## Preliminary Automatics Course

**Chapter 25. Continuous control
**

**Chapter**

**25.1 Introductio**

**n****Fig. 25-1**

**Fig. 25-1a
**The most general control system-

**G(s)**with the negative feedback.

You see:

– set point

**x(t)**

– output signal

**y(t)**(other name -process varable)

**– disturbance**

**z(t)**

– control error

**e(t)**

**Fig. 25-1b**

The

**G(s)**was divided into

**ON-OFF**controller and a object

**Go(s)**. The control signal

**s(t)**is added here.

**Fig. 25-1c**

The

**ON-OFF**controller was replaced by the continuos type controller for example

**P, PI, PD**or

**PID**. The control signal

**s(t)**doesn’t

oscillate in steady state (as

**ON-OFF**control) and this is an advantage of this type control.

**Chapter 25.2 General and typical form of the controlled object Go(s)**

**Fig. 25-2
Fig. 25-2a
**The most general

**Go(s)**form. The nominator

**L(s)**and denominator

**M(s)**are the polynomians.

**Fig. 25-2b**

The particular

**Go(s)**form.

**Fig. 25-2c**

The

**L(s)**is a single real number mostly i.e. polynomian

**n=0**,here–>

**10**

**Fig. 25-2d**

But

**L(s)=1**more often than not. The

**M(s)**is a product

**M(s)=(1+sT1)*(1+sT2)*(1+sT3)**often as for example .

**Fig. 25-3**

The **Go(s) **object input is a controlled signal **s(t)** range **0…+10V. **The **G(s) **object output is a measuring converter signal **y(t) **with the same range** y(t)=0…+10V**. It isn’t difficult to fulfil this requirement and it’s an automatic design engineer job. It means that **k=1 **for objects as **Fig. 25-2d**. This is very comfortable for people who tune **PID **controllers because **Kp**, **Td **and **Ti **controller parameters

involves all–>”controller+object”.

**About** measuring converter a bit.

The liquid temperature range in **Fig. 25-3** is **0…+70ºC **for example. It will be measured by the platinum resistance thermometer **Pt100 **range **0…+100ºC**. The **measuring converter** consists of the platinium **coil wire** in the metalic package and an **electronic
converter**. Why expensive platinum? Because its characteric is repetitive. The

**platinium coil**transforms temperature changes

**ΔT**for resistance

**changes. The**

**ΔR****electronic converter**changes

**for voltage changes**

**ΔR****. The effect is that the liquid temperature**

**ΔV****0…+100ºC**is transformed for voltage

**0…+10 V**. The

**electronic converter**as a part of the thermometer is protruding outside the tank .

**Chapter 25.3 Static and astatic Go(s) objects
**

**Chapter 25.3.1**

**Introduction**

The most general dynamic object division is:

– Static objects

– Dynamic objects

**Chapter 25.3.1 Static Go(s) objects
**Call Desktop/PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/01_czlony_statyczne.zcos

**Fig. 25-4
**You test

**3**typical

**static**objects simultaneously.

Click “start”

**Fig. 25-5**

The steady state step response

**y(t)=1**is constant here, in other words

**static**. Note that nominators

**M(s)**doesn’t contain roots

**s=0**, as

**astatic**objects–>see next.

**Chapter 25.3.2 Astatic Go(s) objects**

Call Desktop/PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/02_czlony_astatyczne.zcos

**Fig. 25-6
**You test

**3**typical

**astatic**objects simultaneously. The nominators

**M(s)**contain roots

**s=0**, double roots

**s=0**even–>

**G2(s)**!

Click “start”

**Fig. 25-7**

The steady state step response

**y(t)=1**isn’t

**static**here, in other words is

**as**

**tatic**. -it arises up to infinity! The typical attribute is that there where situations when

**y(t)**

**≠0**but

**x(t)=0**. It’s not seen at

**Fig. 25-7**but is seen at

**Fig.8-17 chapter 8**.

**Chapter 25.3.3 Static and Astatic objects with negative feedback**

**Chapter 25.3.3.1 Introduction
**

**Astatic**objects are more difficult to control than

**static**. Let’s check it.

**Chapter 25.3.3.2 Static**

Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/03_statyczny_-_sprzezenie.zcos

**Fig. 25-8
**Typical static object

**n=2**degree with the negative feedback.

Call “start”

**Fig. 25-9**

There is a steady state

**y=+0.91**with the steady error

**e=0.09**. This is a typical

**static**objects attribute-

**non-zero**steady error

**e**. This is a weakness of course and the bigger is the

**K**the lower is a steady error

**e**. But the big amplification

**K**means big oscillations and

**instabilities**sometimes even.

**Instability**when polynomian

**M(s)**degree

**n>2**follows

**Nyquist**or

**Hurwitz**benchmark –>this system is stable for all

**K**(n=2).

**Chapter 25.3.3.3 Astatic**

Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/04_astatyczny_-_sprzezenie_a.zcos

**Fig. 25-10
**There is

**s=0**root in the

**M(s)**polynomian–>

**Go(s)**object is static. There will be a very delicate and short

**x(t)**input pulse in

**5 sec**.

I know the result so I changed the oscilloscope

**y**range from

**0…+1.2**to

**-150…+150**.

Call “start”

**Fig. 25-11**

Adversity! Oscillations up to infinity. Classical

**instabilty**! How to do this system

**stable**? There are many ways. The simplest is to drop the amplification

**K**. We are very radical so

**K=0.25**now (

**K=10**before).

Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/05_astatyczny_-_sprzezenie_b.zcos

**Fig. 25-12**

Click “start”

**Fig. 25-13**

The main goal-stability was accomplished-system is

**stable**. But it’s very slowly now. Please compare

**Fig. 25-9**.

**Chapter 25.3.3.4 Static and Astatic in the common graph**

Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/06_3_4_5_-_sprzezenie_porown.zcos

**Fig. 25-14**

Click “start”

Common **x(t)** signal for **3 **inputs:

– **G1(s)** static

– **G2(s)** astatic- big **K=10** –>instable

– **G3(s)** astatic -small **K=0.25**–>stable

**Fig. 25-15
**

**Yelliow**lines-oscillations up to

**+/-infinity**for astatic

**G2(s)**. Note that oscilloscope cuts the negative signals.

**Conclusions**:

–

**Astatic**are more susceptible to

**instabilty**

–

**Null**steady control error for

**Astatic**. (provided that is

**stable**of course.)

–

**Static**is faster but exists steady control

**error**

**Chapter 25.4 Man as a continuos controller**

**Chapter 25.4.1 Introduction
**We are automatic company and our Client ordered control system project for

**Fig. 25-3**. He is a poor Client so he will stand in the

controller. The Manual Control is applied here in other words.

**Design criteria**

– The liquid doesn’t evaporate and freeze. Ok. It’s easier for us.

– Continuos type of liquid temperature control in range

**+10ºC…+70ºC**

– The change from

**+30ºC**to

**+50ºC**and

**reverse from**

**+50ºC**to

**+30ºC**for example, must be quick. This last wish requires active cooling too. The heater converts to fridge! Our resistance heater is a so-known

**Peltier**

**element**. The positive voltage heats and negative cools!

Why is it continuos and not cheap

**ON-FF**control? Client doesn’t wish temperature oscillations in steady state.

**Chapter 25.4.2 Man as a continuos controller without negative feedback
**We proposed first our Client the control without negative feedback or in other word-open loop control.

**Fig. 25-16
**The figure is similar to

**Fig. 25-3**. The only difference is a control potentiometer. There is a control range

**s(t)=-20V…+20V**instead of

**s(t)=0V…+10V**. Why? You will know it later.

**Instruction manual**

This manual is prepaired especially for our Client

**Potentiometer**

**-20V…+20V**is:

–

**400**

**mm**long

–

**Scaled**in

**ºC**–> High =

**+200**

**ºC=+20V**Middle=

**0**

**ºC**

**=0V**Down=

**-200**

**ºC=-20V 1mm=1ºC**

Please note that the

**s(t)**range

**-20V…+20V**is

**4**times bigger than

**y(t)**range

**0…+10V**but the static object parameter as

**k=y/s=1**is the same, because input

**Δs(t)=+1V**increase cause static output

**Δy(t)=+1V**increase too! We have only possibility to set bigger input

**s(t)**range.

I remind that the amplifier and heater are so choosed by the automatic design engineer that:

The assumed ambient temperature

**=0**

**ºC****High=+200**

**ºC=+20V**potentiometer position gives

**+200**liquid temperature in a steady state

**ºC****Middle=0**

**ºC=0V**position gives

**0**liquid temperature in a steady state

**ºC****Low =-200**

**ºC=-20V**potentiometer position gives

**-200**liquid temperature in a steady state.

**ºC**I remind that:

– our liquid doesn’t evaporate and freeze

-our

**heater/cooler resistance**is a

**Peltier element**–>+votlage haeats and -voltage cools–>see

**Chapter 25.4.1**end.

**Client** tries to set **y(t)=+50ºC** output temperature.- **First** atempt

The ambient temperature=**0 ºC**

He is very careful at the begining and he tries to set liquid temperature

**y(t)=+50ºC**. What does he do? He simply set the potentiometer slider at

**50 mm–>+50**and waits. The liquid

**ºC****y(t)**temperature will attain

**+50**after a long time. It’s

**ºC****absolute**open loop control! Client doesn’t observe the liquid temperature

**y(t)**! He trust me as as a automatic project designer that

**50 mm**to the left slider position causes

**+50**. In other words. The output

**ºC****y(t)**has no influence for input

**x(t)**!

**Client** tries to set **y(t)=+50ºC** output temperature.- **S****econd** atempt**
**He observes this

**y(t)**temperature at the converter

**meter**.

Client set the potentiometer slider

**s(t)=+50**and he is sure that the steady temperature will be

**ºC****y(t)=+50ºC**after some time. But what is this? The output temperature is

**y(t)=+63ºC**! He says to me. Mr automatic. Your system is down the tube. I will not pay. But the system is ok. There is

**y(t)=+63ºC**instead of

**y(t)=+50ºC**because

– The ambient temperature is

**+12ºC**now and not

**0ºC**as by the first atempt.

– There are probably other reasons for examply:

– there is less liquid in the tank now

– humidity

– the liquid has a little changed physical parameters here

– the physical model of the object isn’t absolutely accurate

– …

We can multiply the reasons but the main conclusion is one only.

The process operator must observe the output temperature on the measuring converter meter and make appropriate decisions

This decision are the appropriate slider potentiometer movements. The output **y(t)** has a influence for the control input **s(t)**. So this is a an example of the system with the negative feedback! This control loop is closed through Mr Operator! Lets’s analyze this problem in the next subchapter.

**Chapter 25.4.3 Man as a continuos controller with negative feedback**

**Fig. 25-17
**Man as controller

**Fig. 25-17a**

Operator observes all the time the output

**y(t)**and the set point

**x(t)=+50ºC=+5V**and tries to keep the state

**x(t)=y(t)**up. He compares

**x(t)**and

**y(t)**in mind, calculates the error

**e(t)=x(t)-y(t)**and makes decisions.

For example

**e(t)>0**–>

**s(t)=+200ºC=+20V**

e(t)<0–>

e(t)<0

**s(t)=-200ºC=-20V**

Operator is an

**ON-OFF**controller now–>see previous chapter. This control is simple, some say primitive and the effect are

**y(t)**oscillations round

**+50ºC**.

How to dump the oscillations? Let’s do

**continuos**control. We assume

**0ºC**ambient temperature.

There is a initial “middle” potentiometer position

**x(t)=0ºC**. The initial temperature is y

**(t)=0ºC**too. What’s the decision? Do nothing. By the way. Every healthy man likes this type of activity. The set point

**x(t)**jumps up to

**x(t)=+70ºC**

**x(t)=0ºC**suddenly. Operators reaction is

**s(**

**t)=+20V=+200ºC**. Why not

**s(**

**t)=+7V=+70ºC**? The ambient temperature is

**0ºC**so there will be

**y(**

**t)=+70ºC**after some time. But we want to make procees more quick or more dynamic in other words. The decision is simply. We give all the disposal power

**s(t)=+200ºC=+20V at**the beginning. The output

**y(t)**increases very quickly now, and we decrease the

**s(t)**gradually. We observe the output

**y(t)**and try to reach the temperature

**yt)=+70ºC**.

The main conclusions to control the objects are:

– set point

**x(t)**and output

**y(t)**ranges are mostly equal here

**0…+100ºC**

– the control

**s(t)**range is mostly bigger than

**x(t)**and

**y(t)**range. It enables more dynamic control. The lower

**s(t)**range than

**0…+100ºC**is forbidden because some

**y(t)**states are unavailable!

**Fig. 25-17b**

The man is replaced by the continuos controller. The control algorithm is hidden in the operator mind! It’s possible to “copy” this

algorithm to microprocessor controller memory!

**Fig. 25-17c**

The above mentioned algorithm may be surprisingly simply. For example

**proportional**controller.

**1 step**Calculate control error

**e(t)=x(t)-y(t)**

**2 step**Calculate control signal

**s(t)=Kp*e(t)**

3 stepReturn to

3 step

**1 step**

**Algorithm ensures the output signal**

**y(t)**almost the same as

**x**

**(t)**after some time. The

**PI**and

**PID**algorithms are a little more complicated but this is not a problem for microprocessor controllers.

**Chapter 25.5 You as a continuos controller without disturbances****
**Do you remember

**Chapter 24.2 ON-OFF control hand type**? The problem is similar but you will be a

**continuos**controller now. You observe the set point signal

**x(t)**and you swing potentiometer

**s(t)**slider to assure the state

**x(t)=y(t)**. It ‘s impossible to fulfil this condition absolutely exactly but try to do it!

Call Desktop PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/07_Ty_jako_regulator_ciagly.zcos

Fig. 25-18

**Fig. 25-18a**

Full diagram

**Fig. 25-18b**

The simplified diagram with you as Mr Controller.

Click “start”

**Fig. 25-19**

There are opened **3 **windows:

– **main **with the diagram. You can observe the digital meters

– **oscilloscope**

– **control potentiometer** **Tk scale**

**Note:**

The one window may override the others sometimes. Move these windows to see all and the digital meters. You have **20 sec **up to **x(t) 0 –> +70ºC** step.

Observe

**x(t)**and

**y(t)**and try to control leftmouse

**s(t)**that

**y(t)**will follow

**x(t)**–>

**x(t)=y(t)**. Be prepaired that

**x(t)**will drop

**in**

**+70ºC**–>**+30ºC****70 sec**! Are you a wonderful online gamer? You will have no problem to control as in

**Fig. 25-20**

**Fig. 25-20**

You give big as posible

**s(t)=**control signal in

**+200ºC****20 sec**. Note that

**s(t)**is much bigger than

**s(t)=**(+7V) which assures steady

**+70ºC****y(t)=**temperature but after a long time! You decrease

**+70ºC****s(t)**then, but you try to keep

**s(t)**a little over the

**x(t)=**. Your

**+70ºC****s(t)**is

**+/-**near the set point

**x(t)=**. Why near and not exactly

**+70ºC****x(t)=y(t)**? Because you aren’t a superman–> the

**red**

**y(t)**and

**black**

**x(t)**aren’t common lines*–>see

**NOTE**. This big

**s(t)**control signal assure shorter control time!

There is

**x**

**(t)**drop of the signal

**–>**

**+70ºC****in**

**+30ºC****70 sec**. Your reaction is decreasing

**s(t)=**(active freezing!) and inrcesing it slowly up to near

**-200ºC****s(t)=**. Do you see anlaogy to situation in

**+30ºC****20 sec**?

**Conclusions
**

**1.**Set poin

**x(t)**and output

**y(t)**signals ranges are the same! In our example

**x(t)=y(t)=0…+100**

**ºC**=0…+10V**2.**Control signal

**s(t)**range should be

**wider**than

**x(t)**and

**y(t)**signals ranges. It enables better system dynamic. These conclusions are valid for real full automatic control systems.

***NOTE
**The

**red**

**y(t)**and

**black**

**x(t)**are common lines. This is an ideal situation

**x(t)=**

**y(t)**-impossible for

**P**control type. This problem will be described later. But how to reduce control error

**e(t)=x(t)-y(t)**to

**zero**in hand control in

**60 sec**when is steady state?

Imagine that you have

**error meter**and we have possiblity to increase/deacrease the

**s(t)**by the

**additional device**.

You increase/deacrease

**s(t)**by

**+**or

**– push button**as voice

**push button**in

**TV pilot**type. The next imagination is that

**x(t)=**

**all the time. ( not as in**

**+70ºC****Fig. 25-20)**

Your algorithm is simple in this steady state:

**step 1**

if e(t)>0–>

if e(t)>0

**+**

**push button**for a short moment

**if e(t)<0**–>

**-push button**for a short moment

**step 2**

wait for a new

**y(t)**steady state

step 3

step 3

**if**

**|e(t)|<+0.01**

**ºC**go to END else return to step 1**End**

**|**

**e(t)|<+0.01**This algorithm enables to realize at choice small error. Theoretically

**ºC**

**e(t)=0!**

Your algorithm is Integral type now!!!

You work as a **P **type controller up to **60 ****sec **. You work as **I **controller after **60 sec**! Your work is similar to **PI **cotroller type for all the time range–>see **chapter 29**.

**Chapter 25.6 You as a continuos controller with disturbances
Chapter 25.6.1 Positive disturbance**

**The additional heater is a source of the**

**z(t)=**

**disturbance. Your reaction will is obviously–>the control signal**

**+20ºC****s(t)**decreasing. You observe the

**y(t)**signal and you try to assure

**x(t)=y(t)**state, regardless of the disturbance. You are a real Mr Controller!

Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/08_Ty_jako_regulator_ciagly_+_zakl.zcos

**Fig. 25-21**

**Fig. 25-21a**

Full diagram

**Fig. 25-21b**

The simplified diagram with you as Mr Controller.

Click “start”

**Fig. 25-22
**Control and positive disturbance

**z(t)=+20°C.**

There aren’t disturbances up to

**10**

**0 sec**and the diagram is similar to

**Fig. 25-20**. The positive disturbance

**z(t)=+20°C**appeares in the

**100 sec**and your reaction-imput power

**s(t)**decreasing is obviously. Note that

**s(t)**decreasing circa

**-20°C**is (almost) compensating the

**z(t)=+20°C**disturbance in steady state and your reaction for disturbance

**z(t)**is more gentle than for set point

**x(t)**in

**20 sec.**Why? Compare the reasons of your reactions in both cases-the control error

**e(t)**!

**Chapter 25.6.2 Negative disturbance
**The additional cooler (coil with cool medium or

**Peltier**element) is a source of the

**z(t)=**

**disturbance. Your reaction will be obviously–>the control signal**

**-20ºC****s(t)**increasing. You observe the

**y(t)**signal and you try to assure

**x(t)=y(t)**state, regardless of the disturbance.

Call Desktop PID/14_regulacja_ogolnie/9_Ty_jako_regulator_ciagly_-_zakl

**Fig. 25-23**

The

**z(t)=**

**-20°C**(coil with the cool flow or

**Peltier**element) is a disturbance source.

Click “start”

**Fig. 25-24
**Control and negative disturbance

**z(t)=-20°C.**

The negative disturbance

**z(t)=-20°C**appeares in the

**100 sec**and your reaction-imput power

**s(t)**increasing is obviously. Note that

**s(t)**increasing circa

**+20°C**is (almost) compensating the

**z(t)=-20°C**disturbance in steady state and your reaction for disturbance

**z(t)**is more gentle than for set point

**x(t)**in

**20 sec.**