Advanced Control Engineering - Chapter 7 potx

//SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 198 ± [198±231/34] 9.8.2001 2:33PM 7 Digital control system design 7.1 Microprocessor control As a result of developments in microprocessor technology, the implementation of control algorithms is now invariably through the use of embedded microcontrollers rather than employing analogue devices. A typical system using microprocessor control is shown in Figure 7.1. In Figure 7.1 . RAM is Random Access Memory and is used for general purpose working space during computation and data transfer. . ROM, PROM, EPROM is Read Only Memory, Programmable Read Only Mem- ory and Erasable Programmable Read Only Memory and are used for rapid sources of information that seldom, or never need to be modified. . A/D Converter converts analogue signals from sensors into digital form at a given sampling period T seconds and given resolution (8 bits, 16 bits, 24 bits, etc.) . D/A Converter converts digital signals into analogue signals suitable for driving actuators and other devices. The elements of a microprocessor controller (microcontroller) are shown in Figure 7.2. Figure 7.2 shows a Central Processing Unit (CPU) which consists of . the Arithmetic Logic Unit (ALU) which performs arithmetic and logical oper- ations on the data and a number of registers, typically . Program Counter ± incremented each time an instruction is executed . Accumulator(s) ± can undertake arithmetic operations . Instruction register ± holds current instruction . Data address register ± holds memory address of data Control algorithms are implemented in either high level or low level language. The lowest level of code is executable machine code, which is a sequence of binary words that is understood by the CPU. A higher level of language is an assembler, which employs meaningful mnemonics and names for data addresses. Programs written in assembler are rapid in execution. At a higher level still are languages //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 199 ± [198±231/34] 9.8.2001 2:33PM such as C and C, which are rapidly becoming industry standard for control software. The advantages of microprocessor control are . Versatility ± programs may easily be changed . Sophistication ± advanced control laws can be implemented. Microprocessor System () rkT () ckT ukT () () ct · RAM Memory ROM PROM EPROM Memory Microprocessor Controller A/D Converter D/A Converter Plant Sensor ut () Fig. 7.1 Microprocessor control of a plant. program counter address bus accumulator(s) instruction register data bus data address register CPU clock ALU RAM ROM PROM EPROM Fig. 7.2 Elements of a microprocessor controller. Digital control system design 199 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 200 ± [198±231/34] 9.8.2001 2:33PM The disadvantages of microprocessor control are . Works in discrete time ± only snap-shots of the system output through the A/D converter are available. Hence, to ensure that all relevant data is available, the frequency of sampling is very important. 7.2 Shannon's sampling theorem Shannon's sampling theorem states that `A function f (t) that has a bandwidth ! b is uniquely determined by a discrete set of sample values provided that the sampling frequency is greater than 2! b '. The sampling frequency 2! b is called the Nyquist frequency. It is rare in practise to work near to the limit given by Shannon's theorem. A useful rule of thumb is to sample the signal at about ten times higher than the highest frequency thought to be present. If a signal is sampled below Shannon's limit, then a lower frequency signal, called an alias may be constructed as shown in Figure 7.3. To ensure that aliasing does not take place, it is common practice to place an anti- aliasing filter before the A/D converter. This is an analogue low-pass filter with a break-frequency of 0:5! s where ! s is the sampling frequency (! s > 10! b ). The higher ! s is in comparison to ! b , the more closely the digital system resembles an analogue one and as a result, the more applicable are the design methods described in Chapters 5 and 6. ft () –1.5 –1 –0.5 0 0.5 1 1.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Original Signal Alias t Fig. 7.3 Construction of an alias due to undersampling. 200 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 201 ± [198±231/34] 9.8.2001 2:33PM 7.3 Ideal sampling An ideal sample f à (t) of a continuous signal f (t) is a series of zero width impulses spaced at sampling time T seconds apart as shown in Figure 7.4. The sampled signal is represented by equation (7.1). f à (t)   I kÀI f (kT)(t À kT)(7:1) where (t ÀkT) is the unit impulse function occurring at t  kT. A sampler (i.e. an A/D converter) is represented by a switch symbol as shown in Figure 7.5. It is possible to reconstruct f (t) approximately from f à (t) by the use of a hold device, the most common of which is the zero-order hold (D/A converter) as shown in Figure 7.6. From Figure 7.6 it can be seen that a zero-order hold converts a series of impulses into a series of pulses of width T. Hence a unit impulse at time t is converted into a pulse of width T, which may be created by a positive unit step at time t, followed by a negative unit step at time (t À T), i.e. delayed by T. The transfer function for a zero-order hold is l[ f (t)]  1 s À 1 s e ÀTs G h (s)  1 À e ÀTs s (7:2) ( ) f* t TfT (6 ) (k ) fT (a) Continuous Signal (b) Sampled Signal () ft t 0 T 2 T 3 T 4 T 5 T 6 T k Tt Fig. 7.4 The sampling process. ft ( ) ft *( ) T Fig. 7.5 A sampler. Digital control system design 201 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 202 ± [198±231/34] 9.8.2001 2:33PM 7.4 The z -transform The z-transform is the principal analytical tool for single-input±single-output dis- crete-time systems, and is analogous to the Laplace transform for continuous systems. Conceptually, the symbol z can be associated with discrete time shifting in a difference equation in the same way that s can be associated with differentiation in a differential equation. Taking Laplace transforms of equation (7.1), which is the ideal sampled signal, gives F à (s)  l[ f à (t)]   I k0 f (kT)e ÀkTs (7:3) or F à (s)   I k0 f (kT)e sT ÀÁ Àk (7:4) Define z as z  e sT (7:5) then F(z)   I k0 f (kT)z Àk  Z[ f (t)] (7:6) In `long-hand' form equation (7.6) is written as F(z)  f (0) f (T)z À1  f (2T)z À2 ÁÁÁf (kT)z Àk (7:7) Example 7.1 Find the z-transform of the unit step function f (t)  1. * () ft f(t) tt (a) Discrete Time Signal (b) Continous Time Signal T T Fig. 7.6 Construction of a continuous signal using a zero-order hold. 202 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 203 ± [198±231/34] 9.8.2001 2:33PM Solution From equations (7.6) and (7.7) Z[1(t)]   I k0 1(kT)z Àk (7:8) or F(z)  1  z À1  z À2  FFF z Àk (7:9) Figure 7.7 shows a graphical representation of equation (7.9). Equation (7.9) can be written in `closed form' as Z[1(t)]  z z À 1  1 1 À z À1 (7:10) Equations (7.9) and (7.10) can be shown to be the same by long division 1  z À1  z À2 ÁÁÁ z À 1  z 00 z À 1 0  1 1 À z À1 0  z À1 z À1 À z À2 (7:11) Table 7.1 gives Laplace and z-transforms of common functions. z-transform Theorems: (a) Linearity Z[ f 1 (t) Æ f 2 (t)]  F 1 (z) ÆF 2 (z)(7:12) *( ) ft 1.0 t 0 T 2 T 3 T 4 T Fig. 7.7 z-Transform of a sampled unit step function. Digital control system design 203 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 204 ± [198±231/34] 9.8.2001 2:33PM (b) Initial Value Theorem f (0)  lim z3I F(z) (7:13) (c) Final Value Theorem f (I) lim z31 z À 1 z  F(z) ! (7:14) 7.4.1 Inverse transformation The discrete time response can be found using a number of methods. (a) Infinite power series method Example 7.2 A sampled-data system has a transfer function G(s)  1 s  1 Table 7.1 Common Laplace and z-transforms f (t)orf (kT) F(s) F(z) 1 (t)11 2 (t À kT)e ÀkTs z Àk 31(t) 1 s z z À 1 4 t 1 s 2 Tz (z À 1) 2 5e Àat 1 (s  a) z z À e ÀaT 61À e Àat a s(s  a) z(1 À e ÀaT ) (z À 1)(z À e ÀaT ) 7 1 a (at À 1  e Àat ) a s 2 (s  a) zf(aT À 1  e ÀaT )z  (1 À e ÀaT À aTe ÀaT )g a(z À 1) 2 (z À e ÀaT ) 8 sin !t ! s 2  ! 2 z sin !T z 2 À 2z cos !T 1 9 cos !t s s 2  ! 2 z(z À cos !T) z 2 À 2z cos !T 1 10 e Àat sin !t ! (s  a) 2  ! 2 ze ÀaT sin !T z 2 À 2ze ÀaT cos !T  e À2aT 11 e Àat cos !t (s  a) (s  a) 2  ! 2 z 2 À ze ÀaT cos !T z 2 À 2ze ÀaT cos !T  e À2aT 204 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 205 ± [198±231/34] 9.8.2001 2:33PM If the sampling time is one second and the system is subject to a unit step input function, determine the discrete time response. (N.B. normally, a zero-order hold would be included, but, in the interest of simplicity, has been omitted.) Now X o (z)  G(z)X i (z)(7:15) from Table 7.1 X o (z)  z z À e ÀT  z z À 1  (7:16) for T  1 second X o (z)  z z À 0:368  z z À 1   z 2 z 2 À 1:368z  0:368 (7:17) By long division 1  1:368z À1  1:503z À2 ÁÁÁ z 2 À 1:368z  0:368  z 2 000 z 2 À 1:368z  0:368 0  1:368z À 0:368 1:368z À1:871 0:503z À1 0  1:503 À 0:503z À1 1:503 À 2:056z À1  0:553z À2 (7:18) Thus x o (0)  1 x o (1)  1:368 x o (2)  1:503 (b) Difference equation method Consider a system of the form X o X i (z)  b 0  b 1 z À1  b 2 z À2 ÁÁÁ 1  a 1 z À1  a 2 z À2 ÁÁÁ (7:19) Thus (1  a 1 z À1  a 2 z À2 ÁÁÁ)X o (z)  (b 0  b 1 z À1  b 2 z À2 ÁÁÁ)X i (z)(7:20) or X o (z)  (Àa 1 z À1 À a 2 z À2 ÀÁÁÁ)X o (z)  (b 0  b 1 z À1  b 2 z À2 ÁÁÁ)X i (z)(7:21) Equation (7.21) can be expressed as a difference equation of the form x o (kT) Àa 1 x o (k À1)T À a 2 x o (k À 2)T ÀÁÁÁ  b 0 x i (kT)  b 1 x i (k À 1)T  b 2 x i (k À 2)T ÁÁÁ (7:22) Digital control system design 205 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 206 ± [198±231/34] 9.8.2001 2:33PM In Example 7.2 X o X i (s)  1 1  s  z z À e ÀT  z z À 0:368 (7:23) Equation (7.23) can be written as X o X i (z)  1 1 À 0:368z À1 (7:24) Equation (7.24) is in the same form as equation (7.19). Hence (1 À 0:368z À1 )X o (z)  X i (z) or X o (z)  0:368z À1 X o (z) X i (z)(7:25) Equation (7.25) can be expressed as a difference equation x o (kT)  0:368x o (k À1)T  x i (kT)(7: 26) Assume that x o (À1)  0 and x i (kT)  1, then from equation (7.26) x o (0)  0  1  1, k  0 x o (1)  (0:368  1)  1  1:368, k  1 x o (2)  (0:368  1:368)  1  1:503, k  2 etc: These results are the same as with the power series method, but difference equations are more suited to digital computation. 7.4.2 The pulse transfer function Consider the block diagrams shown in Figure 7.8. In Figure 7:8(a) U à (s) is a sampled input to G(s) which gives a continuous output X o (s), which when sampled by a () Us *( ) Us (a) () Uz (b) T T () Gs () Gz Xs o () Xs o *( ) Xz () o Fig. 7.8 Relationship between G(s)andG(z). 206 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC07.3D ± 207 ± [198±231/34] 9.8.2001 2:33PM synchronized sampler becomes X à o (s). Figure 7.8(b) shows the pulse transfer function where U(z) is equivalent to U à (s) and X o (z) is equivalent to X à o (s). From Figure 7.8(b) the pulse transfer function is X o U (z)  G(z)(7:27) Blocks in Cascade: In Figure 7.9(a) there are synchronized samplers either side of blocks G 1 (s) and G 2 (s). The pulse transfer function is therefore X o U (z)  G 1 (z)G 2 (z)(7:28) In Figure 7.9(b) there is no sampler between G 1 (s) and G 2 (s) so they can be combined to give G 1 (s)G 2 (s), or G 1 G 2 (s). Hence the output X o (z) is given by X o (z)  ZfG 1 G 2 (s)gU(z)(7:29) and the pulse transfer function is X o U (z)  G 1 G 2 (z)(7:30) Note that G 1 (z)G 2 (z) T G 1 G 2 (z). Example 7.3 (See also Appendix 1, examp73.m) A first-order sampled-data system is shown in Figure 7.10. Find the pulse transfer function and hence calculate the response to a unit step and unit ramp. T  0:5 seconds. Compare the results with the continuous system response x o (t). The system is of the type shown in Figure 7.9(b) and therefore G(s)  G 1 G 2 (s) Inserting values G(s)  (1 À e ÀTs ) 1 s(s  1) &' (7:31) (a) (b) () Us *( ) Us () Xs *( ) Xs T T T () Gs 1 () Gs 2 () Xs o () Us *( ) Us T () Gs 1 () Xs () Gs 2 () Xs o T Xs o *( ) Xs o *( ) Fig. 7.9 Blocks in cascade. Digital control system design 207 [...]... (7. 75) and (7. 73) K ˆ 9:58 and 105:23 (7: 85) Unit circle crossover: Inserting K ˆ 9:58 into the characteristic equation (7. 82) gives z2 À 0:487z ‡ 1 ˆ 0 (7: 86) z ˆ 0:244 Æ j0: 97 (7: 87) z ˆ 1€ Æ 75 :9 ˆ 1€ Æ 1:33 rad (7: 88) The roots of equation (7. 86) are or //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 220 ± [198±231/34] 9.8.2001 2:33PM 220 Advanced Control Engineering Since from equation (7. 63)... Figure 7. 24 (e) Mapping closed-loop poles from s to z-plane jzj ˆ eT inserting values jzj ˆ eÀ0:5Â0:5 ˆ 0 :77 9 (7: 116) €z ˆ !T ˆ 0:866  0:5 ˆ 0:433 rad (7: 1 17) ˆ 24:88 Converting from polar to cartesian co-ordinates gives the closed-loop poles in the zplane z ˆ 0 :70 7 Æ j0:3 27 (7: 118) which provides a z-plane characteristic equation z2 À 1:414z ‡ 0:6 07 ˆ 0 (7: 119) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D... equation (7. 69), when n ˆ 2 (À1)2 Q(À1) ˆ f1 À (0:092K À 1:368) ‡ (0:368 ‡ 0:066K)g > 0 (7: 72) Equation (7. 72) simplifies to give 2 :73 6 À 0:026K > 0 or K< 2 :73 6 ˆ 105:23 0:026 (7: 73) Im K = 9.58 z-plane 75 .90 K = 60 K=1 K = 105.23 x –2 –1.5 –1 Fig 7. 20 Root locus diagram for Example 7. 4 –0.5 x 0.5 K = 0 .78 1.0 Re //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 218 ± [198±231/34] 9.8.2001 2:33PM 218 Advanced. .. À 1 ˆ À1:414 Add (7: 124) 0: 270 6K À b ˆ 0:6 07 (7: 125) 0:5902K À 1 ˆ À0:8 07 or 0:5902K ˆ 0:193 K ˆ 0:3 27 (7: 126) Inserting equation (7. 126) into (7. 125) (0: 270 6  0:3 27) À 0:6 07 ˆ b b ˆ À0:519 (7: 1 27) Thus the required compensator is D(z) ˆ U 0:3 27( z À 0:6065) (z) ˆ E (z À 0:519) (7: 128) Figure 7. 25 shows that the continuous and discrete responses are identical, both with  ˆ 0:5 The control algorithm... //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 219 ± [198±231/34] 9.8.2001 2:33PM Digital control system design 219 Solution From equation (7. 43) '  & eÀTs 1 G(s) ˆ K 1 À s(s ‡ 2) s and from equation (7. 53), given that T ˆ 0:5 seconds   0:092z ‡ 0:066 G(z) ˆ K 2 z À 1:368z ‡ 0:368 (7: 77) (7: 78) Open-loop poles z2 À 1:368z ‡ 0:368 ˆ 0 (7: 79) z ˆ 0:684 Æ 0:316 ˆ 1 and 0:368 (7: 80) Open-loop zeros 0:092z... 4 ωs 8 π 4 r =1 σ s-plane Re z-plane Fig 7. 18 Mapping constant ! from s to z-plane jω Im x 8 x ω 9 22 x 5 x ω 6 s 8 7 x x 10 x 7 x 1 x 2 x 3 x 4 6 x x 5 σ x 10 x 9 x 8 5 x 5 x 6 x 7 x 8 x 9 x 10 s-plane x 1 x x 2 x 3 x 4 Re x ´ 6 x 7 z-plane Fig 7. 19 Corresponding pole locations on both s and z-planes Figure 7. 19 shows corresponding pole locations on both the s-plane and z-plane 7. 6.2 The Jury stability... 0:368) (7: 68) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 2 17 ± [198±231/34] 9.8.2001 2:33PM Digital control system design 2 17 or Q(z) ˆ z2 ‡ (0:092K À 1:368)z ‡ (0:368 ‡ 0:066K) ˆ 0 (7: 69) The first row of Jury's array is j z0 (0:368 ‡ 0:066K) z1 (0:092K À 1:368) z2 1 (7: 70) Condition 1: Q(1) > 0 From equation (7. 69) Q(1) ˆ f1 ‡ (0:092K À 1:368) ‡ (0:368 ‡ 0:066K)g > 0 (7: 71) From equation (7. 71),...//SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 208 ± [198±231/34] 9.8.2001 2:33PM 208 Advanced Control Engineering Xi(S) –Ts 1 s+1 1–e S Xo(S) T Fig 7. 10 First-order sampled-data system Taking z-transforms using Table 7. 1 G(z) ˆ (1 À zÀ1 ) or  G(z) ˆ zÀ1 z & & which gives G(z) ˆ For T ˆ 0:5 seconds z(1 À eÀT ) (z À 1)(z À eÀT ) z(1 À eÀT ) (z À 1)(z À eÀT ) 0:393 z À 0:6 07 (7: 32) '   1 À eÀT z À eÀT  G(z) ˆ ' (7: 33) (7: 34)... 6 7 8 7. 4.3 xi (kT) xo (kT) xo (t) À 0.5 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 k 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 0 0 0.304 0. 577 0.940 1.3 57 1.805 2. 275 2 .75 7 0 0 0.1 07 0.368 0 .72 3 1.135 1.582 2.050 2.530 3.018 The closed-loop pulse transfer function Consider the error sampled system shown in Figure 7. 11 Since there is no sampler between G(s) and H(s) in the closed-loop system shown in Figure 7. 11,... locus in the z-plane Solution (a) poles zeros z ˆ 1, 0:6 07 z ˆ À0:8 47 z2 ‡ (0:426K À 1:6 07) z ‡ (0:361K ‡ 0:6 07) ˆ 0 breakaway points (b) K ˆ 1:06, 47: 9 (c) radius ˆ 1:6 07, z ˆ 0 :79 5, À2:5 centre ˆ À0:8 47, 0 Example 7. 12 A unity feedback continuous control system has a forward-path transfer function G(s) ˆ K s(s ‡ 5) (a) Find the value of K to give the closed-loop system a damping ratio of 0 .7 The above . e –Ts S Fig. 7. 10 First-order sampled-data system. 208 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D ± 209 ± [198±231/34] 9.8.2001 2:33PM 7. 4.3 The closed-loop pulse. +σ σ s -plane ω ω = s 2 ω ω = – s 2 ω ω = s 4 –ω s 2 ω ω = – s 4 Fig. 7. 17 Mapping constant  from s to z-plane. 214 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 07. 3D. 0 00 000 1 0.5 0.5 0 0.1 07 2 1.0 1.0 0.304 0.368 3 1.5 1.5 0. 577 0 .72 3 4 2.0 2.0 0.940 1.135 5 2.5 2.5 1.3 57 1.582 6 3.0 3.0 1.805 2.050 7 3.5 3.5 2. 275 2.530 8 4.0 4.0 2 .75 7 3.018 () Rs Es ( ) *(