Control 2. Proportional and Integral control

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1 Control 2 Proportional and Integral control 1

2 Disturbance rejection in Proportional Control Θ i =5 + _ Controller K P =20 Motor K=2.45 Θ o Consider first the case where the motor steadystate gain = 2.45 o i

3 Θ i =5 + _ 0.1 Controller K P =20 Motor K= o i i and ss

4 Θ i =5 + _ 0.1 Controller K P =20 Motor K=2.45 Θ o = Now supposing the motor comes under load This has the effect of lowering the motor gain to 2.20 and the motor speed drops

5 Θ i =5 + _ 0.11 Controller K P =20 Motor K= Θ o The CLTF becomes o i ss 0. 11

6 Θ i =5 + _ 0.11 Controller K P =20 Motor K=2.20 Θ o = For the motor to reach its former speed, a new controller setting (K P ) is required. It s easy enough to calculate. Before the motor came under load K K 20 P

7 Θ i =5 + _ 0.11 Controller K P =20 Motor K=2.20 Θ o =4.89 Under load, K=2.20 therefore K P 2.2 The proportional controller gain has to be changed when the load on the motor changes. Either way a steady state error remains when proportional control is used on its own.

8 Integral Control With integral control, the output of the controller is proportional to the integral of the error signal ε with time. Output t K i dt 0

9 Supposing, for example the error had a constant value of 0.1 as in the previous example. 0.1 The Integral controller output would be Output t K 0.1dt i 0 K 0. 1t i The corrective action of the controller is increasing with a slope = K i 0.1

10 It s much easier to consider the integral controller in the Laplace domain The transform of the integral of a function between t=0 and t=t is 1/s times the Laplace transform of the function L t 0 f ( t) dt 1 s F ( s)

11 Recall this example from Systems 4

12 There is a steady-state error of 0.5

13 Now we introduce an integral controller with transfer function 1 and an integral gain K i =10 s

14 The steady-state error has gone but there is some oscillation Let s try a lower gain of K i =5

15 There is still some oscillation and the settling time is still about 10 seconds Let s try a lower gain of K i =1

16 One overshoot ( 15%) and no steady-state error

17 Θ i (s) + _ Controller K i s Motor Example 1: The motor above has a transfer function G( s) 1 s( s 1) And is used with integral control. Use Simulink to investigate the system response to a step and a ramp input. Compare the response to that of a proportional controller.

18 Proportional control, Step response, K P =10

19 Proportional control, Ramp response, K P =10 Note the steady-state error

20 Integral control, Step response, K i =10

21 Integral control, Ramp response, K i =10

22 The system seems to have gone unstable with the integral controller The closed-loop transfer function is s 2 ( s Ki 1) K i s 3 K s i 2 K i There is no coefficient of s, so the system is at best marginally stable

23 The Routh array of coefficients is K i The complete Routh array is K i 0 K i So the system is unstable for all positive values of K i

24 Proportional plus Integral Control Integral control is almost always used in conjunction with proportional control. K i Θ i (s) s + Process Θ o (s) + _ + K P

25 Controller output The diagram below shows how the controller output changes when there is an abrupt change to a constant error t There is a proportional controller output (P) which remains constant since the error does not change t I P Added to this is a steadily increasing integral controller output (I)

26 Controller output Now let s look at the case where the error signal increases from zero and then decreases back to it again. t First we will look at the proportional action alone. The controller mirrors the change t

27 Controller output Now let s look at the case where the error signal increases from zero and then decreases back to it again. t t Now lets see the action of the integral controller acting alone The controller acts in proportion to the area under the error curve Some integral controller action remains even when the error signal has gone to 0

28 Controller output Now let s look at the case where the error signal increases from zero and then decreases back to it again. t The result of the combined action is got by adding the two graphs We get a combination of swift reaction and no steady state error t

29 We ll apply proportional plus integral control to the previous example with the gains shown 29

30 The system is stable with no steady-state error 30

31 If we increase the proportional gain from 2 to 10 The settling time has decreased 31

32 Now we ll return the proportional gain to 2 and increase the integral gain to 1,5 The effect is to increase the oscillations and the settling time 32

33 If we further increase the integral gain to 2 The oscillations are further increased and the system tends towards instability. 33

34 CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR Kp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate The table summarises the effect of increase of the proportional and integral gains on system response 34

35 This sort of control system (P + I) copes well with large changes in the process The process changes must not be too rapid otherwise the integral action will cause oscillations.

36 When the input changes, for example at system start-up or after a step change, the error signal becomes temporarily very large. The integrator output goes immediately to 100% (saturates) and causes large overshoots before the system settles down. This phenomenon, known as integral wind up is avoided by only switching in the integrator when the error has reduced to a small steady state value

37 Integrator switched in when steady-state error becomes small Θ i (s) + _ K i s + + s 2 Process 1 10s 20 Θ o (s) K P

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