# Control Systems/Sampled Data Systems

## Ideal Sampler

In this chapter, we are going to introduce the ideal sampler and the
**Star Transform**. First, we need to introduce (or review) the
**Geometric Series** infinite sum. The results of this sum will be very
useful in calculating the Star Transform, later.

Consider a sampler device that operates as follows: every *T* seconds,
the sampler reads the current value of the input signal at that exact
moment. The sampler then holds that value on the output for *T* seconds,
before taking the next sample. We have a generic input to this system,
*f(t)*, and our sampled output will be denoted *f\*(t)*. We can then
show the following relationship between the two signals:

$$f^{\,*}(t)=f(0)\big(\mathrm{u}(t\,-\,0)\,-\,\mathrm{u}(t\,-\,T)\big)\,+\,f(T)\big(\mathrm{u}(t\,-\,T)\,-\,\mathrm{u}(t\,-\,2T)\big)\,+\;\cdots\;+\,f(nT)\big(\mathrm{u}(t\,-\,nT)\,-\,\mathrm{u}(t\,-\,(n\,+\,1)T)\big)\,+\;\cdots$$

Note that the value of *f^\*^* at time *t* = 1.5 *T* is the same as at
time *t = T*. This relationship works for any fractional value.

Taking the Laplace Transform of this infinite sequence will yield us
with a special result called the **Star Transform**. The Star Transform
is also occasionally called the \"Starred Transform\" in some texts.

## Geometric Series

Before we talk about the Star Transform or even the Z-Transform, it is
useful for us to review the mathematical background behind solving
infinite series. Specifically, because of the nature of these
transforms, we are going to look at methods to solve for the sum of a
**geometric series**.

A geometric series is a sum of values with increasing exponents, as
such:

$$\sum_{k=0}^{n} ar^k = ar^0+ar^1+ar^2+ar^3+\cdots+ar^n \,$$

In the equation above, notice that each term in the series has a
coefficient value, a. We can optionally factor out this coefficient, if
the resulting equation is easier to work with:

$$a \sum_{k=0}^{n} r^k = a \left(r^0+r^1+r^2+r^3+\cdots+r^n \,\right)$$

Once we have an infinite series in either of these formats, we can
conveniently solve for the total sum of this series using the following
equation:

$$a \sum_{k=0}^{n} r^k = a\frac{1-r^{n+1}}{1-r}$$

Let\'s say that we start our series off at a number that isn\'t zero.
Let\'s say for instance that we start our series off at *n = 1* or *n =
100*. Let\'s see:

$$\sum_{k=m}^{n} ar^k = ar^m+ar^{m+1}+ar^{m+2}+ar^{m+3}+\cdots+ar^n \,$$

We can generalize the sum to this series as follows:

$$\sum_{k=m}^n ar^k=\frac{a(r^m-r^{n+1})}{1-r}$$

With that result out of the way, now we need to worry about making this
series converge. In the above sum, we know that n is approaching
infinity (because this is an *infinite sum*). Therefore, any term that
contains the variable n is a matter of worry when we are trying to make
this series converge. If we examine the above equation, we see that
there is one term in the entire result with an *n* in it, and from that,
we can set a fundamental inequality to govern the geometric series.

$$r^{n+1} < \infty$$

To satisfy this equation, we must satisfy the following condition:

$$r \le 1$$ {{-}} Therefore, we come to the final result: **The
geometric series converges if and only if the value of *r* is less than
one.**

## The Star Transform

The **Star Transform** is defined as such:

$$F^*(s) = \mathcal{L}^*[f(t)] = \sum_{k = 0}^\infty f(kT)e^{-skT}$$

The Star Transform depends on the sampling time *T* and is different for
a single signal depending on the frequency at which the signal is
sampled. Since the Star Transform is defined as an infinite series, it
is important to note that some inputs to the Star Transform will not
converge, and therefore some functions do not have a valid Star
Transform. Also, it is important to note that the Star Transform may
only be valid under a particular **region of convergence**. We will
cover this topic more when we discuss the Z-transform.

### Star ↔ Laplace

The Laplace Transform and the Star Transform are clearly related,
because we obtained the Star Transform by using the Laplace Transform on
a time-domain signal. However, the method to convert between the two
results can be a slightly difficult one. To find the Star Transform of a
Laplace function, we must take the residues of the Laplace equation, as
such:

$$X^*(s) = \sum \bigg[ \text{residues of } X(\lambda)\frac{1}{1-e^{-T(s-\lambda)}}\bigg]_{\text{at poles of E}(\lambda)}$$

This math is advanced for most readers, so we can also use an alternate
method, as follows:

$$X^*(s)=\frac{1}{T}\sum_{n=-\infty}^\infty X(s+jm\omega_s)+\frac{x(0)}{2}$$

Neither one of these methods are particularly easy, however, and
therefore we will not discuss the relationship between the Laplace
transform and the Star Transform any more than is absolutely necessary
in this book. Suffice it to say, however, that the Laplace transform and
the Star Transform *are related* mathematically.

### Star + Laplace

In some systems, we may have components that are both continuous and
discrete in nature. For instance, if our feedback loop consists of an
Analog-To-Digital converter, followed by a computer (for processing),
and then a Digital-To-Analog converter. In this case, the computer is
acting on a digital signal, but the rest of the system is acting on
continuous signals. Star transforms can interact with Laplace transforms
in some of the following ways:

### Convergence of the Star Transform

The Star Transform is defined as being an infinite series, so it is
critically important that the series converge (not reach infinity), or
else the result will be nonsensical. Since the Star Transform is a
geometic series (for many input signals), we can use geometric series
analysis to show whether the series converges, and even under what
particular conditions the series converges. The restrictions on the star
transform that allow it to converge are known as the **region of
convergence** (ROC) of the transform. Typically a transform must be
accompanied by the explicit mention of the ROC.

## The Z-Transform

Let us say now that we have a discrete data set that is sampled at
regular intervals. We can call this set *x\[n\]*:

`x[n] = [ x[0] x[1] x[2] x[3] x[4] ... ]`

we can utilize a special transform, called the Z-transform, to make
dealing with this set more easy:

$$X(z) = \mathcal{Z}\left\{x[n]\right\} = \sum_{n = -\infty}^\infty x[n] z^{-n}$$

Like the Star Transform the Z Transform is defined as an infinite series
and therefore we need to worry about convergence. In fact, there are a
number of instances that have identical Z-Transforms, but different
regions of convergence (ROC). Therefore, when talking about the Z
transform, you must include the ROC, or you are missing valuable
information.

{{-}}

### Z Transfer Functions

Like the Laplace Transform, in the Z-domain we can use the input-output
relationship of the system to define a **transfer function**.

![](Z_Block.svg "Z_Block.svg"){width="400"}

The transfer function in the Z domain operates exactly the same as the
transfer function in the S Domain:

$$H(z) = \frac{Y(z)}{X(z)}$$

$$\mathcal{Z}\{h[n]\} = H(z)$$

Similarly, the value *h\[n\]* which represents the response of the
digital system is known as the **impulse response** of the system. It is
important to note, however, that the definition of an \"impulse\" is
different in the analog and digital domains.

### Inverse Z Transform

The **inverse Z Transform** is defined by the following path integral:

$$x[n] = Z^{-1} \{X(z) \}= \frac{1}{2 \pi j} \oint_{C} X(z) z^{n-1} dz \$$

Where *C* is a counterclockwise closed path encircling the origin and
entirely in the region of convergence (ROC). The contour or path, *C*,
must encircle all of the poles of *X(z)*.

This math is relatively advanced compared to some other material in this
book, and therefore little or no further attention will be paid to
solving the inverse Z-Transform in this manner. Z transform pairs are
heavily tabulated in reference texts, so many readers can consider that
to be the primary method of solving for inverse Z transforms. There are
a number of Z-transform pairs available in table form in **The
Appendix**.

### Final Value Theorem

Like the Laplace Transform, the Z Transform also has an associated final
value theorem:

$$\lim_{ n\to \infty} x[n] = \lim_{z \to 1} (z - 1) X(z)$$

This equation can be used to find the steady-state response of a system,
and also to calculate the steady-state error of the system.

## Star ↔ Z

The Z transform is related to the Star transform though the following
change of variables:

$$z = e^{sT}$$

Notice that in the Z domain, we don\'t maintain any information on the
sampling period, so converting to the Z domain from a Star Transformed
signal loses that information. When converting back to the star domain
however, the value for *T* can be re-insterted into the equation, if it
is still available.

Also of some importance is the fact that the Z transform is bilinear,
while the Star Transform is unilinear. This means that we can only
convert between the two transforms if the sampled signal is zero for all
values of *n \< 0*.

Because the two transforms are so closely related, it can be said that
the Z transform is simply a notational convenience for the Star
Transform. With that said, this book could easily use the Star Transform
for all problems, and ignore the added burden of Z transform notation
entirely. A common example of this is Richard Hamming\'s book
\"Numerical Methods for Scientists and Engineers\" which uses the
Fourier Transform for all problems, considering the Laplace, Star, and
Z-Transforms to be merely notational conveniences. However, the Control
Systems wikibook is under the impression that the correct utilization of
different transforms can make problems more easy to solve, and we will
therefore use a multi-transform approach.

### Z plane

*z* is a complex variable with a real part and an imaginary part. In
other words, we can define *z* as such:

$$z = \operatorname{Re}(z) + j\operatorname{Im}(z)$$

Since *z* can be broken down into two independent components, it often
makes sense to graph the variable *z* on the **Z-plane**. In the
Z-plane, the horizontal axis represents the real part of *z*, and the
vertical axis represents the magnitude of the imaginary part of *z*.

Notice also that if we define *z* in terms of the star-transform
relation:

$$z = e^{sT}$$

we can separate out *s* into real and imaginary parts:

$$s = \sigma + j\omega$$

We can plug this into our equation for *z*:

$$z = e^{(\sigma + j\omega)T} = e^{\sigma T} e^{j\omega T}$$

Through **Euler\'s formula**, we can separate out the complex
exponential as such:

$$z = e^{\sigma T} (\cos(\omega T) + j\sin(\omega T))$$

If we define two new variables, *M* and φ:

$$M = e^{\sigma T}$$

$$\phi = \omega T$$

We can write *z* in terms of *M* and φ. Notice that it is Euler\'s
equation:

$$z = M\cos(\phi) + jM\sin(\phi)$$

Which is clearly a polar representation of *z*, with the magnitude of
the polar function (*M*) based on the real-part of *s*, and the angle of
the polar function (φ) is based on the imaginary part of *s*.

### Region of Convergence

To best teach the region of convergance (ROC) for the Z-transform, we
will do a quick example.

{1-r}

`       = 1 \frac{1 - ((e^2z)^{-1})^{n+1}}{1 - (e^2z)^{-1}}``</math>`{=html}

Again, we know that to make this series converge, we need to make the r
value less than 1:

$$|(e^2z)^{-1}| = \left|\frac{1}{e^2z}\right| \le 1$$

$$|e^2z| \ge 1$$

And finally we obtain the region of convergance for this Z-transform:

$$|z| \ge \frac{1}{e^2}$$}}

### Laplace ↔ Z

There are no easy, direct ways to convert between the Laplace transform
and the Z transform directly. Nearly all methods of conversions
reproduce some aspects of the original equation faithfully, and
incorrectly reproduce other aspects. For some of the main mapping
techniques between the two, see the Z Transform Mappings
Appendix.

However, there are some topics that we need to discuss. First and
foremost, conversions between the Laplace domain and the Z domain *are
not linear*, this leads to some of the following problems:

1.  $\mathcal{L}[G(z)H(z)] \ne G(s)H(s)$
2.  $\mathcal{Z}[G(s)H(s)] \ne G(z)H(z)$

This means that when we combine two functions in one domain
multiplicatively, we must find a combined transform in the other domain.
Here is how we denote this combined transform:

$$\mathcal{Z}[G(s)H(s)] = \overline{GH}(z)$$

Notice that we use a horizontal bar over top of the multiplied
functions, to denote that we took the transform of the product, not of
the individual pieces. However, if we have a system that incorporates a
sampler, we can show a simple result. If we have the following format:

$$Y(s) = X^*(s)H(s)$$

Then we can put everything in terms of the Star Transform:

$$Y^*(s) = X^*(s)H^*(s)$$

and once we are in the star domain, we can do a direct change of
variables to reach the Z domain:

$$Y^*(s) = X^*(s)H^*(s) \to Y(z) = X(z)H(z)$$

Note that we can only make this equivalence relationship if the system
incorporates an ideal sampler, and therefore one of the multiplicative
terms is in the star domain.

### Example

## Z ↔ Fourier

By substituting variables, we can relate the Star transform to the
Fourier Transform as well:

$$e^{sT} = e^{j \omega}$$

$$e^{(\sigma + j \omega)T} = e^{j \omega}$$

If we assume that *T = 1*, we can relate the two equations together by
setting the real part of *s* to zero. Notice that the relationship
between the Laplace and Fourier transforms is mirrored here, where the
Fourier transform is the Laplace transform with no real-part to the
transform variable.

There are a number of discrete-time variants to the Fourier transform as
well, which are not discussed in this book. For more information about
these variants, see Digital Signal
Processing.

## Reconstruction

Some of the easiest reconstruction circuits are called \"Holding
circuits\". Once a signal has been transformed using the Star Transform
(passed through an ideal sampler), the signal must be \"reconstructed\"
using one of these hold systems (or an equivalent) before it can be
analyzed in a Laplace-domain system.

If we have a sampled signal denoted by the Star Transform $X^*(s)$, we
want to **reconstruct** that signal into a continuous-time waveform, so
that we can manipulate it using Laplace-transform techniques.

Let\'s say that we have the sampled input signal, a reconstruction
circuit denoted *G(s)*, and an output denoted with the Laplace-transform
variable *Y(s)*. We can show the relationship as follows:

$$Y(s) = X^*(s)G(s)$$

Reconstruction circuits then, are physical devices that we can use to
convert a digital, sampled signal into a continuous-time domain, so that
we can take the Laplace transform of the output signal.

### Zero order Hold

!Zero-Order Hold impulse
response

A **zero-order hold** circuit is a circuit that essentially inverts the
sampling process: The value of the sampled signal at time *t* is held on
the output for *T* time. The output waveform of a zero-order hold
circuit therefore looks like a staircase approximation to the original
waveform.

The transfer function for a zero-order hold circuit, in the Laplace
domain, is written as such:

$$G_{h0} = \frac{1 - e^{-Ts}}{s}$$

The Zero-order hold is the simplest reconstruction circuit, and (like
the rest of the circuits on this page) assumes zero processing delay in
converting between digital to analog.

center\|framed\|A continuous input signal (gray) and the sampled signal
with a zero-order hold (red)

### First Order Hold

!Impulse response of a first-order
hold.

The zero-order hold creates a step output waveform, but this isn\'t
always the best way to reconstruct the circuit. Instead, the
**First-Order Hold** circuit takes the derivative of the waveform at the
time *t*, and uses that derivative to make a guess as to where the
output waveform is going to be at time *(t + T)*. The first-order hold
circuit then \"draws a line\" from the current position to the expected
future position, as the output of the waveform.

$$G_{h1} = \frac{1 + Ts}{T} \left[ \frac{1 - e^{-Ts}}{s}\right]^2$$

Keep in mind, however, that the next value of the signal will probably
not be the same as the expected value of the next data point, and
therefore the first-order hold may have a number of discontinuities.

center\|framed\|An input signal (grey) and the first-order hold circuit
output (red)

### Fractional Order Hold

The Zero-Order hold outputs the current value onto the output, and keeps
it level throughout the entire bit time. The first-order hold uses the
function derivative to predict the next value, and produces a series of
ramp outputs to produce a fluctuating waveform. Sometimes however,
neither of these solutions are desired, and therefore we have a
compromise: **Fractional-Order Hold**. Fractional order hold acts like a
mixture of the other two holding circuits, and takes a fractional number
*k* as an argument. Notice that *k* must be between 0 and 1 for this
circuit to work correctly.

$$G_{hk} = (1 - ke^{-Ts}) \frac{1 - e^{-Ts}}{s} + \frac {k}{Ts^2} (1 - e^{-Ts})^2$$

This circuit is more complicated than either of the other hold circuits,
but sometimes added complexity is worth it if we get better performance
from our reconstruction circuit.

### Other Reconstruction Circuits

!Impulse response to a linear-approximation
circuit.

Another type of circuit that can be used is a **linear approximation**
circuit.

center\|framed\|An input signal (grey) and the output signal through a
linear approximation
circuit

## Further reading

-   Hamming, Richard. \"Numerical Methods for Scientists and Engineers\"
    
-   Digital Signal Processing/Z
    Transform
-   Complex Analysis/Residue
    Theory
-   Analog and Digital
    Conversion

