Preliminary Automatics Course
Chapter 29. PI control
Chapter 29.1 Introduction
The previous chapter conclusion (see p. 28.8) is:
–The P and especially PD control enables fast reaction for x(t) step type but it doesn’t enable null steady control error e(t)=0
– The I control behaviour is inverse. The reaction is slow but it enables null steady control error e(t)=0
The PI controller conception is obvious. Let’s combine P and I controllers!
Chapter 29.2 PI unit
Chapter 29.2.1 PI unit Kp=1 Ti= 1 sec
Call Desktop/PID/15_regulacja_typu_PI/01_czlon_PI_Kp1_I1.zcos
Fig. 29-1
This is not PI controller yet. It isn’t equipped with the most important controller device-the subtractor node!
The input signal x(t) is amplified first (here gain Kp=1) and divided for 2 components
– proportional P (“naked wire”)
– integrating I (here Ti=1 sec)
Click “start”
Fig. 29-2
Kp=1 Ti=1 sec
The input signal is x(t) step type. You see the yp(t) proportional component and yi(t) integral component of the y(t) output signal. The yp(t)=x(t)=1 because Kp=1 here. The yi(t) is a ramp type and the both components yp(t)=yi(t) are equal after Ti=1sec.
For all PI controllers both components yp(t)=yi(t) are equal after so-called reset time Ti. The other name of Ti is a integral time.
Note
The y(t) speed depends on the on the Kp and Ti parameters. But Ti parameter doesn’t depend on Kp. Please analyse Fig. 29-2 when Kp=0.5 for example.
Chapter 29.2.2 PI unit Kp=1 Ti= 10 sec
Call Desktop/PID/15_regulacja_typu_PI/02_czlon_PI_Kp1_I10.zcos
Fig. 29-3
Kp=1 Ti=10 sec
Click “start”
Fig. 29-4
The yi(t) speed is 10 times slower now, but y(t) output signal was doubled after Ti=10 sec too.
Chapter 29.2.3 PI unit Kp=3 Ti=10 sec
Call Desktop/PID/15_regulacja_typu_PI/03_czlon_PI_Kp3_I10.zcos
Fig. 29-5
Kp=3 Ti=10 sec
Click “start”
Fig. 29-6
The integration is 3 times faster now, but y(t) was doubled after Ti=10 sec too. Conlusion-Kp has no influence for Ti.
Chapter 29.3 Control System with the PI didactic controller
Chapter 29.3.1 PI didactic controller
I realised the PI didactic controller. It demonstrates that the component :
– proportional P assures shorter setting time
– integral I assures null steady state control error e(t)=0
– PI didactic controller starts as P controller and ends as a I controller when x(t) is a step type. The P–>I transition is in a one specific moment. The real PI controller P–>I transition from P is continuos. You will be convinced later.
Call Desktop/PID/15_regulacja_typu_PI/04_didactic_PI_Kp3_I35_od_0.zcos
Fig. 29-7
The controllers type depends on the contact position:
Fig. 29-7a – contact “high” – P controller
Fig. 29-7b – contact “low” – I controller
Fig. 29-7c – didactic PI as a Xcos model. The Xcos enables the complicated multiblock replacing for the one block.
Click “Start”
Fig. 29-8
It’s P controller up to 25 sec because contact is “high”and y(t)=Kp*e(t)=3*1=3. The contact is “low” then and the controler changes to I type now. The component I starts integration process from the initial state s(t)=0 as every moral integration unit! The I and P components are equal after Ti=35 sec as in the classical PI controller but this discrete s(t) drop is strange a little. The controller developed the s(t) signal based on the earlier e(t) process after all. The conclusion is only one. The integration must start from the last P controler state that is from the s(t)=3. Let’s improwe the PI didactic controller!
Call/Desktop/ PID/15_regulacja_typu_PI/05_didactic_PI_Kp3_I35_od_3.zcos
Fig.29-9
The controller structure is the same as Fig. 29-7. The difference is in the algorthm. The s(t) starts from the last P state when is the change P–>I. It means s(t) starts from s(t)=3 and not from s(t)=0 as Fig. 26-7)
Click “Start” and all will be clear
Fig. 29-10
The controller transition P–>I appears in 25 sec. If input e(t) jumped immediately in 40 sec up to 1.5 (for example), the s(t) speed will increase as red hash line. There isn’t s(t) jump because the I component is active only. The P is off.
Chapter 29.3.2 Control System with the didactic PI controller
The experiment demonstrates that:
– P component is accountable for fast y(t) value achievement. We know that this value is near the set point x(t).
– I (integral) component is accountable for steady state x(t)=y(t) achievement. In short. It assures null control error. The every control engineer goal.
Call Desktop/ PID/15_regulacja_typu_PI/06_didactic_PI_Kp10_I50_z_inercyjnym
Fig. 29-11
PI didactic controller Kp=10 and Ti=50 sec.
Fig. 29-11a
PI before transition, that is before 13 sec. The P component is active only, contact is “high”. Controller changes for I then, contact is “low”.
Fig. 29-11b
as Fig. 29-11a but Xcos model
Click “Start”
Fig. 29-12
Fig. 22-12a
It’s difficult to see that SP(t) and SI(t) components are time separated.
Fig. 22-12b
There are separate SP(t) and SI(t) processes and one common SPI(t) controll signal.
It’s clearly visible that:
– P component is accountable for fast y(t) growing before 13 sec., but P control can’t make state x(t)=y(t).
– I component is accountable for steady state x(t)=y(t) achievement after 13 sec.
Chapter 29.3.3 PI Didactic and classical P controller comparison
See Rys. 29-12a.
P controller is a PI Didactic special case without changing P–>I. This same y(t)=0.91 signal will be up to Doomsday. –>no null controll error e(t).
PI Didactic controller enables y(t)=x(t) steady state–>null controll error e(t)=0.
Chapter 29.3.4 PI Didactic and classical I controller comparison
Call Desktop/PID/15_regulacja_typu_PI/07_porown_didactic_PI_1T_KP10_I145_I16.zcos
Fig. 29-13
These same inertial objects are controlled by:
– PI didactic controller
– Classical I controller
Who is the winner? It seems like a rhetorical question.
Click “start”
Fig. 29-14
Score 5:0 for PI Didactical controller. The y1(t) setting time is lower than y2(t)and the oscillations are lower too. The PI Didactical controller starts as P and and quickly reaches the steady otput value y1(t)=0.91. It can’t do more–>to do y1(t)=x(t). There is a job for I component now. The controller changes for I type in 13 sec and finish the job–>reduce the control error up to 0–>y1(t)=x(t). This job is easy because the start point is y1(t)=0.91 and not 0! The start point for I controller is 0–>The y2(t) setting time is much longer. Score 5:0.
The real PI controller is better than PI Didactical controller yet!
Chapter. 29.4 PI controller with the one-inertial object
Chapter 29.4.1 Introduction
Let’s go to the real PI controller. These are p.29.4 , p.29.5 and p.29.6 subchapters
Chapter 29.4.2 One-inertial object in open loop
Call Desktop/PID/15_regulacja_typu_PI/08_1T_otwarty.zcos
Fig. 29-15
T=10 sec K=1.
Click “start”
Fig. 29-16
No comment
Chapter 29.4.3 PI controller Kp=3 Integration OFF
I.e we have typical P controller
Call Desktop/PID/15_regulacja_typu_PI/09_1T_Kp3_bez_calkowania.zcos
Fig. 29-17
We test P control to compare it with the PI control. You will know advantage of the PI then
Click “start”
Fig. 29-18
Typical P controll process. The non null steady error e(t)=0.25. This is a weakness of course. I remind you the main rule of the control system by the way. The steady state is achieved when s(t)=y(t). The input object signal=output object signal. Control signal s(t) is a object input signal here.
Chapter 29.4.4 PI controller PI Kp=3 Ti=12 sec
Call Desktop/PID/15_regulacja_typu_PI/10_1T_Kp3_I12.zcos
Fig. 29-19
Kp=3 Ti=12 sec
Click “start”
Fig. 29-20
Yes, yes, yes!!! Steady state y(t)=x(t) after 40 sec. Null control error. The Kp and Ti parameters are “shy” slightly. Do the more “aggressive” I integration. Will be the process better?
Chapter 29.4.5 More detailed analysis of the PI controller
Call Desktop/PID/15_regulacja_typu_PI/11_1T_Kp3_I12_P_I.zcos
Fig. 29-21
Kp=3 Ti=12 sek
The previous control system but we analyse proportional sP(t) and integral sI(t) components additionally.
Click “start”
Fig. 29-22
We are interested in sP(t) and sI(t) components only, therefore the main controller signal sPI(t) has a less pronounced yellow colour.
sP(t)=3 and sI(t)=0 in t=3sec when x(t) started–>sPI(t)=sP(t)
sP(t)=0 and sPI(t)=0 after t=55 sec (steady state)–>sI(t)=sPI(t)
Conclusion
PI controller starts as P controller and ends as a I controller. It’s similar to PI Didactical controller in the present case. But betweentimes? The sP(t) proportional components drops continuosly from 3 to 0 and sI(t) arises from 0 to 1. It means that PI controller changes continuosly its character from P controller to I controller. It’ s less similar to the PI Didactical controler. This one changes character in one moment in 13 sec –>Fig. 29-12.
Chapter 29.4.6 PI controller PI Kp=3 Ti=4 sec
Call Desktop/PID/15_regulacja_typu_PI/12_1T_Kp3_I4_opt.zcos
Fig. 29-23
Kp=3 Ti=4 sec
Integration I is more “aggresive” now. Will be better?
Click “start”
Fig. 29-24
There is a small small initial 10% over-regulation but it’s incontestably better. So try to be more aggresive and give Ti=2 sec.
Chapter 29.4.7 PI controller PI Kp=3 Ti=2 sec
Call Desktop/PID/15_regulacja_typu_PI/13_1T_Kp3_I2.zcos
Fig. 29-25
Kp=3 Ti=2 sec
Click “start”
Fig. 29-26
More oscillations and a longer setting time. Previous Ti=4sec is better.
Let’s give further Kp=10.
Chapter 29.4.8 PI controller Kp=10 Integration OFF
Call Desktop/PID/15_regulacja_typu_PI/14_1T_Kp10_bez_calkowania.zcos
Fig. 29-27
Kp=10 integration off–>P controller
Click “start”
Fig. 29-28
This is P controller now. This is under theory that Kp increament causes system faster and steady error isn’t null. The bigger is Kp the lower is steady error e(t)–>compare Kp=3 in Fig. 29-18. The system system is faster too.
Chapter 29.4.9 PI controller PI Kp=10 Ti=15 sec
Calll Desktop/PID/15_regulacja_typu_PI/15_1T_Kp10_I15.zcos
Fig. 29-29
Kp=10 Ti=15 sec
Click “start”
Fig. 29-30
It’s better than Kp=3 in Fig. 29-20. Let’s increase the integration speed.
Chapter 29.4.10 PI controller PI Kp=10 Ti=5 sec
Call Desktop/PID/15_regulacja_typu_PI/16_1T_Kp10_I5_opt.zcos
Fig. 29-31
Kp=10 Ti=5 sek
Click “start”
Fig. 29-32
The small initial over-regulation but the system is much faster. Let’s go this way and make integration more intensive.
Chapter 29.4.11 PI controller PI Kp=10 Ti=1.75 sec
Call Desktop/PID/15_regulacja_typu_PI/17_1T_Kp10_I.75.zcos
Fig. 29-33
Kp=10 Ti=1.75 sec
Click “start”
Fig. 29-34
The response is very fast but all depends of our Client decision. If this intial over-regulation doesn’t hurt Client technology then the parameters Kp=10 and Ti=1.75 sec are optimal for the Fig. 29-15 object.
Chapter. 29.5 PI controller with the two-inertial object
Chapter 29.5.1 Two-inertial object in open loop
Call Desktop/PID/15_regulacja_typu_PI/18_obiekt_2T.zcos
Fig. 29-35
T1=3 sek, T1=5 sek, K=1.
Click “start”
Fig. 29-36
No comments
Chapter 29.5.2 PI controller Kp=3 Integration OFF
Call Desktop/PID/15_regulacja_typu_PI/19_2T_Kp3_bez_calkowania.zcos
Fig. 29-37
Kp=3 Integration OFF
Click “start”
Fig. 29-38
Typical P control with non steady error e(t)=0.25. The response is with oscillations because the object is more complicated than in the Fig. 29-18
Chapter 29.5.3 PI controller PI Kp=3 Ti=32 sec
Call Desktop/PID/15_regulacja_typu_PI/20_2T_Kp3_I32.zcos
Fig. 29-39
Kp=3 Ti= 32 sec
We start with a very subtle integration as usual.
Click”start”
Fig. 29-40
Fig. 29-40a
The integration is to shy and there isn’t steady error e(t)=0 before 60 sec yet. But this state will be achieved later. See Fig. 29-40b
Fig. 29-40b
The experiment time is 180 sec now. The steady error e(t)=0 was achieved finally. We were very shy with the integration so be brave and do Ti=8sec.
Chapter 29.5.4 PI controller PI Kp=3 Ti=8 sec
Call Desktop/PID/15_regulacja_typu_PI/21_2T_Kp3_I8_opt.
Fig. 29-41
Kp=3 Ti= 8 sec
Wciśnij “start”
Fig. 29-42
The courage pays off. The response is much better. Let’s go on this way.
Chapter 29.5.5 PI controller PI Kp=3 Ti=5 sec
Call PID/15_regulacja_typu_PI/22_2T_Kp3_I5.zcos
Fig. 29-43
Kp=3 Ti= 5 sec
Click “start”
Fig. 29-44
It’s worse than before.
Chapter 29.5.6 PI controller Kp=10 Integration OFF
Call Desktop/PID/15_regulacja_typu_PI/23_2T_Kp10_bez_calkowania.zcos
Fig. 29-45
Kp=10 Integration OFF–>P type control
Click “start”
Fig. 29-46
There are oscillations but the steady non null control error is consistent with the theory–>P type control. We expect troubles with I component.
Chapter 29.5.7 PI controller Kp=10 Ti=20 sec
Call Desktop/PID/15_regulacja_typu_PI/24_2T_Kp10_I20.zcos
Fig. 29-47
Kp=10 Ti=20 sec
Click “start”
Fig. 29-48
We started carefully with Ti=20 sec (low integration speed) but there are oscillations and the setting time is long, more than 60 sec. The steady error is e(t)=0. This a I component job!
By the way. The s(t) oscillations are bigger than y(t) oscillations and it’s typical for control systems. The driver steering wheel movements are bigger than a car course oscillations. The car is going by the “almost” straight line but driver steering wheel movements are quite visible.
Chapter 29.5.7 PI controller Kp=10 Ti=10 sec
Call Desktop/PID/15_regulacja_typu_PI/25_2T_Kp10_I10opt.zcos
Fig. 29-49
Kp=10 Ti=10 sec
Click “start”
Fig. 29-50
The process is decently, but the parameters Kp=3 and Ti=8 sec are better–>Fig. 29-42! The life is full of surprises. The higher Kp was better by now. Let’s decrease Ti. Will be better?
Chapter 29.5.9 PI controller Kp=10 Ti=5 sec
Call Desktop/PID/15_regulacja_typu_PI/26_2T_Kp10_I5.zcos
Fig. 29-51
Kp=10 Ti=5 sec
Click “start”
Fig. 29-52
It’s worse.
Chapter. 29.6 PI controller with the three-inertial object
Chapter 29.6.1 Three-inertial object in open loop
Call Desktop/PID/15_regulacja_typu_PI/27_obiekt_3T.zcos
Fig. 29-53
K=1, T1=0.5 sec T2=3 sec and T3=5 sec
Click “start”
Fig. 29-54
Compare with the two-inertial–> Fig. 29-36. The different isn’ very distinct. But the more “multi” is a multi-inertial unit, the more distinct is To delay parameter.
Chapter 29.6.2 PI controller Kp=3 Integration OFF
Call Desktop/PID/15_regulacja_typu_PI/28_3T_Kp3_bez_calkowania.zcos
Fig. 29-55
Kp=3 Integration OFF
Click “start”
Fig. 29-56
Compare with the Fig. 29-38. The same P controller but the object is simpler-two-inertial. The steady error e(t)=0.25. But the over-regulation (first oscillation) is bigger here.
Chapter 29.6.3 PI controller Kp=3 Ti=16 sec
Call Desktop/PID/15_regulacja_typu_PI/29_3T_Kp3_I16.zcos
Fig. 29-57
Kp=3 Ti=16 sec
Click “start”
Fig. 29-58
Component I made steady error e(t)=0 after 60 sec. But it’s very slowly. Let’s make I more aggresive Ti=10 sec.
Chapter 29.6.4 PI controller Kp=3 Ti=10 sec
Call Desktop/PID/15_regulacja_typu_PI/30_3T_Kp3_I10opt.zcos
Fig. 29-59
Kp=3 Ti=10 sec
Click “start”
Fig. 29-60
Good job! May be better? Let’s minimize on Ti.
Chapter 29.6.5 PI controller Kp=3 Ti=4 sec
Call Desktop/PID/15_regulacja_typu_PI/31_3T_Kp3_I4.zcos
Fig. 29-61
Kp=3 Ti=4 sec
Click “start”
Fig. 29-62
It’s worse. And what about Kp=10?
Chapter 29.6.6 PI controller Kp=10 Integration OFF
Call PID/15_regulacja_typu_PI/32_3T_Kp10_bez_calkowania.zcos
Fig. 29-63
Kp=10 Integration OFF
Click “start”
Fig. 29-64
The steady error is e(t)=0.09 and is lower than for Kp=3. But long setting time and oscillations are terrible.
Chapter 29.6.7 PI controller Kp=10 Ti=16 sec
Call Desktop/PID/15_regulacja_typu_PI/33_3T_Kp10_I16opt.zcos
Fig. 29-65
Kp=10 Ti=16 sec
Click “start”
Fig. 29-66
It’s worse!
Chapter 29.6.8 PI controller Kp=10 Ti=8 sec
Call Desktop/PID/15_regulacja_typu_PI/34_3T_Kp10_I8.zcos
Fig. 29-67
Kp=10 Ti=8 sec
Click “start”
Fig. 29-68
It’s much worse. The parameters Kp=3 i Ti=10 sec for three-inertial object are better! But we are determinded and we will increase integration speed for Ti=2.5 sec.
Chapter 29.6.9 PI controller Kp=10 Ti=2.5 sec
Call Desktop/PID/15_regulacja_typu_PI/35_3T_Kp10_I2.5_niestabilny.zcos
Fig. 29-69
Kp=10 Ti=2.5 sec
Why is there Dirac (pseudo Dirac-strictly) instead of the softie x(t) step type pule as usually?
Click “start”
Fig. 29-70
The short hammer-Dirac caused oscillations up to +/- infinity. By the way. The instability doesn’t mean immobility always. This system is instable but it still up to 3 sec. As a vertical positioned pencil on the table.
Chapter 29.7 How does PI controller suppress the disturbances?
Chapter 29.7.1 Introduction
The one, two and three-inertial objects are used as before. The additional disturbance z(t)=+0.5 or z(t)=-0.5 occures at their inputs in 70 sec. The optimal Kp and Ti=5 parameters are set. They are optimal on the grounds of the x(t) input, not z(t) input. The z(t) response will be better if they are choosed on the grounds of the z(t) disturbance!
Chapter 29.7.2 One-inertial object, Kp=10 Ti=5 sec and positive disturbance z(t)=+0.5
Call Desktop/PID/15_regulacja_typu_PI/36_1T_Kp10_I5opt_zakl+.zcos
Fig. 29-71
Disturbance z(t)=+0.5 occures in 70 sec.
Click “start”
Fig. 29-72
The process is the same as in Fig. 29-32 up to 70 sec. Remember that there are different oscilloscope time scales. The z(t)=+0.5 (additinal heating for example) disturbance is compensated by the controller output Δs=-0.5 drop. (oven power drop for example). There is a steady controll error e(t). The main goal of the control system. This is a good job of the I controller component!
Chapter 29.7.3 One-inertial object, Kp=10 Ti=5 sec and negative disturbance z(t)=-0.5
Call Desktop/PID/15_regulacja_typu_PI/37_1T_Kp10_I5opt_zakl-.zcos
Fig. 29-73
Disturbance z(t)=-0.5 occures in 70 sec.
Click “start”
Fig. 29-74
The z(t)=-0.5 (additinal cooling for example) disturbance is compensated by the controller output Δs=+0.5 growth. (oven power growth for example).
Chapter 29.7.4 Two-inertial object, Kp=3 Ti=8 sec and positive disturbance z(t)=+0.5
Call Deskto/PID/15_regulacja_typu_PI/38_2T_Kp3_I8opt_zakl+.zcos
Fig. 29-75
Disturbance z(t)=+0.5 occures in 70 sec.
Click “start”
Fig. 29-76
Object is more complicated than before and you see the effects.
Chapter 29.7.5 Two-inertial object, Kp=3 Ti=8 sec and negative disturbance z(t)=-0.5
Call Desktop/PID/15_regulacja_typu_PI/39_2T_Kp3_I8opt_zakl-.zcos
Fig. 29-77
Disturbance z(t)=-0.5 occures in 70 sec.
Click “start”
Fig. 29-78
No comments
Chapter 29.7.6 Three-inertial object, Kp=3 Ti=8 sec and positive disturbance z(t)=+0.5
Call Desktop/PID/15_regulacja_typu_PI/40_3T_Kp3_I10opt_zakl+.zcos
Fig. 29-79
Disturbance z(t)=+0.5 occures in 70 sec.
Click “start”
Fig. 29-80
More complicated object yet and the dynamic response parameters are worse–>bigger oscillations and setting time.
Chapter 29.7.7 Three-inertial object, Kp=3 Ti=8 sec and negative disturbance z(t)=-0.5
Call Desktop/PID/18_regulacja_typu_PI/41_3T_Kp3_I10opt_zakl-.zcos
Fig. 29-81
Disturbance z(t)=-0.5 occures in 70 sec.
Click “start”
Fig. 29-82
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