10 Historical noteThe first application of PID controller was in 1922 by Minorsky on ship steering.Minorsky (1922) “Directional stability of automatically steered bodies”, J. Am. Soc. Naval Eng., 34, p.284.This was the first mathematical treatment of the type of controller that is now used to control almost all industrial processes.
11 The current situationDespite the abundance of sophisticated tools, including advanced controller design techniques, PID controllers are still the most widely used controller structure in modern industry, controlling more that 95% of closed-loop industrial processes.Different PID controllers differ in the way how their parameters be tuned, manually, or automatically.Most of the DCS systems have built-in routines to perform auto-tuning of PID controllers based on the loop characteristics. They are often called: auto-tuners.
12 The PID AlgorithmThe PID algorithm is the most popular feedback controller algorithm used. It is a robust easily understood algorithm that can provide excellent control performance despite the varied dynamic characteristics of processes.As the name suggests, the PID algorithm consists of three basic modes:the Proportional mode,the Integral mode& the Derivative mode.
14 P, PI or PID ControllerWhen utilizing the PID algorithm, it is necessary to decide which modes are to be used (P, I or D) and then specify the parameters (or settings) for each mode used.Generally, three basic algorithms are used: P, PI or PID.Controllers are designed to eliminate the need for continuous operator attention. Cruise control in a car and a house thermostatare common examples of how controllers are used toautomatically adjust some variable to hold a measurement(or process variable) to a desired variable (or set-point)
15 Controller OutputThe variable being controlled is the output of the controller (and the input of the plant):The output of the controller will change in response to a change in measurement or set-point (that said a change in the tracking error)provides excitation to the plantsystem to be controlled
16 PID ControllerIn the s-domain, the PID controller may be represented as:In the time domain:proportional gainintegral gainderivative gain
17 PID Controller In the time domain: The signal u(t) will be sent to the plant, and a new output y(t) will be obtained. This new output y(t) will be sent back to the sensor again to find the new error signal e(t). The controllers takes this new error signal and computes its derivative and its integral gain. This process goes on and on.
18 Definitions In the time domain: integral time constant derivative time constantderivative gainproportional gainintegral gain
19 PID structures Standard PID controllers have the following structures: Proportional only:Proportional plus Integral:Proportional plus derivative:Proportional, integral and derivative:
20 What is an Op-Amp?An Operational Amplifier is an electronic device used to perform mathematical operations in a circuit – they are generally abbreviated as “Op-Amps”Op-Amps are high gain devices that amplify a signal using an external power supplyThey are composed of multiple transistors, resistors, and capacitorsCommon types of op-amps:InvertingNon-InvertingIntegratingDifferentialSumming
21 What is an Op-Amp? +Vcc V- V- Vout V- V+ Vout V+ V+ -Vcc All op-amps use a voltage supply (Vcc) to amplify the signalThe supply voltages can either have equal value but opposite signs, or the low side is grounded and the high side has a value of twice the voltage inputSome common applications of op-amps:Low Pass FiltersStrain GaugesPID Controllers+VccV-Inverting InputV-V+VoutV-VoutVoutV+Non-Inverting InputV+-Vcc
23 Amplifier Gain V- Vout = K (V+ - V-) Vout V+ Op-Amp Av = Vout / Vin All op-amps can be represented by the formula:Where K is the gain, and is a property of the individual op-ampThis gain should be distinguished from the gain of the op-amp circuit which is generally denoted by AvA potential source of confusion comes from failing to properly distinguish between the op-amp and the op-amp circuitV-V+VoutVout = K (V+ - V-)Op-AmpAv = Vout / VinOp-Amp Circuit
24 Comparator A comparator is an example of an open-loop op-amp i- = 0A +Vcc-VccVout+-V-V+Vin-Vin+VoutVin+ - Vin-+ Vsat- VsatIf V+ > V- Vout = Vsat ≈ VccIf V+ < V- Vout = -Vsat ≈ - Vcc
25 Summing Op-AmpApplication of a non-inverting op-amp and Millman‘s theorem(1) Millman’s theorem:if RA = RB = RC = R(1)R2(2) Non-inverting Op-Amp:(2)R1V-+-VoutVinARAV+VinBRBTo control the output voltageV’VinCRCVout = VinA + VinB + VinC25
26 Substraction (Differential) Op-Amp Applying Kirchhoff’s Rules and Op-Amp Calculation Rules yields:Vout = VinA - VinBVinBVout+-R4V-V+VinAR1R2R3if R1 = R2 = R3 = R4To control the output voltageNote:if R1 = R3 = R and R2 = R4 = a R26
27 Derivative Op-Amp R C Vin V- - Vout + V+ To control the output voltageApplying Kirchhoff’s Rules and Op-Amp Calculation Rules yields:27
28 Integrating Op-Amp C V- Vin R - Vout + V+ To control the output voltageApplying Kirchhoff’s Rules and Op-Amp Calculation Rules yields:28
29 PID Controller – System Block Diagram VOUTVSETVERROROutput ProcessDVSENSORSensorGoal is to have VSET = VOUTRemember that VERROR = VSET – VSENSOROutput Process uses VERROR from the PID controller to adjust Vout such that it is ~VSET
30 Applications PID Controller – System Circuit Diagram Signal conditioning allows you to introduce a time delay which could account for things like inertiaSystem to controlCalculates VERROR = -(VSET + VSENSOR)-VSENSOR
48 Controller EffectsA proportional controller (P) reduces error responses to disturbances, but still allows a steady-state error.When the controller includes a term proportional to the integral of the error (I), then the steady state error to a constant input is eliminated, although typically at the cost of deterioration in the dynamic response.A derivative control typically makes the system better damped and more stable.
49 Closed-loop Response Rise time Maximum overshoot Settling time Steady-state errorPDecreaseIncreaseSmall changeIEliminateDNote that these correlations may not be exactly accurate, because P, I and D gains are dependent of each other.
50 Example problem of PIDSuppose we have a simple mass, spring, damper problem.The dynamic model is such as:Taking the Laplace Transform, we obtain:The Transfer function is then given by:
51 Example problem (cont’d) LetBy plugging these values in the transfer function:The goal of this problem is to show you how each ofcontribute to obtain:fast rise time,minimum overshoot,no steady-state error.
52 Ex (cont’d): No controller The (open) loop transfer function is given by:The steady-state value for the output is:
53 Ex (cont’d): Open-loop step response 1/20=0.05 is the final value of the output to an unit step input.This corresponds to a steady-state error of 95%, quite large!The settling time is about 1.5 sec.
54 Ex (cont’d): Proportional Controller The closed loop transfer function is given by:
55 Ex (cont’d): Proportional control LetThe above plot shows that the proportional controller reduced both the rise time and the steady-state error, increased the overshoot, and decreased the settling time by small amount.
56 Ex (cont’d): PD Controller The closed loop transfer function is given by:
57 Ex (cont’d): PD control LetThis plot shows that the proportional derivative controller reduced both the overshoot and the settling time, and had small effect on the rise time and the steady-state error.
58 Ex (cont’d): PI Controller The closed loop transfer function is given by:
59 Ex (cont’d): PI Controller LetWe have reduced the proportional gain because the integral controller also reduces the rise time and increases the overshoot as the proportional controller does (double effect).The above response shows that the integral controller eliminated the steady-state error.
60 Ex (cont’d): PID Controller The closed loop transfer function is given by:
61 Ex (cont’d): PID Controller LetNow, we have obtained the system with no overshoot, fast rise time, and no steady-state error.
63 PID Controller Functions Output feedback from Proportional actioncompare output with set-pointEliminate steady-state offset (=error) from Integral actionapply constant control even when error is zeroAnticipation From Derivative actionreact to rapid rate of change before errors grows too big
64 Effect of Proportional, Integral & Derivative Gains on the Dynamic Response
65 Proportional Controller Pure gain (or attenuation) since:the controller input is errorthe controller output is a proportional gain
66 Change in gain in P controller Increase in gain: Upgrade both steady-state and transientresponses Reduce steady-stateerror Reduce stability!