Feedback Control

Feedback Control - ME375 Handouts Introduction to Feedback...

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Unformatted text preview: ME375 Handouts Introduction to Feedback Control • Control System Design System Design – Why Control? – Open-Loop vs Closed-Loop (Feedback) OpenClosed– Why Use Feedback Control? • Closed-Loop Control System Structure Control System Structure – Elements of a Feedback Control System – Closed-Loop Transfer Functions (CLTF) Closed- • Performance Specifications – Steady State Specifications – Transient (Dynamic) Specifications School of Mechanical Engineering Purdue University ME375 Feedback Control - 1 1 ME375 Handouts Control System Design Control: Control: verb, 1. To exercise authority or dominating influence over; direct; direct; regulate. regulate. 2. To hold in restraint. restraint. Control is the process of causing a system to behave in a prescribed manner. manner. Specifically, control system design is the process of causing a system variable (output) to conform to some desired input (reference). Reference Input Input R (s) U (s) System (Plant) GP(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. School of Mechanical Engineering Purdue University ME375 Feedback Control - 2 2 ME375 Handouts OpenOpen-Loop vs Closed-Loop Closed• Open-Loop Control enThe 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). input). Reference Input Input (Command) R(s) System Output C(s) U(s) Control Input GP(s) Y(s) Plant or System Q: Ideally, if we want Y(s) to follow R(s) (i.e. want Y(s) = R(s)), how would you design the controller to how C(s) ? School of Mechanical Engineering Purdue University ME375 Feedback Control - 3 3 ME375 Handouts OpenOpen-Loop Control Example Static Cruise Control W The vehicle speed model can be approximated by a static gain between the throttle angle (input) and the vehicle speed (output). From experiment, on (output). level road, at 55 mph, 1o of throttle angle causes 10 mph change in speed. When speed. the road grade changes by 1%, 1o of throttle angle will only change vehicle speed by 5 mph. Design an open-loop mph. opencruise controller for this vehicle. vehicle. R U Y W : reference speed, mph : throttle angle, degree : actual speed, mph : road grade, % R U − Y + Cruise Controller Speed Model Model Q: What are potential problems with this cruise control ? School of Mechanical Engineering Purdue University ME375 Feedback Control - 4 4 ME375 Handouts OpenOpen-Loop vs Closed-Loop Closed• Closed-Loop (Feedback) Control Co 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: classroom: Disturbance D(s) Heater Actuator School of Mechanical Engineering Purdue University Room Temperature Room Y(s) Plant or System ME375 Feedback Control - 5 5 ME375 Handouts ClosedClosed-Loop Control Example Static Cruise Control Same vehicle system as the previous example. The vehicle speed is measured and fed back. example. back. Design a closed-loop cruise control that uses the measured vehicle speed and the reference closedspeed. speed. W R − U Y + Speed Model Q: How would road grade, plant uncertainty affect the closed-loop performance ? closedQ: How is the steady state performance ? Will you have any steady state error ? School of Mechanical Engineering Purdue University ME375 Feedback Control - 6 6 ME375 Handouts ClosedClosed-Loop Control Example Static Cruise Control (Closed-Loop Control) (Closed(a) Find the actual vehicle speed when the reference speed is 50 mph and the road grade is 1% and 10%, respectively. 10% respectively. (b) If the actual vehicle speed model is 1o of throttle angle corresponds to 9 mph change in speed, what is the actual vehicle speed with the same cruise controller. 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. desired. School of Mechanical Engineering Purdue University ME375 Feedback Control - 7 7 ME375 Handouts Why Feedback ? Using feedback, we can change the closed-loop system’s dynamic behavior (the ClosedClosed-Loop Transfer Function (CLTF) will be different from the original system’s (open-loop) transfer function). By using feedback to change the CLTF, (openfunction). we can achieve the following: 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 School of Mechanical Engineering Purdue University ME375 Feedback Control - 8 8 ME375 Handouts Example More Realistic Cruise Control Problem The relationship between a vehicle’s speed y and the throttle angle u is described by sec. a first order system with a steady state gain KC and a time constant of 3 sec. The gain KC is affected by various operating conditions like the temperature and humidity. humidity. Due to these effects, the actual value of KC is between 5 and 15. The 15. objective of the cruise control is to design a control law (strategy) to 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. Cruise Controller R Speed Model U Y School of Mechanical Engineering Purdue University Use a simple “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 error. constant KP is usually called the feedback gain or the proportional gain. ME375 Feedback Control - 9 9 ME375 Handouts Example Calculate Closed-Loop Transfer Function (CLTF): Closed- Select an appropriate feedback gain KP to satisfy the performance specification : Q: Will this proportional control law work for attenuating external disturbances ? School of Mechanical Engineering Purdue University ME375 Feedback Control - 10 10 ME375 Handouts Elements of Feedback Control Elements of a Feedback Control System: • Plant (Process) GP(s) -- The plant is the system (process) whose output is to be controlled, e.g., the room in the room temperature control example. 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. 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. thermocouple. • Controller GC (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. example. Controller Reference Input Input Disturbance D(s) Output GC (s) R(s) Heater Actuator Sensor GP(s) Y(s) Plant (Process) H(s) School of Mechanical Engineering Purdue University ME375 Feedback Control - 11 11 ME375 Handouts ClosedClosed-Loop Transfer Function Disturbance D(s) Control Input U(s) Output GP(s) Y(s) Plant Plant Equation (Transfer function model that we all know how to obtain ?!): ?! Control Law (Algorithm) (we will try to learn how to design): School of Mechanical Engineering Purdue University ME375 Feedback Control - 12 12 ME375 Handouts ClosedClosed-Loop Transfer Function Disturbance D(s) Reference Input R(s) + Error − E(s) Control Input + GC (s) + U(s) Output GP(s) Y(s) Plant H(s) Y (s) = ⋅ R( s ) + 14444 244444 4 3 GYR ( s ) ⋅ D( s ) 1 4444 244444 4 3 School of Mechanical Engineering Purdue University GYD ( s ) ME375 Feedback Control - 13 13 ME375 Handouts ClosedClosed-Loop Transfer Function The closed-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: are: Y (s) = GYR ( s ) 13 2 ⋅ R( s ) + Closed-Loop Transfer Function From R ( s ) to Y ( s ) ⋅ D( s ) GYD ( s ) 123 Closed-Loop Transfer Function From R ( s ) to Y ( s ) The objective of control system design is to design a controller GC (s), such that certain performance (design) specifications are met. For example: • 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 range. , • 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 occur. to specifying that School of Mechanical Engineering Purdue University ME375 Feedback Control - 14 14 ME375 Handouts Performance Specifications Given an input/output representation, GCL (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) Output GCL (s) r(t) Y(s) y(t) Time School of Mechanical Engineering Purdue University Time ME375 Feedback Control - 15 15 ME375 Handouts Unit Step Response 1.6 yMAX 1.4 Unit Step Response OS 1.2 ± X% 1 0.8 0.6 0.4 0.2 0 tP Time tS tr School of Mechanical Engineering Purdue University ME375 Feedback Control - 16 16 ME375 Handouts 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 e. specified as the steady state response, y(∞) (or ySS(t)), to be within an X% bound of the reference input r(t), i.e., the steady state error eSS(t) = r(t) − ySS(t) should be within a certain percent. For example: percent. example: r ( t ) − y SS ( t ) ≤ 2% = 0.02 r (t ) ⇔ y SS ( t ) ≥ 98% = 0.98 r (t ) ⇔ To find the steady state value of the output, ySS(t): – Sinusoidal references: use frequency response, i.e. – General references: use FVT, provided that School of Mechanical Engineering Purdue University is stable, ... ME375 Feedback Control - 17 17 ME375 Handouts 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: system. are: – Settling Time (tS): Specifies the time required for the response to reach and stay within a specific percent of the final (steady-state) value. Some typical (steadyvalue. settling time specifications are: 5%, 2% and 1%. For 2nd order systems, the are: specification is usually: ⎧ 4 2% ⎪ ζ ω for 2 % b oun d ⎪n ⎨ ⎪ 5 for 1% bound ⎪ ζω n ⎩ ≤ D esired Settling Time (t S ) ⇒ – % Overshoot (%OS):(2nd order systems) Overshoot (% order systems) %OS = 100e −π ζω n ωn 1−ζ 2 = 100e −π ζ 1−ζ 2 ≤ X% Q: How can we link this performance specification to the closed-loop transfer function? closed(Hint) What system characteristics affect the system performance ? School of Mechanical Engineering Purdue University ME375 Feedback Control - 18 18 ME375 Handouts Performance Specifications Transient Performance Specifications and CLTF Characteristic Poles Recall that the positions of the system characteristic poles directly affect the system output. output. For example, assume that the closed-loop transfer function of a feedback closedKω n2 control system is: is: GCL ( s ) = 2 s + 2ζω ns + ω n2 The characteristic poles are: s1 , 2 = − ζ ω n ± jω 1 − ζ 2 = −ζω n n ± jω d =− ± j⋅ Settling Time (2%): → Puts constraint on the real part of the dominating closed-loop poles. closedtS (2 % ) = 4 ζω = 4 n %OS: → Puts constraint on the imaginary part of the dominating closed-loop poles. closed- %OS = 100e − πζ 1−ζ 2 = 100e − π⋅ ζω n ω n 1−ζ 2 = 100e − π⋅ School of Mechanical Engineering Purdue University ME375 Feedback Control - 19 19 ME375 Handouts Performance Specification → CL Pole Positions CL Transient Performance Specifications and CLTF Pole Positions Transient performance specifications can be interpreted as constraints on the positions of the poles of the closed-loop transfer function. Let a pair of closedclosedfunction. closedloop poles be represented as: p 1 , 2 = − σ ± j ω as: Img. Transient Performance Specifications: jω −σ + jω – Settling Time (2 %) ≤ TS tS (2 % ) = 4 σ – %OS ≤ X % %OS = 100e − π⋅ σ ω ≤ TS ⇒ σ≥ 4 TS Real σ π⋅ 100 100 = πσ ≤ X % ⇒ e ω ≥ ω X e −σ − jω ⇒ − jω School of Mechanical Engineering Purdue University ME375 Feedback Control - 20 20 ME375 Handouts Example A DC motor driven positioning system can be modeled by a second order transfer function: on: GP ( s ) = Find closed-loop transfer function: closed- 3 s ( s + 6) A proportional feedback control is proposed and the proportional gain is chosen to be 16/3. 16/ Find the closed-loop transfer function, as well closedas the 2% settling time and the percent overshoot of the closed loop system when given a step input. input. Draw block diagram: School of Mechanical Engineering Purdue University ME375 Feedback Control - 21 21 ME375 Handouts Example Find closed-loop poles: 2% settling time: ti %OS: School of Mechanical Engineering Purdue University ME375 Feedback Control - 22 22 ME375 Handouts Example A DC motor driven positioning system can be modeled by a second order transfer function: on: GP ( s ) = Find closed-loop transfer function: closed- 3 s ( s + 6) A proportional feedback control is proposed. proposed. It is desired that: 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% 10%. Find (1) the condition on the proportional gain such that the steady state performance is satisfied; satisfied; (2) the allowable region in the complex plane for the closed-loop poles. closedpoles. Write down the performance specifications: School of Mechanical Engineering Purdue University ME375 Feedback Control - 23 23 ME375 Handouts Example Steady state performance constraint: Percent Overshoot (%OS) Img. jω Transient performance constraint: 2% Settling Time Real − jω School of Mechanical Engineering Purdue University ME375 Feedback Control - 24 24 ME375 Handouts 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 performed. function GP(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). necessary). (3) Draw the closed-loop block diagram, which includes the plant, sensor, actuator and closedfunctions. controller GC (s) transfer functions. (4) Obtain the closed-loop transfer function GCL (s). closed(5) Based on the performance specifications, find the conditions that the CLTF, GCL (s), has to satisfy. satisfy. (6) Choose controller structure GC (s) and substitute it into the CLTF GCL (s). (7) Select the controller parameters (e.g. the proportional feedback gain of a proportional (e. control law) so that the design constraints established in (5) are satisfied. (8) Verify your design via computer simulation (MATLAB) and actual implementation. School of Mechanical Engineering Purdue University ME375 Feedback Control - 25 25 ME375 Handouts 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 10-20 to 10-23). 1010-23) During the latest design review, the following plant parameters are obtained: obtained: 10 LA = 10 mH RA = 10 Ω 10 KT = 0.06 Nm/A 6.5 JE = 6.5 × 10-6 Kg m2 BE = 1.4 × 10-5 Nm/(rad/sec) 1.4 The drive roller angular position is sensed by a rotational potentiometer with a static sensitivity of KS = 0.03 V/deg. The design (performance) V/deg. specifications for the paper positioning system are: are: – The steady state position for a step input should be within 5% of the desired position. 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: specifications: School of Mechanical Engineering Purdue University ME375 Feedback Control - 26 26 ME375 Handouts In Class Exercise (1) Model the physical system (plant) that we want to control and obtain its I/O transfer function GP(s). (Sometimes, certain model simplification should be performed.) performed. DC Motor + eRa − + ei(t) _ R + eLa − iA A LA + Eemf τm θ, ω N2 θL, ωL JL JA N1 BL B From previous example, the DC motor driven paper positioning system can be modeled by + − 2 1 LA s + RA KT 1 JE s + BE ⎛1⎞ JE = JA + ⎜ ⎟ ⋅ JL ⎝N⎠ 2 ⎛1⎞ BE = B + ⎜ ⎟ ⋅ BL ⎝N⎠ Kb School of Mechanical Engineering Purdue University ME375 Feedback Control - 27 27 ME375 Handouts In Class Exercise The plant transfer function GP(s) can be derived to be: can GP ( s ) = θ (s) Ei ( s ) = KT s ( LA J E s + ( BE LA + RA J E ) s + ( RA BE + K b KT )) 2 As discussed in the previous example, we can further simplify the plant model by neglecting the electrical subsystem dynamics (i.e., by letting LA = 0 ): GP ( s ) = = θ (s) Ei ( s ) = KT = s ( RA J E s + ( RA BE + K b KT )) KM s (τ M s + 1) Substituting in the numerical values, we have our plant transfer function: GP ( s ) = KM = s (τ M s + 1) School of Mechanical Engineering Purdue University ME375 Feedback Control - 28 28 ME375 Handouts In Class Exercise (2) Determine sensor dynamics (transfer function of the measurement system) H(s) and actuator dynamics (if necessary). necessary). (3) Draw the closed-loop block diagram, which includes the plant, sensor, actuator and closedcontroller GC (s) transfer functions. Input Ei (s) School of Mechanical Engineering Purdue University GP(s) Output θ (s) ME375 Feedback Control - 29 29 ME375 Handouts In Class Exercise (4) Obtain the closed-loop transfer function GCL (s). closed- School of Mechanical Engineering Purdue University ME375 Feedback Control - 30 30 ME375 Handouts In Class Exercise (5) Based on the performance specifications, find the conditions that GCL (s) has to satisfy. satisfy. Steady State specification: Imag. jω Transient Specifications: Settling Time Constraint: Real Overshoot Constraint: − jω School of Mechanical Engineering Purdue University ME375 Feedback Control - 31 31 ME375 Handouts In Class Exercise (6) Choose controller structure GC (s) and substitute it into the CLTF GCL (s). Let’s try a simple proportional control, where the control input to the plant is proportional to the current position error: ei (t ) = K P ⋅ eθV (t ) = K P ⋅ (Vθ d (t ) − Vθ (t )) In s-domain (Laplace domain), this control law can be written as: s- Substitute the controller transfer function into GCL (s): School of Mechanical Engineering Purdue University ME375 Feedback Control - 32 32 ME375 Handouts In Class Exercise (7) Select the controller parameters (e.g., the proportional feedback gain of a proportional (e. control law) so that the design constraints established in (5) are satisfied. satisfied. Steady State Constraint: Transient Constraints: To satisfy transient performance specifications, we need to choose KP such that the closedclosedloop poles are within the allowable region on the complex plane. To do this, we first need to plane. find an expression for the closed-loop poles: closed- School of Mechanical Engineering Purdue University ME375 Feedback Control - 33 33 ME375 Handouts In Class Exercise Img. Axis For every KP , there will be two closed-loop poles (closed-loop characteristic roots). It’s closedclosedobvious that the two closed-loop poles change with the selection of different KP . For closedexample: → p1,2 = KP = 30 → p1,2 = KP = 20 KP = → p1,2 = KP = → p1,2 = 10 KP = → p1,2 = 0 → p1,2 = KP = By inspecting the root-locus, we can find root-10 that if -20 then the closed-loop poles will be in the closedallowable region and the performance specifications will be satisfied. -30 -60 -50 School of Mechanical Engineering Purdue University -40 -30 -20 Real Axis -10 0 ME375 Feedback Control - 34 34 ME375 Handouts 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); 1 Unit Step Response >> >> >> >> >> KP = 100 KP = 40 0.8 KP = 29.93 0.6 KP = 15 0.4 0.2 0 0 0.05 0.1 0.15 0.2 0.25 Time (sec) School of Mechanical Engineering Purdue University ME375 Feedback Control - 35 35 ME375 Handouts In Class Exercise (9) Check the Bode Plots of the open loop and closed loop systems: 10 0 KP = 100 Phase (deg); Magnitude (dB) -20 KP = 40 -40 KP = 29.93 -60 KP = 15 -80 Open Loop Loop 0 KP = 100 -45 KP = 40 -90 KP = 29.93 KP = 15 -135 Open Loop -180 10 -1 10 0 1 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University 10 2 10 3 ME375 Feedback Control - 36 36 ...
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