2.010 Fall 2000 Solution of Homework Assignment 8


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1 2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s) _ θ t e θt e θt M(s) θ t Figure 1: Block Diagram of Controlled Servomechanism. The openloop transfer function is and the closedloop transfer function is OLT F(s) =C(s)P (s)m(s) C(s)P (s) CLTF(s) = 1+OLTF(s) In this problem we consider two plant models: the system with an ideal servovalve, for which we found in Problem Set 7, 2π 2 P 1 (s) = s(s +2π) rad/amp and the system with servovalve dynamics, for which we found in Problem Set 6, 2π 2 P 2 (s) = s(s +2π) (s + λ v ) rad/amp 2 It is given that λ v =25Hzor5π rad/sec. We also consider two controllers: the simple proportional control for which λ 2 v C 1 (s) =G amps/volt and the proportional plus derivative controller (PD control) for which C 2 (s) =G(1 + K d s) amps/volt In all cases the measurement operator M(s) is simply the constant M =1. volt/radian. 1
2 (a) When P (s) =P 1 (s) and C(s) =C 1 (s) =G in Fig. 1 the openloop transfer function is OLTF(s) =GM 2π2 s(s +2π) The openloop poles are at s = and s = 2π. The closedloop poles are the roots of the quadratic s 2 +2πs +2π 2 GM = The root locus could be obtained by brute force by simply solving this quadratic and plotting the roots as G is varied from G =tog =. The general nature of this locus can be deduced from the five rootlocus rules. The locus has real segments in the range between s = 2π and s = (this is the region to the left of one pole). The locus has complex branches, which are symmetric about the real axis, and which, because there are two poles and no zeros in the OLTF(s), have two straightline asymptotes as G that emanate from c =( 2π +)/2 = π and make the angles 18(1/2) = 9 degrees and 18(3/2) = 27 degrees with the real axis. Imag c s = 2π s = Figure 2: segment and asymptotes determined by rootlocus rules. The sketch in Fig. 2 indicates the real line segment and the asymptotes. The five rules do not fix the point at which the root locus breaks out from the real axis, or the precise shape of the locus close to the break out. In the present case the quadratic equation for the root locus is easily solved to give s = π ± π 1 2GM which, when M =1. volts/rad, indicates that for small G there are two real poles until G =.5 when there is a double root at s = c = π, and that for all 2
3 larger values of G the poles are complex with real part always equal to c = π. The root locus thus has the shape shown in Fig. 3. Imag c Figure 3: Root locus for system with P 1 (s)andc 1 (s). (b) For the system with C(s) =C 1 (s) and P (s) =P 2 (s) the openloop transfer function is ( ) OLTF(s) =G 2π2 5π 2 s(s +2π) s +5π when M = 1. volt/rad. The openloop poles are at s = and s = 2π and a double pole at s = 5π. There are no openloop zeros so there are four asymptotes at angles 45 degrees, 135 degrees, 225 degrees and 315 degrees, which emanate from the point on the real axis where 2π ++2( 5π) s = c = = 25.5π 4 The only segment of the real axis which belongs to the root locus is the segment 2π <s<. A sketch of a root locus which is consistent with these rules is shown in Fig. 4. Note how the dynamics of the actuator degrades the stability of the control system. With no actuator dynamics the closedloop system of figure 3 is stable for all positive gains. When the actuator dynamics is included, the closedloop system of Fig. 4 is unstable for sufficiently high gains. (c) An accurate root locus is obtained in the following MATLAB session, in which the openloop transfer function with unity gain is entered, and the root locus for the closed loop transfer function is requested: 3
4 Imag c Figure 4: Sketch of root locus for system with C 1 (s) and P 2 (s). sysc1p2 = tf(5*pi^4, [ 1 12*pi 27*pi^2 5*pi^3 ]) Transfer function: 4.87e s^ s^ e4 s^ e5 s rlocus(sysc1p2) The resulting root locus is displayed in Fig. 5 From Fig. 5 we see that that the root locus crosses from the left halfplane (stable) to the right halfplane (unstable) where the imaginary part of the root is approximately ±2 rad/sec. To estimate the gain G at the onset of instability, we estimate the magnitude of the OLTF(s) when s = j2, as shown in Fig. 6. The magnitude of the openloop transfer function for unity gain (G = 1. amp/volt) is 5π 4 OLTF(j2) = j2 2π + j2 5π + j2 =.465 At the onset of instability the magitude of the OLTF is unity, which requires that.465g =1. This occurs when G =21.5 amps/volt which is 27 db greater than G =1. amp/volt. 4
5 Imag Axis Axis Figure 5: MATLAB root locus for system with C 1 (s) and P 2 (s). Imag c Figure 6: Estimating magnitude of OLTF(s). (d) To check these estimates, return to MATLAB and request the gain margin by typing in margin(sysc1p2) The Bode plot displayed in Fig. 7 is returned with the information that a gain increase of db (G = 26. amps/volt) from a gain of unity will produce borderline stability at a frequency of rad/sec. This indicates that the 5
6 estimates obtained above are about 1% too low for the frequency at the onset of instability and about 1.3 db too low for the gain value at the onset of instability. Bode Diagrams Gm= db (at rad/sec), Pm= deg. (at rad/sec) 5 Phase (deg); Magnitude (db) Frequency (rad/sec) Figure 7: MATLAB Bode plot for system with C 1 (s) and P 2 (s). (e) For the system with C(s) =C 2 (s) and P (s) =P 1 (s) the openloop transfer function is OLTF(s) =GK d M 2π2 (s + z) s(s +2π) where z =1/K d is the zero introduced by the derivative action. We are told to choose K d so that z =.9 Hzorz =.9(2π) =5.65 rad/sec. This requires that K d =.1768 sec. With M =1. volts/rad, the OLTF(s) is OLTF(s) =.1768G 2π2 (s +5.65) s(s +2π) 3.49(s +5.65) = G s(s +6.28) which has two poles and one zero. The root locus rules tell us that there is only a single asymptote emanating from c =[ 6.28 ( 5.65)]/1 =.628 rad/sec along the real axis toward. The root locus lies on the real axis in the range 5.65 <s< and in the range <s< A root locus which is consistent with these rules is sketched in Fig. 8. 6
7 Imag c Figure 8: Sketch of root locus for system with C 2 (s) and P 1 (s). (f) For the system with C(s) =C 2 (s) and with P (s) =P 2 (S) the openloop transfer function is ( ) 3.49(s +5.65) 5π 2 86, 1(s +5.65) OLTF(s) =G = G s(s +2π) s +5π s(s +2π)(s +5π) 2 There are four openloop poles at s =,s = 2π, s = 5π, and s = 5π again, and one zero at s = The root locus has real segments in the ranges 5.65 <s<, 5π <s<2π and <s< 5π. There are three highgain asymptotes at angles 6 degrees, 18 degrees, and 3 degrees with respect to the real asxis. The asymptotes emanate from c =[ 2π 2(5π) ( 5.65)]/3 = 14.9 rad/sec. A root locus which is consistent with these rules is sketched in Fig. 9. Note how the actuator dynamics degrades stability and performance. With no actuator dynamics, the closedloop transient responses corresponding to figure 8 exhibit no oscillation for any gain value. When the actuator dynamics is included, the closedloop transient responses corresponding to figure 9 become oscillatory and ultimately unstable for sufficiently high gains. (g) An accurate locus is obtained from the following MATLAB session: sysc2p2 = tf([ ],[1 12*pi 27*pi^2 5*pi^3 ]) Transfer function: 861 s s^ s^ e4 s^ e5 s rlocus(sysc2p2) 7
8 Imag c Figure 9: Sketch of root locus for system with C 2 (s) and P 1 (s) Imag Axis Axis Figure 1: MATLAB root locus for system with C 2 (s) and P 1 (s). which produces the root locus displayed in Fig. 1. From Fig. 1 we see that at the onset of instability, where the root locus crosses from the left halfplane 8
9 to the right halfplane, the imaginary part of the root is approximately ±16 rad/sec. To estimate the gain at the onset of instability we make a rough estimate of the magnitude of OLTF(s) when s = j16 rad/sec as indicated in Fig j16 OLTF(j16) =86, 1G j16 2π + j16 5π + j16 =.17G 2 Imag c Figure 11: Estimating gain at onset of instability. Now the magnitude of the openloop transfer function is unity at the onset of instability, so the gain G required to reach the onset of instability is G =93.4 amps/volt which is 39.4 db greater than 1. amp/volt. (h) To check these estimates, ask MATLAB to calculate the gain margin by entering margin(sysc2p2) MATLAB replies with the Bode plot displayed in Fig. 12 and the information that the gain margin is 39.2 db at the frequency rad/sec. The estimates of 39.4 db and 16 rad/sec obtained in (g) above are fairly accurate in this case. 9
10 Bode Diagrams Gm= db (at rad/sec), Pm=9.171 deg. (at rad/sec) 22 Phase (deg); Magnitude (db) Frequency (rad/sec) Figure 12: MATLAB Bode plot for system with C 2 (s) and P 2 (s). 2. Control of PiezoElectric Machine Tool Head. In this problem we consider a single plant model and three different controllers. The transfer function P (s) for the plant was derived in Assignment 7 (see Eq.(2) in Solution of Problem 2)..368 P (s) = s 2 + 2e4s +2.64e9 meters/volt The roots of the denominator are the physical system poles s = 1, ± j5, 4 rad/sec. The controllers considered are: (i) pure integral action, e a (s) =(K i /s)[e r (s) e x (s)]; (ii) proportional plus integral, e a (s) =G(1 + K i /s)[e r (s) e x (s)]; (iii) proportional plus integral plus derivative, e a (s) =G(1 + K i /s + K d s)[e r (s) e x (s)]. The openloop transfer function has the form OLTF(s) =C(s)P (s)m x (s) with M x (s) = 1 5 functions volts/meter in every case, and with the three control transfer C 1 (s) = K i s 1
11 C 2 (s) = G(1 + K i s ) C 3 (s) = G(1 + K i s + K ds) (a) For Case (i) the openloop transfer function is OLTF(s) =K i 36, 8 s(s 2 + 2e4s +2.64e9) (1) which has three poles and no zeros. The root locus covers all of the negative real axis (to the left of the single pole at the origin). There are three asymptotes, making angles of 6 degrees, 18 degrees, and 3 degrees with the real axis. The asymptote center is at s = c =( 1, 1, )/3 = 6, 67 rad/sec. A root locus compatible with these requirements is sketched in Fig. 13 Imag Figure 13: Sketch of Root Locus for Integral Control. (b) An accurate root locus is obtained by inputting to MATLAB the openloop transfer function with unity gain and requesting the root locus for the closedloop transfer functiion as shown in the following MATLAB session: sys1 = tf(368,[ ]) Transfer function:
12 s^3 + 2 s^ e9 s rlocus(sys1) The resulting plot is displayed in Fig,14. 6 x Imag Axis Axis Figure 14: Root Locus for Integral Control Produced by MATLAB. From Fig. 14 we see that at the onset of instability the frequency is approximately 52, rad/sec. The gain at the onset of instability is estimated by estimating the magnitude of the OLTF(s) when s = j52, K i 36, 8 j52, j52, ( 1, + j5, 4) j52, ( 1, j5, 4) = K i 36, 8 (52, )(1, 13)(12, 9) = K i For this openloop transfer function to have a magnitude of unity at the onset of instability, the gain K i would have to have the value K i =1.47! 1, which is db greater than K i =1.. (c) Integral action introduces a free integrator, which eliminates steadystate error, but also introduces asymptotes which extend into the right halfplane, which means that sufficiently high gain will cause instability. (d) For the PI controller C 2 (s), the openloop transfer function is OLTF(s) =G 36, 8(s + K i) s(s 2 + 2e4s +2.64e9) (2) 12
13 Compared with pure integral control (1), PI control has added a zero at s = K i to the OLTF(s). We are told to locate the zero midway between the origin and the real part of the physical system poles s = 1, ± j5, 4 rad/sec. This is done by setting K i = 5, rad/sec. The portion of the real axis on which the root locus can exist is the segment 5, <s< to the left of the pole at the origin. Because there are three poles and one zero, there are two asymptotes for the root locus, making angles of 9 degrees and 27 degrees with the real axis, and emanating from the asymptote center at s = c = [2( 1, ) + ]/2] = 1, rad/sec. A root locus consistent with these requirements is sketched in Fig. 15. Imag c Figure 15: Sketch of Root Locus for PI Control. (e) An accurate root locus is obtained from the following MATLAB session: sys2 = tf([ ],[ ]) Transfer function: 368 s e s^3 + 2 s^ e9 s rlocus(sys2) The plot produced by MATLAB is displayed in Fig
14 1 x Imag Axis Axis Figure 16: Root Locus for PI Control Produced by MATLAB. (f) Because the asymptotes for the root locus remain in the left halfplane, it can be seen that the root locus for PI control always remains in the left halfplane. The controller is stable for all values of G. (g) For the PID controller C 3 (s), the openloop transfer function is 36, 8(s 2 + s/k d + K i /K d ) OLTF(s) =GK d s(s 2 + 2e4s +2.64e9) The roots of the quadratic in the numerator are the two zeros 1 z 1 = ( K i K d ) 2K d 1 z 2 = (1 1 4K i K d ) 2K d We are told to locate these zeros at z 1 = 5, rad/sec and z 2 = 15, rad/sec. This requires solving the preceding simultaneous equations for K d and K i. If we add the equations we obtain z 1 + z 2 = 155, = 1 K d or K d =6.45e6 seconds (3) 14
15 If we subtract the two equations we obtain z 1 z 2 = 145, = 155, 1 4K i K d or 4K i K d =1 from which we obtain K i =4, 84 rad/sec. (h) An accurate root locus is obtained from the following MATLAB session: ( ) sys3 = tf( 6.45E6*368*[ *155], [ ]) Transfer function:.2374 s^ e4 s e s^3 + 2 s^ e9 s rlocus(sys3) 2 x Imag Axis Axis Figure 17: Root Locus for PID Control Produced by MATLAB. x 1 5 The resulting MATLAB plot is displayed in Fig
16 (i) In Homework Assignment 7 it was found that a gain of G = 287, was required to produce a steadystate error of 2 %. If we insert that gain into the closedloop transfer function e x (s) e r (s) = = OLTF(s) 1+OLTF(s) 287, (.2374s e4 s e8) (s 3 + 2s e9 s) + 287, (.2374s e4 s e8) = = 6.74e4 s e1 s +5.11e13 s e4 s e1 s +5.11e13 we can ask MATLAB to plot the closedloop poles and unitstep response by entering the following commands: sys4 = tf([6.74e4 1.56E1 5.11E13], [1 8.81E4 1.32E1 5.11E13]) Transfer function: 674 s^ e1 s e s^ s^ e1 s e13 pzmap(sys4) step(sys4) The resulting plot of the poles and zero is displayed in Fig. 18 and the unitstep response is displayed in Fig. 19. Note the rapid risetime followed by a decaying oscillation due to the pair of complex poles, which decays more quickly than the firstorder response due to the real pole close to the slow zero. This problem illustrates the general properties of integral, proportional and derivative action. Integral action eliminates steadystate error but induces closed loop instability for sufficiently high gains. Proportional plus integral action introduces a zero that improves stability but does not reduce resonant transient behavior. Derivative action introduces a second zero that may be used to improve the transient, reducing resonance. However, if both zeros are realvalued, the pole introduced by the integral action is confined to be slower than the slower zero, and that results in the slow approach to zero steadystate error evident in figure
17 1.5 x 15 Polezero map 1.5 Imag Axis Axis x 1 4 Figure 18: PoleZero Map for ClosedLoop PID Transfer Function. Step Response 1.8 Amplitude Time (sec.) x 13 Figure 19: UnitStep Response of PiezoElectric Actuator with PID Control. 17
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