Professor Walter W. Olson Department of Mechanical, Industrial and Manufacturing Engineering University of Toledo Loop Shaping Outline of Today’s Lecture Review PID Theory Integrator Windup Noise Improvement Static Error Constants (Review) Loop Shaping Loop Shaping with the Bode Plot Lead and Lag Compensators Lead design with Bode plot Lead design with root locus Lag design with Bode plot PID: A Little Theory Consider a 1st order function where the 1st method of Ziegler Nichols applies The general transfer function for this system is P(s) K sa The term e st e st 0 K 1 a s 1 a e st 0 Ka 1 Ta s 1 e st 0 is the transport lag and delays the action for t0 seconds. Therefore L t 0 The term Ta is the time constant for the system. T measured on the graph is an estimate of this. 0 PID: A Little Theory The method 1 PI controller applied to the loop equation is 1 T 1 1 st 0 k p 1 P ( s ) 0.9 1 K e L s a Ta s 1 Ti s K L 0.3 assum ing that L t 0 , K Ka and T T a 1 T s 0.333 sL 0.9 T s 0.3 1 sL k p 1 e 0.9 e P(s) Ts 1 T s L s L s T s 1 i L 1 0.9 e sL F s L T s 0.333 Ls Ts 1 e 1 f t L 1 t L sL 0.9 T s 0.333 s Ts 1 2 tim e shifted by L PID: A Little Theory In Method 2, the gain was increased until the system was nearly a perfect oscillatory system. Since the gain changes the oscillatory patterns, the lowest order system that this could represent would by a 3rd order system. G (s) K s as bs K 3 2 For this system to oscillate, there must be a solution of the characteristic function for K real and positive where s=±wi s as bs K 0 3 2 w i a w bw i K 0 for s w i 3 2 w i a w bw i K 0 for s w i 3 2 K a w K cr a w 2 2 and Pcr 2 w PID: A Little Theory Applying the PI Controller: C P I ( s ) P ( s ) 0.45 K cr 1.2 K 2 1 3 0.45 a w 2 Pcr s s as bs K 2 1.2 w aw 1 3 2 2 2 s s as bs a w 2 aw s 0.191 w s 0.191 w 2 4 C P I ( s ) P ( s ) 0.45 a w 0.45 a w 3 3 2 2 2 2 s s as bs a w s s as bs a w 2 Integrator Windup We have tacitly assumed that the controlled devices could meet the demands of the controls that we designed. However real devices have limitations that may prevent the system from responding adequately to the control signal When this occurs with an integrating controller, the error which is used to amplify the control signal may build up and saturate the controller. We refer to this as “integrator windup”: the system can’t respond and the integrator signal is extremely large (often maxed out on a real controller) the result is an uncontrolled system that can not return to normal operating conditions until the controller is reset Integrator Windup To avoid windup, a possible solution is to provide a correcting error from the actuator by adding another loop: (the actuator has to be extracted from the plant) kp ++ ++ + kd s ki 1 ++ s -1 A( s ) -1 ++ P(s) Derivative Noise Improvement A major problem with using the derivative part of the PID controller that the derivative has the effect of amplifying the high frequency components which, for most systems, is likely to be noise. Without PID With PID Derivative Noise Improvement One way to improve the noise rejection at higher frequencies is to apply a second order filter that passes low frequency and rejects high frequency The natural frequency of the filter should be chosen as wn Nk p kd with N chosen to give the controller the bandwidth necessary, usually in the range of 2 to 20 The controller then has the design 2 ki wn C P ID ( s ) k p kd s 2 2 s s 2w n s w n Static Error Constants If the system is of type 0 at low frequencies will be level. A type 0 system, (that is, a system without a pole at the origin,) will have a static position error, Kp, equal to lim G ( jw ) K K p w0 If the system is of type 1 (a single pole at the origin) it will have a slope of -20 dB/dec at low frequencies A type 1 system will have a static velocity error, Kv, equal to the value of the -20 dB/dec line where it crosses 1 radian per second If the system is of type 2 ( a double pole at the origin) it will have a slope of -40 dB/dec at low frequencies A type 2 system has a static acceleration error,Ka, equal to the value of the -40 dB/dec line where it crosses 1 radian per second Static Error Constants Kv(dB) G (s) 2( s 0.1) s ( s 0.1 s 4) 2 Error signal E(s) Input r(s) Loop Shaping ++ Controller C(s) Open Loop Signal B(s) Plant P(s) Output y(s) Sensor -1 We have seen that the open loop transfer function, B ( s ) C ( s ) P ( s ),has profound influences on the closed loop response The key concept in loop shaping designs is that there is some ideal open loop transfer (B(s)) that will provide the design specifications that we require of our closed loop system Loop shaping is a trial and error process: Everything is connected and nothing is independent What we gain in one area may (usually?) causes loss in other areas Often times, out best controller is a compromise between demands To perform loop shaping we can used either the root locus plots or the Bode plots depending on the type of response that we wish to achieve We have already considered an important form of loop shaping as the PID controller Loop Shaping with the Bode Plot The open loop Bode plot is the natural design tool when designing in the frequency domain. For the frequency domain, the common specifications are bandwidth, gain cross over frequency, gain margin, resonant frequency, resonant frequency gain, phase margin, static errors and high frequency roll off. -3 db Roll off Rate dB/dec Resonant peak gain, dB Bandwidth rps Resonant peak frequency rps Gain cross over frequency rps Loop Shaping with the Bode Plot Increase of gain also increases bandwidth and resonant gain Poles bend the magnitude and phase down Zeros bend the magnitude and the phase up Break frequency corresponds to the component pole or zero Lead and Lag Compensators The compensator with a transfer function C (s) K sa sb is called a lead compensator if a<b and a lag compensator if b>a The lead and the lag compensator can be used together Note: the compensator sa does add a steady state gain of that needs to be accounted for in the final design sb a b There are analytical methods for designing these compensators (See Ogata or Franklin and Powell) Lead Compensator The lead compensator is used to improve stability and to improve transient characteristics. The lead compensator can be designed using either frequency response or root locus methods Usually, the transient characteristics are better addressed using the root locus methods Addressing excessive phase lag is better addressed using the frequency methods The pole of the system is usually limited by physical limitations of the components use to implement the compensator In the lead compensator, the zero and pole are usually separated in frequency from about .4 decades to 1.5 decades depending on the design Lead Compensator (Frequency Design) Note: 1) the lead compensator opens up the high frequency region which could cause noise problems 2) The Lead compensator adds phase C (s) sa sb ab a b 1 sin f f wm b1 1 sin f a wm b Mechanical Lead Compensator b2 a xi x0 k y Example An aircraft has a pitch rate control as shown. Design a lead compensator for this system for a static velocity error of 4/sec, and a phase margin of 40 degrees. Aircraft Pitch Rate Dynamics Compensator R ++ 2 s .0 5 C(s) s ( s 0 .1 s 4 ) 2 -1 Y Example Current System: 33 4 0.0224 K v 10 20 0.0224 N o G ain M argin -33dB When design a lead compensator first adjust the gain to meet the static error condition In this case the gain needs the be increase by 180 or 20Log10180= 45.1 dB added Example Gain Adjusted System 12 dB Then noting where the phase currently is, that is the desired location for the peak of the lead phase f spec – Pm=40-0.153 + a small safety = 55 deg a b 1 sin f 1 sin f 0.1808 0.0994 1.8192 a w m 19.1 0.0994 6.0222 rps b a 6.0222 60.5775 rps 0.0994 Finally adjust them as necessary Example Initial design: Phase good but Kv not Gain needs to be increased by about 20 dB Final design: 60.58 s 6.022 C L ead ( s ) 180 6.022 s 60.58 C L ead ( s ) 1810 s 6.022 s 60.58 Example An aircraft has a pitch rate control as shown. The response of the pitch control is under damped, sluggish has an objectionable transient vibration mode. Design a lead compensator that provides a damping ratio of between 0.45 to 0.50 and 5% settling time of 150 seconds which reduces the vibration mode as much as possible. R ++ 2 s .0 5 C(s) s ( s 0 .1 s 4 ) 2 -1 Y Example The root locus from the complex poles has very little damping and causes the vibration seen in the response. There is a pole at -0.01 on the real axis that is dominant and causes the sluggish behavior. Strategy: Use a lead compensator to bend the curves to the left and into the 0.6 damping region. The zero of the compensator should counteract the vibrational mode Example The initial design with the pole at -5 and the zero at -1 had the desired effect of bending the root locus to the left and removing most of the vibration. However the pole is still too close to the origin such that 0.6 damping can not be achieved. Example We achieved the specifications once the pole of the compensator was moved out to -9 and we adjusted the gain for the 0.6 damping. Final com pensator design: C (s)=16 s 1 s9 Lag Compensator Lag compensators are used to improve steady state characteristics where the transient characteristics are adequate and to attenuate high frequency noise In order to not change the transient characteristics, the zero and pole are located near the origin on the root locus plot The starting point for the design on a root locus is to start with a pole location at about s = -0.001 and then locate the pole as needed for the desired effect In order to not give up too much phase, the zero and pole are located away from the phase margin frequency Lag Compensator Mechanical Lag Compensator b2 xi k b1 b a x0 Note that the lag compensator causes a drop in the magnitude and phase This could be useful in reducing bandwidth, and improving gain margin; however it might reduce phase margin Example A linear motor has an open loop rate transfer function of G (s) 24 s s 2s 6 It is desired that the system have a static velocity error constant greater than 20/sec, a phase margin of 45 degrees plus or minus 5 degrees and gain crossover frequency of 1 radian/sec. Design a lag compensator for this system. Linear Motor Compensator Rate Dynamics R 24 ++ s s 2s 6 C(s) -1 Y Example 4.9dB 4.9 K v 10 20 1.7579 Km 20 11.3 use 12 1.7579 Current System: Phase margin is low and the static velocity error constant must be improved. K 12 * 24 288 20 log 10 288 49.18 Start by correcting the static velocity error constant Example Gain Adjusted: Gain and Phase Margin problems Need to shape the curve like this Need to move PM to here try placing a pole at -0.01 rps and adjust the zero Example Final Design .0 1 s .2 0 C ( s) 12 .2 0 s 0 .0 1 C ( s ) 0 .6 s .2 0 s 0 .0 1 Summary Static Error Constants (Review) Loop Shaping Loop Shaping with the Bode Plot Lead and Lag Compensators Lead design with Bode plot Lead design with root locus Lag design with Bode plot Next: Sensitivity Analysis