Preliminary Automatics Course
Chapter 25. Continuous control
Chapter 25.1 Introduction
The most general control system-G(s) with the negative feedback.
– set point x(t)
– output signal y(t) (other name -process varable)
– disturbance z(t)
– control error e(t)
The G(s) was divided into ON-OFF controller and a object Go(s). The control signal s(t) is added here.
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)
The most general Go(s) form. The nominator L(s) and denominator M(s) are the polynomians.
The particular Go(s) form.
The L(s) is a single real number mostly i.e. polynomian n=0,here–>10
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 .
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
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 ΔR changes. The electronic converter changes ΔR for voltage changes ΔV. The effect is that the liquid temperature 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
You test 3 typical static objects simultaneously.
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
You test 3 typical astatic objects simultaneously. The nominators M(s) contain roots s=0, double roots s=0 even–>G2(s)!
The steady state step response y(t)=1 isn’t static here, in other words is astatic. -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 22.214.171.124 Introduction
Astatic objects are more difficult to control than static . Let’s check it.
Chapter 126.96.36.199 Static
Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/03_statyczny_-_sprzezenie.zcos
Typical static object n=2 degree with the negative feedback.
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 188.8.131.52 Astatic
Call PID/11_Obiekt i regulator oraz człowiek jako regulator ciągły/04_astatyczny_-_sprzezenie_a.zcos
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.
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
The main goal-stability was accomplished-system is stable. But it’s very slowly now. Please compare Fig. 25-9.
Chapter 184.108.40.206 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
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
Yelliow lines-oscillations up to +/-infinity for astatic G2(s). Note that oscilloscope cuts the negative signals.
– 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.
– 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.
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.
This manual is prepaired especially for our Client
Potentiometer -20V…+20V is:
– 400 mm long
– Scaled in ºC–> High =+200ºC=+20V Middle=0ºC=0V Down=-200ºC=-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=0ºC
High=+200ºC=+20V potentiometer position gives +200ºC liquid temperature in a steady state
Middle=0ºC=0V position gives 0ºC liquid temperature in a steady state
Low =-200ºC=-20V potentiometer position gives -200ºC liquid temperature in a steady state.
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=0ºC
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–>+50ºC and waits. The liquid y(t) temperature will attain +50ºC after a long time. It’s 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 +50ºC. In other words. The output y(t) has no influence for input x(t)!
Client tries to set y(t)=+50ºC output temperature.- Second atempt
He observes this y(t) temperature at the converter meter.
Client set the potentiometer slider s(t)=+50ºC and he is sure that the steady temperature will be 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
– 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
Man as controller
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.
e(t)>0 –> s(t)=+200ºC=+20V
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!
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!
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 step Return to 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
The simplified diagram with you as Mr Controller.
There are opened 3 windows:
– main with the diagram. You can observe the digital meters
– control potentiometer Tk scale
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 +70ºC –> +30ºC in 70 sec! Are you a wonderful online gamer? You will have no problem to control as in Fig. 25-20
You give big as posible s(t)=+200ºC control signal in 20 sec. Note that s(t) is much bigger than s(t)=+70ºC (+7V) which assures steady y(t)=+70ºC temperature but after a long time! You decrease s(t) then, but you try to keep s(t) a little over the x(t)=+70ºC. Your s(t) is +/- near the set point x(t)=+70ºC. Why near and not exactly 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–>+30ºC in 70 sec. Your reaction is decreasing s(t)=-200ºC (active freezing!) and inrcesing it slowly up to near s(t)=+30ºC. Do you see anlaogy to situation in 20 sec?
1. Set poin x(t) and output y(t) signals ranges are the same! In our example x(t)=y(t)=0…+100ºC=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.
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)=+70ºC all the time. ( not as in Fig. 25-20)
Your algorithm is simple in this steady state:
if e(t)>0–>+push button for a short moment
if e(t)<0–>-push button for a short moment
wait for a new y(t) steady state
if |e(t)|<+0.01ºC go to END else return to step 1
This algorithm enables to realize at choice small error. Theoretically 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)=+20ºC disturbance. Your reaction will is obviously–>the control signal 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
The simplified diagram with you as Mr Controller.
Control and positive disturbance z(t)=+20°C.
There aren’t disturbances up to 100 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)=-20ºC disturbance. Your reaction will be obviously–>the control signal 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
The z(t)=-20°C (coil with the cool flow or Peltier element) is a disturbance source.
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.