Introduction to Feedback Control


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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 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 a system to behave in a prescribed manner. ner. Specifically, control system design is the process of causing a system variable (output)) to conform to some desired input (reference( reference). Reference Input R (s) Input U (s) System (Plant) G P (s) 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 a set of reference inputs r. ME375 Feedback Control  2 1
2 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 o to make it conform to the desired output (reference input). Reference Input (Command) R(s) C(s) U(s) Control Input System Output G P (s) Y(s) Plant or System Q: Ideally, if we want Y(s) to follow R(s) R ) (i.e.( want Y(s) Y ) = R(s)), how would you design the controller C(s)? ME375 Feedback Control  3 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 2
3 OpenLoop vs ClosedLoop ClosedLoop (Feedback) Control The control input u(t) ) (or U(s)) is synthesized based on the a priori 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 ClosedLoop Control Example Static Cruise Control Same vehicle system as the previous example. The vehicle speed is measured and fed back. Design a closedloop loop 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 performance? Q: How is the steady state performance? Will you have any steady state error? ME375 Feedback Control  6 3
4 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 9 mph 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 Why Feedback? Using feedback, we can change the closedloop loop system s s dynamic behavior (the ClosedLoop Transfer Function (CLTF) will be different from the original system s 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) 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 4
5 Example More Realistic Cruise Control Problem The relationship between a vehicle s s speed y and the throttle angle u is described by a first order system with a steady state gain K C and a time constant of 3 sec. The gain K C is affected by various operating conditions like the temperature e and humidity. Due to these effects, the actual value of K C is between 5 and 15. The objective of the cruise control is to design a control law (strategy) tegy) to determine the throttle angle u such that the vehicle s 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 a simple proportional feedback control, i.e. the control input u(t) u 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 is usually called the feedback gain or the proportional gain. ME375 Feedback Control  9 Example Calculate ClosedLoop Transfer Function (CLTF): Select an appropriate feedback gain to satisfy the performance specification : Q: Will this proportional control law work for attenuating external disturbances? ME375 Feedback Control
6 Elements of Feedback Control Elements of a Feedback Control System: Plant (Process) G 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 a device 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 measures the system output,, e.g., a thermocouple. Controller G C (s)  The controller is the device that generates the controlled input t 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 ClosedLoop Transfer Function Disturbance D(s) Control Input U(s) G P (s) Plant Output Y(s) Plant Equation (Transfer Transfer function model that we all know how to obtain?!): Control Law (Algorithm) (we( we will try to learn how to design): ME375 Feedback Control
7 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) Y() s = R() s + D() s G ( s) G ( s) YR YD ME375 Feedback Control  13 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 R( s) to Y( s) From R( s) to Y( s) The objective of control system design is to design a controller G C (s), such that certain performance (design) specifications are met. For example: e: we want the output y(t) ) to follow the 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
8 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 R(s) G CL (s) Output Y(s) r(t) y(t) Time Time ME375 Feedback Control  15 Unit Step Response Unit Step Response 1.6 y MAX OS ± X% t P Time t S t r ME375 Feedback Control
9 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: r() t y () SS () t yss t 2% = % = 0.98 r() t r() t 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 Performance Specifications Transient Performance (Transient Response) Transient performance of a system is usually specified using the unit step response of the system. Some typical transient response specifications are: a Settling Time (t S ): Specifies the time required for the response to reach and stay within a specific 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 for 2% bound ω ζ n 5 for 1% bound ζω n % Overshoot (%OS OS):(2nd order systems) ζωn ζ π π 2 2 ωn 1 ζ 1 ζ %OS = 100e = 100 e X% Desired Settling Time ( t ) 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 S 9
10 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 closedloop loop transfer function of a feedback 2 control system is: K ω n GCL( s) = 2 2 s + 2ζω ns + ω n The characteristic poles are: 2 s12, = ζω n ± jω n 1 ζ = ζω n ± jω d = ± j Settling Time (2%): Puts constraint on the real part of the dominating closedloop loop poles. 4 4 t S (2% ) = = ζ ω n %OS: Puts constraint on the imaginary part of the dominating closedloop loop poles. πζ π ζωn 2 1 ζ ωn 2 1 ζ %OS = 100e = 100e = 100e π ME375 Feedback Control  19 Performance Specification CL Pole Positions Transient Performance Specifications and CLTF Pole Positions Transient performance specifications can be interpreted as constraints on the positions of the poles of the closedloop loop transfer function. Let a pair of closed loop poles be represented as: p 1, 2 = σ ± jω Img. Transient Performance Specifications: jω Settling Time (2 %) T σ + jω S t S 4 4 (2% ) = TS σ T σ %OS X % σ π 100 π 100 ω ω %OS = 100 e = π X% e e σ ω X S σ σ jω jω Real ME375 Feedback Control
11 Example A DC 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 feedback 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 a step input. Find closedloop loop transfer function: Draw block diagram: ME375 Feedback Control  21 Example Find closedloop loop poles: 2% settling time: %OS: ME375 Feedback Control
12 Example A DC 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 Example Steady state performance constraint: Percent Overshoot (%OS) Img. Transient performance constraint: 2% Settling Time jω Real jω ME375 Feedback Control
13 Feedback Control Design Process A typical feedback controller design process involves the following steps: (1) Model the physical 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 a 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 t 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) Select the controller parameters (e.g. the proportional feedback gain of a proportional control law) so that the design constraints established in (5) are a satisfied. (8) Verify your design via computer simulation (MATLAB) and actual implementation. i ME375 Feedback Control  25 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 1023). During the latest design review, the following plant parameters are obtained: L A = 10 mh R A = 10 Ω K T = 0.06 Nm/A 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 = 0.03 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
14 (1) Model the physical system (plant) that we want to control and obtain its I/O transfer function G P (s). (Sometimes, certain model simplification should be performed.) DC Motor θ L, ω L N + e Ra + e La 2 θ, ω + e i (t) _ R A i A L A + E emf _ τ m B From previous example, the DC motor driven paper positioning system can be modeled by + 1 L A s + R A K b K T J A 1 J E s + B E N 1 J L B L 1 J = J + J N 2 1 BE = B+ BL N E A L 2 ME375 Feedback Control  27 The plant transfer function G P (s) can be derived to be: θ () s K GP() s = = 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 e plant model by neglecting the electrical subsystem dynamics (i.e., by letting L A = 0 ): θ () s KT GP() s = = = E () s s( R J s+ ( R B + K K )) i A E A E b T KM = s( τ s+ 1) M Substituting in the numerical values, we have our plant transfer function: KM GP() s = = s( τ s+ 1) M ME375 Feedback Control
15 (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 a controller G C (s) transfer functions. Input E i (s) G P (s) Output θ (s) ME375 Feedback Control  29 (4) Obtain the closedloop loop transfer function G CL (s). ME375 Feedback Control
16 (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: jω Real Overshoot Constraint: jω ME375 Feedback Control  31 (6) Choose controller structure G C (s)) and substitute it into the CLTF G CL (s). Let s s try a simple proportional control, where the control input to the plant is proportional to the current position error: ei() t = KP eθ () t = K ( () ()) V P Vθd t Vθ t In sdomain s (Laplace domain),, this control law can be written as: Substitute the controller transfer function into G CL (s): ME375 Feedback Control
17 (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 a satisfied. Steady State Constraint: Transient Constraints: To satisfy transient performance specifications, we need to choose 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 For every, there will be two closedloop loop poles (closedlooploop characteristic roots).. It s obvious that the two closedloop loop poles change with the selection of different. For example: = p 1,2 = 30 = p 1,2 = = p 1,2 = 20 = p 1,2 = 10 = p 1,2 = = p 1,2 = 0 By inspecting the rootlocus locus,, we can find 10 that if Img. Axis 20 then the closedloop loop poles will be in the allowable region and the performance specifications will be satisfied Real Axis ME375 Feedback Control
18 (8) Verify your design via computer simulation (MATLAB) and actual implementation. i >> num = 16*Ks*Kp; >> den = [taum 1 16*Ks*Kp]; >> T = (0:0.0002:0.25) ; >> y = step(num,den,t); >> plot(t,y); Unit Step Response = 100 = 40 = = Time (sec) ME375 Feedback Control  35 (9) Check the Bode Plots of the open loop and closed loop systems: Phase (deg); Magnitude (db) Frequency (rad/sec) = 100 = 40 = = 15 Open Loop = 100 = 40 = = 15 Open Loop ME375 Feedback Control
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