Essentials of Process Control phần 10 ppt

cttAt?‘EK IS: Stability Analysis of Sampled-Data Systems 525 hods. .s the value - gain 1 for ristic From i #r z in 5.35) 1 : Fig. L ) On the CR contour, z = Re’” where R + 00 and 8 goes from 7r through 0 to 7~. Substituting into Eq. ( 15.32) gives ~GWzP(z) = &K,>(l - b) Rei8 - b (15.36) As R -+ m, f-G@(,) -+ 0. Therefore, the infinite circle in the z plane maps into the origin in the ffG,+,(,)D(,) plane. The Ci, contour is just the reverse of the C,,, contour, going from the origin out along the negative real axis. The C- contour is just the reflection of the C+ contour over the real axis. So, just as in a continuous system, we only have to plot the C+ contour. If it goes around the (- 1,O) point, this sampled-data system is closedloop unstable since P = 0. The farther the curve is from the (- 1,O) point, the more stable the system. We can use exactly the same frequency-domain specifications we used for continuous sys- tems: phase margin, gain margin, and maximum closedloop log modulus. The last is ob- tained by plotting the function HGM(iw,Dcio,l( 1 + HG M(iu)D(io)). For this process (with 70 = K, = 1) with a proportional sampled-data controller and a sampling period of 0.5 minutes, the controller gain that gives a phase margin of 45” is K, = 3.43. The controller gain that gives a +2-dB maximum closedloop log modulus is Kc = 2.28. The ultimate gain is K, = 4.08. n EX A M PL E 15.6. If a deadtime of one sampling period is added to the process considered in the previous example, ~Gd4z) = K&(1 - b) z(z - b) (15.37) We substitute eiwTs for z in Eq. (15.37) and let o go from 0 to 0,/2. At o = 0, where z = + 1, the Nyquist plot starts at KcKp on the positive real axis. At o = 0,/2, where z = - 1, the curve ends at HGM(iw)D(iw) = K&(1 - b) = Kc&,(1 - b) -1(-l - b) l+b (15.38) This is on the positive real axis. Figure 15.6~ shows the complete curve in the HGM~D plane. At some frequency the curve crosses the negative real axis. This occurs when the real part of HG ~(i~&,) is equal to -0.394 for the numerical case considered in the previous example. Thus the ultimate gain is i K, = l/O.394 = 2.54 Note that this is smaller than the ultimate gain for the process with no deadtime. The controller gain that gives a +2-dB maximum closedloop log modulus is Kc = 1.36. The final phase angle (at 0,/2) for the first-order lag process with no deadtime was - 180”. For the process with a deadtime of one sampling period, it was -360’. If we had a deadtime that was equal to two sampling periods, the final phase angle would be -540”. Every multiple of the sampling period subtracts 180” from the final phase angle. Remember that in a continuous system, the presence of deadtime made the phase angle go to co as u went to ~0. So the effect of deadtime on the Nyquist plots of sampled- data systems is different than its effect in continuous systems. n EXAMPLE 15.7. As our last example, let’s consider the second-order two-heated-tank process studied in Example 15.4. GMts) = - 2.315 (s + l)(5s + I) E 526 PAW FIVE: Sampled-Data Systems i Using a zero-order hold, a proportional sampled-data controller, and a sampling period of T, = 0.5 minutes gives h~Guz, = O.O479K,(z + 0.8133) (z - 0.607)(z - 0.905) We substitute eiwrs for z and let o vary from 0 to wJ2. (15.39) 2T.r 271. - 4= us = - = - - T,f 0.5 At o = 0, the Nyquist plot starts at 2.315&. This is the same starting point that the continuous system would have. At o = 0,/2 = 27r, where z = - 1, the Nyquist plot ends on the negative real axis at 0.0479&(- 1 + 0.8133) (- 1 - 0.607)(- 1 - 0.905) = -o 0029K . ’ The entire curve is given in Fig. 15.M. It crosses the negative real axis at -0.087&. So the ultimate gain is K, = l/O.087 = 11.6, which is the same result we obtained from the root locus analysis. The controller gain that gives a phase margin of 45“ is KC = 2.88. The controller gain that gives a +2-dB maximum closedloop log modulus is K, = 2.68. n 15.2.3 Approximate Method To generate the HG M+) Nyquist plots discussed above, the z transform of the appro- priate transfer functions must first be obtained. Then e jwTs is substituted for z, and o is varied from 0 to 0,/2. There is an alternative method that is often more convenient to use, particularly in high-order systems. Equation (15.40) gives a doubly infinite series representation of HG~ci~j. 1 +m ffGM(iu) = T >: H(iw+inw,)G~(iw+ino,) (15.40) S n=-CC where Ht,, and GM($) are the transfer functions of the original continuous elements before z transforming. If the series in Eq. (15.40) converges in a reasonable number of terms, we can approximate H GMtiw) with a few terms in the series. Usually two or three are all that are required. 1 +m ffGM(ico) = 7 >: H(iw +inw,)Gn/r(iw +irwy) (15.41) S n=-cc 1 ffG~~(iw) 2: Ti’H(iw)GM(iw) + H(icx+-iwr)G~(iw+iwJ) + H(iw-iwS)GM(iw-iw,) S + H(i,+i2w,,)G~(iw+i2ws) + H(iw-i2u,)G~(iw-i20,)1 (15.42) This series approximation can be easily generated on a digital computer. The big advantage of this method is that the analytical step of taking the z trans- formation is eliminated. You just deal with the original continuous transfer functions. For complex, high-order systems, this ca;r elintinate a lot of messy algebra. iod 3% the jlot so om Her m ro- lo znt lite w nts :an hat $1) 42) ns- ns. C’~IAWI~K IS, Stability Analysis of Sampled-Data Sysrems 327 15.2.4 Use of MATLAB The frequency response of sampled-data systems can be easily calculated using MATLAB software. Table 15.2 gives a program that generates a Nyquist plot for the first-order process with a deadtime of one sampling period. After parameter values are specified (K, = 3, T, = 0.2, r0 = 1, and K, = i), the frequency range is specified from o = 0.01 to o = wJ2. The vector of complex variables z for each frequency is calculated. Then the complex function HGM(iW&,) is calculated in two steps: hgmd= kc*kp*(l -b)./(z-6); hgmd = hgmd .I z; Term-by-term division is specified by the use of the “I” operator. TABLE 15.2 MATLAB program for discrete frequency response 70 Program “dfreq.m” generates Nyquist plot 70 for jirst-order process with deadtime = sampling period % using a sampled-data P controller and zero-order hold % GM(s) = Kp*exp( -Ts*s)/(tauo*s+l) % 950 Give parameter values kc=3; ts=0.2; tauo=l; kp=l; b=exp(-ts/tauo); % Calculate sampling frequency ws=2*pi/ts; 940 Specify frequency range from 0.01 up to ws/2 w=[0.01:0.01:ws/2]; % Calculate vector of 2 values i=sqrt(-1); z=exp(i*w*ts); 70 % Calculate value of HGM(iw)*D(iw) at each frequency 70 hgmd=kc*kp*( 1 -b) ./ (z-b); hgmd=hgmd ./ z; % elf plot(hgmd) grid title( ‘Polar Plot for HGM(iw)*D(iw)‘) axis( ‘square ‘) clxis([-I 3.5 -2.5 21); xlabel( ‘Real HGM*D’) yiabel( ‘[mag HGM*D ‘) / . i t text(O.S. -0.5, ‘Ts=O.Z. Kc=3’) pause 528 PART FIVE: Sampled-Data Systems 15.3 PHYSICAL REALIZABILITY In a digital computer control system the feedback controller DtZ) has a pulse transfer function. What we need is an equation or algorithm that can be programmed into the digital computer. At the sampling time for a given loop, the computer looks at the current process output yu), compares it with a setpoint, and calculates a current value of the error e(,). This error plus some old values of the error and old values of the controller output or manipulated variable that have been stored in computer memory are then used to calculate a new value of the controller output m(,). These algorithms are basically difference equations that relate the current value of m to the current value of e and old values of m and e. These difference equations can be derived from the pulse transfer function &I. Suppose the current moment in time is the nth sampling period t = nT,. The current value of the error et,) is e(,Ts). We will call this e,. The value of et,) at the previous sampling time was e,-1. Other old values of error are en-z, en-j, etc. The value of the-controller output m(,) that is computed at the current instant in time t = nT, is rn(,~~) or m,. Old values are m,-1, mn-2, etc. Suppose we have the following difference equation: mn = hoe, + blend1 + b2en-2 + ** * + bMen-M - aim,-] - u2mn-2 - u3mn-3 - ” - - ahIm,-N (15.43) m(nTs) = boe(nl;) + ble(nTs-T,) f he(nT,-2T,) + * * * + b,we(nTS ,w,) - vqnTS-T3) - a2m(nTs-2Ts) - a3m(c-3TS) - . . . - aNm(,TS-NT3) (15.44) Limiting t to some multiple of T,, mw = he(t) + hqt-Ts) + b2+2T,) + . . . + h@-,t4Ts) (15.45) - alm(f-T,) - a2m(j-2Ts) - a3m(t-3T,) - '. . - aNfl(f-NTJ If each of these time functions is impulse sampled and z transformed, Eq. (15.45) becomes Mcz) = boEc,) + blz-‘EcZ, + b2z-2EtZj + b3z-3Et,j + . . . - q-7 -‘hf(,) - u2z-2k!(z) - . ’ - - uNzmNhrl(,) (15.46) Putting this in terms of a pulse transfer function gives M,,, _ b. + b,z-’ + b2z-2 + *.a + bMz+ D(,) = - - 4,) 1 + alz-’ + a2ze2 + *. . + aNzeN (15.47) A sampled-data controller is a ratio of polynomials in either positive or negative powers of z. It can be directly converted into a difference equation for programming into the computer. Continuous transfer functions are physically realizable if the order of the poly- nomial in s of the numerator is less than or equal to the order of the polynomial in s of the denominator. (*IIAP~EK IS: Stability Analysis of Sampled-Data Systems 529 sfer into s at rent iues uter ilue .ons The the The t= Gng .43) ,45) .45) .46) .47) tive ling Ay- in s The physical realizability of pulsed transfer functions uses the basic criterion that the current output of a device (digital computer) cannot depend upon future information about the input. We cannot build a gadget that can predict the future. If DtZJ is expressed as a polynomial in negative powers of Z, as in Eq. (15.47), the requirement for physical realizability is that there must be a “1” term in the denominator. If DtZ) is expressed as a polynomial in positive powers of z, as shown in Eq. (15.48) below, the requirement for physical realizability is that the order of the numerator polynomial in z must be,less than or equal to the order of the denominator polynomial in z. These two ways of expressing physical realizability are completely equivalent, but since the second is analogous to continuous transfer functions in s, it is probably used more often. M(z) D($ = - = bOzM + b,z”-’ + b2zM-2 + *. . + bM E(z) ZN -k alZN-’ + a2zNp2 +“’ + aN (15.48) Multiplying numerator and denominator by zeN and converting to difference equa- tion form give the current value of the output m,: ml = b@,+M-N + b,e,+,+N-1 + ‘. . + bMen-N - am-1 - a2mn-2 - ” ’ - aNmn-N (15.49) If the order of the numerator M is greater than the order of the denominator N in, Eq. (15.48), the calculation of m, requires future values of error. For example, if M - N = 1, Eq. (15.49) tells us that we need to know e,+l or e(,+~~) to calculate m, or m(,). Since we do not know at time t what the error e(,+TX) will be one sampling period in the future, this calculation is physically impossible. 15.4 MINIMAL-PROTOTYPE DESIGN One of the most interesting and unique approaches to the design of sampled-data controllers is called minimal-prototype design. It is one of the earliest examples of model-based or direct-synthesis controllers. The basic idea is to specify the desired response of the system to a particular type of disturbance and then, knowing the model of the process, back-calculate the controller required. There is no guarantee that the minimal-prototype controller is physically realizable for the given process and the specified response. Therefore, the specified response may have to be modified to make the controller realizable. Let us consider the closedloop response of an arbitrary system with a sampled- data controller. Y(z) = D(~,HGM(~, ’ G&z) 1 + D(z,HGM(z) Yg; + 1 + &,HGM(,, If we consider for the moment only changes in setpoint, (15.50) &.I= D(,)HGM(z) Yg l + &$GM(z) (15.51) 530 I+~T PINE: Sampled-Data Systems If we specify the form of the input YF$ and the desired form of the output Ytzj, and if the process and hold transfer functions are known, we can rearrange Eq. (I 5.5 I) to give the required controller designed for setpoint changes QT(~ (15.52) j If all of the terms on the right side of this equation have been specified, the controller can be calculated. EXAMPLE 15.9. The first-order lag process with a zero-order hold has a pulse transfer function HGnr(,, = K,(l - b) z-b (15.53) where b E eeTsJ7~~. Suppose we want to derive a minimal-prototype controller for step changes in setpoint. (15.54) We know that it is impossible to have the output of the process respond instanta- neously to the change in setpoint. Therefore, the best possible response that we could expect from the process would be to drive the output YCZ, up the setpoint in one sampling period. This is sketched in Fig. 15.7~. Remember, we are specifying only the values of the variables at the sampling times. The output at t = 0 is zero. At t = T,, the output should be 1 and should stay at’1 for all subsequent sampling times. Therefore, the desired YtEJ is y(z) = y(0) + y(T$-’ + y(2Ts)z-2 + y(3Ts,z-3 + . ” = 0 + z-1 + z-2 + z-3 + . . * (15.55) i Z -I Y(z) = 1 1-z-l =- z-l Plugging these specified functions for YQ) and Y$ into Eq. (15.52) gives 1 (15.56) h(z) = Y (2) z-l = HGm,tY;Z”; - Y(z)) (fGqz,) 5 - !- z-1 (15.57) D 1 ‘(,-) = HGM(<)(z - 1) Now, for this first-order process, Eq. (15.53) gives HG MCZJ. Plugging this into Eq. (15.57) gives the minimal-prototype controller. D 1 1 s(z) = HGMc,,(z - 1) = K,>( 1 - b) z-b K,,( I - b)(z - 1) (15.58) This sampled-data controller is physically realizable since the order of the polynomial in the numerator is equal to the order of the polynomial in the denominaror. Therefore, the desired setpoint response is achievable for this process. CHAPTER IS: Stability Analysis of Sampled-Data Systems 531 :r :r l- d g bf 1 I I l Yli l l l l . 2 h Only can specify values of y(,) 2 at sampling times Discontinuity in slope at t = TfT (b) *t 2T, 3T, 4T, 5T, 6T, 1 T, 2T, 3T, 4T, 5T, 6T, IT, 8T, 9T, (4 _ / I I I *t T, 2T, 3T, 4T, . FIGURE 15.7 Minimal-prototype responses. (a) Desired response to unit step change in setpoint. (b) Response of first-order process. (c) Response of second-order system when driven to setpoint in one sampling period. (d) Modified response of second-order system to take two sampling periods to reach setpoint without rippling. 532 PART FIVE: Sampled-Data Systems Before we leave this example, let’s look at the closedloop characteristic equation of the system. 1 + DScZ)HGM(z) = 0 If we substitute Eqs. (I 5.58) and (15.53), we get I+ Z-h K,U-6) =. &,(I -b)(z- 1) z-b I+ I -=o j z=o z-l Thus, the closedloop root is located at the origin. This corresponds to a critically damped closedloop system ([ = 1). The specified response in the output was for no overshoot, so this damping coefficient is to be expected. n EXAMPLE 15.10. If we have a first-order lag process with a deadtime equal to one sam- pling period, the process transfer function becomes HGMI(~) = Kp(l - b) z(z - b) (15.59) Suppose we specified the same kind of response for a step change in setpoint as in Ex- ample 15.9: the output is driven to the setpoint in one sampling period. Substituting our new process transfer function into Eq. (15.58) gives D 1 1 S(z) = HGM,,,(z - 1) = K,(l - 6) z(z - b) = K,(l - b)(z - 1) (15.60) z(z - 6) (z - 1) This controller is WC physically realizable because the order of the numerator is higher (second) than the order of the denominator (first). Therefore, we cannot achieve the re- sponse specified. This result should really be no surprise. The deadtime does not let the output even begin to change during the first sampling period, and we cannot drive the output up to its setpoint instantaneously at t = TX. Let us back off on the specified output and allow two sampling periods to drive the output to the setpoint. Y(z) = Y(0) + Y(T,)z-’ + y(*T,)z-* + y(J&)z-3 + . . . = o+(o)~-‘+~-*+z-3+ Y(z) = Z-* 1 1 - z-’ = z(z - 1) Now the minimal-prototype controller for step changes in setpoint is 1 D Y S(z) = (z) HG,w(~,(Y;',~: - Yc,I) = PW - b) z(z - 1) I[ z 1 ___ - Z-1 z(z - 1) (15.61) (15.62) - I (15.63) z(z - 6) = K,(l -6)(22-i) ~mwrts IS: Stability Analysis of Satnpled~Data Systems 533 led 06 n m- i9) 1X- ur 10) ler *e- he he he 1) 2) 3) Using long division to see the values of VZ(,T,~) gives The controller is physically realizable since N = 2 and M = 2. Note that there are two poles, one at z = + 1 and the other at z = - I, and there are two zeros (at z = 0 and z = b). n A first-order process can be driven to the setpoint in one sampling period and held right on the setpoint even between sampling periods. This is possible be- cause we can change the slope of a first-order process response curve, as shown in Fig. 15.76. If the process is second or higher order, we are not able to make a discontin- uous change in the slope of the response curve. Consequently, we would expect a second-order process to overshoot the setpoint if we forced it to reach the setpoint in one sampling period. The output would oscillate between sampling periods, and the manipulated variable would change at each sampling period. This is called rippling and is illustrated in Fig. 15.7~. Rippling is undesirable since we do not want to keep wiggling the control valve. We may want to modify the specified output response to eliminate rippling. Allowing two sampling periods for the process to come up to the setpoint gives us two switches of the manipulated variable and should let us bring a second-order process up to the setpoint without rippling. This is illustrated in Example 15.11. In general, an Nth- order process must be given N sampling periods to come up to the setpoint if the response is to be completely ripple free. Since we know only the values of the output ytn~,) at the sampling times, we cannot use Yt,) to see if there are ripples. We can see what the manipulated variable rn(,~~) is doing at each sampling period. If it is changing, rippling is occurring. SO we choose Ytz, such that Mt,, does not ripple. If the controller is designed for setpoint changes [Eq. (15.52)], D(z) = Y(z) HGdY(Z) set - Y(z)> Y(z) M(z) = HGMM(,) Let’s check the first-order system from Example 15.9. &Al - b) 1 HG M(z) = z-b and Ytz) = - z-l 1 M Y(z) Z-l z-b (‘) = HGM(Z, = K,(l - b) = I$( 1 - b)(z - 1) z-b 1 M -___ 1 -I 1 -2, It-‘+ (‘) = K,,(l-b)+K,Z +Gz KP (15.64) (15.65) (15.66) (15.67) 534 PART FIVE : Sampled-Data Systems Thus, the manipulated variable holds constant after the first sampling period, indi- cating no rippling. E X A M P L, E t s. t I . The second-order process considered in Example 15.4 has the follow- ing openloop transfer function: Using a zero-order hold gives an openloop transfer function ~GI(,) = K,ao(z - ZI> (z - PINZ - P2) (15.69) We want to design a minimal-prototype controller for a unit step setpoint change. The output is supposed to come up to the new setpoint in one sampling period. Substituting Eq. (15.69) into Eq. (15.52) gives 1 1 D S(z) = = fG,&. - 1) K,aok - ZI) cz _ 1> (2 - Pl>(Z - P2) (15.70) (z - Pl>(Z - P2) = K,,ao(z - ZI)(Z - 1) This controller is physically realizable. Therefore, minimal-prototype control should be attainable. But what about intersample rippling? Let us check the manipulated variable. 1 yw (z-pd(z-P2) ’ - = M(z) = HGM(,, z-1 K,ao(z - zr > = ao(z - l)(z - Zl> (15.71) (z - PINZ - P2) . Long division shows that the manipulated variable changes at each sampling period, SO rippling occurs. For a specific numerical case (Kp = rol = 1; 7,~ = 5; T, = 0.2) the parameter values are p1 = 0.8187, p2 = 0.9608, aa = 0.0037, and ZI = -0.923. Mcz) = 270 -4602-l + 4272 -2 - 392~-~ + 364~-~ - 334z-’ + . . . (15.72) This system exhibits rippling. To prevent rippling, we.modify our desired output response to give the system two sampling periods to come up to the setpoint. The value of yCr) at the first sampling period, the yl shown in Fig. 15.711, is unspecified at this point. The output YcZ) is now Y(z) = VIZ -I +z-2+z-3+ = y, z-1 + z-2(1 + z-1 + z-2 + z-3 -t . . .) = y,z-’ + z-2& Y(z) = Ylz+ . 1 -yt z(z - I) - (15.73) The setpoint disturbance is still a unit step: [...]... response of the process at the sampling times if a controller gain of 0.722 is used 15.9 Make a root locus plot for the process considered in Problem 15.7 in the z plane iat IO- 15 .10 A first-order lag process with a zero-order hold is controlled by a proportional sampled-data controller (a) What value of gain gives a critically damped closedloop system? (b) What is the gain margin when this value of gain... Use sampling times of 0.1 and I minute (S +81)3 536 15.2 Repeat Problem 15.1 using a sampled-data PI controller Use r/ values of 0.5 and 2 minutes 15.3 Make Nyquist plots for the process of Problem 15.1 and find the value of gain that gives the following specifications: l l l Gain margin of 2 Phase margin of 45” Maximum closedloop log modulus of +2 dB 15.4 Make a root locus plot of the system in Problem... integrator process with a proportional sampled-data controller (b) Use the Nyquist stability criterion to find the ultimate gain (c) What is the phase margin of the system if Kc = l/T,7? 15.25 A pure integrating process with unity gain is controlled by a PI controller with reset 71 = 1 minute If the controller is analog, what value of controller gain gives a closedloop damping coefficient of 0.3? Now... use MI to control Y (a) Using a digital proportional controller D1 and zero-order hold with sampling period T, = 0.5, find the value of controller gain K, that gives a closedloop damping coefficient of 0.5 However,, using manipulated variable Ml is more expensive than using M2 because M2 is cheaper Therefore, we want to use a “valve position controller” (VPC), a simple type of optimizing control, that... What is the closedloop characteristic equation of the entire system with the DI controller tuned using the gain determined in part (Al)? (d) Solve for the roots of the closedloop characteristic equation as functions of the gain K, of the 112 controller 540 fm7‘ I:fvti: San~plcd-Data Systems 15.23 A process has an opcnloop transfer function relating the controlled variable Y and the manipulated variable... response of the system when this controller is used and a unit step change in the load input occurs 15.16 An openloop-unstable, first-order process has the transfer function KP G M(s) = ~ 7,s - 1 A discrete approximation of a PI controller is used (a) Sketch a root locus plot for this system Show the effect of changing the reset time ~1 from very large to very small values (b) Find the maximum value of controller... important This means that an accurate value of the steadystate gain (the o = 0 point) is not required Figure 16.1 illustrates the point Suppose we have two models of a process, model A and model B, and we know exactly what the true transfer function of the process is The step responses of the alternative models are compared with the response of the real process in Fig 16 la Model A fits the real plant... duration of the input pulse We need to keep the width of the pulse fairly small to prevent its “frequency content” from becoming too low at higher frequencies A good rule of thumb, is to keep the width of the pulse less than about half the smallest time constant of interest If the dynamics of the process are completely unknown, it takes a few trials to establish a reasonable pulse width If the width of the... HGM~;) (6) Calculate the value of controller gain that puts the system right at the limit of closedloop stability (c) Calculate the controller gain that gives a closedloop damping coefficient of 0.3 CIJA~~EK IS: Stability Analysis of Sampled-Data Systems 537 15.8 A process controlled by a proportional digital controller with zero-order hold and sampling period T, = 0.25 has the openloop pulse transfer... setpoint changes, the process should be brought up to the setpoint in one sampling period For load changes, the process should be driven back to its initial steady-state value in two sampling periods Calculate the controller outputs for both controllers to check for rippling _ (IIAI~I B.K IS: Stability Analysis of Sampled-Data Systems 15.20, Dmign a minimal-pl-vtvtyp~ aamplcd-data controller process rtmsf’er . integrating process with unity gain is controlled by a PI controller with reset 71 = 1 minute. If the controller is analog, what value of controller gain gives a closedloop damp- ing coefficient of 0.3? Now. direct-synthesis controllers. The basic idea is to specify the desired response of the system to a particular type of disturbance and then, knowing the model of the process, back-calculate the controller. PI controller. Use r/ values of 0.5 and 2 minutes. 15.3. Make Nyquist plots for the process of Problem 15.1 and find the value of gain that gives the following specifications: l Gain margin of