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Computer Controlled System ELE3105/70520

Computer Controlled System ELE3105/70520. 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.

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Computer Controlled System ELE3105/70520

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  1. 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

  2. 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.

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

  4. 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.

  5. Assessment

  6. Questions?

  7. Linear systems & control Numerical computing Computer Controlled Systems Signal processing Robotics & machine vision Subject Overview Computer engineering I & II

  8. Output; Response Input; Stimulus Control System Desired response Actual response 1.1 Introduction Control system definition & Control system application.

  9. 1.1 Introduction • Advantages of control systems • Power amplification • Remote control • Convenience • Compensation for disturbance

  10. M(s) Y(s) E(s) R(s) GP(s) Gc(s) Controller Plant 1.1 Introduction Control system

  11. GHP(z) M(z) Y(z) E(z) R(z) GP(s) ZOH Gc(z) D/A A/D Computer system Plant 1.1 Introduction Computer controlled system

  12. 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.

  13. 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).

  14. 1.2 Digital control loop:Sequence 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)

  15. 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

  16. 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

  17. f(kT) f(t) Sampling time Time kT f(kT) f(t) A/D time Time kT 1.3 ADC and DAC

  18. f(kT) f(kT) D/A Time kT Time kT 1.3 ADC and DAC D/A is used as a ZOH.

  19. Output Output Input ADC  Digital Analog Input 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

  20. 1.3 ADC and DACADC

  21. 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)

  22. 20 2N-1 Bit 0 Bit N-1  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.

  23. 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

  24. 1.3 ADC and DACADC Decimal to binary Exercise: For N=10, repeat the above example

  25. Input Output Output DAC  Analog Input Digital 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

  26. AD MUX  Digital signal  Status Control Analog signal 1.3 ADC and DAC Multi-channel A/D converter

  27. Control … Analog signals MUX DA   Digital signals 1.3 ADC and DAC Multi-channel D/A converter

  28. Output  ADC Errors  Input 1.4 Errors

  29. 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.

  30. Output Input ADC  1.4 Errors Gain error Output Digital Analog Input

  31. f(kT) f(t) Sampling time Time kT f(kT) f(t) A/D time Time kT 1.5 Sampling theorem

  32. 1.5 Sampling theorem

  33. 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.

  34. 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 sis 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.

  35. 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.

  36. f(t) F() Fourier Transform  0 -0 1.6 The proven of sampling theorem

  37. Fourier Transform  Fs() -2s -s -0 0 s 2s 1.6 The proven of sampling theorem 2. For a sampled signal fs(t), we have

  38. F() Fourier Transform  Fs()  0 -0 -2s -s -0 0 s 2s 1.6 The proven of sampling theorem 3. The relationship between f(t) and fs(t), and F() and Fs().

  39. Fs()  Fs() s -2s -s 0 2s -0 -2s -s -0 0 s 2s 1.6 The proven of sampling theorem 4. If we change the sampling frequency, what will happen with fs(t) and Fs().

  40. Fs() Fs() s -2s -s 0 2s -0 Fs() -0 -2s -s 0 s 2s -2s -s -0 0 s 2s 1.6 The proven of sampling theorem

  41. 1.6 The proven of sampling theorem 5. Conclusions If our sampling frequency s is faster enough, that is s>2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.

  42. 1.7 Aliasing 1. Aliasing problem

  43. 1.7 Aliasing Ambiguity: alias

  44. 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 f0=90Hz & fs=100, find alias. Solution: =2f, fn=fs/2=50Hz, f=| (f0+ fn)mod(fs) - fn|=|(90+50)mod(100)-50| =|40-50|=10Hz

  45. 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

  46. 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?

  47. 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.

  48. 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.

  49. F()  Krad/s -8 -4 4 8 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.

  50. Hints The relationship between f(t) and fs(t), and F() and Fs(). F() Fourier Transform  0 -0 Fs()  -0-s -0-2s -0 0 0+s 0+2s -0+s 0-2s 0-s -0+2s

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