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 open-loop transfer function and conclude that the closed-loop 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 closed-loop system is stable with a phase margin (PM) greater than 5 o and a gain margin (GM) greater than 0 db.. The low-frequency gain of the compensated open-loop system is equal to that of the uncompensated system. 3. The bandwidth of the compensated open-loop system is twice that of the uncompensated system. Solution: Part A: Stability Analysis of the Uncompensated System: The magnitude and phase of the uncompensated open-loop 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 open-loop transfer function are shown in solid line on the next page. The closed-loop system is unstable because the open-loop 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: Stabilizing-Network 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 open-loop 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 open-loop 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 low-frequency approximation of the compensated open-loop 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 open-loop transfer function is equal to the zero-db crossover frequency, that is ω B = / rad/s. The new open-loop bandwidth should be twice that of the uncompensated bandwith, that is, ω B = = rad/s As a rule of thumb, we want the compensated open-loop transfer function to cross the zero-db line with a slope of -6 db/oct (equivalently, -0 db/dec). Using the previous Bode plots, the phase margin of the uncompensated open-loop 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 open-loop 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 open-loop 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 open-loop 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 open-loop 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 straight-line 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