Computer Controlled System ELE310570520

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Computer Controlled SystemELE3105/70520

• Examiner Dr Paul Wen
• Faculty of Engineering and Surveying
• The University of Southern Queensland
• Toowoomba, Qld 4350
• Contact Information
• Email PengWen_at_usq.edu.au
• Phone [phone]
• Office Z408

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Study Materials

• Study Book Computer controlled systems
• Textbook Discrete time control system,
  Katsuhiko Ogata, 2nd ed. Prentice-Hall, Inc.,
  1995.
• Support Materials Matlab ( 5.3/later Control
  toolbox)
• References
• Astrom K. J. Wittenmark B., Computer
  controlled systems theory and design, 3rd
  edition, Prentice Hall, New York, 1996.
• Nagle H. Philips C., Digital control system
  analysis design, Prentice Hall.
• Norman S. Nise, Control system engineering
  Third edition, John Wiley Sons, Inc., New York,
  2000.

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Timetable

• Note Consulting in non-consulting time
• Quick questions (5 minutes) Any time
• Others By Appointments

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Assessment

• Students attention is drawn to the student
  related policies and the unit specification 2002,
  which outline the Universitys Assessment Policy
  and the requirements.
• For unit 70520, Computer controlled systems, you
  must obtain 50 from each of three assessments
  and 50 all over.

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Assessment

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Questions?
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Subject Overview
Computer engineering I II
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1.1 Introduction

• Control system definition Control system
  application.

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1.1 Introduction

• Advantages of control systems
• Power amplification
• Remote control
• Convenience
• Compensation for disturbance

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1.1 Introduction
Control system
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1.1 Introduction
Computer controlled system
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1.2 Digital control loop Components

• GHP(z) is the transfer function of control object
  ZOH, where z indicates the discrete time domain
• GC(z) is a controller implemented in computer
  languages.
• A/D is the Analog-to-Digital converter (Voltage
  ?Binary number).
• D/A is the Digital-to-Analog converter (Binary
  number ? Voltage).
• The little switch indicate a sampling operation.

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1.2 Digital control loopSignals

• Discrete time domain
• R(z) is the desired output
• E(z) is the error signal
• M(z) is the controller output/control action
• C(z) is the actual output
• Continuous time domain
• In continuous time domain, R(z), E(z), M(z) and
  C(z) are corresponding to r(t), e(t) m(t) and
  c(t).

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1.2 Digital control loopSequence of events

• Get desired output r(t) at this instant in time
• Measure actual output c(t)
• Calculate error e(t)r(t)-c(t)
• Derive control signal m(t) based on proper
  control algorithm
• Output this control signal m(t) to controlled
  object
• Save previous history of error and output for
  later use
• Repeat step 1 to 6 (go to 1)

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1.2 Digital control loopForms of signals

• Computer cannot sample while calculating, so
  there is a sample frequency 1/T for data
  acquisition through a A/D, where T is sampling
  interval.
• The data of a signal are recorded and represented
  as a sequence of number in memory.
• Based on these numbers, a control signal is
  derived and then conveyed to controlled object
  through a D/A

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1.2 Digital control loopForms of signals

• In between sample instants, the input is supposed
  as constant and the output is held as a constant
  by a device termed as zero-order-hold (ZOH).
• The reconstruction of a signal will be a
  stair-step, and a low-pass filter is employed to
  smooth out the rough edges

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1.3 ADC and DAC
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1.3 ADC and DAC
D/A is used as a ZOH.
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1.3 ADC and DACADC

• Have a discrete number of quantization levels
• Number of levels L2N, where N is the number of
  bits
• eg N3 bits, L238 levels

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1.3 ADC and DACADC
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1.3 ADC and DACADC

• More bits more accuracy. The commonly used ADC
  has
• 8-bits L28256 (coarse)
• 10-bits L2101024 (adequate)
• 12-bits L2124096 (works well)
• 16-bits L21665536 (almost overkill)

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1.3 ADC and DACADC

• Distances between sequential levels are the same.
  eg 5v/280.0195v
• The weight of each bit is different. The most
  significant bit is the most left bit and the
  least significant bit is the most right bit.

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1.3 ADC and DACADC
Example For N8, find the number range of the
ADC in binary, decimal and hexadecimal numbers.
If the input signal is from 0 to 5 voltage for
the above number range, what will be the number
for a 2 voltage signal in decimal and binary
numbers? Solution In binary 00000000B?
11111111B In decimal 0 ? 2726252423222120
255 In hexadecimal 0 ? F15 00H ?FFH
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1.3 ADC and DACADC
Decimal to binary Exercise For N10,
repeat the above example
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1.3 ADC and DACDAC
Example For N8 and the signal is from 0 to
5, find the output value for the number
145. Solution5/255x/145, x5145/2552.84312.84
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1.3 ADC and DAC
Multi-channel A/D converter
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1.3 ADC and DAC
Multi-channel D/A converter
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1.4 Errors

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1.4 Errors

• The quantization error or resolution error is the
  difference between the analog input value and the
  equivalent digital value. On average it is one
  half of the LSB.
• Linearity error the maximum deviation in step
  size from ideal step size, expressed as a
  percentage of full scale.
• Settling time the time it takes for the output
  to reach within /- half of the step size of the
  final output.

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1.4 Errors

• Gain error

Output
Digital
Analog
Input
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1.5 Sampling theorem
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1.5 Sampling theorem
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1.5 Sampling theorem
If we need the sampled data to keep all the
features of the original signal, what is the
minimum sampling frequency? Or what conditions
should we meet if we wish that the sampled data
can represent the original data exactly? The
answer to the above question forms the Sampling
theorem/Shannons sampling theorem/Shannons
theorem.
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1.5 Sampling theorem
A continuous-time signal f(t) with a finite
bandwidth ?0 (the highest frequency component in
the signal, or the Nyquist frequency) can be
uniquely described by the sampled signal
f(kT)k,-1,0,1., when the sampling frequency
?s is greater than 2?0. In other words, if a
signal is sampled twice faster than its highest
frequency component, the sampled date can
represent all the features of this signal.
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1.6 The proven of sampling theorem

• The proven is based on Fourier Transform
• Fourier transform A transformation from time
  domain to frequency domain f(t) ? F(?), where t
  is time and ? is frequency.
• For a continuous time function f(t), we can
  uniquely find F(?). If given F(?), we can also
  unique determine f(t).
• It means that f(t) and F(?) are equivalent.

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1.6 The proven of sampling theorem
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1.6 The proven of sampling theorem
2. For a sampled signal fs(t), we have
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1.6 The proven of sampling theorem
3. The relationship between f(t) and fs(t), and
F(?) and Fs(?).
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1.6 The proven of sampling theorem
4. If we change the sampling frequency, what will
happen with fs(t) and Fs(?).
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1.6 The proven of sampling theorem
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1.6 The proven of sampling theorem

• 5. Conclusions
• If our sampling frequency ?s is faster enough,
  that is ?sgt2?0, there will be gaps between the
  shifting F(?) in Fs(?). We can always put a
  filter to figure out F(?) from Fs(?). Otherwise
  if the repeating F(?) figures overlap in Fs(?),
  we cannot put a filter to figure out F(?) from
  Fs(?). The turning point from possible to
  impossible is ?s 2?0, where ?0 is the highest
  frequency component or Nyquist Frequency of the
  signal.

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1.7 Aliasing

• 1. Aliasing problem

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1.7 Aliasing

• Ambiguity alias

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1.7 Aliasing

• 2. Finding aliases
• The fundamental alias frequency is given by
• ? (?0 ?n)mod(?s) - ?n
• where mod() means the remainder of an division
  operation, ?0 is signal bandwidth, ?n Nyquist
  frequency, and ?s sampling frequency
• Example For f090Hz fs100, find alias.
• Solution ?2?f, fnfs/250Hz,
• f (f0 fn)mod(fs) - fn(9050)mod(100)-50
• 40-5010Hz

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1.7 Aliasing

• 3. Preventing aliases
• Make sure your sampling frequency is greater than
  twice of the highest frequency component of the
  signal
• Pre-filtering
• Set your sampling frequency to the maximum if
  possible

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1.7 Aliasing

• Suppose that the Nyquist frequency of a signal is
  100Hz. If we use an 8-bit ADC to sample this
  signal at the frequency of 200Hz, can the sample
  data represents this signal exactly? Why?

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1.7 Aliasing

• Theoretically, as long as the sampling frequency
  is greater than or equal to twice the Nyquist
  frequency, aliases will not happen. However,
  because of the conversion/quantisation error, the
  practical sampling frequency is much higher than
  that (5 to 10 times of the Nyquist frequency).
  Fortunately, most of the time the speeds of ADC
  and computer are also much greater than signals
  Nyquist frequency.

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Reading

• Study book
• Module 1 The digital control loop
• Textbook
• Chapter 1 Introduction to discrete time
  control system
• Chapter 3 pages 90-92 96-98.

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Exercise

• Exercise 1 The frequency spectrum of a
  continuous-time signal is shown below.
• What is the minimum sampling frequency for this
  signal to be sampled without aliasing.
• If the above process were to be sampled at 10
  Krad/s, sketch the resulting spectrum from 20
  Krad/s to 20 Krad/s.

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Hints
The relationship between f(t) and fs(t), and F(?)
and Fs(?).
F(?)
Fourier Transform ?
?0
-?0
Fs(?)
?
-?0-?s
?0
-?0
-?0-2?s
?0?s
?02?s
-?0?s
?0-?s
?0-2?s
-?02?s
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Answers
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Tutorial
Solution 1) From the spectrum, we can see that
the bandwidth of the continuous signal is 8
Krad/s. The Sampling Theorem says that the
sampling frequency must be at least twice the
highest frequency component of the signal.
Therefore, the minimum sampling frequency for
this signal is 2816 Krad/s.
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Tutorial
2) Spectrum of the sampled signal is formed by
shifting up and down the spectrum of the original
signal along the frequency axis at i times of
sampling frequency. As ?s10 Krad/s, for i 0, we
have the figure in bold line. For i1, we have
the figure in bold-dot line.
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Tutorial
For I-1, ?2, we have