Chapter 33.1 Introduction
Control structures doesn’t depend on controller types. It may be PID, ON-OFF or any other non continues PID types.
A smal digression about “non continues PID types” now. You read the ON-OFF controller manual and there is something about Kp, Ti and Td parameters. What is this? These parameters are possibly for continuos controllers only. Really? But imagine that  the s(t) controller output is a square wave  with the fullness factor  proportional to the required value. The step controller is the most known non-continuos controller. It has only 3 output valuses. Ther are +max, 0 and  -max and is mainly used for valves positioning.  This positioning is realised by the characteristic steps–> the name step controller.
But revisit our main subject – Control Systems Structures.
We will test systems structures:
1– Opened loop
2– Opened loop with the noise compensation
3– Closed loop
4– Closed loop with the noise compensation
5– Cascade control
6– Ratio control

Chapter 33.2 Opened loop

Fig. 33-1
The above-mentioned heat exchanger is an example of the opened loop system. It’s convenient when regulating unit – power amplifier with the heater here, is a part of the object Go(s). Open loop system has some advantages. It’s simple and is stable almost always. But it’s not resistant for disturbances- z(t)=+0.4 additional heater here.
Every serious controller is equipped with the switch A/M – automatic/manual.
Switch  A/M is a A – automatic position–>converts Fig. 33-1 (open loop) to Fig. 32-11–>previous chapter  (closed loop)
Switch  A/M is a M  manual position–>converts  Fig. 32-11 (open loop)–>previous chapter to Fig. 33-1 (open loop). It’s  manual control and is used in the emergency situations or other not typical.  It maybe a common switch but mostly is software switch on the computer monitor.
The main  A/M switch requirement is to assure the is a gentle transition from A to M mode and vice versa. What is this “gentle”? There aren’t control signals s(t) non-continuos steps.
Call Desktop/PID/19_struktury_ukladow_regulacji/01_otw

Fig. 33-2
X-cos open loop model.
Click “start”
Fig. 33-3
Set point x(t) and disturbance z(t) response. This experiment evidences that open loop system isn’t resistant to z(t) disturbances.

Chapter 33.3 Opened loop with the noise compensation or feedforward control-other name.
Chapter 33.3.1 Introduction
The name feedforward  is as a feedback opposite.  It is possible when we can measure the z(t) disturbance.

Chapter 33.3.2  Opened loop with the incomplete noise compensation k=0.7

Fig. 33-4
The diagram is a Fig. 33-1 + additional component-compensator.  It measures the z(t)=+0.4 disturbance  and calculates the control control signal s(t)=x(t)-k*z(t)  given to Go(s) object input. The control  signal s(t) additional component -k*z(t) has opposite direction than a z(t) disturbance. It acts as a compensator.
This controller is curious a little. It does’t used Go(s) output signal y(t) but:
x(t) input signal. It isn’t curious
z(t) disturbance input signal
Call Desktop/PID/19_struktury_ukladow_regulacji/02_otw_niepelna_komp.zcos
Fig. 33-5
Click “start”
Fig. 33-6
Compare with the Fig. 33-3 classical opened loop system. The z(t) disturbance was clearly but partly only suppressed.  How to do full the z(t) disturbance compensation? It’s obviously. Let’s make full k=1 compensator

Chapter 33.3.3  Opened loop with the complete noise compensation k=1

Fig. 33-7
k=1 compensation
Call Desktop/PID/19_struktury_ukladow_regulacji/03_otw_pelna_komp.zcos

Fig. 33-8

Fig. 33-9
The s(t) control signal is in the anaphase  to z(t)=+0.4 and the disturbance was full compensated. The y(t) output stands as a lancer lance regardless of z(t) disturbance!
Two questions  arises now:
1– What is the reason of the Fig. 33-4 experiment with k=0.7? It’s obviously that k=1 is the best.
2– The k=1 disturbance compensation is ok. We even don’t see the z(t) disturbance effect! Why isn’t this method universal?
Answer for the questions
We are in good situation. We know the object Go(s) and compesation k=1 is clear as a day. But imagine that you are in the refinery site. You don’t see any transfer function Go(s) but columns, pipelines etcc…  The z(t) disturbance isn’t so distinct as in the Fig. 33-4. Even you have the controller potentiometer with the range k=0…1.5 the ideal k value is possible by experiments or by exactly mathematical object model Go(s). These methods aren’t so obviously as in the Fig. 33-4!

Chapter 33.3.4  Opened loop with the noise compensation in “middle” of the Go(s) object

Fig. 33-10
The “disturbance heater” is located near the tank no. pipe coil. I hope you understand the word “middle” in the chapter title.  We expect the disturbance suppresion.  But will be as good as Fig. 33-9 with the disturbance in the Go(s) input?
Call Desktop/PID/19_struktury_ukladow_regulacji/04_otw_pelna_komp_srodek.zcos

Fig. 33-11
Xcos model of the Fig. 33-10.
Click “start”
Fig. 33-12
The compensator reaction was right. The disturbance was compensated after 65 sec.  But the Fig. 33-9 was much better because there were no temporary state here.  How do we improve the disturbance response? If the Gzk(s) transfer function between z(t) and compensation signal sk(t) was Gzk(s)=1–> the compensation will be ideal as in the Fig. 33-9!

Chapter 33.3.5 The “inverse” transfer function
How to transform every G(s) for G(s)=1?

Fig. 33-13
I hope that the idea is simple. Let’s try transform the inertial unit for proportional unit G(s)=1.
Call Desktop/PID/19_struktury_ukladow_regulacji/05_odwrotn_G(s).zcos
Fig. 33-14
G(s)=(1+3*s)/(1+3*s)=1  But  Xcos doesn’t realise  the pure 1+3*s block. It realises almost this same block but with a smal inertia 1/(1+0.01*s) as in the above figure.   The G(s) will be almost G(s)=1, almost proportional unit!
Click “start”

Fig. 33-15
Fig. 33-15a
y1(t) inertial unit response
y2(t) almost  “proportional” unit  G(s)=1 as a inertial unit and its “inverse” product.
Look at the y2(t) in the Fig. 33-15a. It seems like y2(t)=x(t). The “not ideal” is the begining of  the y2(t) nearby 1 sec. There is a small influence  of the inertia 1/(1+0.01*s). But we transformed the inertia 1/(1+3*s) for proportional unit G(s)=1. Almost!
Fig. 33-15b
It shows the reason of the fast y2(t) compared to slow y1(t). The reason is a strong differentiation of the signal x(t) good seen by the other oscilloscope parameters.
The Go(s) transformation to proportional Go(s)=1 requires strong differentiation and big momentary power. See  blue xp(t)  signal-especially Fig. 33-15b.

Chapter 33.3.6  Opened loop with the noise “inverse” compensation in the “middle” of the Go(s) object.
The chapter title is complicated a little, but we know “what’s the matter”.

Fig. 33-16
Call Desktop/PID/19_struktury_ukladow_regulacji/06_otw_pelna_komp_zakl_srodek_odwr.zcos

Fig. 33-17
It’s Fig 33-16 Xcos model.  The compensator denominator (1 + 0.01*s) is a one difference only here. Why?  See Fig. 33-14.  We constructed the (almost) inverse transfer function.
Click “start”

Fig. 33-18
Fig. 33-18a
The disturbance suppresion was  like in the Fig. 33-10 because the Gzk(s) transfer function between z(t) and compensation signal sk(t) was almost proportional Gzk(s)=1! It means that the sk(t)-pipe coil temperature in the tank no. 2 can react for z(t)=+0.4 disturbance immediately! It’s possible to reduce disturbance influence which is good seen in the Fig. 33-12. But the controller reaction s(t) signal is very rapid now, especially it’s differenttiation component D.
Fig. 33-18b
You see all  the control signal s(t) compared to x(t)=1! Very,  very  big. It compensates the 1/(1+3*s) inertia. The rest of the object – tank no. 2 dynamics “thinks” that the +step disturbance z(t)=+0.4 is (almost) ideally compensated Tc1 temperature  Δ=-0.4 step drop.
The s(t) and sk(t) analysis is difficult because these signals are partly covered.  The next figure shows them separately

Fig. 33-19
s(t), sk(t) and  z(t) signals shown separatly.
Fig. 33-19a
There is a very strong s(t) controller reaction  for disturbance z(t)=+0.4 in the 40 sec! All the signal is seen  in the Fig. 33-18a. I remind  that our electrical heater in the tank no. is a Peltier element. Posive voltage heats and negative cools! Momentary powerful cooling s(t)=-120. It corresponds to -12000°C! We know that the minimal temperature in the world is -273.15°C! But we agreed that -12000°C temperature exists in this course. This cooling causes the immediately sk(t) drop int the Fig. 33-19and (almost) ideal z(t)=+0.4 disturbance compensation.

Chapter 33.4 Closed loop
The most popular structure in the control system theory.  The others are only its modifications.

Fig. 33-20
An closed loop system example. Almost all this course are closed loop systems and the subject is finished now .

Chapter 33.5 Closed loop with the noise compensation-Other name–>Closed-opened system
Chapter 33.5.1 Introduction
The Opened loop with the noise compensation enables noise suppresion provided that this noise-disturbance is possible to measure. But:
– there are othere disturbances possible
– the static and dynamic Go(s) characteristic doesn’t change.
The problem-solving is obviously. It’s necessary to close the loop with the aid of the PID controller. All other disturbances should be suppressed.
The different combinations will be tested:
-Full compensation k=1 and the z(t) disturbance located in the object input–>p. 33.5.2
-Incomplete compensation k=0.7 and the z(t) disturbance located in the object input–> p. 33.5.3
-Full compensation k=1 and the z(t) disturbance located in the “middle” of the object –>p. 33.5.3
-Full compensation k=1 and the z(t) disturbance located in the object input and aditional disturbance located in the “middle” of the object –>p. 33.5.5

Chapter 33.5.2 Full compensation k=1 and the z(t) disturbance located in the object input

Fig. 33-21
The Kp, Ti and Td parameters enables optimal set point x(t) response.
Call Desktop/PID/19_struktury_ukladow_regulacji/07_zamk_komp_1.zcos

Fig. 33-22
Wciśnij “start”
Fig. 33-23
The y(t) response is identical as for pure Closed loop–>Fig. 32-13 chapter 32 up to 40 sec.
The z(t)=+0.4 disturbance in 40 sec causes immediate compensator  -0.4 reaction–>The disturbance is absolutely compensated! The y(t) doesn’t change. Classical PID controller (e.i. PID without compesator part) doesn’t react! It even didn’t notice the z(t)! All the job was made by the compensator part of the controller.

Chapter 33.5.3 Incomplete compensation k=0.7 and the z(t) disturbance located in the object
It isn’t easy to do full disturbance compensation. The disturbance may be hard to meaure. The Go(s) static and dynamic parameters aren’t good identified… etc. This situation corrresponds to k=0.7.

Fig. 33-24
Almost the Fig. 33-21 copy. The difference is  k=0.7  and not  k=1. We expect some z(t) influence now.
Call Desktop/PID/19_struktury_ukladow_regulacji/08_zamk_komp_0.7.zcos

Fig. 33-25
Click “start”

Fig. 33-26
The compensator k=0.7 does what is possible only. It compensates a part of the z(t) disturbanse only. The rest of the z(t) disturbanse is suppresed by the classical PID controller. Other words-by the closed loop.

Chapter 33.5.4 Full compensation k=1 and the z(t) disturbance located in the “middle” of the object

Fig. 33-27
Call Desktop/PID/19_struktury_ukladow_regulacji/09_zamk_pelna_komp_zakl_srodek_odwr.zcos

Fig. 33-28
Click “start”
Fig. 33-29
The z(t)=+0.4 disturbance is compensated by the compensator part of the PID controller. The PID part didn’t react, it even didn’t notice the disturbance. Other words-PID controller part  was at a loose end.

Chapter 33.5.5 Full compensation k=1, the z1(t) disturbance located in the object input and the additional z2(t) disturbance in the “middle” of the object
The z1(t)=+0.4 disturbance  is a heating near tank no. 1 pipe coil in 40 sec. It will be full commpensated.
The additional z2(t)=-0.6 disturbance  is a cooling- negative voltage on the Peltier element  near tank no. 2 pipe coil in 55 sec.

Fig. 33-30
Technological diagram with 2 disturbances.
Call Desktop PID/19_struktury_ukladow_regulacji/10_zamk_komp_2_zaklocenia.zcos

Fig. 33-31
Click “start”
Fig. 33-32
The z1(t)=+0.4 heating in 40 sec  was compensated only by the controller compensator part-the immediate s(t) signal drop -0.4.
The z2(t)=-0.6  cooling was compensated by the controller PID part-the s(t) signal +0.6 increase. Other words-it was normal closed loop feedback action.

Chapter 33.5.6 Conclusions
1. Closed loop with the noise compensation system enables ideally suppression the disturbances provided that:
– accurate disturbance measurement
– appropriate compensator transfer function–>chapters 33.5.2 and 33.5.3
2. There is a big control quality improvement even a/m conditions aren’t ideally fulfilled.
3. There are s in the compensator transfer function nominator–>differentiation. It causes some obviously problems.
4. The other not compensated disturbances are suppresessed by PID with the normal feedback.

Chapter 33.6 Cascade Control
Chapter 33.6.1 Introduction
There is a better control quality than a classical Closed Loop and a little worse than a Closed loop with the noise compensation. But as distinct from the last,  it doesn’t require the disturbance measurment and the object Go(s) good knowledge.
The action principle is similar to the corporation. The corporation president rules all the corporation and the manager rules a department only.
The president assign the task to manager and doesn’t care on. Strictly-president doesn’t care about department disturbances. The worker Mr Brown is drunken or some VAT invoices are missing. This is department menager job because he is near the disturbances than a president. The menager is ideal when president doesn’t  care about trivia. The president is a superior PID controller and the menager is a lowly controller, mostly P type.

Chapter 33.6.2 Cascade Control with one z(t) disturbance

Fig. 33-33
P controller compares Tank no. 1 liquid temperature Tc1 with the sPID(t) control signal. There is an internal loop and Tc1 tries to follow the sPID(t) signal.  PID controller is a president of the all corporation – all the object with 2 tanks and P is a  manager of the tank no. 1 only.  PID is a superior controller and P  is a lowly controller.
There are 2 advantages of the cascade control.
1. The internal loop causes that the inertia of the tank no 1 is reduced. All the object is easier to control.
2. The z1(t)=+0.4 is faster suppresed than with the classical closed loop system–>Fig. 32-11 chapter 32.  The lowly P controller suppresses the disturbance so fast that the superior PID almost doesn’t observe the z(t) influence for the y(t) signal!
Let’s check it.
Call Desktop/PID/19_struktury_ukladow_regulacji/11_kaskadowy_10_7_2.zcos

Fig. 33-34
Click “start”
Fig. 33-35
Compare red y(t) with the classical closed loop–>Fig. 32-13 chapter 32.  The small z(t) influence exist but isn’t visible here! But the set point x(t)response is much worse! Why? The answer is simply. All the object inertia is lower now, because internal loop “lowered” the tnak no. 1 inertia. It means that there are other optimal PID parameters now!

Chapter 33.6.3 Cascade Control with one z(t) disturbance and with more optimal PID parameters
The new PID parameters Kp=10 Ti=5 sec i Td=0.55 sek were set by the trial-and-error method.  Will be better?
Call Desktop/PID/19_struktury_ukladow_regulacji/12_kaskadowy_10_5_0.55.zcos

Fig. 33-36
New Kp,Ti,Td parameters
Click “start”

Fig. 33-37
Shock. But megashock when compared to Fig. 32-13 chapter 32!
But one comment please. Look at Fig. 33-32. The steady state s(t)=y(t) is here–>y(t) e.i. Go(s) output is equal the control signal s(t). We accustommed that “it’s always”. Look at Fig. 33-37–> sPID(t)≠y(t)! Why?
General steady state rule is that:
is a control signal and k is Go(s) object static gain.
The object gain was  k=1 so far. And what’s the cascade control situation. The internal loop gain is 0.909 now! It means that sPID(t)=1.1*y(t) when z(t)=0 e.i. before 40 sec. The experiment confirms it.

Chapter 33.6.4 Cascade Control with two disturbances

Fig. 33-38
Disturbance z1(t)=+0.4
is a additional heating in the inernal loop. It will be strong suppressed mainly by the lowly P controller.
Disturbance z2(t)=-0.6 is a additional cooling (minus on the Peltier element!) of the tank no. 2 pipe coil. It should suppressed too, but not so fast as z1(t) disturbance. This is a job for superior PID controller. It’s responsible for all the other disturbances too.
Call Desktop/PID/19_struktury_ukladow_regulacji/14_kaskadowy_10_5_0.55_2_zaklocenia.zcos

Fig. 33-39
Fig. 33-40
The system behaviour identical as  Fig. 33-37 is up to 55 sec, e.i. up to z2(t) appearance. The PID controller reaction is right after 55 sec. The z2(t) was suppressed too, but it’s seen the z2(t) influence.
Has lowly P controller positive effect for z2(t) suppresion? Yes! The P controller lowered the internal loop inertia–>all the object inertia is lower now. We can set more aggresive PID parameters now.

Chapter 33.7 Ratio control
Chapter 33.7.1 Introduction
The common control system tries to follow input signal x(t) that the steady state is y(t)=x(t). The Ratio control system does similar job but the steady state is y(t)=k*x(t). Parameter k is a ratio coeficient.
It may be for example the paints mixing plant–>chapter 33.7.2
The other is a gas steam boiler–>p. 33.7.2 There is one combustion efficiency. The one optimal parameter coefficient k=gas/air must be fulfilled. When gas is too much–>not all gas will be fired, When air is too much–>there is unnecessary boiler cooling.

Chapter 33.7.2 Paints mixing plant
The appropriate colour is the appropriate 3 basic colour mixture.
By the way. The german poet Johan Wolfgang Goethe is a colours circle author.  And the “Sorrows of  Young Werther” were for him less valuable than the colour theory.
The appropriate coulour is 2 colours mixture K=green/red=1/3.  The tank will be filled by red and green pipes with the flows Fz, Fcz and K=Fz/ Fcz=1/3.

Fig. 33-41
Paint mixing plant
How does paint mixing plant work?
The set point x(t) flow range suits 0…+10V and measured Fcz(t) suits this same range 0…+10V of course. P controller compares set point x(t) with the Fcz(t) flow  from the flowmeter PP and calculates control signal to the valve. The valve shaft and electrical actuator SE form the integrating unit with transfer function 1/5*s. How to convince you about integrating unit? Imagine that the controller gives small + voltage. The schaft is slowly moved right and the valve is slowly opening. The twice bigger + voltage will double the valve opening speed. The 0 voltage means that x(t)=Fcz(t) and the valve is in the stillness state. The – voltage will cause the valve closing. Do you believe me that valve+electrical actuator is an integrating unit? The integrating unit 1/10*s is double slower than the unit 1/5*s.  The valve with the flowmeter forms inertial unit 1/(1+1*s).
The “right” green paint part control system part action is similar.
The steady state is Fz=Fcz/3
Let’s check it.
Call Desktop/PID/19_struktury_ukladow_regulacji/15_reg_stos_mieszalnia.zcos
Fig. 33-42
There are to I type control systems.
Click “Start”
Fig. 33-43
The ratio control goal is achieved. There are steady state flows and K=Fz/ Fcz=1/3.

Chapter 33.7.3 The gas steam boiler
This is less trivial ratio control example than the paints mixing plant

Fig. 33-44
Gas steam boiler
This is boiler drum  filled with the water and supplied by the natural gas. The boiler supplies many devices not shown in the figure.  Our goal is to assure steady steam pressure cp(t) regardless of disturbances as on-off different steam receivers.
The gas boilers and the thermal engineering aren’t my business generally. The optimal is gas flow/air flow=k=10 reportedly. But due to the didactic reason the optimal is k=1.5.

Gkoc(s)-Steam boiler
Inputgas flow Fg(t) and  k=1.5 bigger air flow Fp(t)*
Outputsteam pressure Cp(t)- more strictly the voltage  0…+10V corresponding tp 0…10 MPa pressure.
The phisics and control system is very simplified here. We assume  that there is a level control system (not seen here) and the water level in the boiler is steady regardless of steam flow changes. The Fg(t)-natural gas flow inrease causes temoperature increase and Cp(t) steam pressure inrease. This is an inertial unit with T=120 sec with the PI control loop.
*The air flow Fp(t) exists only by the higher figure near boiler.  There isn’t in the lower block diagram as a input. It’s right because  air flow is proportional to gas flow Fp(t)=k*Fg(t).

Gas control unit Gg(s)
InputPI control signal s1(t)
OutputFg(t) gas flow
SE-electrical actuator- It’s proportional unit as distinct from SE-integral unit in the paint mixing plant. It’s equipped with the internal shaft position feedback-not shown in the figure.
PP – flowmeter changes the Fg(t)–> 0…+10V.
Valve – It changes shaft position–>valve flow surface change–>flow changes. We assume that Gg(s) is a inertial unit with T=5sec.

PI steam pressure controller
Compares the Cp(t) pressure with the set point x(t).
Cp(t) steam pressure increase causes Fg(t) gas flow decrease–>water temperature decrease–>Cp(t) steam pressure decrease and vice versa. The Cp(t) steam pressure is stabilized as a result of the above mentioned actions. The PI control assures null steady controll error and not too shabby dynamics.

Air control unit Gp(s)

InputPI control signal s2(t)
Output– air flow Fp(t)
The action is similar to Gas control unit Gg(s)

PI air flow controller
The setpoint is k*Fg(t)–>the ratio control is assured–>Fp(t)=k*Fg(t). The air flow is proportional to the gas flow and the optimal combustion is assured.
Let’s test it
Call Desktop/PID/19_struktury_ukladow_regulacji/16_reg_stos_kociol.zcos

Fig. 33-45
The Xcos model of the gas steam boiler
Click “start”
Fig. 33-46
The ratio control goal is attained. The air flow is proportional to gas flow. I emphasize that ratio control is only a auxiliary cotrol here. The main goal is to attain steady Cp(t) steam pressure regardless of the steam consumption devices.

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