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INTRODUCTION TODISCRETE CONTROL
SYSTEM
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UPON COMPLETION OF THIS CHAPTER STUDENTSSHOULD BE ABLE T0:
DESCRIBE THE ADVANTAGES AND DISADVANTAGES OF
DIGITAL CONTROL
DIFFERENTIATE BETWEEN ANALOGUE AND DIGITAL SYSTEM
EXPLAIN THE SAMPLING PROCESS AND SIGNAL COVERSION
FROM ANALOG TO DIGITAL SIGNAL
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INTRODUCTION
The use of a digital computer as a compensator (controller) device has
grown during the past three decades as the price and reliability of digitalcomputers have improved dramatically .
The cost of microcontrollers and digital signal processors has reduced
and you can buy an 8-bit micro controller for a few RM.
A digital control system uses digital signals and a digital computer tocontrol a process.
The measurement data are converter from analog form to digital form by
means of the analog-to-digital converter shown in Figure 1.1.
Figure 1.1: A block diagram of a computer control system
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Digital control systems are used in many applications: for
machine tools, metalworking processes, chemical processes,
aircraft control, and automobile traffic control and others.
Digital Control Problem
Usually trying to control a continuous time system, but using a
digital controller to control it.
Need to be able to convert from continuous time signal to discrete
time signal (A/D converter) this is just sampling Need to be able to convert from discrete time to continuous time
signal (D/A converter) may ways to do this
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Flexibility.With analogue circuitry it would be difficult to suddenly change a
resistor to modify the integral gain while the plant is operating! So it
allows operators to, modify, to tune, to experiment with the controller.
With a good user interface any parameter inside that piece of control
software can be adjusted.
Improved measurement sensitivity
the use of digitally coded signals, digital sensors and transducers, and
microprocessors; reduced sensitivity to signal noise; and the
capability to easily reconfigure the control algorithm in software.
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Both algorithmic and decision-making control in one
package
Usually Smaller/lighter
Usually needs Less power
Often More precise Can Re-program
It's easy for parameters to be passed around in a program, far easier than
hardware signals.
More complex control can be realised
It allows algorithms such as vector control of machines involving
flux modelling and reference frame transformation to be carried
out.
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Computation is performed serially,
Instructions take a finite time, albeit in the region of nano-
seconds, but if sever hundred instructions are required to carry
out a task then this time can add up. This can limit the
complexity, but each year faster processors are launched and this
problem becomes less significant for all but the cheapest micros.
New theory involved
There is generally a resistance by industry to adopt new ideas,
although this is becoming less of a problem.
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Sampled Data Systems
The basic difference between these controllers is that the digital
system operates on discrete signals (or samples of the sensed
signal) rather than on continuous signals.
A typical sampled data control system is shown in Figure 1.2.
Figure 1.2
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Sampled Data Systems
The digital computer performs the compensation function within
the system.
The A/D converter converts the error signal, which is a continuous
signal, into digital form so that it can be processed by the computer.
At the computer output the D/A converter converts the digital output
of the computer into a form which can be used to drive the plant.
A sampler is basically a switch that closes every T seconds, as shown
in Figure 1.3.
Figure 1.3
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The ideal sampling process can be considered as the
multiplication of a pulse train with a continuous signal, i.e.
Figure 1.4
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where P(t) is the delta pulse train and expressed as:
Taking the Laplace transform gives:
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A D/A converter converts the sampled signal into a continuous signal y(t).
The D/A can be approximated by a zero-order hold (ZOH) circuit as shown in
Figure 1.5.
This circuit remembers the last information until a new sample is obtained, i.e. the
zero-order hold takes the value r(nT) and holds it constant for and the value r(nT)
is used during the sampling period.
*( )r t
Figure 1.5: sampler and zero-order hold
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The transfer function of a zero-order hold is given by
A sampler and zero-order hold can accurately follow the input
signal if the sampling time T is small compared to the
transient changes in the signal.
Where H(t) is the step function, and taking the Laplace transform yields
Figure 1.6: waveforms after the sampler and zero-order hold
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Example 1.1
Figure 1.7 shows an ideal sampler followed by a zero-order hold. Assuming the
input signal r(t) is as shown in the figure. Show the waveforms after the
sampler and also after the zero-order hold.
Solution
Figure 1.7
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Systems The z-transform can be defined through the following equation:
Its used in sampled data systems just as the Laplace transformation
is
used in continuous-time systems.
The z-transform of the function r(t) is Z[r(t)]=R(z) which is given by:
The response of a sampled data system can be determined easily by
finding the z-transform of the output and then calculating the inversez-transform, just like the Laplace transform techniques used in
continuous-time systems.
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EXPONENTIAL FUNCTION
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SINE FUNCTION
so that
FromPreviously we already knowthe z-transform of an
exponential function is:
)
2cos
2sin
jx jx
jx jx
From
e e x
e e x
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COSINE FUNCTION
Previously we already know
the z-transform of an
exponential function is:
2cos
2sin
jx jx
jx jx
e e x
e e x
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THE Z TRANSFORM OF COMMON DISCRETE TIME FUNCTIONS
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THE Z TRANSFORM OF COMMON DISCRETE TIME FUNCTIONS (cont..)
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Determine G(z) for the system with transfer function:
Example 1.2
Solution:
G(s) can be express as a sum of its partial fractions:
The inverse Laplace transform of this equation is :
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In z-transforms we can write this equation as:
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Properties of z-Transforms
1. Linearity property
Suppose that the z-transform of f (nT ) is F(z) and the z-transform of g(nT ) is
G(z). Then
and for any scalar a
2. Left-shifting property
Suppose that the z-transform of f (nT ) is F(z) and let y(nT ) = f (nT+mT ).
Then
If the initial conditions are all zero, i.e. f (iT ) =0, i = 0,1,2,..., m1, then,
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3. Right-shifting property
Suppose that the z-transform of f (nT ) is F(z) and let y(nT ) = f (nTmT).
Then
If f(nT ) = 0 for k < 0, then the theorem simplifies to
4. Attenuation property
Suppose that the z-transform of f (nT ) is F(z). Then,
This result states that if a function is multiplied by the exponential eanT
then in the z-transform of this function z is replaced by zeaT
.
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5. Initial value theorem
Suppose that the z-transform of f (nT ) is F(z). Then the initial value of the
time response is given by
6. Final value theorem
Suppose that the z-transform of f (nT ) is F(z). Then the final value of the
time response is given by
Note that this theorem is valid if the poles of (1 z 1)F(z) are inside the
unit circle or at z =1.
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Example 1.3
The z-transform of a unit ramp function r(nT ) is
Find the z-transform of the function 5r(nT ).
Solution
Using the linearity property of z-transforms,
Example 1.4
The z-transform of trigonometric function r(nT) =sinnwT is
find the z-transform of the function y(nT) = e 2T sin nW T .
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Example 1.5