Preliminary Automatics Course
Chapter 34. Nonlinearity influence for regulation
Chapter 34.1 Introduction
The most important influence for the control process is a saturation of the power amplifier. The more general name of the power amplifier is a final control element or actuator. This element transforms control signal for power signal. Please note that all the remaining signals e.i. set point x(t), measured output signał y(t) or control signal y(t) doesn’t require much energy. And the nonlinearity of the power amplifier is a saturation. Thus the chapter title should be more unambiguous “The final control element saturation influence for the control process”
I propose the reminder about transfer functions now. When you use the transfer it means that you are living in the linear objects world. The y(t) output is proportional to x(t) input in steady states.
The linear object characteristic in steady state.
But non steady state? There are ruling linear differential equations now. All the coefficients are constant, for example
Linear differential equation
Non-linear differential equation example. The red circles are the bugs. The reason that differential equation isn’t linear.
The linear differential equation Fig. 34-2 and a/m G(s) transfer function are equivalent. I propose you to associate differential equation coefficients Fig. 34-2 with transfer function parameters. Have you any problems? Go to Fig. 19-8 chapter 19.
We used mainly the linear objects in this course. And only the transfer function Go(s) makes a sense for these objects. The linear differential equation theory is deeply and good established now. It enables problems solving as stability, PID controllers adjusting etc. But there is one condition only. The object must be linear and we forgot it often. The one nonlinear element can break down–> all the system is nonlinear.
Chapter 34.2 Nonlinear elements
Chapter 34.2.1 Introduction
The most every real object nolinearity is saturation. It’s +/-15V for amplifiers or maksimal valve flow. Sorry for unscientific names of the typical elements.
Fig. 34-5a – “progressively growth with the saturation”
Rys. 34-5b – “linear with the saturation”
Rys. 34-5c – “more and more slowly with the saturation”
Rys. 34-5d – histeresis
Rys. 34-5e – “linear with the dead band and saturation”
We will test only 2 elements. The input is a black sine wave x(t) and the output is a red y(t).
Chapter 34.2.2 The linear element with the saturation
x(t) is a sine wave
The saturation when x(t)=+/-0.5.
It acts as a amplifier k=1 before saturation and y(t)=x(t)
Chapter 34.2.3 The linear element with the dead band and saturation
x(t) is a sine wave.
The dead band is for x(t) -0.5….+0.5.
Chapter 34.3 How does the ideal control system work?
E.i the linear control system. It’s most popular experiment in this course but it does no harm repetition.
Typical PID control system with 2-inertial object.
We expect steady state y(t)=x(t)=+1.
It may be temperature control system and +1 means +100°C.
The integral I component made null control error, as usual. The process is very fast because of the big sPID(t) control signal amplitudes. The driver uses strongly gas pedal and brake pedal. The effect is a dynamic drive. The controller job is the same. The gas pedal is a big positive sPID(t) and the brake pedal is a big negative sPID(t).
Let’s change the oscilloscope parameters and observe the real sPID(t) (not “cutted”) signals.
The block diagram is the same but there are other oscilloscope parameters. You will see all the sPID(t) signal.
The main x(t), y(t) signals are small and hardly visible now but you see all the sPID(t) control signal majesty! Compare please x(t), y(t) circa 0…+1 range and sPID(t) range -5…+160. This is a reason of a very good y(t) dynamic–>”brake and gas” control algorithm.
The real control system has power limits. Any design engineer will not give 160 kW power when steady state process requires 1 kW only! The negative power (cooling for example) is also rarely used.
There is a difference between theory and practise*. The real system have the control signals limits. There are saturations 0…+1.5 instead of -5…+160. The range is even more narrow 0.2…+1.5! The power off causes y(t)temperature decrease up to ambient temperature +20°C and not to 0°C.
The linear systems are smart, relatively easy and there are answers for many general questions. Stability, Hurwitz etc… The real systems are different. The problem is how big is this difference.
*I have seen the poster in the worthy control systems company
Practitioner- All is operating but he he don’t know why
Theoretician-Nothing is operating but he knows why
Our company combines practice and theory. Nothing is operating and we don’t know why. I took to them quickly.
Chapter 34.4 How does the real control system work?
Chapter 34.4.1 Introduction
I try to argue that the main nonlinearity in the real systems is a power amplifier-other names final element or actuator.
We will analyse power amplifier, PID controller and the object.
Chapter 34.4.2 Power amplifier with the saturation
The power amplifier is just a difference between Fig. 34-10. This is a linear unit with the saturation Fig. 34-5b. It makes that all the object is nonlinear. We can’t calculate the closed loop transfer function for example!
Chapter 34.4.3 PID controller
This almost a linear unit and its transfer function is:
The appropriate Fig.34-13 controllers parameters are:
Ti =7 sec
Note This is a real differentiation, not ideal!
Comment to word “almost a linear unit”
The PID controllers are mostly the microprocessor type devices. They can calculate every sPID(t) control signal digital value–>there are no limits here. Some people will complain that this value is quantum timed. But these quantum times are very small compared to object inertia and PID is for us a pure tranfer function Gpid(s).
Chapter 34.4.4 G(s) Object
Object is a furnace, rectifying column or rocket. There are weaker linearity arguments than PID. The most popular static object
characteristic is Fig. 34-5c type–> “more and more slowly with the saturation”.
What’s the conclusion?
The G(s) nominator e.i. unit static gain k=1 only for null working point, then is lower. The dynamic parameters T time constants also may change due to working point. It means that the real parameters are “fuzzy”. They change in the some range.
Fig. 34-15a– the ideal G(s) 2-inertial object with parameters k=1, T1=10 sec and T2=5 sec.
Fig. 34-15b-The real parammeters are variable in the ranges. These parameters depend on working point and our identification methods. The real parameter values are inside the ranges k=0.95…1.05, T1=9.5…10.5 sec and T2 4.5…5.5 sec.
Many objects, especially multi-inertial may be approximate as Substitute Transfer Functions with K, T and To parameters.
The first question. What for? The better is exactly than approximately model. It’s better to be rich, young and healthy than poor, old and ill.
Firstly– It’s easier to experimentally determine K, T and To parameters than the real multi-inertial object parameters.
Secondly– There are ready optimal PID parameters dependend of Substitute Transfer Functions K, T and To parameters–>chapter 31
The Substitute Transfer Function was discussed in chapter 31. I remind the step response.
Substitute Transfer Function step response.
Chapter 34.4.5 Conclusions
– PID controller is a linear unit Gpid(s)–>Fig. 34-14.
– Object is a linear object too but with some toleration dose.
– Power amplifier is a nonlinear unit–>linear with the saturation. This unit (other names- final control element, actuator). This element “perverts” all the closed system unit, makes it nonlinear.
The actuator exists always (almost) in closed loop systems. It means that all (almost!) closed loop systems are nonlinear and all the theory (G(s) units configuration,Nyquist, Hurwitz…) isn’t necessary. Fortunately the situation isn’t so bad.
Chapter 34.5 Ideal and Real control systems comparison.
Chapter 34.5.1 Introduction
We will test:
– step responses x(t)=+1
– disturbances z(t)=+0.4 and z(t)=-0.4 responses
The saturations are 0…+1.5 and 0…+5
The common sense says that:
– the lower are input signals x(t) and z(t) the more is a system similar to ideal
– the wider is saturation range the more is a system similar to ideal. 0…+5 is better than 0…+1.5.
Chapter 34.5.2 Saturation 0…+1.5, x(t)=+1, z(t)=+0.4 (heating)
We have 2 identical tanks with identical liquid. Controllers parameters are identical too. The different are the heater power limits-saturations.
– Ideal System – unlimited heater power. The heating power +100 000 kW and cooling power -1000 000kW are possibly when required.
– Real System – limited heater power 0…+1.5 kW
The additional heater +0.4 kW is a disturbance.
It’s certainly the ideal is better than a real system. But how better? 100:0 or 5:0 only.
The ideal system hasn’t saturations. It means that all control signals are possible, the negative (cooling) values too.
The real control signal sPID(t) enables temperatures 0…+150°C when ambient temperature is 0°C.
The set point x(t)=1 step is in 3 sec and the disturbance z(t)=+0.4 is in 60 sec.
IDEAL SYSTEM without saturations.
The PID has control signal with -infinity…+infinity amplitudes at disposal. The step yi(t) response is fast and assures null control error. It’s possible beacuse big sPIDi(t) control signal has big amplitudes and “brake and gas” control algorithm is used here–>see Fig. 34-12–>blue sPIDi(t)–>“brake=-5 and “gas=+160“.
REAL SYSTEM with saturations 0…+1.5
The PID has control signal with 0…+1.5 amplitudes at disposal. It means that “brake and gas” effect is weaker than for ideal system–>real–>green sPIDr(t)–>“brake=0.9 and “gas=+1.5“. The set point x(t)=1 response isn’t fast as ideal. There is even 30% overregulation here! On the other hand. The “brake and gas” effect is circa 100 times weaker than for ideal system. But the setting time isn’t 100 times worse! And the most inportant. There is null steady error.
The main conclusion. The nonlinearity (saturation here) impairs the quality of the control systems but “not to bad”.
Note that z(t) disturbance supressions are the same for ideal and real systems! The real green sPIDr(t) control signal doesn’t exceed the saturations now and that’s a reason.
Chapter 34.5.3 Saturation 0…+1.5, x(t)=+1, z(t)=-0.4 (cooling)
The block diagram is similar to Fig. 34-17 but z(t)=-0.4 (cooling)
The set poin x(t)=1 response is the same and it’s obviously. The z(t)=-0.4 (cooling) was compensating by the green sPIDr(t) control signal increase (heating)
Chapter 34.5.4 Saturation 0…+5, x(t)=+1, z(t)=+0.4 (heating)
This and the next block diagram differs in saturation only. There is 0…+5 instead 0…+1.5. We expect that x(t)=+1 step response will be better because system is “more ideal” now. Our power amplifier-actuator has more power now–> it’s expensiver.
The set point x(t) response is faster (better) than by 0…+1.5 saturation. The z(t)=+0.4 disturbance suppresion is the same as ideal because sPIDr(t) doesn’t exceed 0 saturation value.
Chapter 34.5.5 Saturation 0…+5, x(t)=+1, z(t)=-0.4 (cooling)
The diffrence is in the z(t)=-0.4 disturbance suppresion.
Chapter 34.5.6 Conclusions
It’s obviously that the ideal system is better than real system with final element power limits. It’s obviously too that the “more ideal” is the system-the more powerfull is the actuator–>the more expensiver it’s. But the real system isn’t to bad as come from saturations. Let’s go to football analogy. The difference between ideal and real system is more similar to Germany-Austria than to Germany-Liechtenstein.
Chapter 34.6 The set point x(t) exceeds the saturations levels.
Chapter 34.6.1 The first approach – signals x(t), yi(t) and yr(t) only
Let’s assume that PID “wishes” to heat the liquid up to +105 °C. But the actuator (or other name power amplifier) has a maksimal power +0.95 kW and it enables +95 °C only. The PID can’t do this job. It’s impossible as a leader who demands from his employee more than employee professional competence.
We compare ideal and real systems behaviours. The set point x(t) is multistep type signal and x(t) exceeds the high+0.95 real PID saturation level sometimes. Can you predict the yi(t) and yr(t) processes?
Period 1 x(t)=+0.3
The x(t)=+0.3 is lower than +0.95 saturation. The feedback will be active and the behaviours are discussed in the Chapter 34.5. The real yr(t) response is slower than ideal yi(t) and with the overregulation but null control error condition is fulfilled.
Period 2 x(t)=+1.05
The ideal yi(t) is obvious because there are no PID control signal limits. The real yr(t) can’t attain x(t)=1.05 because it exceeds the saturation 0.95 level of the real PID. The feedback doesn’t work and the response is as for open loop system.
Period 3 x(t)=+0.5
The setpoint x(t) falls rapidly up to +0.5. The ideal system reacted immediately and the steady state yi(t)=x(t) is quickly attained. The real steady state yr(t)=x(t) is slowly attained. The feedback works again because the sPIDr(t) control signal is in the inside of the saturation range 0…+0.95 again. But what’s about the “dead” time? It wasn’t in the period 1. I propose to test sPIDi(t) and sPIDr(t) control signals.
Chapter 34.6.2 The second more accurate approach
We will observe the sPIDi(t) and sPIDr(t) control signals of the PID. Do we discover the “dead” time reason?
The ideal system is obvious.
The sPIDr(t) of the real system in the period 2 is limited by the saturation and sPIDr(t) =x(t)=0.95. The feedback doesn’t act now the yr(t) is an opened loop type now. But the “dead” time in the period 3 remains mysterious. I propose to observe sPIDir(t) signal before power amplifier.
Chapter 34.6.3 The third even more so accurate approach
The yellow sPIDi(t) control signal (fuzzy colour!) is obvious and doesn’t require the comment.
We observe the green sPIDir(t) before the power amplifier. The i letter in the name sPIDir(t) suggests that the signal is after the ideal part of the real PID cpontroller.
Period 1 x(t)=+0.3
The green sPIDir(t) is bigger than saturation +0.95, therefore the blue sPIDr(t) control signal will be cutted. There isn’t feedback now and yr(t) arises up to +0.95 as an opened loop system. The feedback shows after 6.5 sec and the yr(t) attains steady state yr(t)=x(t)=0.3 after circa 25 sec.
Period 2 x(t)=+0.95
The sPIDir(t)=+0.95 all the period 2 and yr(t) is going to +0.95 as opened loop system.
Period 3 x(t)=+0.5
Something is clear with the “dead time”. The control signal blue PIDr(t)=+0.95=saturation during the “dead time”! It’s steady and it means that feedback doesn’t work here! It works after the “dead time” and the steady state yr(t)=x(t)=0.5 is attained after circa 110 sec. Is the “dead time” secret hidden in the green sPIDir(t) signal? Let’s look at this not cutted by the oscilloscope signal.
Chapter 34.6.4 This same experimen but the bigger oscilloscope range.
This same block diagram but other oscilloscope range -100…+140, before was -1…+2. All the green sPIDir(t) (before power amplifier) will be seen.
The signals x(t), sPIDr(t), yi(t), yr(t) are very small, almost unvisibile. But you see all the sPIDi(t) i sPIDir(t) control signals.
Whoa! I see the green sPIDir(t) real control signal before the power amplifier. But where is the sPIDi(t) ideal control signal? It’s covered by the green sPIDir(t).
Period 1 x(t)=+0.3
See Fig. 34-28. All the real blue sPIDr(t), output yr(t) and a piece of the sPIDir(t) are seen here. You haven’t seen the falling sPIDir(t) up to 6.5 sec. The feedback works after 6.5 sec and output yr(t) attained steady state yr(t)=x(t)=0.3. Nothing new and interesting in this period.
Period 2 x(t)=+0.95
The feedback doesn’t work (see Fig. 34-28) and the ideal PID part “thins” that it’s opened loop. It causes the typical arising of the green sPIDir(t) signal. The typical PID step response. But the blue sPIDr(t)=0.95 is steady because of the saturation=0.95. We can say that this green sPIDir(t) signal is unnecessary beacuse it doesn’t cause output yr(t) reaction.
Period 3 x(t)=+0.5
See Fig. 34-28 again. You see a piece of the falling green sPIDir(t) signal. The ideal part of the real PID controller “thinks” that is opened loop (saturation=0.05) and it’s reaction for the x(t) 0.95/0.5 signal drop (negative step) is typical. It falls up to +0.95 value and this is a reason of the “dead time”. ( Fig. 34-28) The feedback is again after the “dead time” and the steady state yr(t)=x(t)=+0.5 is attained atfter the 115 sec.
Chapter 34.6.5 Conclusions
1. The PID control is possible when the set point x(t) signal is inside the of the power amplifier (other name is final control element or actuator) range. It’s obviously. It’s impossible to heat a bath of water up to +100°C when you dispose a small electrical immersion “glass” heater.
2. When the et point x(t) signal is inside the range of the power amplifier –> the integral I of PID is able to assure null control error.
3. The wider is power amplifier output signal range–>the more similar to ideal is a real system.
4. The “dead time” of the PID control reason is a exiting from the PID saturation state. This is an harmful effect because there are time delays.* How to counteract? The PID “feels” that it’s in the saturation state (The signal from valve end contact or output y(t) signal is steady when PID control signal arises) and changes its structure to PD controller. The integration doesn’t act.
5. The smaller is the set point x(t) signal the more are the real and ideal systems similar.
6. The real and ideal systems disturbances suppresions are similar often. The similarities are closer when the z(t) disturbances are smaller.
* The time delay To is always harmfull in the control systems. An example. There is To=1 sec delay between steering wheel and the wheels. How is it made? No important. Is this a secure car?