School of Mechanical Engineering Purdue University. ME375 Feedback Control  1


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1 Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System ClosedLoop Transfer Functions (CLTF) Performance Specifications Steady State Specifications Transient (Dynamic) Specifications ME375 Feedback Control  1
2 Control System Design Control: verb, 1. To exercise authority or dominating influence over; direct; regulate. 2. To hold in restraint. Control is the process of causing asystem to behave in aprescribed manner. Specifically, control system design is the process of causing asystem variable (output output) to conform to some desired input (reference reference). Reference Input R (s) Input U (s) System (Plant) G P (s) Output Y(s) The objective of the control system is to control the output y by using the input u, such that the output y follows aset of reference inputs r. ME375 Feedback Control  2
3 OpenLoop vs ClosedLoop OpenLoop Control The control input u(t) (or U(s)) is synthesized based on the a priori knowledge of the system (plant) and the reference input r(t) (or R(s)) )). The control system does not measure the output, and there is no comparison of the output to make it conform to the desired output (reference input). Reference Input (Command) R(s) C(s) U(s) Control Input G P (s) Plant or System System Output Y(s) Q: Ideally, if we want Y(s) to follow R(s) ) (i.e. want Y(s) ) = R(s)), how would you design the controller C(s)? ME375 Feedback Control  3
4 OpenLoop Control Example Static Cruise Control The vehicle speed model can be approximated by a static gain between the throttle angle (input) and the vehicle speed (output). From experiment, on level road, at 55 mph, 1 o of throttle angle causes 10 mph change in speed. When the road grade changes by 1%, 1 o of throttle angle will only change vehicle speed by 5 mph. Design an openloop cruise controller for this vehicle. R Cruise Controller U W + Speed Model Y R : reference speed, mph U : throttle angle, degree Y : actual speed, mph W : road grade, % Q: What are potential problems with this cruise control? ME375 Feedback Control  4
5 OpenLoop vs ClosedLoop ClosedLoop (Feedback) Control The control input u(t) (or U(s)) is synthesized based on the a priori i knowledge of the system (plant), the reference input r(t) (or R(s)) and the measurement of the actual output y(t) (or Y(s)) )). For example the temperature control of this classroom: Disturbance D(s) Heater Actuator Room Plant or System Room Temperature Y(s) ME375 Feedback Control  5
6 ClosedLoop Control Example Static Cruise Control Same vehicle system as the previous example. The vehicle speed is measured and fed back. Design aclosedlooploop cruise control that uses the measured vehicle speed and the reference speed. W R U + Speed Model Y Q: How would road grade, plant uncertainty affect the closedloop loop pp performance? Q: How is the steady state performance? Will you have any steady state error? ME375 Feedback Control  6
7 ClosedLoop Control Example Static Cruise Control (ClosedLoop Control) (a) Find the actual vehicle speed when the reference speed is 50 mph and the road grade is 1% and 10%, respectively. (b) If the actual vehicle speed model is 1 o of throttle angle corresponds to 9mph change in speed, what is the actual vehicle speed with the same cruise controller. (c) When there is no grade and the vehicle speed model is accurate, what is the actual output speed when a reference speed of 50 mph is desired. ME375 Feedback Control  7
8 Why Feedback? Using feedback, we can change the closedlooploop system s dynamic behavior (the ClosedLoop Transfer Function (CLTF) will be different from the original system s (openloop) transfer function). By using feedback to change the CLTF, we can achieve the following: Stabilize Unstable Systems For example, unstable plants such as inverted pendulum and DC motor positioning systems can be stabilized using feedback. Improve System Performance (Achieve Performance Specifications) i Steady State Performance  For example, reduce steady state error... Transient Performance  For example, reduce rise time, reduce settling time, reduce overshoot Reduce (attenuate) the effect of modeling uncertainty (error) and external disturbances ME375 Feedback Control  8
9 Example More Realistic Cruise Control Problem The relationship between a vehicle s speed y and the throttle angle u is described by afirst order system with asteady state gain K C and a time constant of 3sec. The gain K C is affected by various operating conditions like the temperature and humidity. Due to these effects, the actual value of K C is between 5and 15. The objective of the cruise control is to design acontrol law (strategy) (t t )t to dt determine the throttle angle u such that the vehicle s steady state speed will stay within 2% of the desired reference speed set by the driver. R Cruise Controller U Speed Model Y Use asimple proportional feedback control, i.e. the control input u(t) is proportional to the regulation error e(t) = r(t) y(t). The control design parameter is the proportional constant between the input and the error. This constant K P is usually called the feedback gain or the proportional gain. ME375 Feedback Control  9
10 Example Calculate ClosedLoop Transfer Function (CLTF): Select an appropriate feedback gain K P to satisfy the performance specification : Q: Will this proportional control law work for attenuating external disturbances? ME375 Feedback Control  10
11 Elements of Feedback Control Elements of a Feedback Control System: Plant (Process) G P P( (s)  The plant is the system (process) whose output is to be controlled, e.g., the room in the room temperature control example. Actuator  An actuator is adevice that can influence the input to the plant, e.g. the heater (furnace) in the room temperature control example. Disturbance d(t)  Disturbances are uncontrollable signals to the plant that tend to adversely affect the output of the system, e.g., opening the windows in the room temperature control example. Sensor (Measurement System) H(s)  The transfer function (frequency response function) of the device (system) that t measures the system output, t e.g., a thermocouple. Controller G C (s)  The controller is the device that generates the controlled input that is to affect the system output, e.g., the thermostat in the room temperature control example. Reference Input R(s) Sensor Controller G C (s) H(s) Disturbance D(s) () Heater Actuator G P (s) Plant (Process) Output Y(s) ME375 Feedback Control  11
12 ClosedLoop Transfer Function Disturbance D(s) Control Input U(s) G P (s) Plant Output Y(s) Plant Equation (Transfer function model that we all know how to obtain?!): Control Law (Algorithm) (we will try to learn how to design): ME375 Feedback Control  12
13 ClosedLoop Transfer Function Disturbance D(s) Reference Input R(s) + Error E(s) G C (s) Control Input U(s) + + G P (s) Plant Output Y(s) H(s) Ys () Rs () Ds () G () s G () s YR YD ME375 Feedback Control  13
14 ClosedLoop Transfer Function The closedloop loop transfer functions relating the output y(t) (or Y(s)) to the reference input r(t) (or R(s)) and the disturbance d(t) (or D(s)) are: Y() s GYR() s R() s GYD() s D() s ClosedLoop Transfer Function ClosedLoop Transfer Function From Rs ( ) to Ys ( ) From Rs ( ) to Ys ( ) The objective of control system design is to design a controller G C (s), such that certain performance (design) specifications are met. For example: we want the teoutput y(t) to follow ow the teeee reference input r(t), i.e.,, for certain frequency range. This is equivalent to specifying that we want the disturbance d(t) to have very little effect on the output y(t) within the frequency range where disturbances are most likely to occur. This is equivalent to specifying that ME375 Feedback Control  14
15 Performance Specifications Given an input/output representation, G CL (s), for which the output of the system should follow the input, what specifications should you make to guarantee that the system will behave in a manner that will satisfy its functional requirements? Input G CL (s) Output R(s) ) Y(s) ) r(t) y(t) Time Time ME375 Feedback Control  15
16 Unit Step Response 1.6 y MAX it Step Respon nse OS X% Un t P Time t S t r ME375 Feedback Control  16
17 Performance Specifications Steady State Performance Steady State Gain of the Transfer Function Specifies the tracking performance of the system at steady state. Often it is specified as the steady state response, y() (or y SS (t)), to be within an X% bound of the reference input r(t), i.e., the steady state error e SS (t) = r(t) y SS (t) should be within a certain percent. For example: rt () y () () SS t yss t 2% % 0.98 rt () rt () To find the steady state value of the output, y SS (t): Sinusoidal references: : use frequency response, i.e. General references: : use FVT, provided that is stable,... ME375 Feedback Control  17
18 Performance Specifications Transient Performance (Transient Response) Transient performance of asystem is usually specified using the unit step response of the system. Some typical transient response specifications are: Settling Time (t S ): Specifies the time required for the response to reach and stay within aspecific percent of the final (steadystate) state) value. Some typical settling time specifications are: 5%, 2% and 1%. For 2nd order systems, the specification is usually: 4 n 5 n for 2% bound for 1% bound % Overshoot (%OS):(2nd order systems) n 2 2 n 1 1 %OS 100e 100 e X% t S Desired Settling Time ( ) Q: How can we link this performance specification to the closedloop loop transfer function? (Hint) What system characteristics affect the system performance? ME375 Feedback Control  18
19 Performance Specifications Transient Performance Specifications and CLTF Characteristic Poles Recall that the positions of the system characteristic poles directly affect the system output. For example, assume that the closedlooploop transfer function of a feedback 2 control system is: K n GCL() s 2 2 s 2 ns n The characteristic poles are: 2 s12, j 1 j j n n n d Settling Time (2%): Puts constraint t on the real part of the dominating closedloop loop poles. 4 4 t S (2%) n %OS: Puts constraint on the imaginary i part of the dominating i closedloop loop poles. n n 1 %OS e e e ME375 Feedback Control  19
20 Performance Specification CL Pole Positions Transient Performance Specifications and CLTF Pole Positions Transient performance specifications can beinterpreted as constraints on the positions of the poles of the closedloop loop transfer function. Let apair of closed loop poles be represented as: p12, j Img. Transient Performance Specifications: Settling Time (2 %) T S t S 4 4 (2%) TS T %OS X X % %OS 100 e X% e X e S j j j j Real ME375 Feedback Control  20
21 Example ADC motor driven positioning system can be modeled by a second order transfer function: 3 GP() s ss ( 6) A proportional feedback control is proposed and the proportional gain is chosen to be 16/3. Find the closedloop loop transfer function, as well as the 2% settling time and the percent overshoot of the closed loop system when given astep input. Find closedloop loop transfer function: Draw block diagram: ME375 Feedback Control  21
22 Example Find closedloop loop poles: 2% settling time: %OS: ME375 Feedback Control  22
23 Example ADC motor driven positioning system can be modeled by a second order transfer function: 3 GP() s ss ( 6) A proportional feedback control is proposed. It is desired that: for a unit step response, the steady state position should be within 2% of the desired position, the 2% settling time should be less than 2 sec, and the percent overshoot should be less than 10%. Find (1) the condition on the proportional gain such that the steady state performance is satisfied; (2) the allowable region in the complex plane for the closedloop loop poles. Find closedloop loop transfer function: Write down the performance specifications: ME375 Feedback Control  23
24 Example Steady state performance constraint: Percent Overshoot (%OS) Img. Transient performance constraint: 2% Settling Time j Real j ME375 Feedback Control  24
25 Feedback Control Design Process A typical feedback controller design process involves the following steps: (1) Model the physical py system (plant) that we want to control and obtain its I/O transfer function G P (s). (Sometimes, certain model simplification should be performed.) (2) Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator dynamics (if necessary). (3) Draw the closedloop loop block diagram, which includes the plant, sensor, actuator and controller G C (s) transfer functions. (4) Obtain the closedloop loop transfer function G CL (s). (5) Based on the performance specifications, find the conditions that the CLTF, G CL (s), has to satisfy. (6) Choose controller structure G C (s) and substitute it into the CLTF G CL (s). (7) Sl Select tth thecontroller parameters (e.g. the proportional feedback gain of a proportional control law) so that the design constraints established in (5) are satisfied. (8) Verify your design via computer simulation (MATLAB) and actual implementation. ME375 Feedback Control  25
26 In Class Exercise You are the young engineer that is in charge of designing the control system for the next generation inkjet printer (refer the example discussed in lecture notes to ). During the latest design review, the following plant parameters are obtained: L A = 10 mh R A = 10 K = 006Nm/A 0.06 K T J E = Kg m 2 B E = Nm/(rad/sec) The drive roller angular position is sensed by a rotational potentiometer with a static sensitivity of K S = V/deg. The design (performance) specifications for the paper positioning system are: The steady state position for a step input should be within 5% of the desired position. The 2% settling time should be less than 200 msec, and the percent overshoot should be less than 5%. You are to design a controller that satisfies the above specifications: ME375 Feedback Control  26
27 In Class Exercise (1) Model the physical system (plant) that we want to control and obtain its I/O transfer function G P P( (s). (Sometimes, certain model simplification should be performed.) DC Motor L L N + e Ra + e La 2 + R i A L A + A e i (t) _ E emf _ m B From previous example, the DC motor driven paper positioning system can be modeled by J A N 1 J L B L 1 L A s+r A K T 1 J E s+b E 1 J J J N E A L 2 K b 1 BE B B N 2 L ME375 Feedback Control  27
28 In Class Exercise The plant transfer function G P (s) can be derived to be: G P () s () s K E s s L J s B L R J s R B K K T 2 i() ( A E ( E A A E) ( A E b T)) As discussed in the previous example, we can further simplify the plant model by neglecting the electrical subsystem dynamics (i.e., by letting L A = 0): G P () s KT () s E () s s ( R J s ( R B K K )) i A E A E b T KM s( s1) M Substituting in the numerical values, we have our plant transfer function: KM GP() s s ( s 1) M ME375 Feedback Control  28
29 In Class Exercise (2) Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator dynamics (if necessary). (3) Draw the closedloop loop block diagram, which includes the plant, sensor, actuator and controller G C (s) transfer functions. Input E i (s) G P (s) Output (s) ME375 Feedback Control  29
30 In Class Exercise (4) Obtain the closedloop loop transfer function G CL (s). ME375 Feedback Control  30
31 In Class Exercise (5) Based on the performance specifications, find the conditions that G CL (s) has to satisfy. Steady State specification: Imag. Transient Specifications: Settling Time Constraint: j Real Overshoot Constraint: j ME375 Feedback Control  31
32 In Class Exercise (6) Choose controller structure G C (s) and substitute it into the CLTF G CL (s). Let s try a simple proportional control, where the control input to the plant is proportional to the current position error: e() t K e () t K ( V () t V ()) t i P P d In sdomain (Laplace p domain), ) this control law can be written as: V Substitute the controller transfer function into G CL (s): ME375 Feedback Control  32
33 In Class Exercise (7) Select the controller parameters (e.g., the proportional feedback gain of a proportional control law) so that the design constraints established in (5) are satisfied. Steady State Constraint: Transient Constraints: To satisfy transient performance specifications, we need to choose K P such that the closed loop poles are within the allowable region on the complex plane. To do this, we first need to find an expression for the closedloop loop poles: ME375 Feedback Control  33
34 In Class Exercise For every K P, theree will be two closedlooploop poles (closedlooploop characteristic roots). obvious that the two closedloop loop poles change with the selection of different K P. example: K P = p 1,2 = K P = p 1,2 = K P = p 1,2 = K P = p 1,2 = K P = p 1,2 = K P = p 1,2 = By inspecting the rootlocus locus, we can find that if Img. Axis It s For then the closedloop loop poles will be in the allowable region and the performance specifications will be satisfied Real Axis ME375 Feedback Control  34
35 In Class Exercise (8) Verify your design via computer simulation (MATLAB) and actual implementation. >> num = 16*Ks*Kp; >> den = [taum 1 16*Ks*Kp]; >> T = (0:0.0002:0.25) ; >> y = step(num,den,t); >> plot(t,y); t Step Respons se Unit K P = 100 K P = 40 K P = K P = Time (sec) ME375 Feedback Control  35
36 In Class Exercise (9) Check the Bode Plots of the open loop and closed loop systems: (db) g); Magnitude Phase (de 10 0 K P = K P = K P = K P = Open Loop 0 K P = K P = K P = K P = Open Loop Frequency (rad/sec) ME375 Feedback Control  36
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