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@usq.edu.au

Phone: (07)46312586

Office: Z408

Study Materials

Study Book: Computer controlled systemsTextbook: “Discrete time control system”, Katsuhiko Ogata, 2nd

ed. Prentice-Hall, Inc., 1995.Support Materials: Matlab ( 5.3/later + Control toolbox)References1. Astrom K. J. & Wittenmark B., “Computer controlled

systems theory and design”, 3rd edition, Prentice Hall, New York, 1996.

2. Nagle H. & Philips C., “Digital control system analysis & design”, Prentice Hall.

3. Norman S. Nise, “Control system engineering” Third edition, John Wiley & Sons, Inc., New York, 2000.

Timetable

Note: Consulting in non-consulting time1) Quick questions (5 minutes): Any time2) Others: By Appointments

Lecture Tutorial Consulting

Mon. 9 -10 am

Z313

10 – 11 am

Z313/office

Wed. 8 -10 am

Z313

4 –5 pm

Z313/office

Fri. 10 -12 am

Z313

Assessment

Students’ attention is drawn to the student related policies and the unit specification 2002, which outline the University’s Assessment Policy and the requirements.

For unit 70520, Computer controlled systems, you must obtain 50% from each of three assessments and 50% all over.

Assessment

Weight Due Penalty Return Pass

Ass. 1 20% 28 March

10% 3 weeks 50%

Ass. 2 20% 31 May 10% 3 weeks 50%

Exam. 60% 50%

Result 100% 50%*

Questions?

Subject Overview

Computer engineering I & II

Linear systems & control

Numerical computing

Computer Controlled Systems

Signal processing Robotics & machine vision

1.1 Introduction

Control system definition & Control system application.

Control SystemInput; Stimulus

Desired response

Output; Response

Actual response

1.1 Introduction

Advantages of control systems

1. Power amplification

2. Remote control

3. Convenience

4. Compensation for disturbance

1.1 Introduction

Control system

Gc(s) GP(s)R(s) E(s) M(s) Y(s)

Controller Plant

1.1 Introduction

Computer controlled system

Gc(z) ZOH GP(s)R(z)E(z)

M(z)

GHP(z)

Computer system

Y(z)

Plant

A/D D/A

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.

1.2 Digital control loop:Signals

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).

1.2 Digital control loop:Sequence of events

1. Get desired output r(t) at this instant in time2. Measure actual output c(t)3. Calculate error e(t)=r(t)-c(t)4. Derive control signal m(t) based on proper control

algorithm5. Output this control signal m(t) to controlled object6. Save previous history of error and output for later

use7. Repeat step 1 to 6 (go to 1)

1.2 Digital control loop:Forms 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

1.2 Digital control loop:Forms 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

1.3 ADC and DAC

time

f(t)

Time kT

f(kT)

Sampling

time

f(t)

Time kT

f(kT)

A/D

1.3 ADC and DAC

D/A is used as a ZOH.Time kT

f(kT)

D/A

Time kT

f(kT)

1.3 ADC and DACADC

• Have a discrete number of quantization levels

• Number of levels L=2N, where N is the number of bits

• eg N=3 bits, L=23=8 levels

Output

Input

ADC

Input

Output

AnalogDigital

1.3 ADC and DACADC

Bits Level Signal Error

1 2 5 5/2=2.5

2 4 5 5/4=1.25

3 8 5 5/8=0.625

4 16 5 5/16=0.3125

1.3 ADC and DACADC

More bits more accuracy. The commonly used ADC has

• 8-bits: L=28=256 (coarse)

• 10-bits: L=210=1024 (adequate)

• 12-bits: L=212=4096 (works well)

• 16-bits: L=216=65536 (almost overkill)

1.3 ADC and DACADC

• Distances between sequential levels are the same. eg 5v/28=0.0195v

• 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.

Bit 0Bit N-1

202N-1

1.3 ADC and DACADC

Example: For N=8, 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 27+26+25+24+23+22+21+20= 255

In hexadecimal: 0 F=15; 00H FFH

1.3 ADC and DACADC

Decimal to binary

Exercise: For N=10, repeat the above example66H ;1100110

0

1

0

12 1

1

2

32 3

0

6

62 6

0

12

122 12

1

24

252 25

1

50

512 51

0

102

1022

1025

255225525

2

255

5

B

xxx

1.3 ADC and DACDAC

Example: For N=8 and the signal is from 0 to 5, find the output value for the number 145.

Solution:5/255=x/145, x=5*145/255=2.8431=2.84

Output

Input

DAC

InputOutput

DigitalAnalog

1.3 ADC and DAC

Multi-channel A/D converter

MUX ADDigital signal Status Control

…

Analog signal

1.3 ADC and DAC

Multi-channel D/A converter

DA MUXAnalogsignals

Control…

Digital signals

1.4 Errors

Input

ADC Output

Errors

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.

1.4 Errors

Gain error

Output

Input

ADC

Input

Output

AnalogDigital

1.5 Sampling theorem

time

f(t)

Time kT

f(kT)

Sampling

time

f(t)

Time kT

f(kT)

A/D

1.5 Sampling theorem

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

0 1 2 3 4 5 6 7 8 9 100.5

1

1.5

2

2.5

3

3.5

4

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/Shannon’s sampling theorem/Shannon’s theorem.

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 20.

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.

1.6 The proven of sampling theorem

The proven is based on Fourier Transform

1. 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.

1.6 The proven of sampling theorem

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

FourierTransform

F()

0-0

f(t)

1.6 The proven of sampling theorem

2. For a sampled signal fs(t), we have

FourierTransform

0 1 2 3 4 5 6 7 8 9 10

-5

-4

-3

-2

-1

0

1

2

3

4

5

Fs()

0-0-s s 2s-2s

1.6 The proven of sampling theorem

3. The relationship between f(t) and fs(t), and F() and Fs().

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5

FourierTransform

F()

0-0

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5 Fs()

0-0-s s 2s-2s

1.6 The proven of sampling theorem

4. If we change the sampling frequency, what will happen with fs(t) and Fs().

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5 Fs()

0-0-s s 2s-2s

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

Fs()

0-0-ss 2s-2s

1.6 The proven of sampling theorem

0 1 2 3 4 5 6 7 8 9 10

-5

-4

-3

-2

-1

0

1

2

3

4

5 Fs()

0-0-s s 2s-2s

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

Fs()

0-0-ss 2s-2s

0 1 2 3 4 5 6 7 8 9 100.5

1

1.5

2

2.5

3

3.5

4

Fs()

0-0-s s2s-2s

1.6 The proven of sampling theorem

5. Conclusions

If our sampling frequency s is faster enough, that is s>20, 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 =20, where 0 is the highest frequency component or Nyquist Frequency of the signal.

1.7 Aliasing

1. Aliasing problem

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.7 Aliasing

Ambiguity: alias

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50 60 70 80 90 100-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.7 Aliasing2. Finding aliasesThe 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 f0=90Hz & fs=100, find alias.

Solution: =2f, fn=fs/2=50Hz,

f=| (f0+ fn)mod(fs) - fn|=|(90+50)mod(100)-50| =|40-50|=10Hz

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

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?

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 signal’s Nyquist frequency.

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.

Exercise

Exercise 1: The frequency spectrum of a continuous-time signal is shown below.

1) What is the minimum sampling frequency for this signal to be sampled without aliasing.

2) If the above process were to be sampled at 10 Krad/s, sketch the resulting spectrum from –20 Krad/s to 20 Krad/s.

-8 -4 4 8

Krad/s

F()

Hints

The relationship between f(t) and fs(t), and F() and Fs().

0 1 2 3 4 5 6 7 8 9 10-5

-4

-3

-2

-1

0

1

2

3

4

5 FourierTransform

F()

0-0

0 1 2 3 4 5 6 7 8 9 10

-5

-4

-3

-2

-1

0

1

2

3

4

5 Fs()

0-0

0-s

-0-s

0-2s

-0-2s-0+s

0+s

-0+2s

0+2s

Answers

sKradf s /168*2

4 8

Krad/s

122 6 14 181610

F()

TutorialSolution: 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 2*8=16 Krad/s.

-8 -4 4 8

Krad/s

F()

Tutorial2) 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 s=10 Krad/s, for i =0, we have the figure in bold line. For i=1, we have the figure in bold-dot line.

4 8

Krad/s

F()

122 6 14 181610-4-8

TutorialFor I=-1, 2,… we have

4 8

Krad/s

122 6 14 181610

-18

F()

4 8

Krad/s

122 6 14 181610-2-4-6-8-14