FEEDBACK CONTROL SYSTEMS


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1 FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control Systems. Control System Design. Establishing system goals (objectives), e.g. to control the velocity of a motor accurately What do you want to control? e.g. speed, position, level, temperature,... What do you want to achieve? e.g. fast response, less vibration,... How do you translate them into control terms? e.g. overshoot, rising time,.... Obtaining a model of the process, the actuator, and the sensor. E.g. G k/(ts) Differential equations Transfer functions Block diagrams State space equations 3. Designing a controller (PID, root locus, frequency response, state space) Proportional (P) control Integral (I) control ProportionalDerivative (PD) control ProportionalIntegral (PI) control ProportionalDerivativeIntegral (PID) control State space based control 4. Simulation (MATLAB) or experimental test.. Open and ClosedLoop Control Systems Design requirements and specification: An excellent transient response in terms of risetime, size of overshoot and settlingtime; An excellent steadystate response in terms of steadystate error; Acceptable stability margins; Robustness in terms of disturbance rejection; and Robustness in terms of sensitivity to parameter changes.
2 Openloop control system: utilises an actuating device to control the process (plant) directly without using feedback. Input Actuating device Process or plant Output Closedloop control system: uses a measurement of the output and feedback of this signal to compare it with the desired input (reference or command). E U G c _ G H Notation: : Laplace transform of the input signal; U: Laplace transform of the control or actuating signal; : Laplace transform of the output signal; E: Laplace transform of the error signal; G c : Transfer function of the controller; G: Transfer function of the process or plant; H: Transfer function of the sensor.
3 3. Why ClosedLoop Control? An advantage of the closedloop control system is the fact that the use of feedback makes the system response relatively insensitive to external disturbances (e.g. temperature and pressure) and internal variations in system parameters (e.g. component tolerances) which are not known or predicted. 3.. Sensitivity of control systems to parameter variations Suppose that G changes to G G due to the environment, ageing, etc. For small G << G, changes to. This can be illustrated by expanding to a firstorder Taylor series: d Y(G G) Y(G) G 443 dg Openloop: G G G Closedloop: G/[GH] G/[GH] _ G H The system sensitivity is defined as: the ratio of the percentage change in the system transfer function T to the percentage change of the process transfer function G: Openloop: S T G Closedloop: S T G GH dt / T S T G dg / G G T dt dg A system with zero sensitivity is ideal (i.e. changes in the system parameters have no effect on the transfer function). Exercise: A temperature control system of a chemical fluid in a tank is shown, where R is the resired temperature, Y the actual temperature and 'a' is a constant related to the mass in the tank. Determine the openloop and closedloop sensitivity of this sytem to changes in s a parameter 'a' due to chemical reactions an ageing. How would you reduce the sensitivity at low frequencies (i.e. )? 3
4 3.. Disturbance rejection Suppose that there is an input disturbance D applied between the controller and plant: D Openloop: U G c G c GGD G c U G Closedloop: U G c [ H] GcG G GH c G D G GH c G c D U H G Exercise: For the systems shown above, the process transfer function G /(0.5s), the controller G c 99, R 0, a stepunit disturbance D /s, and H. Determine the response to the disturbance for both the openloop and closedloop systems and sketch the outputs. 4
5 3.3. Transient response The transient response is the response of a system as a function of time. A motor speed control system: Objective: the actual speed y approaches the desired speed r quickly Openloop: G c k, G k /(Ts) Assume, for example, kk, T 0 Closedloop with a proportional control term k, H Assume, for example, kk 0, T 0. Unitstep response of openloop control Unitstep response of closedloop control Amplitude Amplitude Time (sec.) Time (sec.) 3.4. Steadystate error The steadystate error is the error after the transient response has decayed, leaving only the continuous response, that is for a unityfeedback system: e( ) lim[r(t) y(t)] t E . Considering a unitstep input, /s: Openloop: G, E [G]. By the finalvalue theorem, e( ) lim e(t) lim se G(0) t Closedloop: G/[GH]. For H, E /[G] By the finalvalue theorem, e( ) lim e(t) limse /[ G(0)] t 5
6 Exercise: Find the open and closedloop steadystate errors for the following system for unitstep and unitramp inputs. E 5 s Summary The advantages of closedloop control: Reduce effects of process (plant) disturbances Make system insensitive to process variations Stabilise an unstable system Create welldefined relations between output and reference The disadvantages: Risk of instability Complexity in analysis and implementation, and expensive 6
7 4. Case Study  Speed Control of a DC Motor The purpose of this case study is: to compare the openloop and closedloop control, and to provide a guide to the solution of a firstorder control design problem using a proportional (P) controller. Techniques used in the solution include: Mathematical modelling of physical systems Transfer function analysis Laplace transforms Feedback control 4.. DC motor  problem description R L i a (t) v a _ Armature circuit v b b ω _ J θ T m Bω T d For the armature controlled DC motor, we will use the following parameter values: R armature resistance Ω L armature inductance 0.5 H m motortorque constant 0 Nm/A b back emf constant 0. Vs/rad J moment of inertia of the motor kgm B viscousfriction coefficient of the motor 0.5 Nms/rad The motor torque, T m, is related to the armature current, i a, by a constant factor m : T m (t) m i a (t) The back emf, v b, is related to the rotational velocity, ω dθ/dt, by the following equation: v b (t) b ω(t) 4.. Design objectives (goals) The objective is to design a simple proportional controller (selecting the gain ) to achieve a fast response to the step input, a small steadystate error, and a reduction of the effect of disturbance. These requirements can be represented, for a unitstep input and unitstep disturbance, as: Steadystate error e ss 0.0 to a unitstep input The effect of a unitstep disturbance < rad/s Time constant 0. s. Can we meet these requirements using an openloop controller? If not, we will try the closedloop controller. 7
8 4.3. Mathematical model of the DC motor In this study, zero initial conditions are used. Time relation Laplace transform The motor torque: T m (t) m i a (t) T m m I a The back emf voltage: Electrical circuit equation: v b b ω(t) V b b Ω di a (t) va (t) bω(t) Ri a (t) L V a b Ω (slr)i a () dt The torque relation: T (t) T m (t) T d (t) T T m T d Mechanical system (rotor): dω(t) J Bω(t) T dt m (t) T d (t) (JsB)Ω m I a T d () Equations () and () are used to obtain the following block diagram: T d V a _ m Ls R T m T l Js B Ω b 4.4. Simplified model Transfer function from V a to Ω: Transfer function from T d to Ω: m G (Ls R)(Js B) G d (Ls R) (Ls R)(Js B) m m b b Neglecting L (assumed very small), we have the simplified firstorder model: m G RJs RB, R G d RJs RB m b m b Substituting the numerical values into G and G d : 0 G, s.5 G d s.5 8
9 T d V a _ 0 T m T l s 0.5 Ω 0. The block diagram can be simplified by moving the summing point ahead of the block with the constant gain of 0: T d /0 V a _ 0 s 0.5 Ω 0. This can be further reduced by representing the feedback loop with an equivalent block: /0 T d V a 0 s.5 Ω 4.5. Simplicity versus accuracy We must make a compromise between the simplicity of the model and the accuracy of the results of the analysis. In deriving a reasonably simplified mathematical model, we frequently find it necessary to ignore certain parameters Openloop control Introducing the constant gain (proportional),, block into the system (R desired velocity): T d Motor dynamics /0 V a 0 s.5 Ω 9
10 Steadystate error: T d 0, /s, 0 E  Ω s s.5 0 ess lim se. By choosing 3/0, e ss 0, O. s 0.5 Disturbance rejection: 0, T d /s, Ω s(s.5) Taking the inverse Laplace transform: ω(t)(/3)[ e 0.75t ] ω(t ) /3 rad/s, not O! Time constant: /.5 4/3 s, again not O! A simple openloop controller can not achieve the required goals Position and velocity sensors in control systems Potentiometers Converts mechanical energy to electrical energy. Input displacement can either be linear or rotational. Three terminal device: two fixed where voltage is applied, third is variable with the movable shaft to provide a voltage (with respect to ground) proportional to the displacement: e(t) e(t) Ly(t) Linear Rθ(t) Rotational Fixed terminals V s y(t) Variable terminal e(t) y(t) Rotary type available for single and multiturn rotational motion. Rotary potentiometer Linear potentiometer 0
11 4.7.. Incremental optical encoder Converts linear or rotary displacement of shaft into a digital pulse signal for position and velocity sensing. Consists of a rotary glass or plastic disk with evenly spaced opaque and transparent region each region represents a displacement increment. For example, the angular resolution of 000 lines disk is θ π/ radians per increment. Light passes through transparent lines as the disk rotates with the movement of the shaft, giving a sinusoidal or triangular waveform at the photodetector, followed by an amplifier/comparator to produce the pulse signal. Position information is obtained from counting the number of pulses using a counter; speed information is obtained from the frequency of the pulse train. Light source Photodetector Squaring Shaft Transparent Opaque For direction of rotation information (clockwise or anticlockwise), a second track of opaque/transparent lines at a different radius of the disk, that is offset 90 o with respect to the first track (quadrature tracks), is included. The pulse trains from the two channels are then compared for lead/lag to determine the direction of rotation Tachometer Works as a voltage generator with a DC voltage proportional to the angular velocity of the input shaft. Used in velocity indicators, velocity feedback control and stabilisation. dθ(t) Mathematical model: e(t) t tω(t) dt t is given as a parameter by the manufacturer in volts/000 rpm. E The transfer function is: Ω t
12 4.8. Closedloop control A tachometer is attached directly to the motor that measures the angular velocity and produces a proportional voltage that is fedback and compared with the reference voltage corresponding to the desired speed. e ref Controller DC motor amplifier M Load t ω Tachometer T ω 5V V REF R R Gain control Driver M ω R V REF t ω Summing amplifier t ω T
13 Assuming that t for simplicity: /0 T d V a 0 s.5 Ω t Steadystate error: T d 0, /s, s.5 E  Ω s s e ss limse (s ). By choosing 5, e ss < 0.0, O. s Disturbance rejection: 0, T d /s, Ω s(s.5 0) Taking the inverse Laplace transform (or from finalvalue theorem): ω( ) /(.50), by choosing 0, ω( ) < 0.005, O. Time constant: /(.50), by choosing, T 0.7 s, O. From the above, we choose a suitable value 0. From the above study, it can be seen that the larger, the better performance for a simple firstorder system. Can we choose very large? 4.9. Problems with a very large gain V a [  Ω] E, in the time domain v a (0) e(0), the small error e(0) at the beginning will lead to a large v a (0) since is large. When is big, the inductance L (which we ignored for simplicity) can not be neglected any more. It will lead to a problem. Assume L mh, then the transfer function from V a to Ω is: m (Ls R)(Js B) s s.5 G m b The closedloop transfer function with the proportional term is: 3
14 T 0.00s 0 s.5 0 For 0, two real poles: s 4 and s 886 (ζ > ) For 500, two complex poles: s 500j500 and s 500j500 (ζ < ) A compromise should be found between the need to increase and the need to avoid the above problems Summary Feedback (closedloop) control can be used to stabilise systems, speed up the transient response, improve the steadystate characteristics, provide disturbance rejection, and decrease the sensitivity to parameter variations. Proportional feedback control reduces errors and improves transient responses, but the high gain (large ) may lead to many problems. 4
15 5. SteadyState Errors in Unity Feedback Control Systems When a command input (desired output) is applied to a control system it is generally hoped that after any transient effects have died away the system output will settle down to the command value. The error with any system is the difference between the required (desired) output signal, i.e. the reference input signal which specifies what is required, and the actual output signal. For the unityfeedback control system shown in the figure, the error is: E G _ E G Using the finalvalue theorem (assuming the system is stable), the steadystate error is: e ss s e( ) lim e(t) limse lim t G A general representation of G is: G (s b s m m m N n n s (s a ns L b s b L a s a where is a constant and m and n are integers and neither a 0 nor b 0 is zero. N is an integer, the value of which is called the type of the system. Thus if N 0 then the system is said to be type 0, if N then type, if N then type and so on. The type number is thus the number of /s factors in the openloop transfer function G. Since /s represents integration, the type number is the number of integrators in the openloop transfer function. Exercise: What are the type numbers for the systems shown in the following figures? 0 0 ) ) (s ) (s )(s 6) (s ) s(s )(s 6) 5
16 5.. Static error constants Just as we defined the damping ratio, the natural frequency, settling time, percentage overshoot etc. as performance specifications for the transient response of a system, we can use static error constants to specify the steadystate error characteristics of control systems Static position error constant po The steadystate error of the unityfeedback control system for a unitstep input is: e ss s e( ) lim e(t) limse lim t s( G) G(0) The static position error constant po is defined by: lim G G(0) po s 0 Thus, the steadystate error in terms of the static position error constant po is; ess For a type 0 system, po C where C is a constant. For a type system, po. Hence, for a type 0 system, the static position error constant po is finite, while for a type or higher system, po is infinite. This is, e ss /(C) for a type 0 system, and e ss 0 for a type or higher system. Exercise: Find the static position error constants and steadystate errors, respectively, of the systems shown in the following figures for a unitstep input. po s s s s(s ) 6
17 5... Static velocity error constant v The steadystate error of a system with a unitramp (velocity) input is: e ss s e( ) lim e(t) lim se lim t s ( G) lim sg The static velocity error constant v is defined by: limsg v s 0 e ss For a type 0 system, limsg 0. For a type system, limsg C. For a type or higher system, v s 0 limsg. v s 0 v s Static acceleration error constant a The steadystate error of a system with a unitparabolic (acceleration) input, r(t) t /, is: s e( ) lim e(t) lim se lim lim t 3 s ( G) s G ess The static acceleration error constant a is defined by: a lims G s 0 e ss For a type 0 system, lims G 0. For a type system, lims G 0. For a type system, a s 0 lims G C a s 0 a s 0 lims G s 0.. For a type 3 or higher system, Exercise: A unity feedback system has the following forward transfer function: 000(s 8) G (s 7)(s 9) (a) Evaluate the system type, po, v, and a. (b) Use your answers to (a) to find the steadystate errors for the unitstep, unitramp, and unitparabolic inputs. 7
18 Exercise: For the unity feedback system shown, determine the value of to yield a 0% error in the steadystate when the input is r(t). s (s 4)(s 8) 8
19 5.. SteadyState Error for Disturbances D E G c G The error component when D 0 is: E  Y R GcG where the output due to the reference input, Y R, is given by: YR G G The steadystate error value due to is given by the final value theorem: e R s ( ) lim se lim G G c c When 0, the error in this case is given by: E 0  Y D where the output due to the disturbance, Y D, is: Y D G D G G The steadystate error value due to D is found using final value theorem: e D c sgd ( ) lim se lim G G The steadystate error due to the reference and disturbance inputs is: e( ) e R ( ) e D ( ) Exercise: For the block diagram shown below, determine the steadstate error component due to a unitstep disturbance. D c 00 s(s 5) 9
20 TUTORIAL PROBLEM SHEET 5. List the major advantages and disadvantages of closedloop control systems.. For the system shown in the figure, what are the steadystate errors when a unitstep input is applied to the following openloop transfer functions G: (a) (b) (c) 0 G (s )(s ) 6(s 3) G (s 6)(s ) 0 G s(s )(s ) E G 3. What are the type numbers for the systems shown in the following figures? (s ) (s )(s 6) (a) (s ) s(s )(s 6) 4. Determine the steadystate error for the system shown in question () above with: (s ) G s (s ) when subjected to the input r(t) t t. Plot the time response y(t) with MATLAB. (b) 5. Find the static position error constants and steadystate errors, respectively, of the systems shown in the following figures for a unitstep input. s s s s(s ) (a) (b) Plot the time response y(t) for a unitstep input using MATLAB. 0
21 6. For the system shown in the following figure: s(s 5)(s 0) (a) What value of will yield a steadystate error in position of 0.0 for an input of r(t) t/0? (b) What is the value of v for the value of found in (a)? (c) Plot the time response using MATLAB. 7. Find the total steadystate error due to a unitstep input and a unitstep disturbance in the system shown below. D 00 s 5 s 8. Consider the system shown in the following figure with G 5/(s ). What are the values of the gain which will achieve the following design specifications? D G (a) Steadystate error due to a unitstep input to be less than 0.. (b) Steadystate error due to a unitstep change in disturbance D to be less than 0.. (c) Validate the above results using MATLAB, that is to plot the time response for a unitstep input and a unitstep disturbance, respectively. 9. Redo question (8) for 0. s 0s 50 G
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