Chapter  28.1 Introduction
You know the and PD controller already.
28-1
Fig. 28-1
These controllers assure closed loop gain Kz and error gain Ke in steady state. The is a static gain of  all the open loop and it includes controller and object.  The gain of the controlled object Go(s) is Ko=1 mostly–> see Fig. 25-3 in chapter 25.
Conclusion
All the  open loop gain is K=Kp. It’s  very convenient solution to addjust controllers parameters. —>Chapter 31.
Big Kp means Kz≈1 and Ke≈0. It means that y(t)≈x(t) and control error e(t)≈0 in steady state. It’s almost ideal situation! Output signal y(t) tries to follow set point x(t) but the ideal steady state  situation x(t)=y(t) will be never.
The main thing now!
The I control assures
28-2
Fig. 28-2
Easier formulas are hard to imagine!  It means that the output signal y(t)=x(t) or e(t)=0  in steady states. The formulas are the special case of the Fig. 28-1. What’s the steady state of the integral unit  I which is a part of the open loop. See Fig. 28-4. This state is y(t)=infinity! –> The open loop gain K is infinity too. Put the K=infinity to the formulas Fig. 28-1. The result is Fig. 28-2

Chapter 28.2 Integral Unit
Chapter 28.2.1 Introduction
We will test 3 integral units with different speed of integration
“Slow”
“Middle”
“Fast”
These units aren’t I controllers because they haven’t subtractor node. This node calculates control error e(t)=x(t)-y(t)

Chapter 28.2.2  “Slow” I unit Ti=2 sec
Call PID/14_regulacja_typu_I/01_calkujacy_wolny.zcos
28-3a
Fig. 28-3
Ti=2 sec
Clickj “start”
28-4a
Fig. 28-4
The input x(t)=1 is step type and the y(t) speed is steady. Note that x(t)=y(t) after Ti=2sec. Conclusion -The bigger is Ti, the “slower” is the I integration unit. The next units will be faster.

Chapter 28.2.3  “Middle” I unit Ti=1 sec
Call PID/14_regulacja_typu_I/02_calkujacy_taki_sobie.zcos
28-5a
Fig. 28-5
Ti=1 sek
Click “start”
28-6a
Fig. 28-6
Ti=1 sec
The unit is 2 times faster

Chapter 28.2.4  “Fast” I unit Ti=0.5 sec
Call PID/14_regulacja_typu_I/03_calkujacy_szybki.zcos
28-7a
Fig. 28-7
Ti=0.5 sek.
Click “start”
28-8a
Fig. 28-8
The unit is 2 times faster.

Chapter 28.3 How does I control make control error e(t)=0.
Chapter 28.3.1 Introduction
It’s an example that I control (Integration control type) is absolutely accurate–>x(t)=y(t)
28-9a
Fig. 28-9
This servomechanism is shows that I control can make error e=x-y=0! Other words. The potentiometer B x(t) voltage will absolutely equal to potentiometer A y(t) voltage in steady state x(t)=y(t)=0. Other words. The initial state is x(t)=y(t)=0.  We move the potentiometer slider up to +5 V. The voltage +Kp*5V occures immadietaly because y(t)=0V in this moment. The D.C.  motor start rotation and the potentiometer slider will start with the maximal intial speed. Please analize this process. This B slider ascent speed is decreasing because D.C. motor voltage  +Kp*[5V-y(t)] is decreasing (y(t) voltage is growing). When does motor stop? It stops when x(t)=y(t)! Please note that our motor is ideal! Very small voltage, for example 1µeven, causes rotation. The gear is ideal too. I hope you haven’t problems with block diagrams Fig.28-9b and Fig.28-9c. Ks and Kp depends on the electrical and mechanical parameters. Note that d.c. motor is integral I type. I will not sink in this subject deeper.
The main conclusion.
There is K/s*Ti integral unit (D.C. motor!) in the closed loop and it causes control error  e(t)=0. The output signal doesn’t oscilate as in ON-OFF control type!

Chapter 28.3.2 “Slow” servomechanism
Call/Desktop/PID/14_regulacja_typu_I/04_model_serwo_wolny.zcos
28-10a
Fig. 28-10
Ti=4 sec
Click “start”
28-11a
Fig. 28-11
Conlusion
Closed loop integral unit is inertial type.
Note
x(t)=y(t) in steady state!

Chapter 28.3.3 “Fast” servomechanism
Call/Desktop/PID/14_regulacja_typu_I/05_model_serwo_szybki.zcos
28-12a
Fig. 28-12
Ti=1 sec
Wciśnij “start”
28-13a
Fig. 28-13
Make your own conclusions.

Chapter. 28.4 I controller with the inertial object.
Chapter 28.4.1 Introduction
We will test I controller with the Inertial object.

Chapter 28.4.2 Inertial object in open loop

28-14
Fig. 28-14
Inertial unit K=1 T=10 sec
Click “start”
28-15a
Fig. 28-15
No comment

Chapter 28.4.3 I controller Ti=36 sec
Call Desktop/PID/17_regulacja_typu_I/07_1T_I36
28-16a
Fig. 28-16
I
controller has only a a single parameter Ti .  We start with the “slow” Ti=36sec. , because we are afraid of the instability.
Click “start”
28-17
Fig. 28-17
Our concerns were bloated. There aren’t oscillations and the setting time is very long. The steady state x(t)=y(t) isn’t seen in the figure, because steady state is after 120 sec.

Chapter 28.4.4 I controller Ti=16 sec
The previous system was slow. The faster integration Ti=16 should make better.
Call Desktop/PID/14_regulacja_typu_I/08_1T_I16_opt.zcos
28-18a
Fig. 28-18
Ti=16 sec
Call “start”
28-19
Fig. 28-19
The system is faster but there are oscillations now.

Chapter 28.4.5 I controller Ti=8 sec
Will be better?
Call Desktop/PID/14_regulacja_typu_I/09_1T_I8.zcos
28-20a
Fig. 28-20
Ti=8 sek
Click “start”
28-21
Fig. 28-21
To much oscillations. The best parameter is Ti=16 sec if so.

Chapter. 28.5 I controller with the Two-inertial object.
Chapter 28.5.1 Introduction
We will test I controller with the Two-inertial object. This object is more complicated than Inertial. We are almost sure that it’s more difficult to control–>more oscillations and long setting time.

Chapter 28.5.2 Inertial object in open loop
Call Desktop/PID/14_regulacja_typu_I/10_2T_otwarty.zcos
28-22
Fig. 28-22
K=1, T1=3 sec and T2=5 sec
Click “start”
28-23
Fig. 28-23
No comment

Chapter 28.5.3 I controller Ti=25 sec
Call Desktop/PID/14_regulacja_typu_I/11_2T_I25.zcos
28-24a
Fig. 28-24
Ti=25 sek
Click “start”
28-25
Fig. 28-25
Surprice. The response is much better than for a “easier” one-inertial unit as Fig. 28-17!

Chapter 28.5.4 I controller Ti=15 sec
Call Desktop/PID/14_regulacja_typu_I/12_2T_I15_opt.zcos
28-26a
Fig. 28-26
Ti=15 sek
Click “start”
28-27
Fig. 28-27
We are braver and the integration Ti=15 sec faster than previopus Ti=15 sec. The system is faster but there are oscillations now. Which is better? I don’t know.

Chapter 28.5.5 I controller Ti=6 sec
Let’s go baldheaded and give Ti=6 sec
Call Desktop/PID/14_regulacja_typu_I/13_2T_I6.zcos
28-28a
Fig. 28-28
Ti=6 sek
Click “start”
28-29
Fig. 28-29
The oscillations are not to accept. The optimal integration Ti for two-inertial at Fig. 28-22 is Ti=15sec.

Chapter. 28.6 I controller with the Three-inertial object.
Chapter 28.6.1 Introduction
We will test I controller with the thrre-inertial object.

Chapter 28.6.2 Three-inertial object in open loop
Call Desktop/PID/14_regulacja_typu_I/14_3T_otwarty.zcos
28-30
Fig. 28-30
K=1 T1=0.5 sek, T2=3 sek and T3=5 sek
Click “start”
28-31
Fig. 28-31
No comment

Chapter 28.6.3 I controller Ti=30 sec
Call PID/14_regulacja_typu_I/15_3T_I30.zcos
28-32a
Fig. 28-32
Ti=30 sek
Click “start”
28-33
Fig. 28-33
The careful control effect. Long setting time and without oscillations. Control signal s(t) is a little bigger than y(t) only.
As a little demanding teacher-s(t) from the student-y(t).

Chapter 28.6.4 I controller Ti=10 sec
Call Desktop/PID/14_regulacja_typu_I/16_3T_I10_opt.zcos
28-34a
Fig. 28-34
Ti=10 sek
Click “start”
28-35
Fig. 28-35
Better? The answer is difficult.

Chapter 28.6.5 I controller Ti=5 sec
Call Desktop/PID/14_regulacja_typu_I/17_3T_I5.zcos
28-36a
Fig. 28-36
Ti=5 sek
Click “start”<
28-37
Fig. 28-37
Terrible oscillations. The  Ti=10 sec is optimal for three-inertial if so.

Chapter 28.6.6 The integration speed Ti  exaggeration
For example Ti = 1.5 sec
Call Desktop/PID/14_regulacja_typu_I/18_3T_I1.5_niestabilny.zcos
28-38a
Fig. 28-38
Ti=1.5 sec
The oscilloscope parameters are changed to see big signals. Do you conjecture why?
Click “start”
28-39
Fig. 28-39
Beautilful instability! The amplitude is growing up to +/-infinity.

Chapter 28.7 How does I controller suppress the disturbances?
Chapter 26.7.1 Introduction
The one, two and threeinertial objects are used as before. The additional disturbance z(t)=+0.5 or z(t)=-0.5 occures at their inputs in 130 sec. The big patience is required because there is long experiment time-4 min. It’s typical that I control is very slowly and it’s very rarely (never?) used in practice.  It’s didactic important for me only.

Chapter 28.7.2 One-inertial object Ti = 16 sec and positive disturbance z(t)==+0.5
Call Desktop/PID/14_regulacja_typu_I/19_1T_I16_opt_zakl+.zcos
28-40a
Fig. 28-40
Disturbance z(t)=+0.5 appeares in 130 sec.
Click “start”
28-41
Fig. 28-41
The time chart is the same as in Fig. 28-19 up to 130 sec. But please note that there are different oscilloscope scales. The disturbance z(t)=+0.5 caused y(t) growth but the controller I started  decreasing the  s(t)  for value Δs(t)=-0.5 in steady state.
Resultat-The disturbance z(t)=+0.5 was compensated by Δs(t)=-0.5 and the y(t) is the same as before the disturbance appearance! This is a main task of the controller.

Chapter 28.7.3 One-inertial object Ti = 16 sec and negative disturbance z(t)==-0.5
Call Desktop/PID/14_regulacja_typu_I/20_1T_I16_opt_zakl-.zcos
28-42a
Fig. 28-42
Disturbance z(t)=-0.5 appeares in 130 sec.
Click “start”
28-43
Fig. 28-43
The disturbance z(t)=-0.5 was compensated by Δs(t)=+0.5

Chapter 28.7.4 Two-inertial object Ti = 15 sec and positive disturbance z(t)==+0.5
Call Desktop/PID/14_regulacja_typu_I/21_2T_I15_opt_zakl+.zcos
28-44a
Fig. 28-44
Disturbance z(t)=+0.5 appeares in 130 sec.
Click “start”
28-45
Fig. 28-45
The disturbance z(t)=+0.5 was compensated by Δs(t)=-0.5

Chapter 28.7.5 Two-inertial object Ti = 15 sec and negative disturbance z(t)==-0.5
Call Desktop/PID/14_regulacja_typu_I/22_2T_I15_opt_zakl-.zcos
28-46a
Fig. 28-46
Disturbance z(t)=-0.5 appeares in 130 sec.
Click “start”
28-47
Fig. 28-47
The disturbance z(t)=-0.5 was compensated by Δs(t)=+0.5

Chapter 28.7.6 Three-inertial object Ti = 10 sec and positive disturbance z(t)==+0.5
Call Desktop/PID/14_regulacja_typu_I/23_3T_I10_opt_zakl+.zcos
28-48a
Fig. 28-48
Disturbance z(t)=+0.5 appeares in 130 sec.
Click “start”
28-49
Fig. 28-49
Disturbance z(t)=+0.5 appeares in 130 sec.

Chapter 28.7.7 Three-inertial object Ti = 10 sec and negative disturbance z(t)==-0.5
Call Desktop/PID/14_regulacja_typu_I//24_3T_I10_opt_zakl-.zcos
28-50a
Fig. 28-50
Disturbance z(t)=-0.5 appeares in 130 sec.

28-51
Fig. 28-51
The disturbance z(t)=-0.5 was compensated by Δs(t)=+0.5 increase.
Click “start”

Chapter 28.8 I, P and PD controllers comparison
28.8.1 Introduction
These same x(t)=1 and  z(t)=+0.5  signals are acting for PD, P and I control systems.
Compare the outputs y1(t), y2(t) and y(3) signals.  Especially the steady states and setting times.

Chapter 28.8.2 Positive disturbance z(t)=+0.5.
Call Desktop/PID/14_regulacja_typu_I/25_2T_Porown_P_PD_I_zakl+.zcos
28-52a
Fig. 28-52

Click “start”
28-53a
Fig. 28-53
P-blue controller ensures steady error 9%.
PD-green controller ensures steady error 9% too but dynamic (oscillations and setting time) is much better here!
I-red controller ensures steady error 0%. This is its main advantage–>compare P and PD steady errors. But dynamic characheristics? Waste your breath! The I control is so lazy, because doesn’t exist the initial ” peak” type signal as in the P and especially in PD controller type.
Conclusion
The pure I controller is possible of course but is rarely used in practics. The integration I component is used in the PI and PID controllers.  See chapters 29 and  30.

Chapter 28.8.3 Negative disturbance z(t)=-0.5.
Call Desktop/PID/14_regulacja_typu_I/26_2T_Porown_P_PD_I_zakl-.zcos
28-54a
Fig. 28-54
z(t)=-0.5
Click “start”
28-55a
Fig. 28-55
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