### Automatics

**Chapter 30 ****Control Systems Structures**

**Chapter 30.1 Introduction**

The structures do not depend on the type of controller. It can be

**PID**, two-position and any other, e.g.

**discontinuous PID**variants. Here a small digression to “discontinuous PID variants”. You read the specification of the

**two-position controller**, and there is, for example, the

**Kp=3**setting typical for a

**proportiona**l controller. What is it about? After all, he only

**turns**the power

**on**and

**off**to the object. What

**Kp**gain can a

**two-position controller**have? But if it outputs a

**1 Hz**square wave with a duty cycle proportional to the error

**e(t)**, and the object has an

**inertia**of the order of

**minutes**, then we are dealing with

**almost continuous**control! Thus, we have a two-position

**controller**, which acts as a

**proportional**one. Nay. There can be

**two-position PID**controllers with

**Kp, Ti, D**settings. The most typical representative of

**discontinuous PID**s is the so-called

**step controllers**. It is most commonly used in flow control and is used to control the degree of valve opening. Here, the

**controller**reaches the setpoint

**x(t)**value by acting on the actuator with a voltage signal, e.g.

**+max**,

**0, -max**in jumps, otherwise – in steps (increase, stop, decrease). All in all, although the control signal is discontinuous, due to the large inertia of the object, its response is continuous.

But let’s get back to the topic.

We will examine typical structures of control systems:

**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 30.2 Opened loop**

Fig. 30-1

Fig. 30-1

An example is the heat

**exchanger control diagram**above. It is convenient for the analysis when the actuator is part of the exchanger – the

**Go(s)**object. Here the actuator is a power amplifier with a heater. An

**open**system has some advantages-

**simplicity**and it is always

**stable**. On the other hand, it is completely

**vulnerable**to disturbances

**z(t)**. The disturbance

**z(t)=+0.4**is a voltage spike of

**+4V**caused by a short-circuit of the contact on the “disturbance” heater.

Every serious controller has an

**A/M**switch for

**Auto/Manua**l operation.

Automatic operation

**A**is the version

**Fig. 29-7**from

**chapter 29**.

Manual work

**M**is an

**open**system, which is the figure above.

Then in

**Fig. 29-7**from

**chapter 29**the

**negative**input to the

**controller**has been connected to

**0V**, and the

**PID**controller itself will change its

**P**controller structure by

**Kp=1**. In other words, the

**feedback loop**was broken and an open system was created, i.e.

**Fig. 30-1**.

**Manual**work is used during the

**commissioning**of the facility, in

**emergency**and other

**unusua**l situations.

The

**A/M**switch can be a

**regular**switch or just a

**software**one. Then the operator switches

**A/M**from the keyboard. Provides a

**smooth**transition from mode

**M**and vice versa. It is somehow implemented there, but I will not go into details.

What will happen if we move the adjuster slider from

**0 V**to

**+10V**in

**3**seconds, and we give a

**+4V**step to the “disturbance” heater in

**40**seconds?

**Note**

Most controllers are implemented in a

**digital**version. In other words, addition, subtraction, multiplication, division, integration and differentiation are calculated by the microprocessor, not analog.

**Fig. 30-2**It’s like you suddenly moved the slider from

**0V**to

**+10V**, and then in

**40 seconds**gave a step of

**+4V**to the disturbance heater. The response

**y(t)**is obvious. The result of the

**x(t)**step is the state

**y(t)=Tc3=+100ºC**after

**35**

**seconds**and the state

**y(t)=Tc3=+140ºC**after

**75 seconds**.

Let me remind you that in

**chapter 29.3.2**we have “simplified” electrical engineering. The power dissipated in the heater is proportional to the

**voltage**, not the

**square**of the

**voltage**.

**Chapter 30.3 Opened loop with the noise compensation**

**Chapter 30.3.1 Introduction**

What to do so that the

**Open System**does not have “absolute lack of resistance to disturbance”. The solution may be to add

**compensation**. Let’s try to limit the

**influence**of the

**noise**by compensating the

**noise**. Always approach something new like a dog to a hedgehog. Therefore, at the beginning we will give incomplete compensation

**k=0.7**.

**Chapter 30.3.2**

**Opened**loop with the incomplete noise compensation k=0.7

Fig. 30-3

The diagram above is **Fig. 30-1** + additional **compensator** block. It measures the** voltage** (i.e. disturbance **z(t)=+0.4**) and then calculates the control signal **s(t)=x(t)-k*z(t)** which is fed to the input of the object. The control signal **s(t)** due to the subtraction operation has the **opposite** direction to the disturbance **z(t)**. If the disturbance **heats** (as in the diagram), then** s(t)** reduces the power supply to the heater, i.e. it **cools** it down. If the disturbance **cools**, then **s(t)** additionally **heats**. The compensator therefore functions as a controller.

Some people call it a **feedforward** controller. It’s a strange controller. It does not use the output signal **y(t)**, only the input signals, i.e.:

–** x(t)** (which is normal for any controller)

– **disturbance z(t)** (this is the wonder!).

**Fig. 30-4**

The response to **x(t)**, i.e. the time chart of **y(t)** in** 3…40 seconds**, is obvious. In **40 sec** there was a disturbance** z(t)=+0.4** – additional heating. The control signal **s(t)** reacted correctly, it decreased the heating by **0.7*0.4=0.28** to the value of** 0.72**. In total, a power of** 0.72+0.28=1.12** is delivered, which corresponds to a steady state temperature of **y(t)=+112ºC**. Compared to **Fig. 30-2**, where there was no compensation, the disturbance z(t) was partially suppressed. You don’t need to be a genius to predict the effect of **full** compensation. **Full** disturbance suppression **z(t)** will be ensured only by** full** compensation **k=1**.**Chapter 30.3.3 Opened loop with the complete noise compensation k=1**

Fig. 30-5

Compensation factor **k=1.** Can you guess the effect of full compensation?

**Fig. 30-6**

The control signal** s(t)** is now in **anti-phase** to **z(t)** and the disturbance has been fully **compensated**. The output signal **y(t)** despite the disturbance** z(t)=+0.4** (additional heating) will not budge. It stands like a Uhlan lance. There are **2** questions here?**1-** Why didn’t we give the compensation factor **k=1** right away?**2-** Since there is** 100%** disturbance suppression, why is it not a universal method.**Answer no 1**

It was easy for us because we know the object’s transmittance** Go(s)** and **k=1** was **obvious**. But on a real object, e.g. in the Refinery, you don’t see the **transmittance**, only some column, heating pipelines and those not fully worked out “disturbance” signals. Even if you have a compensation potentiometer on the controller, e.g. **k = 0 … 1.5**, you will select the ideal value either experimentally by turning the potentiometer, or theoretically, when you have a thoroughly mathematically worked out object. Returning to the “**k potentiometer**“, in serious systems it appears only as **software**. The operator “turns” it through the keyboard, entering the appropriate numbers into the system.**Note***

Earlier, there were systems where, despite **z(t)** disturbance, the **PID** controller worked so that the** red y(t)** “stood” in place. However, this was only due to the small gain of the oscilloscope. If it was larger, the **red y(t)** would “move”.**Answer to 2**

Actually. So far, we have not encountered such a perfect suppression of disturbance. But the compensation is successful only when we know the object well, the disturbance **z(t)** can be measured and the parameters of the object are constant in time. And that’s not always the case. In addition, the system is not immune to other disturbances! Therefore, in **chapter. 30.5** you will learn about the **Closed-Open Systems**, in which there is a compensator that suppresses the measured disturbance and a simple feedback that suppresses the remaining disturbances.**Chapter 30.3.4 Opened System with “normal” Compensation for disturbances “in the middle” of Go(s)**Disturbance in the “center of the object”, but compensator

**k=1**. As in the diagram below.

**Fig. 30-7**

The “disturbance” heater is right next to the

**tank 2**coil. We expect the disturbance

**z(t)**to be compensated to some extent. The response should be better than the uncompensated system in

**Fig. 30-2**. But is it as good as in

**Fig. 30-6**with disturbance at the beginning of the object?

**Fig. 30-8**During

**3…40 seconds**, the output signal

**y(t)=Tc3**is the same as in

**Fig. 30-3, 6**and

**9**because there was no disturbance yet. In

**40 sec z(t)=+0.4**occurred from the “disturbance” heater. The compensator reacted correctly by lowering the control signal

**s(t)**also to

**0.4**– in antiphase. But this time you can see the impact of the disturbance

**z(t)**. Transitional indeed. Why?

Because the decrease in the inflow of heating power to

**tank 2**did not occur abruptly, but with an inertia of

**3 seconds**, as the temperature

**Tc1**in

**tank 1**equals the temperature of the liquid in the coil of

**tank 2**. After

**65 sec**. the output signal returned to its initial state

**y(t)=+100ºC**. The compensator finally did its job, though not as gloriously as in

**Fig. 30-5**, where

**z(t)**acted directly on the input. Then the disturbance was attenuated perfectly immediately by a decrease in the heating power of the heater in

**tank 1**.

It’s good, but we want it to be even better. How? For example, you can put an additional

**electric heater**with a

**Peltier**element right next to the

**disturbance**heater. Then the

**negative**Peltier voltage will cause

**cooling**, i.e. it will cause an immediate drop in the thermal power delivered to

**tank 2**. This is a solution, but it comes with additional structure and cost, compromising energy

**efficiency**.

The existing heater in

**tank 1**can also be used. Note that the imperfection of disturbance suppression

**z(t)=+0.4**is due to the fact that the disturbance

**transmittance**between

**z(t)**and

**sk(t)**is

**Gzk(s)=1/(1 +sT)**where

**T=3 sec**. So it is an

**inertial**unit with constants

**k=-1**and

**T=3 sec**. What about making it a proportional unit

**Gzk(s)=1**? Then it would be possible to completely suppress the disturbance, as in

**Fig. 30-5.**Just how to do it?

**Chapter 30.3.5 “Reverse” transmittance**

Fig. 30-9

How to make any **G(s)** term proportional **G(s)=1**?

Theoretically, it is very simple. **G(s)** should be preceded by its **inverse**. Just like **2** amplifiers connected in series with gains **k1=0.1** and** k2=10** give a gain **k=k1*k2=1**.

And what does it look like when **G(s)** is e.g. an** inertial** unit?

Fig. 30-10

From the inertial unit **G(s)=1/(1+3*s)** we want to make a **proportional** unit **G(s)=1**. As we have shown in **Fig. 30-9**, it should be connected in series with the inverse **G(s)**, i.e. with the unit **(1 + 3s)**. Right away. and what is it doing here in the denominator **(1 + 0.01s)**? Oh yeah. It is difficult to realize the perfect **(1 + 3s)**. It gives** infinity** as a response to the step** x(t)**, because the speed of the jump increase is also **infinite**.

Therefore, we will put **1 + 0.01s** in the denominator. This corresponds to a series connection **(1 + 3s)** with an **inertial** unit with a very small time constant** T=0.01 sec**, i.e.** 1/(1 + 0.01*s)**. The response **y2(t)** to step **x(t)** should be “almost a step”, not a perfect jump. Let’s check.

**Fig. 30-11****– y1(t)** response of the **inertial** unit**– y2(t)** response of the “almost” proportional term **G(s)=1** as the product of the **inertial** unit and its** reciprocal**.**Note**

In a moment it will turn out to be the** “almost proportional”** unit.

At the** first** moment, **y2(t)** perfectly coincides with the step **x(t)**. But if you look closely, the **red y2(t)** in **1 second** does not coincide with the **x(t)** step. Rather, there is a very rapid increase to **y2(t)** as in the** inertial** unit with a very small time constant. This is where the impact **(1+0.01*s)** in the denominator is! Therefore, **y2(t)** is the answer of the **“nearly proportional” **unit.

And why does **y2(t)** increase very quickly during the step compared to the response of **y1(t)** of the** inertial** unit alone? Then look at how powerful the **xp(t)** signal is at the input of the **inertial** unit. The** xp(t)** signal is cut off by the oscilloscope, but it reaches the value **xp(t)=301**! Then it drops very quickly to** 1**. Such a time chart **xp(t)** causes a very rapid increase in the signal **y2(t)** to the value **y2(t)=1**.**Conclusion**

A **differentiating** unit appeared in the **nominator** of the compensating element, which causes large instantaneous value–> **xp(t)=301**! This requires the provision of large temporary power, which can be quite troublesome. Even stronger differentiations – read **second** and **higher derivatives** – occur with more complex objects than **inertial** ones. Despite this, compensation often improves disturbances suppression, even when supply voltages are limited. Especially with simple dynamic objects.**Chapter 30.3.6 System with “inverse” compensation of disturbances “in the middle” of Go(s)**

Fig. 30-12

Fig. 30-12

Below is the system model. The only difference is the denominator of the compensator

**(1 + 0.01*s)**introducing a small

**inertia**. It prevents

**diracs – infinitely**large pins when jumping. The pin will remain, but of

**finite**value! It does not significantly negatively affect the attenuation of disturbance

**z(t)**. Let’s agree that this is reverse compensation, although formally it is not.

**Fig. 30-13**

Until the disturbance **z(t)=+0.4**, in **40 seconds**, the compensating element has not yet joined the action, i.e. the time charts are the same as in **Fig. 30-11**. It is clearly visible how the** x(t)** step is ahead of **Tc1** and** Tc1** is ahead of **y(t)=Tc3**. Let me remind you that **Tc1** as the temperature in the coil of **tank 2** is the **input** signal to this object, and **y(t)=Tc3** (temperature measured with a thermometer) is the **output** signal.

In** 40 seconds**, a disturbance **z(t)=+0.4** (additional heating) appeared, but **y(t)** did not even move! Great, compare it with **Fig. 30-8** where there was “normal” compensation, otherwise “without reverse transmittance”. A little comment here. However, it “**twitched**” a bit, because this is not a perfect **reverse** compensation. But an **insensitive** oscilloscope doesn’t show it.**Further** analysis is difficult:**Firstly**

The oscilloscope “cuts”. The control signal **s(t)** got a **negative** kick in **40** seconds -> **negative pin** about **-120**! This is the **differentiation** effect of the compensating unit. And only thanks to a small **inertia** (the denominator of the compensator) the pin is not a **dirac** with a minus **infinity** signal. In fact, the pin is slightly stretched in time as in** Fig. 30-11**.**Secondly**

the time charts partially overlap, therefore in **Fig. 30-14** the following are shown **separately**:**– Temperature**in the coil **Tc1 **the input signal to the tank **2****– disturbance z(t)=+0.4**

The control signal **s(t)=-120** in **40 sec** is a temporary powerful **cooling** of the **Peltier** heater-cooler. Now the temperature **Tc1** in the **tank 2** coil will drop almost immediately, thus compensating for the disturbance heater **z(t)=+0.4**.

**Fig. 30-14**

Thanks to the compensator, the input signal **s(t)**, (actually its negative “cooling” pin **Fig. 30-13**), the temperature **Tc1** of the tank** 2** coil will be almost “negative rectangular”. And this will almost perfectly compensate for the disturbance **z(t)=+0.4**. Recall that **Tc1** is the input for** tank 2**. This is all clear from animation **Fig. 30-13**. And even more in **Fig. 30-14** where there are only **Tc1** and **z(t)=+0.4** time charts.

**Chapter 30.4 Closed System**

This is the most commonly used control scheme. You can say that everyone else is just a modification of it.

For example:

– **Open** System is a** Closed** System with a **broken loop**

–** Open **System** with Compensation** is **Closed** System + **Compensation** with a **broken loop****–Cascade** Control is a combination of two **Closed** Controls.

Fig.30-15

Closed System.

This is a copy of the **Fig. 29-10** from **Chapter 29**. The** open** object – **Heat Exchanger** controlled from the **0…+10V** potentiometer, was closed with a **negative** feedback loop with the **PID** controller. This is the end of the topic, because the entire course is about** Closed Systems**.

**Chapter 30.5 Closed System with Compensation – another name Closed-Open System****Chapter 30.5.1 Introduction**An

**Open System**with

**Compensation**can completely suppress the disturbance

**z(t),**which can be measured as in

**Fig. 30-5**. It is, of course, vulnerable to other uncompensated disturbances, such as the “disturbance” heater at the coil in tank

**2**. The solution suggests itself. Close the system with a feedback loop with a

**PID**controller. We expect this to result in perfect disturbance

**suppression**with full

**compensation**and normal

**disturbance suppression**.

We will explore

**different**combinations:

**– Full**

**compensation k=1**with disturbance on the object input –>

**chapter 30.5.2**

**– Incomplete compensation k=0.7**with disturbance at the object input–>

**chapter 30.5.3**

**– Full compensation k=1**with disturbance in the “center” of the object–>

**chapter 30.5.4**

**– Full compensation k=1**with disturbance at the input of the object and with

**additional**disturbance in the “center” of the object–>

**chapter 30.5.5**

**Chap. 30.5.2 Full Compensation k=1 with disturbance at the object input**

**Fig. 30-16**The

**Full Compensated System**from

**Fig. 30-7**after closing the loop with the

**PID**controller.

**Kp, Ti, Td**settings provide optimal response to

**x(t)**step.

**Fig. 30-17**

Up to the disturbance **z(t)=+0.4** in **40 second**, the response to the step **x(t)** is identical to that of the Closed System–>**Fig. 29-11** from **chapter 29**.

In **40 seconds**, the disturbance** z(t)=+0.4** caused an immediate reaction of the **compensator** in the **opposite phase** in the form of a **decrease** in the signal **s(t)** also by **0.4**. The disturbance **z(t)** was thus fully** compensated**. The output signal **y(t)** didn’t even budge (because the **PID** “thinks” there was no **disturbance**!) and the error** e(t)** still remained **zero**.

The control signal** s(t)** is the sum of the **PID->sPID(t)** controller and the compensator **signal**. Only the compensator **signal** reacted to** z(t)=+0.4**.**Chap. 30.5.3 Incomplete Compensation k=0.7 with disturbance at the object input**

We try to make everything perfect, including compensation. But it doesn’t always work, for various reasons. The disturbance **z(t)** may be difficult to measure, for example… This corresponds to the case where, for example, **k=0.7** and not** k=1**.**Fig. 30-18**Almost a copy of

**Fig. 30-16**. The only difference is

**k=0.7**and not

**k=1**. So we expect that there will be some disturbance effect on the output

**y(t)**.

**Fig. 30-19**Up to

**40 second**the response to the

**x(t)**step is of course the same as before. However, from

**40 second**only part of the disturbance

**z(t)=+0.4**ie

**0.7*0.4=0.28**is immediately compensated by the

**compensator**. The rest, i.e.

**0.4-0.28=0.12**, is suppressed by

**PID**and therefore not immediately! It lasts over

**10**seconds, which can be seen in the time chart

**s(t)**. All in all,

**y(t)**is not perfectly damped as with full compensation, but it is clearly better than the system with feedback without compensation –>

**Fig. 29-11**from

**chapter 29**. The response is as if it were a

**feedback system**with

**no compensation**, but with less disturbance.

**z(t)=+0.12**( and not

**z(t)=+0.4**)

**Chapter 30.5.4 Full compensation k=1 with disturbance in the “center” of the object**

It will arise when the open

**System with “Inverse” Compensation**from

**Fig. 30-12**is closed with a feedback loop with the

**PID**controller.

Fig. 30-20

Fig. 30-20

**Closed**system with

**full compensation**and with a

**disturbance**in the

**middle**of the object.

**Fig. 30-21**

Until the disturbance **z(t)=+0.4** in **40 seconds**, the response to the **x(t) step** is the same as for the **Closed System->Fig. 29-11** from **chapter 29**. Just at this time **z(t)=0**, i.e. the “compensating part”, does nothing. In **40 seconds**, **z(t)** appeared at the coil of **tank 2**, which was immediately* compensated by the signal **Tc1**. Therefore, **z(t)=+0.4** was not “noticed” by the **PID** and the response **y(t)** is the same as for a closed system without disturbance!**Note**

In **40 seconds**, a **negative pin s(t)** appears, exactly as in **Fig. 30-13**. This makes the temperature signal almost perfectly rectangular. The oscilloscope didn’t show it.**Chapter 30.5.5 Full Compensation k=1 with disturbance at the input of the object and with additional disturbance in the “center” of the object**

It will be a **Closed System** with **Full Compensation** from **chapter 30.5.2** with an additional **uncompensated disturbance**. A compensated heating disturbance **z1(t)=+0.4** will occur as usual in **40 sec**. and **uncompensated cooling** disturbance **z2(t)=-0.6** in **55 sec**.

Fig. 30-22

The **cooling** disturbance **z2(t)=-0.6** right next to the tank **coil 2** is caused by a **negative** voltage spike of **-6V** in **55 seconds**. I remind you that the heater is the so-called **Peltier** element, where** positive** voltage **heats** and **negative** voltage **cools**.

**Fig. 30-23**Until the disturbance

**z1(t)=+0.4**in

**40 sec**, the response to the

**x(t)**step is identical to that for the

**Closed System**without compensation. It’s just that during this time

**z1(t)=0**, i.e. the “compensating part”, it does nothing, as if it were not there. In

**40 sec**additional heating appeared

**z1(t)=+0.4**, which was immediately compensated by the signal

**s(t)**. Therefore,

**y(t)**itself did not change and the

**PID**controller did not even

**react**. So until then the system behaves as in

**Chap. 30.5.2**.

As for the “cooling” disturbance

**z2(t)=-0.6**, as it was uncompensated, it was suppressed with the usual

**PID**feedback. Except that for the PID itself it was only a disturbance

**z2′(t)=-0.6+0.4=-0.2**! Because the disturbance

**z1(t)=+0.4**was already taken care of by the compensation part of the controller

**k=1**.

**Chapter 30.5.6 Conclusions**

**1. A Closed System with Compensation**can theoretically (!) perfectly suppress a disturbance, provided that:

**– accurate**disturbances measurement

**– selection**of the appropriate transmittance of the

**compensator**. This involves a thorough knowledge of the

**Go(s)**object.

**When**the

**disturbance**is at the input of the object, the transmittance of the compensator is

**k=1**–>

**chapter 30.5.2**

**When**the

**disturbance**is in the “center” of the object, it is the inverse transmittance

**1/Gzk(s)**–>

**chapter 30.5.4**

**2. Even**when it is not possible to accurately measure or select the ideal

**compensator**, the quality of control is significantly

**improved**.–>

**chapter 30.5.3**. This also applies to the inverse transmittance when the disturbance is in the “center” of the object.

**3. In**the

**inverse transmittance**there is an

**s**in the

**numerator**(causing a strong differentiation of the signal. It is

**difficult**to realize (infinite instantaneous power) and therefore the practical results are not as beautiful as in our experiments. But even then there is a clear improvement in

**control quality**.

**4. Other**uncompensated disturbances are suppressed by

**PID**by ordinary feedback–>

**chapter 30.5.5**

**Chap. 30.6 Cascade System****Chap. 30.6.1 Introduction**

The system allows better suppression of a **specific** disturbance than the **classic Closed System** and a little worse than the **Closed System **with **Compensation**. Contrary to the latter, it does not require **measurement** of the **disturbance**, nor such precise knowledge of the **Go(s)** object.

The **principle** is similar to the **organization** of an **enterprise**. The **director** assigns a task to be performed by the **menager** of the subordinate **department** and **ceases** to be interested in it. He does not investigate the disturbances affecting the **ward**, whether **mr, Smith** came drunk or the **Client** did not pay the invoice. For the **director**, the monthly plan is to be completed. **Information** about a** disturbance** reaches the manager** faster** than the **director**, and therefore he has greater possibilities to **suppress** it. He will use the method of the **stick** (threat of dismissal) and the** carrot** (bonus). The** manager** will eliminate the **disturbance** faster than the **director**. In the **Cascade System**, the role of the **director** is performed by the **master controller** – most often **PID**, and the **manager** is the **slave controller** – usually** P** type.**Chap. 30.6.2 Cascade System with one disturbance**

More precisely, with a disturbance or even several affecting a certain part of the object.

Fig. 30-24**Cascade** system with disturbance

The **P** controller compares the temperature **Tc1** of the tank **1** liquid with the **PID** output signal. So an **inner loop** was created – an ordinary **closed** system, in which the **Tc1** output signal “tries” to imitate the signal from the **PID**. It is clearly visible which regulator is the **director** of the entire company – **2** tanks, and which is only the **manager** of tank **1**.

So **PID** is the **master** controller and **P** is the **slave** controller. Sometimes the subordinate is **PI** or even **PID**.

The disturbance **z(t)=+0.4** is the activation of the auxiliary heater located at the control heater of tank **1**. Compare this diagram with the usual **Closed System** in **Fig. 29-10** **chapter 29** . When we break the **internal** feedback loop from **Tc1** and replace the **P** regulator with a **bare** wire, we get an ordinary **Closed System**. From **Fig. 29-11** **chapter 29** it can be seen that the **disturbance** of a normal closed system has been completely **suppressed**.

And what to expect from the **Cascade System**?**Firstly****– The inertia** of tank **1,** which has been covered by a **strong** feedback, will decrease –>**Kp=10**. Therefore, the **inertia** of the entire object will also decrease and it will become easier to **control**. So it should give a **shorter** response time to **x(t)** step than **classical** feedback.**Secondly**– The disturbance attenuation

**z(t)**will improve dramatically. Before the

**PID**reacts, the

**P**controller will do it much faster! After all, it only has to do with one

**inertia**of tank

**1**.

I will say more. The

**P**controller will suppress

**z(t)**so quickly that the

**PID**will barely notice the change in

**y(t)**. The main work of suppressing disturbance

**z(t)**will be done by the

**slave**controller

**P**and the master

**PID**will only finish it.

Let’s check it.

**Fig. 30-25**

The output is **red y(t)**. Compare with the response of a **classic closed system**–>**Fig. 29-11** from** chapter 29**. Suppression of **z(t)** super as if it wasn’t there, but the response to the **x(t) **step is a shame to talk about. It’s easy to explain. Simply, the controller settings **Kp=10 Ti=7 sec. Td=2 se**c. have been selected for the system shown in **Fig. 32-11** from **chapter 32**. And as I wrote in the introduction, the **inertia** of the object in **Fig. 30-24** is smaller, because the inertia of tank **1** covered by the inner loop has almost disappeared. Conclusion -> there will be other **optimal settings**. Let’s check.**Chapter 30.6.3 Cascade with one disturbace and with more optimal settings**

The settings of the **PID** controller, i.e. **Kp=10 Ti=5 sec and Td=0.55 sec**, were selected by trial and error. Although they may not be fully optimal, there should be a clear improvement in time charts.

**Fig. 30-26**Recall that the output is

**y(t)**! Shocking. And if you compare it with the response of a classical closed system

**Fig. 29-11**from

**chapter 29**, it’s a megashock! One more thing. For example, see

**Fig. 30-25**, where the steady state was

**s(t)=y(t)**–> the control signal at the input of the object

**s(t)**equaled the output signal

**y(t)**. We got used to “it’s always like this”. And in

**Fig. 30-26**it is not (here instead of

**s(t)**there is

**sPID(t)**). How to explain it? The matter is simple. In general, it is always the case that in the steady state there is

**s(t)=k*y(t)**, where

**k**is the

**static**gain of the object with the controller. Let me remind you that the

**steady**state is the state of

**equilibrium**found by the controller. When

**s(t)**is too large, the controller will reduce

**y(t)**and vice versa. So far with normal feedback, it’s mostly

**k=1**, and then it’s always

**s(t)=y(t)**?

And what about cascade control in

**Fig. 30-24**? Is the gain of the object between the signals

**sPID(t)**and

**y(t)=TC3**equal to

**k=1**?

NO! because the inner loop around the tank

**1**, closing by the

**P**controller, caused

**Kz=0.909**. Such amplification “sees” the master

**PID**controller.

And this means that in steady state

**y(t)=0.909*sPID(t)**or

**sPID(t)=1.1y(t)**. This is confirmed in

**Fig. 30-26**.

**Chapter 30.6.4 Cascade with two disturbances**

Fig. 30-27**The** disturbance **z1(t)=+0.4** is additional heating entering the **inner** loop. Which is exactly the same as **z(t)=+0.4** in previous experiments. Therefore, we expect it to be very strongly damped as well.**The** disturbance **z2(t)=-0.6** is cooling (minus the Peltier!) of the input coil to tank** 2**. It is subject to ordinary feedback. The disturbance should also be completely suppressed, although not as fast as **z1(t)**.

**Fig. 30-28****Up** to **55 sec**, i.e. until **z2(t)** appears, the response is for obvious reasons identical to **Fig. 30-26**. The **disturbance** suppression **z1(t)** is so strong (so good!) that there is no visible effect on the output signal **y(t)**. Although there is, but the oscilloscope does not show it.**In 55 seconds**, a disturbance** z2(t)=-0.6** occurred – cooling at the **tank 2** coil. The **PID** controller reacted correctly by increasing the power of the **tank 1** heater. The influence of the disturbance** z2(t)** is already visible here, although it was also completely suppressed. Did the **P** slave controller have any (**positive!**) effect on the disturbance suppression **z2(t)**. I think so. After all, the internal feedback involving **tank 1** seems to have reduced its** inertia**. Now almost all the inertia of the **two** tanks is only in **tank 2**. And objects with smaller inertias are easier to control.

**Chap. 30.7 Ratio Control System****Chap. 30.7.1 Introduction****The** output signal **y(t)** of the ordinary **Control System** tries to follow the input signal so that in steady state **y(t)=x(t)**. The **Ratio Control System** does a similar job, just according to the formula **y(t)=k*x(t)**.

It can be used, for example, in the **Paint Mixing Plant**–>**Chapter 30.7.2**, where a **new paint** will be created by mixing **two** different **ones**. It is known that the** color** depends on the ratio of their **mixing**. So it is enough to **fill** the **mixing tank** with **2** flows of **two paints**, e.g.** red** and **green**, whose flows are related by the relationship **red=k*green.**

Another application is the **Steam Boiler**–>**Chapter 30.7.3**. It is known that optimal **combustion** depends on the right **ratio** of **gas** and **air**. It’s bad when there’s **too** much **gas**. Then not only will not everything **burn up** -> l**ow efficiency**, but the environment will be **polluted**. And when there is not enough **gas**? It is true that everything will **burn**, but the **excess air** unnecessarily **cools** the **boiler** and heats the atmosphere, reducing the energy **efficiency** of the boiler. It is therefore necessary to **ensure** an optimal **ratio** of gas and air flows.**Chapter 30.7.2 Paint Mixing Plant**

It is a** good example** of the application of the **Ratio Control System**. The right **color** is the **right** mix of **three colors**, here for simplicity **only** two** red** and **green**. We want a color where **green/red=1/3**. So it is enough to fill the tank with **2** flows from the pipelines – **green** and **red**, for which the **ratio** of flows.**Fig. 30-29**Technological and control diagram of the mixing plant

**first let’s talk about**

But

But

**valves**. You can leave if you know the topic.

The

**Ratio Control System**is most often associated with

**flows**. The

**flows**in the pipelines are controlled by

**valves**. Therefore, to

**begin**with, the absolute minimum of knowledge. Everyone has it at home. A

**closed valve**means

**no flow**, an

**open valve**means

**maximum flow**limited only by the

**pressure**in the network, the hydraulic

**resistance**of the

**piping**and the

**resistance**of the

**fully**open valve itself. By

**turning**the tap, I set the indirect

**flow**, i.e. I have the ability to

**contro**l it.

The most important

**feature**of each

**valve**is the so-called

**Kv**factor. I will not go into details further. I will just say that a small

**tap**in the house is small

**Kv**, and a large valve on the industrial pipeline is large

**Kv**.

The

**second**feature is its characteristics, i.e. the function between the

**shift**of the valve

**stem**(at home it is the

**angle**of rotation of the

**tap**) and the flow

**F**. It would seem that with unscrewing the valve, the “hole”, i.e. the

**surface**through which the liquid flows, should increase proportionally from

**0**(closed) to

**MAX**(open). So that the flow also increases in a

**linear**way, which is the most desirable in control. But it’s not like that! At

**first**, the area increases slowly and then

**faster**and

**faster**. Something like an

**exponential**function. This is the so-called

**equal percentage characteristics**. It turns out that just such

**non-linearity**will provide (approximately) a

**linear characteristic**of the

**valve**, when the

**input**is the

**movement**of the valve stem (and plug) and the

**output**is

**flow**.

**How to explain it?**

At the

**beginning**there is a small “hole”, but the

**liquid**has a terrible desire to

**flow**through it, because there is a

**large pressure drop**between the two sides of the

**“hole”**. Then the

**surface**area

**increases**, but the

**desire**to flow

**decreases**because the

**pressure**drop

**decreases**. Overall, the increase in surface area and the decrease in pressure work against each other and the characteristic becomes more linear.

**From the above**, it follows that when selecting a valve, you must select:

**a-Kv**such that for

**x(t)=1**(+10V on the potentiometer) the valve is

**fully open**->

**maximum flow**

**b-valve**with

**equal percentage characteristics**

**ad.a**

-When

**Kv**too

**large**, it will open only, for example, up to

**10%,**giving the

**maximum flow**limited by the resistance of the

**external piping**– it’s a pity for a

**large**valve when you can give a

**smaller**one, i.e.

**cheaper**.

-When

**Kv**too

**small**, it will open almost immediately to

**100%**, then it works as a permanent

**choke**, although the pressure of the network and piping could give a greater flow.

**ad.b**

-Provides the desired

**linearity**in automatics

How does the mixer work?

How does the mixer work?

With the adjuster –

**potentiometer 0…+10V**, we set the maximum flow

**Fcz**of the

**red paint**, which is measured with the flow

**converter**

**PP**. The maximum flow also corresponds to

**+10V**at the output of the

**PP converter**. The

**P**controller

**compares**the

**setpoint**value

**x(t)**with the

**current**flow and generates an appropriate

**control signal**for the

**SE**electric

**actuator**. This signal moves the

**valve stem**at a

**rate**proportional to the

**voltage**from the

**controller**.

**Positive**voltage

**opens**the

**valve**,

**negative**voltage closes it, and

**zero**voltage disables it. So

**valve+actuator**is a typical

**integrating**unit with a transmittance of

**1/5s**. The number

**5**in the denominator means that at

**+10V**, but without

**feedback,**the valve will open to the max within

**5 sec***. The valve with the

**PP**converter is an

**inertial**element with a time constant

**T=1 sec**. The

**pipeline**with

**red**paint + automatics form the

**I**– i.e.

**integrating**control system. Control

**I**was discussed in detail in

**Chapter 28**. The

**green**part works in a similar way. Here, in turn, the

**set value**is the

**flow**

**1/3*Fcz**. We expect that in the steady state the flow

**Fz**will be

**3**times smaller than z Fcz. Let’s check.

* the

**s**symbol in the

**1/5s**transmittance is not a

**second**, but a complex number

**s**, as it happens in

**transmittances**.

**Fig. 30-30**The goal of the

**Ratio Control**has been achieved. The

**green**

**Fz**flow in steady state is

**one third**of the

**red Fz**.

**Chapter. 30.7.3 Steam Boiler**

It is a less trivial example of a

**Ratio Control**than a

**Paint Mixing Plant**.

**Fig. 30-31**Technological and block diagram of a typical

**Steam Boiler**.

It is a

**drum**filled with

**water**and

**heated**with natural

**gas**. Supplies various devices not shown in the diagram with steam. The

**automatics**tries to maintain a constant steam

**pressure**at the boiler outlet

**cp(t)**regardless of disturbances, which may be, for example, switching

**on**an additional steam

**receiver**.

**Gas**needs

**air**, more precisely

**oxygen**. Optimum combustion ensured at a certain

**ratio**of

**air**flow to

**gas**flow.

If it is

**too small**, not all the gas will be

**burned**–> lower

**efficiency**and environmental

**pollution**.

If it is

**too large**, the unburnt

**air cools**the exhaust gases

**unnecessarily**–> lower

**efficiency**.

Boilers are not my field, but

**k=10**is supposedly

**optimal**. To make the

**graphs**of

**gas**and air

**less**different from each other, for

**didactic**reasons we will assume that

**k=1.5**.

Now let’s look at the schematic from the

**automatics**point of view. We distinguish the following

**dynamic blocks**in

**it.**

**Gkoc(s)-Steam Boiler**

**Input**– gas flow

**Fg(t)**and

**k=1.5**times the air flow

**Fp(t)***

**Output**– pressure

**Cp(t)**on the pipeline. More precisely, it will be the voltage, where

**0…10V**is

**0…10 MPa**.

The physics here is

**very simple**. We assume that by some miracle the

**water level**is kept

**constant**regardless of the

**steam discharge**. This is usually done by a very precise and fast

**level control system**. It provides fluctuations of less than 1 cm! And the

**steam**keeps going!

This is where

**Cascade Control**comes into play. It is also possible, by additionally measuring the water

**inflow**to the

**drum**and the

**steam outflow**, to accurately calculate the

**disturbance**affecting the

**level**. That is

**Closed System with Compensation.**

These were just side notes. So we’re still stuck with the

**constant-level**miracle. An

**increase**in the gas supply

**Fg(t)**and

**Fp(t)**increases the

**temperature**and thus the

**pressure Cp(t)**of the steam. Of course, not immediately, but just like the

**inertial unit**with a time constant

**T=120 sec**, subject to

**negative feedback**.

***Fp(t)**flow is only shown in the top drawing at the

**drum**. However, it is not shown in the lower figure as an input to

**Gkoc(s)**. And rightly so, because the vapor pressure

**Cp(t)**depends only on the flow

**Fp(t)**, assuming that

**Fp(t)=k*Fg(t)**.

**Gg(s)-Gas Control Block**

**Input**– control signal

**s1(t)**from the

**PI**controller

**Output**– gas flow

**Fg(t)**

**SE-Electric Actuator**– Unlike the

**SE**actuator in the

**Mixing Plant**, it is a

**proportional**and not an

**integrating unit**. So it has (not shown in the diagram) internal feedback from the

**position**of the

**stem**.

**Cascade Control**is bowing again -> there will be better

**time charts**.

**PP**–flow transducer converts the flow signal

**Fg(t)**to

**0…10V**.

**Valve**–Stem displacement changes into an increase in the flow surface, i.e. a

**change**in

**flow**. The designer selected the appropriate valve size->

**Kv**and equal percentage characteristics to achieve

**linearity**. We assume that

**Gg(s)**is an

**inertial**unit with a time constant

**T=5 sec**.

**Type PI Pressure Controller**

**It**constantly compares the pressure

**Cp(t)**with the

**setpoint**value

**x(t)**.

**When**

**Cp(t)**increases, the error

**e1(t)**decreases–>

**Fg(t)**decreases, the water

**temperature**and steam pressure

**Cp(t)**decreases.

**When Cp(t)**decreases, the error

**e1(t)**increases–>

**Fg(t)**increases, water

**temperature**and steam pressure

**Cp(t)**increases.

In this way, the

**controller**tries to keep the pressure

**Cp(t)**constant equal to the setpoint value

**x(t)**. The

**PI**control will ensure

**zero error**and good time charts.

**Gp(s)-Air Control Block**

**Input**– control signal s2(t) from the PI controller

**Output**– airflow Fp(t)

Operation and dynamics the same as

**Gg(s)**–

**Gas Control Block**

**PI Flow Controller**

The setpoint is

**k*Fg(t)**. Therefore, the

**controller**tracks the airflow ensuring

**Fp(t)=k*Fg(t)**.

Let’s see how the whole system works.

**Fig. 30-32**The goal of control the has been achieved. In steady state, the

**air flow**is

**1.5 times**greater than the

**gas flow**. I emphasize that the flows of gas

**Fg(t)**and air

**Fg(t)**are only auxiliary signals here. The most important is the steam pressure

**Cp(t)**.