

Use values of 15^o, 30^o, 45^o, . . . up to 180^o, for the angle.
Use values of 1.0, 0.9, 0.8, . . . for the decay factor.
Observe results and see what conclusions you can draw. Make sure that the results make sense to you.

Are the results what you expected?
Notice how the signal is undamped when it is on the unit circle.
Notice how the apparent damping increases as the poles move toward the center (origin) of the unit circle.

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:

For a nintey degree angular rate, the poles are at +90^o and -90^o.
The angle of the poles determines the number of samples in a period of oscillation.
For an angle of 90^o, there are four samples in a period.
With four samples in a period, in this case there is:

one up,
one at zero,
one down, and
one at zero, etc.

We can sum up what you should have obtained from this part of the lesson.

You should be able to relate the distance of a pole from the origin of the z-plane to the decay rate.
You should be able to relate the angle of the pole off the horizontal - measured from the origin - to the number of samples in an oscilation period.

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, y_k, and the transform of that signal sequence is Y[z].

Z(a y_k) = a Y[z] Multiplication by a constant
Z(a y_k+ b w_k) = a Y[z] + b W[z] Linearity

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:

Z(a y_k+ b w_k) = a Y[z] + b W[z] Linearity

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:

Z(a y_k-1) = z^-1 Y[z]

You need to understand what y_k-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 y_k-1 is the same as the signal y_k except that it takes on specific values one sample period later than y_k.

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

Shifting the index by -1 (changing k to k-1) delays the function by one sample point.

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.

The first term in the sum, for k = 0, is y_-1. We will assume that the signals we deal with - including y_k here - all start at zero so that y_-1 = 0.
Noting that, we can let m = k - 1, and substitute that in the sum. Then the sum will run from m = -1 to infinity. Here is that expression.

Now, evalute the sum. That leads to the following expression.

Ultimately, we conclude that the transform of the delayed signal is just z-1 times the transform of the undelayed signal, i.e.:

Z[y_k-1] = z^-1 Y[z] = Y[z]/z

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, y_k.

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:

[y_o - 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

Introduction to Z Transforms
Application To Systems - Transfer Functions

Moving Along - More Advanced Material

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