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: [email protected] Phone: (07)46312586 Office: Z408 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 1. 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 Lecture Mon. Wed. Fri. Tutorial Consulting 10 – 11 am Z313/office 4 –5 pm Z313/office 9 -10 am Z313 8 -10 am Z313 10 -12 am Z313 Note: Consulting in non-consulting time 1) Quick questions (5 minutes): Any time 2) Others: By Appointments 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 28 March 31 May Penalty Return Pass Ass. 1 20% 10% 3 weeks 50% Ass. 2 20% 10% 3 weeks 50% Exam. 60% 50% Result 100% 50%* Questions? Subject Overview Linear systems & control Computer engineering I & II Numerical computing Computer Controlled Systems Signal processing Robotics & machine vision 1.1 Introduction Control system definition & Control system application. Input; Stimulus Desired response Output; Response Control System 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 R(s) M(s) E(s) Gc(s) Controller Y(s) GP(s) Plant 1.1 Introduction Computer controlled system GHP(z) R(z) M(z) E(z) ZOH Gc(z) A/D Y(z) GP(s) D/A Computer system Plant 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. 2. 3. 4. 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 5. Output this control signal m(t) to controlled object 6. Save previous history of error and output for later use 7. 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 zeroorder-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 f(kT) f(t) Sampling time Time kT f(kT) f(t) A/D time Time kT 1.3 ADC and DAC f(kT) f(kT) D/A Time kT D/A is used as a ZOH. Time kT 1.3 ADC and DAC ADC Output Output Input ADC Digital Analog Input • 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 1.3 ADC and DAC ADC 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 DAC ADC 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 DAC ADC • 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. 20 2N-1 Bit N-1 Bit 0 1.3 ADC and DAC ADC 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 DAC ADC Decimal to binary 5 255 2 5 x 2 255 x x 51 2 102 102 0 2 255 102 5 25 6 3 0 12 1 2 51 2 25 2 12 2 6 2 3 2 1 50 24 12 6 2 0 1 1 0 0 1 1 1100110 B ; 66H Exercise: For N=10, repeat the above example 1.3 ADC and DAC DAC Input Digital Output Output DAC Analog Input 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 1.3 ADC and DAC … MUX AD Digital signal Status Analog signal Multi-channel A/D converter Control 1.3 ADC and DAC Control DA … MUX Analog signals Digital signals Multi-channel D/A converter 1.4 Errors ADC Output Errors Input 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 Output Input ADC Digital Analog Input 1.5 Sampling theorem f(kT) f(t) Sampling time Time kT f(kT) f(t) A/D time Time kT 1.5 Sampling theorem 5 5 4 4 3 3 2 2 1 1 0 0 -1 -1 -2 -2 -3 -3 -4 -4 -5 -5 0 1 2 3 4 5 6 7 8 9 10 4 4 3 3.5 0 1 2 0 1 2 3 4 5 6 7 8 9 10 2 3 1 2.5 0 2 -1 -2 1.5 -3 1 -4 -5 0.5 0 1 2 3 4 5 6 7 8 9 10 3 4 5 6 7 8 9 10 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 f(t) F() 5 4 Fourier Transform 3 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 -0 0 1.6 The proven of sampling theorem 2. For a sampled signal fs(t), we have 5 4 Fourier Transform 3 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 Fs() -2s -s -0 0 s 2s 1.6 The proven of sampling theorem 3. The relationship between f(t) and fs(t), and F() and Fs(). F() 5 4 Fourier Transform 3 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 Fs() 5 4 3 2 1 0 0 -0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 -2s -s -0 0 s 2s 1.6 The proven of sampling theorem 4. If we change the sampling frequency, what will happen with fs(t) and Fs(). Fs() 5 4 3 2 1 0 -1 -2 -3 -2s -4 -5 0 1 2 3 4 5 6 7 8 9 10 -s -0 0 s 4 3 Fs() 2 1 0 -1 -2 -2s -s -0 0 s -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 2s 2s 1.6 The proven of sampling theorem 5 4 3 2 1 0 -1 Fs() -2 -2s -s -0 0 s -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 4 3 Fs() 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 4 -2s -s -0 0 s 2s Fs() 3.5 3 2.5 2 1.5 1 0.5 0 1 2 3 4 5 6 7 8 9 10 -2s -s -0 0s 2s 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 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 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 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 0 10 20 30 40 50 60 70 80 90 100 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 0. 8 0. 6 0. 4 0. 2 0 -0. 2 -0. 4 -0. 6 -0. 8 -1 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: =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. F() Krad/s -8 -4 4 8 Hints The relationship between f(t) and fs(t), and F() and Fs(). F() Fourier Transform 5 4 3 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 4 5 6 7 8 9 10 Fs() 5 4 3 2 1 0 -1 -2 -3 -4 -5 0 -0 0 1 2 3 4 5 6 7 8 9 10 -0-s -0-2s -0 0-2s 0-s 0 + 0+2s -0+s 0 s -0+2s Answers f s 2 * 8 16 Krad / s F() 2 4 6 8 10 12 14 16 18 Krad/s 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 2*8=16 Krad/s. F() Krad/s -8 -4 4 8 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 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. F() -8 -4 2 4 6 8 10 12 14 16 18 Krad/s Tutorial For I=-1, 2,… we have -18 -14 -8 -6 -4 -2 F() 2 4 6 8 10 12 14 16 18 Krad/s 2 4 6 8 10 12 14 16 18 Krad/s

Descargar
# Linear Systems and Control