Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain


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1 Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism: Stabilizing Network Servoamplifier + servomotor i s (+0.5s) o Part A: Draw the Bode plots of the uncompensated openloop transfer function and conclude that the closedloop system is unstable. Can this system be stabilized by a network consisting of a preamplifier with gain K? Part B: Design a stabilizing network so that:. The compensated closedloop system is stable with a phase margin (PM) greater than 5 o and a gain margin (GM) greater than 0 db.. The lowfrequency gain of the compensated openloop system is equal to that of the uncompensated system. 3. The bandwidth of the compensated openloop system is twice that of the uncompensated system. Solution: Part A: Stability Analysis of the Uncompensated System: The magnitude and phase of the uncompensated openloop transfer function are given by: G(jω) = = o (jω) = ( + jω/) ω 80 o + (ω/) tan ω/ 0.5 ω + (ω/) 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the uncompensated openloop transfer function are shown in solid line on the next page. The closedloop system is unstable because the openloop transfer function has negative phase margin at the crossover frequency ω c = / rad/s. Because of this negative phase margin, the system can never be stabilized by a preamplifier with a real gain K. The Bode magnitude plot would be shifted upwards and the croossover frequency would increase. Clearly, the phase will always be less than 80 o resulting in a negative phase margin.
2 6 db/oct 0 /8 / 6 db/oct / ω (rad/s) MAGNITUDE (db) 36 db/oct 8 db/oct deg/oct 5 deg/oct /8 / / ω (rad/s) PHASE (deg) 90 5 deg/oct 5 deg/oct deg/oct 70 5 deg/oct
3 Part B: StabilizingNetwork Design Stabilizing Network Servoamplifier + servomotor i + K +st α +s α T 0.5 s (+0.5s) o Preamplifier Lead Network We need to add some phase lead to the uncompensated openloop transfer function in order to obtain positive phase margin. To realize the stabilizing network, we select a preamplifier followed by a lead network. The overall compensated openloop transfer function has the following form: G(s) = K α + st + sαt 0.5 s ( + s/) = K α + s/ω s/ω s ( + s/) The lowfrequency approximation of the compensated openloop transfer function should equal that of the uncompensated system. Therefore, G(s) = K α + st + sαt 0.5 s ( + s/) Therefore, the preamplifier gain K should be selected such that = Kα 0.5 s = 0.5 s = Kα = K = α From the previous Bode plots, the bandwidth of the uncompensated openloop transfer function is equal to the zerodb crossover frequency, that is ω B = / rad/s. The new openloop bandwidth should be twice that of the uncompensated bandwith, that is, ω B = = rad/s As a rule of thumb, we want the compensated openloop transfer function to cross the zerodb line with a slope of 6 db/oct (equivalently, 0 db/dec). Using the previous Bode plots, the phase margin of the uncompensated openloop transfer function at ω c = rad/s is equal to 5 o (that is, it is 5 o below the 80 o line). Since we want a minimum phase margin of 5 o, the phase lead to be added at the crossover frequency ω c = rad/s should be at least 5 o + 5 o = 60 o. Recalling the relationship between the maximum phase of a lead network and α, we should have tan α α 60o α α tan 60o = 3 α 3 α + α α α α α + 0 = α 7 8 = or α = 3.9 Since α <, we choose α = 6 = = K = 6 3
4 The maximum phase occurs at ω c = T α = rad/s = T α = T 6 = = T = ω = = T rad/s ω = = = rad/s αt /6 Finally, the openloop transfer function of the compensated system can be expressed as G(s) = 6 + s 6 + s/ 0.5 s ( + s/) = 0.5( + s) s ( + s/) The magnitude and phase of the compensated openloop transfer function are given by: G(jω) = 0.5( + jω) (jω) ( + jω/) = (ω) tan ω ω 80 o [ + (ω/) ] tan ω/ = ω ω ( + ω /6 tan ω 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the compensated openloop transfer function are shown in dashed line next to the uncompensated Bode plots. Clearly, the system is now stable with a positive phase margin at the crossover frequency ω c = rad/s. We can either use the Bode straightline approximations to compute the gain and phase margins of the compensated system, or we can do exact calculations using a calculator. We will do both below. Phase Margin: Assuming that the crossover frequency is exactly ω c = rad/s, we read an approximate phase margin of PM = 5 o on the graph. An exact computation yields φ(ω c ) = tan ω c 80 o tan ω c = tan 80 o tan PM = 80 o + φ(ω c ) = 80 o 3. o = 7.9 o = 3. o The exact crossover frequency is such that G(ω) = ω ω ( + ω /6) = This expression can be reduced to the following polynomial equation: ω 8 + 3ω ω 56ω 6 = 0 which has ω c = rad/s as a solution. The exact phase margin at this frequency is φ(ω c ) = tan ω c 80 o tan ω c φ(ω c ) = 3.8 o PM = 80 o + φ(ω c ) = 80 o 3.8 o = 8. o = tan o tan
5 Gain Margin: From the Bode plots, the compensated openloop transfer function has a phase of 80 o at ω o = rad/s. The actual phase at ω o = rad/s is φ(ω o ) = tan ω o 80 o tan ω o The approximate gain margin at this frequency is GM (db) = 0 log G(ω o ) = tan 80 o tan = 0 log G(ω o ) = ( ) = db = 83. o Using the fact that there is a maximum error of 6 db at the double pole ω o = rad/s, we can conclude that the gain margin is truly GM = 6 = 8 db An exact computation yields G(ω o ) = ωo ωo( + ωo/6) = ( + = 0.5 /6) GM = 0 log G(ω o ) = 0 log 0.5 = 8.0 db To be more accurate, we need to compute the exact frequency at which the phase is equal to 80 o. Using a calculator, and, by trial and error: By interpolation, ω o ω o (rad/s) φ(ω o ) (deg) 68.5 o 76 o 78 o 79.5 o 80.8 o = 3.7 rad/s. We check that φ(ω o ) = tan ω o 80 o tan ω o φ(ω o ) = o = 80 o = tan o 3.7 tan G(ω o ) = ωo ωo( + ωo/6) = ( = 0.3 /6) GM = 0 log G(ω o ) = 0 log 0.3 = 6.9 db Therefore, all of the criteria have been met and this design is satisfactory. Exercise #: Redo the design using decade straightline approximations. Exercise #: Redo the design using a PD controller of the form G c (s) = K p + K d s = K p ( + sk d /K p ) = ( + sk). (Notice that K p must be equal to using Criterion #.) Design for an approximate gain K that will meet the design objectives. There is no need for a preamplifier in this case. 5
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