## Preliminary Automatics Course

**Chapter 26. P control
**

**Chapter 26.1 Introduction**

Don’t be afraid of the chapters

**26…30**big sizes! Majority of them are repetitive.

Controller

**P**(other name-

**Proportional**controller) is the simplest of the

**PID**controllers family. It has one adjustment only-gain

**Kp**.

We will test controller

**P**with the objects:

–

**one-inertial**

–

**two-inertial**

–

**three-inertial**

**Chapter 26.2 Closed loop transfer functon Gz(s) and steady state gain Kz
**It’s a short reminder of the

**Chapter 20 How does feedback work?**

**Fig. 26-1**

**Fig. 26-1a**-open loop diagram

**Fig. 26-1b**-closed loop diagram

Theory confirms the practice. The steady state gain

**Kz**in particular!

**Note**-there are different oscilloscope parameters–> The real close loop dynamic is much better than you see above!

The formula K*e(t)=y(t) is fulfilled in steady state. It’s very important to understand the feedback principle! All the Chapter 20 is a justification.

**Chapter 26.3 Controller P with the one-inertial object**

**Chapter 26.3.1 Introduction
**We said in the

**20 Chapter**that

**static objects**are equipped with the steady gain

**K=1**mostly. It means that

**Kp**is a steady state gain of all the open loop object (objest+controller). It’s easier to adjust

**Kp**parameter now, more generally-all

**PID**parameters.

**Chapter 26.3.2 One-inertial unit as a open loop **

Call Desktop/PID/12_regulacja_typu_P/01_obiekt_1T.zcos

**Fig. 26-2
K=1 **and

**T=10 sec**

Click “start”

**Fig. 26-3**

The

**K**and

**T**of the

**one-inertial**unit calculation.

**Chapter 26.3.3 Controller P with Kp=10
**Call Desktop/PID/12_regulacja_typu_P/02_1T_Kp10.zcos

**Fig. 26-4**

Controller

**P**executes very simple algorithm. It calculates the control error

**e(t)=x(t)-y(t)**and calculates the control signal

**s(t)**then:

**s(t) = Kp*e(t) = Kp*[x(t) – y(t)]**. Here

**Kp=10**.

Click “start”

**Fig. 26-5**

The RC circuit is in the steady state when

**s(t)=y(t)**. The steady state with the feedback is analogously. All

**static**and

**stable**units with the feedback are going to the state

**K*e(t)=s(t)=y(t)**. The example is

**Fig. 26-5**. The reason of the signal

**y(t)**growth or decline is a

difference

**s(t)-y(t)**! It’s very important to understand the

**feedback**. When you have problems, return to

**chapter 20**!

The diagram confirms the theoretical

**Kz=0.91**.

The oscilloscope cuts the

**y(t)**on the

**2**level. Let’s change the oscilloscope parameters to see the whole

**y(t)**signal

Click “start”

**Fig. 26-6**

The closed loop

**s(t)**signal is much bigger than in open loop! See

**Fig. 26-6**. It’s the reason that

**y(t)**changes faster the in closed loop. The bigger is

**Kp**the more dynamic is the system. The steady state is when

**y(t)=s(t)**.

Let’s compare

**open**and

**closed loop**systems in the common diagram.

Call Desktop/PID/12_regulacja_typu_P/04_1T_Kp10_porownanie_z_otwartym.zcos

**Fig. 26-7**

Let’s compare open

**y1(t)**and closed loop

**y2(t)**diagrams.

Click “start”

**Fig. 26-8**

The closed loop system

**y2(t)**is

**11**times faster than the opened loop system

**y1(t)**. But there is a steady state error

**. I hope you don’t come to splendid conclusion “Open loop system is slowly but it enables steady state error**

**e=1-0.91=0.09****e=0**because

**x=y”**. Ok. You are right but the open loop system isn’t resistant to disturbances!

**Chapter 26.3.4 Controller P with Kp=100**

Call Desktop/PID/15_regulacja_typu_P/05_1T_Kp100.zcos

**Fig. 26-9**

**Kp****=100**

Click “start”

**Fig. 26-10
**System is faster and the steady state error is very small. The red

**y(t)**almost squares with with black

**x(t)**! The theory confirms too–>

**Kz=0.99**i

**Tz=0.1 sec**. The oscilloscope cuts the

**y(t)**on the

**2**level. Let’s change the oscilloscope parameters to see the whole

**y(t)**signal.

Click “Start”

**Fig. 26-11**

You see the whole

**s(t)**control signal. Very, very big! It may be a technical problem. An example. The furnace require

**10 kW**steady power, but

**1MW**at the start! It assures fast

**y(t) response.**The crazy engineer will project this power only. The wise engineer gives

**30 kW**. The system will be not so fast then. We return to this subject in the

**Chapter 34 Nonlinearity influence for regulation**.

**Chapter 26.3.5 Summary P controller with the one-inertial object
**It’s easy to control. Big

**Kp**causes fast responses and small steady error. We can prove (Nyquist or Hurwitz) that

**one-inertial**object with the feedback is always stable. By big

**Kp**too!

**Chapter 26.4 Controller P with the two-inertial object**

**Chapter 26.4.1 Introduction
**We will repeat these same experiments but with the

**two-inertial object**.

**Chapter 26.4.2 Two-inertial unit as a open loop**

Call Desktop/PID/12_regulacja_typu_P/07_obiekt_2T.zcos

**Fig. 26-12**

**K1=K2=1** and **T1=3 sec**, **T2=5 sec**.

Click “start”

**Fig. 26-13
**There is a typical for multiinertial unit so-called point of inflection. It’s easy to obtain steady state gain

**Ku**. But it isn’t so nice with the

**T1**and

**T2**parameters. There are the methods of course but let’s give it a rest.

**Rozdz. 26.4.3 Controller P with Kp=10**

Call PID/12_regulacja_typu_P/08_2T_Kp10.zcos

**Fig. 26-14**

The **two-inertial **object is a only difference of the **Fig. 26-4**

Click “start”

**Fig. 26-15
**There are oscillations now. The steady state is in circa

**t=35 sec**when:

**1. s(t)=y(t)**

2.there is

2.

**“no movement”**(all derivatives are null)

**Note**

There are moments when

**s(t)=y(t)**(circa t=5 sec for example) but there is “movement” here! So it isn’t steady state.

Let’s look the whole

**s(t)**signal.

Call Desktop/PID/12_regulacja_typu_P/09_2T_Kp10_pelen_widok.zcos

This is the same block diagram but there a different oscilloscope parameters.

Click “start”

**Fig. 26-16**

The

**s(t)**control signal is crazy now. There is big steady error

**e(t)=0.09**here and in

**Fig. 26-15**. We can minimize this error (but not

**e(t)=0!)**by

**Kp**increasing.

**Chapter 26.4.4 Control Kp=100**

Call Desktop PID/12_regulacja_typu_P/10_2T_Kp100.zcos

**Fig. 26-17**

Click “Start”

**Fig. 26-18
**There is a steady error

**e(t)=0.01**but the oscillations and long control time (

**t=45 sec**. circa) are indecent!

**Chapter 26.4.5 Summary P controller with the two-inertial object
**It’s more difficult to control than a

**one-inertial**object. There are oscillations now but we can prove (Nyquist or Hurwitz) that

**two-inertial**object with the feedback is always stable. By big

**Kp**too!

**Chapter 26.5 Controller P with the three-inertial object**

**Chapter 26.5.1 Three-inertial unit as a open loop
**Call Desktop/PID/12_regulacja_typu_P/11_obiekt_3T.zcos

**Fig. 26-19
**The one-inertial with a small time constant

**T3=0.5sec**is added to the

**two**

**-inertial**(

**Fig. 26-12**). We have a

**t**

**hree-inertial unit**now. The change is small. I expect that system with the feedback behaviour will be simillar as for

**two**

**-inertial**. Let’s check it.

Click “Start”

**Fig. 26-20**

The response is similar to

**Fig. 26-13**, but a bit more lazy. The

**To**parameter increased.

**Chapter 26.5.2 Control Kp=100**

Call Desktop/PID/12_regulacja_typu_P/12_3T_Kp10.zcos

**Fig. 26-21**

What will be the influence of this small **T3=****0.5 sec**?

Click “start”

**Fig. 26-22
**Terrible oscillations and a long control time! The increased

**Kp**will worse the situation probably.

**Chapter 26.5.3 Control Kp=30**

Call Desktop/PID/12_regulacja_typu_P/13_3T_Kp30_niestabilny.zcos

**Fig. 26-23**

Click”start”

**Fig. 26-24
**The system is instable! By the way. Has a calculated gain

**Kz=0.97**a sense? It has a bit of sense.

It’s a constant component of the oscillations! Let’s look not cutted

**s(t)**control signal yet.

Call Desktop/PID/12_regulacja_typu_P/14_3T_Kp30_niestabilny_pelen_widok.zcos

The block diagram is the same as **Fig. 26-23 **but the oscilloscope parameters are different now.

Click “start”

**Fig. 26-25
**The increasing

**s(t)**and

**y(t)**amplitudes are seen. Note that

**s(t)**amplitudes are bigger than

**y**

**(t)**.

**Chapter 26.5.4 Summary P controller with the three-inertial object
**It’s more difficult object to control than the previously. The

**Amplitude Phase Characteristic**of the

**three-inertial**objects crosses

**3**quadrants–> This object may be instable–>see Nyquist benchmark.

**Chapter 26.6 How does P controller suppress the disturbances?**

**Chapter 26.6.1 Introduction
**The disturbances suppression is the main job of all the controllers. The earlier experiments lasted

**1 min**. The next will be

**2 min**. The step

**x(t)**is the first and the step type disturbance

**z(t)**is the next in

**70 sec**. Imagine that you put the additional heater or cooler to the liquid tank. The “healthy” system should compensate this disturbance by the additional cooling or heating. “Healthy” means that the

**Kp**is good adjusted.

This

**2 min**isn’t too much but some people may be nervous.

**Scilab**and all simulation oriented language have a good option for these nervous people–>time shortening. The

**1 sec**is shortened up to

**0.1 sec**. I don’t recommend this method for automatic beginners. The main advantage of this course is a intuition and time shortening isn’t good for it.

How to shorten the time from

**120 sec**up to

**12 sec**? Make steps

**1,2**and

**3**under.

**Fig. 26-26
**Time shortening.

We will test controller **P **with the objects :

– **one-inertial
**–

**two-inertial**

–

**three-inertial**

The input signals are

–

**x(t)**as usual

–

**z(t)**positive or negative disturbance (heating or cooling)

**Chapter 26.6.2 One-inertial object, Kp=10 and positive disturbance z(t)=+0.2**

Call Desktop/PID/12_regulacja_typu_P/15_1T_Kp10_zakl+.zcos

**Fig. 26-27
**

**z(t)=+0.2**The additional heater is put in the liquid tank.

Click “start”

**Fig. 26-28**

The

**P**controller reaction for heating is right. The

**z(t)=+0.2 increasing**is compensated by the control signal

**s(t) decreasing.**This compensation isn’t ideal–>

**Δs(t)=-0.182**instead

**Δs(t)=-0.2**. It’s typical for

**Proportional Controllers.**The bigger is

**Kp**the better is compensation.

**But the ideal compensation isn’t possible for P controllers**!

**You see the small disturbance**

**z(t)=+0.2**influence for the

**y(t)**signal. There is a addtional

**+0.2 heating**but the otput signal increases only about

**+0.018**! The absolute compensation is possible when

**integration**algorithm is included. We will speak about it later.

**Chapter 26.6.3 One-inertial object, Kp=10 and negative disturbance z(t)=-0.2**

Call Desktop/PID/12_regulacja_typu_P/16_1T_Kp10_zakl-.zcos

**Fig. 26-29**

The negative dosturbance **z(t)=-0.2 **(cooling)

Click “start”

**Fig. 26-30
**The

**P**controller reaction for heating is right. The

**z(t)=-0.2 decreasing**is compensated by the control signal

**s(t) increasing.**The compensation isn’t ideal too. The situation will be better when the

**Kp**parameter is increasing. For example

**Kp=100**.

**Chapter 26.6.4 One-inertial object, Kp=100 and positive disturbance z(t)=+0.2
**Call Desktop/PID/12_regulacja_typu_P/17_1T_Kp100_zakl+.zcos

**Fig. 26-31**

Click “start”

**Fig. 26-32**

Beautifully! The

**y(t)**stands almost a lancers lanca! Regardless of

**z(t)**disturbance! There is a small

**disturbance**

**z(t)**influence, but it’s not seen at these oscilloscope parameters.

**Chapter 26.6.5 One-inertial object, Kp=100 and negative disturbance z(t)=-0.2**

Call Desktop/PID/12_regulacja_typu_P/18_1T_Kp100_zakl-.zcos

**Fig. 26-33**

Click “start”

**Fig. 26-34**

No comments.

**Chapter 26.6.6 Two-inertial object, Kp=10 and positive disturbance z(t)=+0.2**

Call Desktop/PID/12_regulacja_typu_P/19_2T_Kp10_zakl+.zcos

**Fig. 26-35
**Two-inertial object,

**Kp=10**and positive

**z(t) = +0.2**

Click “start”

**Fig. 26-36**

The experiment is an example that response for

**x(t)**is different than for disturbance

**z(t)**! The response for

**z(t)**is better than for

**x(t)**! Many people are adjusting the

**P**(more generally

**PID)**controler when

**x(t)**is an input signal. They are glad that is ok but they forget the above effect!

When

**x(t)**changes often–>the input

**x(t)**is most important, otherwise the disturbance

**z(t)**.

**Chapter 26.6.7 Two-inertial object, Kp=10 and negative disturbance z(t)=-0.2**

Call Desktop/PID/12_regulacja_typu_P/20_2T_Kp10_zakl-.zcos

**Fig. 26-37**

**Fig. 26-38
**The

**P**controller reaction is right. It tries to compensate the negative (cooling)

**z(t)=-0.2**disturbance by the control increase signal (heating)

**s(t)**.

**Chapter 26.6.8 Three-inertial object, Kp=10 and positive disturbance z(t)=+0.2**

Call Desktop/PID/12_regulacja_typu_P/21_3T_Kp10_zakl+.zcos

**Fig. 26-39**

Click”start”

**Fig. 26-40**

No comments

**Chapter 26.6.9 Three-inertial object, Kp=10 and negative disturbance z(t)=-0.2**

Call Desktop/PID/12_regulacja_typu_P/22_3T_Kp10_zakl-.zcos

**Fig. 26-41**

Click “start”

**Fig. 26-42**

No comments

We will not test the **Kp=30** (and more) because the system is instable then (see **Fig****. 26-24** ** **and **26-25**)

**Chapter 26.7 Why the closed loop system is better than the open loop?
**For majority of readers the question is like “Who want you be? Young rich and healthy or old, poor and ill?” So the rest of this subchapter is for a doubtful minority.

Let’s compare the open and closed loop system!

Call Desktop/PID/12_regulacja_typu_P/23_zakl_otw_zamkn_por.zcos

**Fig. 26-43**

These same signals

**x(t)**and

**z(t)**are acting for:

– open loop system (higher object.)

– closed loop system (lower object)

Click “start”

**Fig. 26-44**

Come to your own conclusions.

The fault of the

**P**control type is

**no null steady error**signal and that they aren’t absolutely resistant to

**z(t)**disturbances. The

**PI**and

**PID**controllers are better in this area. We will discuss it later.

** **