FEEDBACK CONTROL SYSTEMS

FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State 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 Proportional-Derivative (PD) control Proportional-Integral (PI) control Proportional-Derivative-Integral (PID) control State space based control 4. Simulation (MATLAB) or experimental test.. Open and Closed-Loop Control Systems Design requirements and specification: An excellent transient response in terms of rise-time, size of overshoot and settling-time; An excellent steady-state response in terms of steady-state error; Acceptable stability margins; Robustness in terms of disturbance rejection; and Robustness in terms of sensitivity to parameter changes.

Open-loop control system: utilises an actuating device to control the process (plant) directly without using feedback. Input Actuating device Process or plant Output Closed-loop 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. Why Closed-Loop Control? An advantage of the closed-loop 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 first-order Taylor series: d Y(G G) Y(G) G 443 dg 443 4 Open-loop: G G G Closed-loop: 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: Open-loop: S T G Closed-loop: 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 open-loop and closed-loop 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

3.. Disturbance rejection Suppose that there is an input disturbance D applied between the controller and plant: D Open-loop: U G c G c GGD G c U G Closed-loop: 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 step-unit disturbance D /s, and H. Determine the response to the disturbance for both the open-loop and closed-loop systems and sketch the outputs. 4

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 Open-loop: G c k, G k /(Ts) Assume, for example, kk, T 0 Closed-loop with a proportional control term k, H Assume, for example, kk 0, T 0. Unit-step response of open-loop control Unit-step response of closed-loop control 0.9 0.9 0.8 0.8 0.7 0.7 Amplitude 0.6 0.5 0.4 Amplitude 0.6 0.5 0.4 0.3 0.3 0. 0. 0. 0. 0 0 5 0 5 Time (sec.) 0 0 5 0 5 Time (sec.) 3.4. Steady-state error The steady-state error is the error after the transient response has decayed, leaving only the continuous response, that is for a unity-feedback system: e( ) lim[r(t) y(t)] t E -. Considering a unit-step input, /s: Open-loop: G, E [-G]. By the final-value theorem, e( ) lim e(t) lim se G(0) t Closed-loop: G/[GH]. For H, E /[G] By the final-value theorem, e( ) lim e(t) limse /[ G(0)] t 5

Exercise: Find the open and closed-loop steady-state errors for the following system for unit-step and unit-ramp inputs. E 5 s 5 3.5. Summary The advantages of closed-loop control: Reduce effects of process (plant) disturbances Make system insensitive to process variations Stabilise an unstable system Create well-defined relations between output and reference The disadvantages: Risk of instability Complexity in analysis and implementation, and expensive 6

4. Case Study --- Speed Control of a DC Motor The purpose of this case study is: to compare the open-loop and closed-loop control, and to provide a guide to the solution of a first-order 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 motor-torque constant 0 N-m/A b back emf constant 0. V-s/rad J moment of inertia of the motor kg-m B viscous-friction coefficient of the motor 0.5 N-m-s/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 steady-state error, and a reduction of the effect of disturbance. These requirements can be represented, for a unit-step input and unit-step disturbance, as: Steady-state error e ss 0.0 to a unit-step input The effect of a unit-step disturbance < 0.005 rad/s Time constant 0. s. Can we meet these requirements using an open-loop controller? If not, we will try the closed-loop controller. 7

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 first-order 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

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. 4.6. Open-loop 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

Steady-state 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 open-loop controller can not achieve the required goals. 4.7. Position and velocity sensors in control systems 4.7.. 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 multi-turn rotational motion. Rotary potentiometer Linear potentiometer 0

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/000 0.006 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 photo-detector, 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 anti-clockwise), 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. 4.7.3. 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

4.8. Closed-loop control A tachometer is attached directly to the motor that measures the angular velocity and produces a proportional voltage that is fed-back 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

Assuming that t for simplicity: /0 T d V a 0 s.5 Ω t Steady-state error: T d 0, /s, s.5 E - Ω s s.5 0.5 e ss limse (s ). By choosing 5, e ss < 0.0, O. s 0.5 0 Disturbance rejection: 0, T d /s, Ω s(s.5 0) Taking the inverse Laplace transform (or from final-value 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) 0 0.00s s.5 G m b The closed-loop transfer function with the proportional term is: 3

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 -500-j500 (ζ < ) A compromise should be found between the need to increase and the need to avoid the above problems. 4.0. Summary Feedback (closed-loop) control can be used to stabilise systems, speed up the transient response, improve the steady-state 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

5. Steady-State 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 final-value theorem (assuming the system is stable), the steady-state 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 open-loop transfer function G. Since /s represents integration, the type number is the number of integrators in the open-loop 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

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 steady-state error characteristics of control systems. 5... Static position error constant po The steady-state error of the unity-feedback control system for a unit-step 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 steady-state 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 steady-state errors, respectively, of the systems shown in the following figures for a unit-step input. po s s s s(s ) 6

5... Static velocity error constant v The steady-state error of a system with a unit-ramp (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 0 5..3. Static acceleration error constant a The steady-state error of a system with a unit-parabolic (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 steady-state errors for the unit-step, unit-ramp, and unitparabolic inputs. 7

Exercise: For the unity feedback system shown, determine the value of to yield a 0% error in the steady-state when the input is r(t). s (s 4)(s 8) 8

5.. Steady-State 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 steady-state 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 steady-state error value due to D is found using final value theorem: e D c sgd ( ) lim se lim G G The steady-state error due to the reference and disturbance inputs is: e( ) e R ( ) e D ( ) Exercise: For the block diagram shown below, determine the stead-state error component due to a unit-step disturbance. D c 00 s(s 5) 9

TUTORIAL PROBLEM SHEET 5. List the major advantages and disadvantages of closed-loop control systems.. For the system shown in the figure, what are the steady-state errors when a unit-step input is applied to the following open-loop 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 steady-state 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 steady-state errors, respectively, of the systems shown in the following figures for a unit-step input. s s s s(s ) (a) (b) Plot the time response y(t) for a unit-step input using MATLAB. 0

6. For the system shown in the following figure: s(s 5)(s 0) (a) What value of will yield a steady-state 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 steady-state error due to a unit-step input and a unit-step 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) Steady-state error due to a unit-step input to be less than 0.. (b) Steady-state error due to a unit-step 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 unit-step input and a unit-step disturbance, respectively. 9. Redo question (8) for 0. s 0s 50 G