Feedback ControlComponents of Engineered Feedback SystemsSlide Number 3Basics in the Frequency DomainSlide Number 5Random Physical DisturbancesSlide Number 7LaPlace vs. Fourier XFMLaPlace Transform and StabilityDecoding the transfer functionExample with a double integrator: e.g., a motor or dynamic positioningSlide Number 12Slide Number 13The PIDSelected Application NotesHeuristic Tuning of PID loops1. Zeigler-Nichols Methods for Tuning of PID ControllersSlide Number 18The 2p Discontinuity in Heading Control3. Integrator Windup4. Sensor Noise & OutliersLinear FilteringMassachusetts Institute of Technology Subject 2.017Feedback ControlMassachusetts Institute of Technology Subject 2.017Components of Engineered Feedback Systems• Plant: the system whose behavior is to be controlled. Examples: vehicle attitude, temperature, chemical process, business accounting, team and personal relationships, global climate• Actuator: systems which alter the behavior of the plant. Examples: motor, heater, valve, law enforcement (!), pump, FET, hydraulic ram, generator, US Mint• Sensor: system which measures certain states of the plant. Examples: thermometer, voltmeter, Geiger counter, opinion poll, balance sheet, financial analyst• Controller: translates sensor output into actuator input. Examples: computer, analog device, human interface, committee• Extreme variability in time scales: – active noise cancellation requires ~100 kiloHertz sensing and actuation– Social Security is assessed and corrected at ~3 nanoHertz (10 years)Massachusetts Institute of Technology Subject 2.017referencemeasurementController SensorerrorinputFederal Reserveanalystsinterest rateseconomic indicatorsshaft positioncontroller encoderPWM duty cyclecruise control tachometerfuel ratespeedPlantworld economymotor & loadengine & carstate____++++Feedback fundamentally creates a new dynamics!Massachusetts Institute of Technology Subject 2.017Basics in the Frequency DomainCP_+reu ye = r – yu = Ce = C(r-y)y = Pu = PCe = PC(r-y)  (PC + 1)y = PCr  y / r = PC / (PC + 1)Similarly, e = r – y = r – PCe  (PC+1)e = r  e / r = 1 / (PC + 1)u = C(r-Pu)  (PC + 1)u = Cr  u / r = C / (PC + 1)Why can we do this? Convolution in time domain = Multiplication in freq. domain!P must roll off at high frequencies – because no physical plant can respond to input at arbitrarily high frequency.• Ideal case: e is a small fraction of r: e/r << 1, equivalent to y/r ~ 1• This implies mag (PC + 1) >> 1 or mag (PC) >> 1.• If plant P is given, then C has to be designed to make PC big.• But mag (u / r) ~ mag(1 / P): HUGE when P gets small at high frequencies  excessive control action which will saturate or break actuators, excite unmodelled plant behavior, etc..  issues of robustnessMassachusetts Institute of Technology Subject 2.01710e/r = 1/(PC+1)y/r = PC/(PC + 1)0log |PC|PerformanceRobustnessGood tracking only possible at low frequencies  leads to a “formula” for design:Make |PC| large at low frequencies, e/r ~ 0, y/r ~ 1;Good regulation and tracking at low frequenciesMake |PC| small at high frequencies, e/r ~ 1, y/r ~ 0, u/r ~ CPoor tracking at high frequencies, but reasonable control actionThe frequency where |PC| = 1 is the crossover frequency c ; Above this point, closed loop t.f. y/r = PC/(PC+1) drops off to zero.So c is about the bandwidth of the closed-loop t.f. ccMassachusetts Institute of Technology Subject 2.017Random Physical DisturbancesCP_+reu yd++e = r – y and u = Ce = C(r-y)y = Pu + d = PCe + d = PC(r-y) + d With r = 0, (PC + 1)y = d  y / d = 1 / (PC + 1) ( = - e / d also)u = C(r- Pu - d) With r = 0, (PC + 1)u = -Cd  u / d = - C / (PC + 1)Because PC+1 is large at low frequencies, y/d will be small at low frequencies; the closed-loop system rejects low-frequency disturbances competing!Massachusetts Institute of Technology Subject 2.017• d is a random input, sometimes white or with frequency content, e.g., ocean waves!• Spectrum of y when system is driven by random noise as in previous analysis:Sy = [y/d]* [y/d] Sd• d can enter either at the plant output (as above), or at the plant input, i.e., it has the same units as control u. (Equations are different.)CP_+reu yd++Massachusetts Institute of Technology Subject 2.017LaPlace vs. Fourier XFMFourier Transform integrates x(t) e –jt over the time range from negative infinity to positive infinityLaplace Transform integrates x(t) e-st over the time range from zero to positive infinityResult: X(j) can describe acausal systems, X(s) describes only causal ones!Many important results of Fourier Transform carry over to LaPlace Transform:L (x(t)) = X(s) (notation)L (ax(t)) = a X(s) (linearity)L (x(t) * y(t)) = X(s)Y(s) (convolution) L (xt (t))  sX(s) (first time derivative)L (xtt (t))  s2X(s) (second and higher time derivatives)L (∫ x(t)dt)  X(s) / s (time integral)L ((t)) = 1 (unit impulse)L (1(t)) = 1/s (unit step)Massachusetts Institute of Technology Subject 2.017LaPlace Transform and Stability• For linear systems, stability of a system refers to whether the impulse response has exponentially growing components. • No pre-determined input can stabilize an unstable system; no pre-determined input can destabilize a stable system.• Some examples you can work out:L (e-t) = 1 / (s + )L (t e-t) = 1 / (s + )2L [ e-t sin(t) ] =  / (s2 + 2s + 2 + 2)L [ d e-nt sin (d t) / (1-2) ] = n2 / (s2 + 2n s + n2)Major observation: stable signal  roots of L denominator have negative real parts: EQUALITY IS TRUE FOR ALL FIRST- AND SECOND-ORDER SYSTEMSMassachusetts Institute of Technology Subject 2.017Decoding the transfer functionNumerator polynomials are a snap:(s + 2)/(s2+s+5) = s/(s2 + s + 5) + 2/(s2+s+5)“input derivative plus two times the input, divided by the denominator”For higher-order polynomials in the denominator: use partial fractions, e.g.,(s+1)/(s+2)(s+3)(s+4) = -0.5/(s+2) + 2/(s+3) -1.5/(s+4) (all real poles)(s+1)/s(s2+s+1) = -s/(s2+s+1) + 1/s (some complex poles)Any high-order transfer function can always be broken down into a sum of transfer functions with factored first- and