Feedback Systems. Many embedded system applications involve the concept of feedback. Sometimes feedback is designed into systems: Actuator
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1 Feedback Systems Many embedded system applications involve the concept of feedback Sometimes feedback is designed into systems: Operator Input CPU Actuator Physical System position velocity temperature current... Sensor Other systems have naturally occuring feedback, dictated by the physical principles that govern their operation EECS46, Lecture 7, updated September 24, 2008
2 Feedback Systems Some examples we will see: opamp motor equations: mechanical motor equations: electrical DC motor: back EMF current controlled amplifier velocity feedback control How many examples of feedback can you think of? EECS46, Lecture 7, updated September 24,
3 Issues with Feedback A feedback loop in a system raises many issues requires a sensor! changes gain reduces effects of parameter uncertainty may alter stability changes both steady state as well as dynamic response introduces phase lag sensitive to computation/communication delay Detailed analysis (and design) of feedback systems is beyond the scope of our course, but we will need to understand these basic issues... EECS46, Lecture 7, updated September 24,
4 Feedback and Gain Using high gain in a feedback system can make output track input: u Σ e K G y feedback response: y = KG + KG u error response: e = + KG u high gain: as K, y u and e 0 open loop gain : KG closed loop gain : KG/( + KG) we can make the output track the input even if we don;t know the exact value of the open loop gain! CAVEAT: only useful if system is stable! for all but very simple systems, use of excessively high gain will tend to destabilize the system! a simple example where dynamics are usually ignored: op amp EECS46, Lecture 7, updated September 24,
5 Operational Amplifier (Op Amp) An op amp [2] is used in many electronics found in embedded systems. Hence it is of interest in its own right, as well as being a simple example of a feedback system v o v v, v 2 : input voltages v 2 + v o : output voltage output voltage depends on difference of input voltages v o = K(v 2 v ) = K(v v 2 ) Typically K , but varies significantly due to manufacturing tolerances Ideal op amp no current flows into input terminals output voltage unaffected by load In reality op amp is a low pass filter with very high bandwidth draws a little current is slightly affected by load we shall assume an ideal op amp EECS46, Lecture 7, updated September 24,
6 Inverting Amplifier, I Q: How to use the op amp as an amplifier given that gain is uncertain? A: Feedback! I 2 R 2 I R v v i 0 + v o currents: I = v i v, I 2 = v v o R R 2 feedback equations: from previous page, v o depends on v : v o = Kv v depends on v o : v = v o + R 2 I 2 R 2 I 2 Σ v K v o EECS46, Lecture 7, updated September 24,
7 Inverting Amplifier, II Feedback diagram: R 2 I 2 Σ v K v o Apply rule for transfer function of feedback system: «K v o = R 2 I 2 + K If K >>, then the feedback equations imply that v o R 2 I 2 It further follows that v = v o + R 2 I 2 0. By assumption that the op amp draws no current, I = I 2, and thus v o = R2 R «v i Feedback allows us to use an op amp to construct an amplifier without knowing the precise value of K! EECS46, Lecture 7, updated September 24,
8 More Complex Feedback Examples to analyze op amp, we ignored dynamics and treated the op amp as a pure gain that was constant with frequency in general, dynamics cannot be ignored transient response stability Two examples where feedback arises from the physics motor dynamics: mechanical motor dynamics: electrical we shall discuss these examples, but we will first consider a simple case: feedback around an integrator EECS46, Lecture 7, updated September 24,
9 Integrator Equations of integrator ẋ = u x(t) = x(0) + Z t 0 u(σ)dσ Examples: u is velocity, x is position u is acceleration, v is velocity R voltage and current through inductor: I = L R V dt voltage and current through capacitor: V = C Idt Integrator is an unstable system the bounded input, u(t) =, yields the unbounded output x(t) = x(0) + t Transfer function of an integrator Z s integrator has infinite gain at DC, s = 0 EECS46, Lecture 7, updated September 24,
10 Feedback Around an Integrator Suppose there is feedback around integrator: u Σ K. x x differential equation of feedback system ẋ = Kx + Ku Transfer function of feedback system: u Σ K s x X(s) = «K/s U(s) = + K/s «K U(s) s + K The system is stable if K > 0. The response to the constant input u(t) = yields x(t) ẋ(t) 0 independently of the value of K EECS46, Lecture 7, updated September 24,
11 Uses of an Integrator sometimes integrators arise from the physics other times they are constructed to perform analog simulation of physical system to add integral control to a system Opamp integrator C 2 R 4 R R 3 v + i + v o Transfer function: v o = R 4 R 3 R C 2 s v i Can also implement integrator on a microprocessor discrete simulations digital control EECS46, Lecture 7, updated September 24, 2008
12 Motor Equations, Mechanical equations of motion for shaft dynamics J Ω = T M BΩ «Ω = T M J «B Ω J Ω: shaft speed, B 0: friction coefficient, J > 0: shaft inertia, T M : motor torque Feedback diagram T M J Ω s B. Ω Transfer function: Ω(s) = sj + B sj T M (s) = /B sj/b + T M(s) Constant torque speed goes to a steady state value: Ω ss = T M /B NOTE: with no friction (B = 0), system is unstable! constant torque implies Ω(t) EECS46, Lecture 7, updated September 24,
13 Motor Equations, Electrical equations of armature winding (ignoring back emf) LI = V RI «I = V L «R I L I: current, R: resistance, J: inductance, V : applied voltage Feedback diagram V R L. Ι s Ι Transfer function: I(s) = sl + R sl V (s) = /R sl/r + V (s) Constant voltage current goes to a steady state value: I ss = V/R EECS46, Lecture 7, updated September 24,
14 First Order Systems Shaft dynamics and circuit dynamics are each examples of a first order systems; i.e., they each have one integrator In general, a first order system may be written in the form ẋ = ax + bu where x is the integrator state, u is the input, and a and b are constants. Feedback diagram: u b Σ a. x x Equivalently u Σ b. x x a/b Transfer function: X(s) = H(s)U(s) ««b H(s) = a s/a + EECS46, Lecture 7, updated September 24,
15 Stability and Time Constant Time response: x(t) = e at x(0) + Z t 0 e a(t σ) bu(σ)dσ Response to a unit step, u(t) =, t 0: x(t) = b a e at The system is stable if a > 0 stability implies that x(t) b a as t Rate of convergence determined by time constant, τ = /a at t = τ, step response achieves 63% of its final value at t = 2τ, step response achieves 87% of its final value at t = 3τ, step response achieves 95% of its final value To easily compare rate of convergence, normalize so that b = a Normalized frequency response: x = H(jω)u, H(jω) = jτω + NOTE: The time constant determines the rate at which the response of the system must be sampled in order to adequately represent it in digital form. «EECS46, Lecture 7, updated September 24,
16 Bandwidth and Response Speed Time constant, τ determines bandwidth of frequency response: 0 frequency response of first order filter, /(jτω+) magnitude, db τ = τ = τ = phase, degrees ω, radians/second speed of response to unit step input, u(t) = : 0.9 step response of first order filter, /(jτω+) τ = 0. τ = τ = time, seconds Plots created with Matlab file first order.m. EECS46, Lecture 7, updated September 24,
17 Complete Motor Model The motor has both electrical and mechanical components, interconnected by the back EMF feedback loop: V VV B R L. Ι s Ι K M T M J Ω s B. Ω V B K V Two integrators a second order system Rules for combining transfer functions Ω(s) = sl+r + K v sl+r KM sj+b KM sj+b V (s) = K M (sj + B)(sL + R) + K v K M V (s) EECS46, Lecture 7, updated September 24,
18 Second Order Systems Question: How to analyze and describe properties of second order systems? stability steady state response transient response Approach : If the system can be decomposed into component first order subsystems, then (perhaps) properties of the overall system can be deduced from those of these subsystems. Example: DC motor Approach 2: General analysis procedure. Roots of characteristic equation Damping coefficient and natural frequency determine response Example: Virtual spring/mass/damper systems We will need to understand the relation between transient response and characteristic roots (natural frequency and damping) in order to design force feedback algorithms in Lab 6! EECS46, Lecture 7, updated September 24,
19 Time Scale Separation For a DC motor, the time constants for each first order subsystem may be very different: electrical subsystem: τ e = L/R = 0.00 mechanical subsystem: τ m = J/B = 0.35 Mechanical subsystem is much slower than the electrical subsystem Response of motor shaft is dominated by the mechanical subsystem On the shaft speed time scale, current appears to be instantaneous Since current and torque are related directly, T M = K M I, torque also responds rapidly response of DC motor to a unit step in voltage input speed torque time, seconds 2 Matlab files motor linear.m and DC motor linear.mdl EECS46, Lecture 7, updated September 24,
20 Second Order Systems Electrical dynamics can be ignored by setting L = response of DC motor to a unit step in voltage input speed, L = 0 torque, L = 0 speed, L = 0.0 torque, L = Detail: time, seconds 0.2 response of DC motor to a unit step in voltage input L = 0 L = time, seconds Will need to model current when we implement torque control 3 Matlab files motor neglect circuit.m and DC motor linear.mdl EECS46, Lecture 7, updated September 24,
21 Second Order Systems Systems with two integrators DC motor system with input and output described by the differential equation ÿ + bẏ + ay = cu u c Σ Σ.. y b. y y a The frequency response function can be written as H(s) = c s 2 + bs + a Example: DC Motor H(s) = s 2 + `BL+JR JL K M JL s + BR+KM K V JL EECS46, Lecture 7, updated September 24,
22 Characteristic Roots Suppose the frequency response is given by H(s) = c s 2 + bs + a Define the characteristic equation: Characteristic roots s 2 + bs + a = 0 s = b ± b 2 4a 2 () Possibilities: (i) b 2 4a > 0 two distinct real roots splane x x (ii) b 2 4a = 0 one repeated real root x+ splane (iii) b 2 4a < 0 two complex conjugate roots x splane x EECS46, Lecture 7, updated September 24,
23 Characteristic Roots and Stability Second order system is stable if the characteristic roots lie in the Open Left Half Plane (OLHP) unstable if the characteristic roots lie in the Closed Right Half Plane (CRHP) (roots on the imaginary axis are sometimes called marginally stable) splane stable unstable EECS46, Lecture 7, updated September 24,
24 Natural Frequency and Damping Parameterize roots of s 2 + bs + a = 0 by s = ζω n ± jω n p ζ 2 (2) where natural frequency, ω n, and damping coefficient, ζ, are defined by (compare (2) with ()) b = 2ζω n, a = ω 2 n roots lie on circle of radius ω n at an angle with the imaginary axis: θ = arctan ζ/ p ζ 2 jω n + θ + Roots are real if ζ 2 complex and stable if 0 < ζ < imaginary if ζ = 0 EECS46, Lecture 7, updated September 24,
25 Frequency and Time Response Natural frequency, ω n and damping ratio, ζ determine 4 bandwidth and peak of frequency response: magnitude, db frequency response of second order filter, /((jω/ω n ) 2 + 2ζ(jω/ω n )+) ζ = ζ = 0.2 ζ = phase, degrees speed and overshoot of unit step response: ω/ω n 2.8 step response of second order filter, /((jω/ω n ) 2 + 2ζ(jω/ω n )+) ζ = ζ = 0.2 ζ = ω n t 4 Plots created with Matlab mfile second order.m. EECS46, Lecture 7, updated September 24,
26 General Systems The characteristic equation of an nth order system will have n roots; these roots are either real, or they occur in complex conjugate pairs. The characteristic polynomial can be factored as N R Y i= (s + p i ) N C /2 Y i= (s 2 + b i s + a i ) Each pair of complex roots may be written as s i± = b i 2 ± q b 2 i 4a i 2 = x i ± jy i and have natural frequency and damping defined from s i± = ζ i ω ni ± jω ni q ζ 2 i Hence ζ and ω n can be computed from the real and imaginary parts as ω ni = q x 2 i + y2 i, ζ i = x i /ω ni Note: It often happens that the response of a high order system is well approximated by one complex pair of characteristic roots. EECS46, Lecture 7, updated September 24,
27 Spring/Mass/Damper System K F B M y Newton s laws: Mÿ + Bẏ + Ky = F ÿ = B M ẏ K M y + F M Second Order System F Σ M.. y B. y y K Transfer Function: Y (s) = M s 2 + B M s + K M F (s) EECS46, Lecture 7, updated September 24,
28 Motor Control Strategies Can conceive of controlling four signals associated with the motor input voltage, V shaft position, Θ shaft velocity, Ω torque, T M (equivalently, current, I) V VV B R L. Ι s Ι K M T L. T M Ω J s B Ω s Θ V B K V Issues: Input (V ) vs. output (Θ, Ω, I) variables Open loop vs. feedback control (i.e., do we use sensors?) Effect of load torque Control algorithm (P, I,...) Motor control results in higher order systems (more than two integrators) Higher order systems Can still define characteristic polynomial and roots Stability dictates that characteristic roots must lie in OLHP Integral control may still be used to obtain zero error (provided that stability is present) More complex control algorithms may be required to obtain stability EECS46, Lecture 7, updated September 24,
29 Voltage Control Apply desired V (either with a linear or a PWM amplifier) Suppose there is a constant load torque, T L. state speed and torque depend on the load: Position Ω = K MV RT L K M K V + RB T M = K M(V B + K V T L ) K M K V + RB Then steady V VV B R L. Ι s Ι K M T L. T M Ω J s B Ω s Θ V B K V Issues: V is an input variable, and usually not as important as T M, Θ, or Ω Suppose we want to command a desired speed (or torque), independently of load or friction * Problem: usually load torque (and often friction) are unknown Suppose we want to command a desired position * Problem: no control at all over position! EECS46, Lecture 7, updated September 24,
30 Position Control, I Suppose we want to control position We can use a sensor (e.g., potentiometer) to produce a voltage proportional to position, and compare that to a commanded position (also in volts). Θ* (volts) Θ error PI VV B sl+r I K M T M Ω sj+b s Θ K V volts potentiometer radians an integral controller cannot stabilize the system. Instead use a proportionalintegral (PI) controller: 0 + s responses of speed, torque, and position due to a unit step command to position speed torque position time, seconds 5 Matlab files motor position FB.m and DC motor position.mdl EECS46, Lecture 7, updated September 24,
31 Position Control, II PI control: if feedback system is stable, then error approaches zero, and position tracks desired value Can implement analog PI control using op amp circuit Control can also be implemented digitally using a microprocessor An encoder can be used instead of a potentiometer to obtain digital measurement PWM can be used instead of linear amplifier EECS46, Lecture 7, updated September 24,
32 Velocity Control, I Using an analog velocity measurement, from a tachometer, and an analog integral controller, allows us to track velocity T L Ω* (volts) Ω error controller V VV B sl+r I K M T M sj+b Ω V B K V volts tachometer radians/second despite the presence of an unknown load torque 6.4 load torque, T L = 0.5 speed torque time, seconds 6 Matlab files motor speed FB.m and DC motor speed.mdl EECS46, Lecture 7, updated September 24,
33 Velocity Control, II microprocessor control use encoder measurement to generate digital velocity estimate compare measured speed with desired speed feed error signal into digital integral controller generate PWM signal proportional to error Note: Performance depends on the controller gain 7. Consider the difference between 0/s and 00/s:.6.4 speed, K = 0 torque, K = 0 speed, K = 00 torque, K = time, seconds Usually,excessively high gain leads to oscillatory response or instability! 7 Matlab files motor speed FB.m and DC motor speed.mdl EECS46, Lecture 7, updated September 24,
34 Torque Control Using a measurement of current and an analog integral controller, allows us to track torqe, which is directly proportional to current: T M = K M I T L I* I error controller V VV B sl+r I K M T M sj+b Ω V B K V despite the presence of an unknown load torque load torque, T L = 0.5 speed torque time, seconds Question: How does our lab setup implement torque control? 8 Matlab files and motor current FB.m and DC motor current.mdl EECS46, Lecture 7, updated September 24,
35 PWM Amplifier, I Copley 422D DC brush servo amplifier with PWM inputs [] Two feedback control modes: velocity control (requires a tachomoter) torque (current) control We use torque control so that we can provide force feedback through our haptic interface VELOCITY MODE ENCODER TORQUE MODE ENCODER PWM 6 J2 DIR 6 J 2 + MOTOR PWM 6 J2 DIR 6 J 2 + MOTOR J2 4 + TACH GND 0 5 GND 0 ENABLE 3 ENABLE 3 POS ENABLE 2 J 4 GND POS ENABLE 2 J 4 GND NEG ENABLE HV NEG ENABLE HV Note: JP on pins 23 ( default ) Notes Note: JP on pins 23 ( default ). All amplifier grounds are common (J3, J4, J22, J27, and J20 ) Amplifier grounds are isolated from case & heatplate.. 2. Jumper JP default position is on pins 23 for ground active /Enable input ( J2 ) For /Inhibit function at J2 ( +5V enables ), move JP to pins 2 3. For best noise immunity, use twisted shielded pair cable for tachometer inputs. Twist motor and power cables and shield to reduce radiated electrical noise from pwm outputs. EECS46, Lecture 7, updated September 24,
36 PWM Amplifier, II onewire mode: 50% duty cycle corresponds to zero requested torque analog integral controller with antiwindup H bridge PWM amplifier 25 khz PWM output LM629 9 PWM MAG 8 PWM SIGN PWM 6 DIR 6 TACH() / GND 2 +/0V MAX Rext + 4 TACH 5 7 GND PWM BOARD PWM RH5 TO +/0v 00 K ANALOG CONVERSION TACH LEAD RH6 CH8 + RH7 00 K TACH AMP Gv = Rin = 47K CH9 RH K + SERVO PREAMP INTEGRATOR (OPEN) CH 00PF * INTEGRATOR RESET SWITCHES TURN ON WHEN AMP IS DISABLED * NOTE: DEFAULT VALUE OF CH IS 0 OHMS FOR TORQUE MODE OPERATION FOR VELOCITY MODE ( BRUSH TACH ) SEE APPLICATIONS SECTION + PEAK CURRENT CONT CURRENT CURRENT LIMIT SECTION RH2 RH3 * JP: PINS 23 FOR /ENABLE AT J2 PINS 2 FOR /INHIBIT AT J2 82K 4.7MEG GREEN = NORMAL RED = FAULT 47K 47K *JP CH LED +5V RH5 CH6 CURRENT ERROR AMP STATUS & CONTROL LOGIC GND 0 ENABLE * POS ENABLE 2 NEG ENABLE 3 4 +FAULT OUTPUT RESET 5 MOMENTARY SWITCH CLOSURE RESETS FAULT WIRE RESET TO GROUND FOR SELFRESET J2 SIGNAL CONNECTOR VALUE DEPENDS ON MODEL SEE "ARMATURE INDUCTANCE" TABLE J MOTOR & POWER CONNECTOR CURRENT REF CURRENT MONITOR V CW RH BALANCE 50K 0 MEG 5V K 4.8 khz FILTER 33NF K 4.8 khz FILTER 33NF PEAK TIME CH4 Voltage gain = +/6V at +/Ipeak 0.47U +/6V for +/Ipeak OUTPUT CURRENT SENSE PWM STAGE MOSFET "H" BRIDGE VOLTAGE GAIN Gv = +HV 0 MOTOR + MOTOR 2 GND 3 GND 4 +HV 5 + MOTOR K +HV +5 AUXILIARY DC OUTPUTS 5 3 0k 0k +5V 5V DC / DC CONVERTER CASE MAY BE GROUNDED FOR SHIELDING CASE GROUND NOT CONNECTED TO CIRCUIT GROUND POWER GROUND AND SIGNAL GROUNDS ARE COMMON EECS46, Lecture 7, updated September 24,
37 References [] Copley Controls. Models 422D, 422D DC brush servo amplifiers with PWM inputs. [2] K. Ogata. Modern Control Engineering. PrenticeHall, 3rd edition, 997. EECS46, Lecture 7, updated September 24,
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