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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_{o}a^{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_{o}a^{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_{o}a^{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_{o}a^{k}
That's what we have been
working with.
The transform is given
by: Y[z] = y_{o}z/(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_{o}a^{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_{o}e^{-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:
Now, you can think of
this as y_{k} = y_{o}a^{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_{k}is
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_{o}z^{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:
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:
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 - p_{1})(z
- p_{2})]
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.
p_{1}
= 0.7
p_{2}
= 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^{-akT}sin(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^{-}^{akT}sin(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^{-}^{akT}sin(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}.
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)]
There are two poles for
this signal. Those poles are at:
z_{1} =
e^{-}^{aT-jbT}
z_{2} =
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^{-}^{akT}sin(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:
z_{1} =
e^{-}^{aT+jbT}
z_{1} =
e^{-}^{aT}[cos(bT)
+ jsin(bT)]
So, the real and imaginary
parts are:
Re(z_{1})
= e^{-}^{aT}cos(bT)
= 0.909
Im(z_{1})
= e^{-}^{aT}sin(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:
z_{1} =
e^{-}^{aT-jbT}
z_{2} =
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^{-}^{akT}sin(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^{-}^{akT}sin(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:
z_{1} =
e^{-}^{aT-jbT}
z_{2} =
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^{-}^{akT}sin(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 y_{2}. 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:
Notice how y_{-1} = 0, and
how every term gets cancelled except the very last. In what is shown
y_{2} is left. In the limit, the "final value" is
left. So, we have the final value theorem as a result.