### Automatics

**Chapter 22 Continuous Control**

**Chapter 22.1 Introduction**

**Fig. 22-1**

**Fig. 22-1a**

The most general control diagram showing:

**–**

**setpoint x(t)**

**– output signal y(t)**–other name process variable

**– disturbance z(t)**

**– control error e(t)**

**Fig. 22-1b**

A more accurate version in which the entire

**G(s) objec**t has been divided into an

**ON/OFF**controller and the proper

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

**s(t)**appeared

**after**the controller.

**Fig. 22-1c**

Instead of a

**ON/OFF**controller, there is a

**P, PI, PD**or

**PID**

**continuous**type controller.

The control

**signal**

**s(t)**, unlike the

**ON/OFF**controller, is

**continuous**.

**Chapter 22.2 General and typical form of the Go(s) object**

From now on, we are only interested in continuous control, i.e. **Fig. 22-1c.** But first, let’s deal with the object itself, which will be controlled by the controller. The **Go(s) **examples are:

Fig. 22-2**Fig. 22-2a**

General form of the** G(s)** transmittance where the numerator **L(s)** and the denominator **M(s)** are polynomials.**Fig. 22-2b**

Specific example where the numerator **L(s)** is a **polynomial** of degree **2** and the denominator **M(s)** of degree **4**.**Fig. 22-2c**

Cases of **Go(s)** object transmittance when **L(s)** is of **degree** greater than **1** are rather rare! A typical numerator **L(s)** is only the gain **K**. Here **K=10**.**Fig. 22-2d**

In turn, most often **K=1** and we will treat this **transmittance** as typical. The denominator **M(s)** is still a** polynomial** of any degree **n**. Often **M(s)** is a product of.. E.g. **M(s)=(1+sT1)(1+sT2)(1+sT3)**** ****Fig. 22-3**An example that

**K=1**is typical as in

**Fig. 22-2d**

The

**Go(s)**

**object**is a water tank with a resistance

**heater**controlled by a

**power amplifier**. The

**water temperature**will be controlled in the range of

**0…+100ºC**. We measure it with a

**Pt100**resistance thermometer with a measuring range of

**0…+100ºC**–>voltage

**0…+10V**. The thermometer consists of a

**platinum**wire in a metal casing-tube and a

**ΔR–>0…10V**converter.

**ΔR**is the increase in resistance

**R**due to an increase in

**temperature**. The tube with the wire is quite delicate and therefore it is placed in a metal

**sheath**. The

**transducer**protrudes outside the tank, from where the signal can be connected with a cable to the

**controller**. In the picture you can see how the

**sheath**enters the tank through the hole.

The

**power amplifier**is tuned so that

**+10V**per input will heat up to

**+100ºC**, but without boiling! Also the temperature transducer will give

**+10V**. And when it is not possible to achieve this value,but for example

**+55ºC**? Now you need to increase the

**boiler**power by increasing the power of the

**amplifier.**

When

**s(t)=0…+10V**then

**s(t)=0…+10V**. That is relationship between

**s(t)**and

**y(t)**is like the

**K=1**amplifier. Of course, with time constants

**T**which are included in the transmittance denominator

**Go(s)**. The degree of the

**M(s)**polynomial is the

**greater**the more

**accurately**we know the mathematical description of the water heating process in the tank. The simplest approximation, is an

**inertial unit**.

**Conclusion**

The

**Go(s)**object is a

**water tank + heater + Ko=1 power amplifier**. And when in front of the object, i.e. in front of the power amplifier, there is also an voltage amplifier (not power!) with a

**gain**of, for example,

**Kp=5**? Later you will find out that this amplifier is a proportional controller or in other words a

**P controller**. Anyway, the whole open circuit is

**K=Kp*Ko=Kp**, because

**Ko=1**So it depends only on the setting of the

**P**controller –

**Kp**gain. And that helps us. controller tuning. Thus, such

**Kp**that provide an appropriate response to the

**x(t)**step.

**Chapter 22.3 Static and astatic Go(s) objects.**

The **automatics engineer** begins the project with a good knowledge of the **Go(s)** objects he is to control, i.e. its **static** and **dynamic** properties.

These are:**-static****-astatic**

**Chapter 22.3.1 Static objects.**

You are examining the simultaneous response to a

**step x(t)**of

**three**typical s

**tatic**objects

**Fig. 22-4**

**G1(s)-Inertial**unit

**K=1 T=2sec**

**G2(s)-Double**

**Inertial**unit

**K=1, T1=2sek, T2=3sec**

**G3(s)-Oscillation**unit

**K=1, q=0.3, T=1sec**–>

**chapter.7 Fig. 7-2**

In

**steady**state, all outputs in response to

**x(t)**are

**constant**, i.e.

**static**. Hence the name

**static objects**. Note that each denominator

**M(s)**of the transmittance

**G(s)**does not have a

**zero root**.

**Chapter 22.3.2 Astatic objects**

Simultaneous response to a

**step x(t)**of three typical

**astatic objects**.

There are

**zero roots**in the denominator

**M(s)**. Even double–>

**G2(s)**.

**Fig. 22-5****G1(s)-Integral****G2(s)-Double Integral****G3(s)-Integral with inertia**

In **steady** state, the outputs **y(t)** keep increasing! **y1(t), y3(t)** at a constant **speed**. For **y2(t)** even better, with the** speed** that is increasing **all the time** – like after the start of the** Apollo rocket**! This is typical for **astatic** objects, i.e. **non-static**. That is, those that, in response to a **step x(t)** in a **steady** state, **move** with a **constant** or with an **increasing speed**. A characteristic feature of these objects is also a **non-zero** output with **zero input** in a **steady** state. This feature ensures precisely **zero control** error is related to the **I** component of the **PID** controller. This will be discussed later.**Interpretation****G1(s)** – an ideal **DC** motor with a gear and a lever. So, in our way, actuator. The **input** is the **voltage** on the motor, the **output** is the angle** α**. This **α** can control, for example, a **valve** in the **pipeline**. I emphasize that there is no **negative feedback** here, i.e. the angle can change theoretically from **–** to **+** **infinity**. In practice, the engine will be switched **off** earlier by the so-called **limit switches**. And even earlier it is turned off or even “turned back” by the regulator providing negative feedback.**G3(s)** – Same thing, just an imperfect actuator. That is, taking into account **inertia** and **friction**. Hence the **acceleration** effect before it **reaches** a **constant** speed.**G2(s)** – Rocket as above, or an **actuator** that **increases speed** all the time!

The latter is a purely **theoretical** case, rather not used in practice. Or maybe they are?

An** astatic** object can only be **stopped** by giving **zero** to the **input**.**Even** more trivial.

With a **non-zero** input **x(t)**:**Static** in a** steady** state, it does not move. **Astatic** in a** stead**y state it moves.

**Chapter 22.4 Comparison of static and astatic objects due to stability****Chapter 22.4.1 Introduction**

It is probably easier to control something that **does not move** in a steady state than something **that moves**.

Therefore, **static objects** are **“easier”** than astatic ones. Let’s check it.**Chap. 22.4.2 Static**

A typical **2nd** degree static object with **negative** feedback.

**Fig. 22-6**The system reached a

**steady**state

**y=0.91**. It is the fixed

**error**, here

**e=0.09**, the

**smaller**the greater the gain

**K**, is typical for static systems with

**negative**feedback. If you want to

**reduce**the error, you need to increase

**K**. However, this will cause

**greater oscillation**and sometimes even

**instability**. It can be proved, for example, from the

**Nyquist**criterion that systems of the

**two-inertial**type will never become

**unstable**, even with a large

**K**. Because their

**amplitude-phase**characteristics only pass through

**2 quadrants**. Only those

**transmittances**whose denominator

**degree**

**M(s)**is greater than

**2**can enter

**instability**.

**Chap. 22.4.3 Astatic**

**Astatic**ones are more

**difficult**to

**control**because they become

**unstable**more

**easily**when closed with

**negative**feedback.

**Fig. 22-7**The

**zero**root

**s=0**in the denominator

**M(s)**of transmittance appeared. A typical feature of

**astatic**systems. Let’s give

**5 seconds**for the input “click”

**x(t)**. I already know what will happen. Therefore, I radically changed the scale of the oscilloscope. Previously it was

**0…+1.2**, now

**-150…+150**. Misfortune. Instability, oscillations from

**–**to

**+ infinity**. And only the small letter

**s**in the transmittance in

**Fig.**

**22-7**did it.

And how to make the system

**stable**? Generally, it does different things. The first thought is to reduce

**K**.

It will be quite a

**radical**reduction from

**K=10**to

**K=0.25**.

**Fig. 22-8**The most important is that the answer is stable. But how slow! Compare with

**Fig. 22-6**, where the steady state occurred after approx.

**10 seconds**, here after

**55 seconds**. But there is one nice thing about it.

**After 45 sec**. we have the state

**y(t)=x(t)**. So

**zero error**! And that’s regardless of

**K**! For example,

**K=0.1**, the fixed error would also be zero. Unfortunately, at the expense of slow time charts,

That is,

**astatic**objects with negative feedback are more

**susceptible**to

**oscillations**and

**instability**. They are also

**slower**to reach a steady state. However, they can

**reduce**the

**error**to

**zero**, which is of course their main advantage.

**Chapter 22.4.4 Comparison of static and astatic with feedback in a common graph**

We will compare the respones with:

**-Fig. 22-6**(static stable)

**-Fig.**

**22-7**(astatic unstable)

**-Fig. 27-8**(astatic stable).

In these figures, there were different

**oscilloscope**settings for time

**t**and

**y(t)**. Therefore, we will examine all

**3**transmittances again, but this time

**simultaneously**.

**Press start**

Fig. 22-9

Fig. 22-9

The same step

**x(t)**enters all

**3 objects**:

–

**G1(s)**static

–

**G2(s)**astatic – high gain

**K=10**therefore unstable

–

**G3(s)**astatic – low gain

**K=0.25**therefore stable

**Fig. 22-10**

These vertical yellow lines are increasing oscillations of the object G2(s) with negative feedback. In** Fig. 22-7** there was an answer to x(t)=dirac and now x(t)=step. **Conclusions**:**– Astatic** systems are more **susceptible** to **instability** than static ones**– Stable** static system **G1(s**) clearly reaches a **steady** state** faster** than **stable astatic G3(s)****– Stable astatic** system **G1(s)** may have a **zero error**! Of course, provided that it is stable, which is more difficult for astatic ones. It is precisely this zero **control error e(t)** in the steady state that is the basic advantage of** astatism**.**Static systems**, unlike astatic ones, provide** zero error** only for the gain **K=infinity**. This rather theoretical condition can cause large oscillations and even instability.

**Chapter 22.5 Man as a Continuous Controller****Chapter 22.5.1 Introduction**

A customer contacted us about the design of the** control system** for the facility shown in** Fig. 22-3**.

He is a poor investor and cannot afford a **controller**. He will do it for him himself, i.e. he will control it manually

About the facility itself, he only said:

– it is a **tank** with a **liquid** z which does not change state at all. This assumption will facilitate our analysis.

– wishes to **control** the **temperature continuously** in the range of **0ºC…+50ºC**

He also cares about the **fastest** possible transition from one state to another, e.g. from **+20ºC to +40ºC**, or vice versa from **+40ºC to +20ºC**. The latter is associated with active cooling, i.e. the heater will turn into a refrigerator! Hence **the cooling Peltier** element.

And why should the controller be **continuous** and not **ON-OFF**? Because he knows that in a steady state for **ON-OFF** control, the output signal **y(t)** oscillates around a certain average value all the time. And this has a very **bad** effect on the quality of the customer’s **product**!**Chapter 22.5.2 Man as a Continuous controller without feedback**

We proposed **open-loop** control, or if you prefer control **without feedback**.

**Fig. 22-11**It is almost repeated

**Fig. 22-3**with a man as a

**controller**and a

**potentiometer**as a direct

**control device**. We already know about the object itself that changes in the voltage

**s(t)=0…+10V**at its input correspond to the same changes

**y(t)=0…+10V**at the output, but in a steady state! Strictly the same changes but only for

**s(t)=0…+5V**. We just assumed that the output temperature will

**never**exceed

**+50ºC**and there is no point in giving the thermometer range

**0… +100ºC**.

So the whole complicated object with a

**boiler, liquid and thermometer**has been

**reduced**, even

**degraded**to an ordinary

**amplifier**with a gain of

**K=1**with certain

**T**time constants. Because if there is e.g.

**+3V**on the potentiometer, there will also be

**+3V**on the

**steady-state**output of the

**thermometer**(meausuring converter)! This approach greatly facilitates the selection of a continuous controller

**PID**type, in which the entire

**controller**dynamics of the

**closed**system will depend only on the

**Kp**,

**Ti**and

**Td**settings of this controller. This will be discussed in the following chapters.

**User manual**

Our

**Client**not only cannot afford the

**controller**, but he is also an orphan–> he needs to be taught how to control it, i.e. give him an

**instruction manual**.

**1.**It will control the potentiometer providing the

**voltage**in the range

**-10V…+10V**. It is

**200 mm**long and is calibrated in units of

**0.1V<=>1ºC**or

**0.1mm<=>1ºC**. It is

**0ºC**in the middle,

**+100ºC**at the

**top**and

**-100ºC**at the

**bottom**.

One more thing. This is an

**automatics**course, not

**physics**. Therefore, the liquid does not change state -> does not

**freeze**or

**evaporate**.

In addition, our electric

**boiler**heats with

**positive voltages**and

**cools**with

**negative voltages**! There is a freak known since the

**19th**century – the

**Peltier**element.

The

**Client**reads the manual with mixed feelings

He has at his disposal:

**-control device**–> potentiometer with a range of

**-10V…+10V,**otherwise

**-100ºC…+100ºC**

–

**temperature converter**with a range of

**0…+50ºC**, otherwise

**0…+5V**

The Client is surprised. Should I control the temperature of the liquid in the range of

**0…+50ºC**? Why redundant “heating/cooling” power, e.g.

**-10 kW…+10kW**, if

**0…+5 kW**is enough for

**0ºC…+50ºC**? Or maybe the

**designer**wants to let me in on costs, like a

**taxi driver**driving not the shortest route? Why do I need control enabling temperatures of

**-100ºC…+100ºC**and the

**temperature transmitter**range is only

**0…+50ºC**?

And so, with doubts, he proceeded to the

**first**attempt to heat the liquid to

**+30ºC**.

**The first attempt to heat the liquid to +30ºC**.

He approaches the problem as carefully as a dog to a hedgehog. Therefore, he moved the slider up by

**30 mm**to

**s(t)=+3V**or, if you prefer, to

**+30ºC**and waits for the effect. The

**Client**blindly believes in the

**s(t)**control slider and does not even look at the

**thermometer**! I will add that the ambient temperature is

**0ºC**. The client’s faith had a basis. After some time, the temperature increased to

**+30ºC**as shown below.

**Fig. 22-12**You can see in the upper left corner how the

**Client**moved the potentiometer slider from

**0V to +3V**after about

**5**seconds, which corresponds to the temperature

**step**from

**0ºC to +30ºC**. The

**step**wasn’t perfect. Our

**Client**is getting old and the paw is shaking a bit. Anyway, you can see how beautifully the temperature

**y(t)**goes from

**0ºC to +30ºC**.

It was an example of absolutely

**open**control. i.e. The

**Client**didn’t even look at the

**thermometer**, just trusted the

**control slider**completely. The output signal

**y(t)**had no effect on the control signal

**s(t)**from the slider! And the

**time chart**itself is a classic

**inertial unit**with

**K=1 and T=10sec**

**Chapter 22.5.3 Man as a Continuous controller with feedback.**

It would seem that

**open-loop control**is great! The

**Client**set the

**slider**to

**s(t)=+3V=+30mm=+30ºC**and after some time the temperature reached

**+30ºC**What more could you want?

But

**First**

The ambient temperature was

**0ºC**. If it were otherwise, e.g.

**+13ºC**, then in steady state it would be

**y(t)=+43ºC**(actually a little less) and not

**y(t)=+30ºC**!

**Second**

We assumed that the

**object**the object has been identified

**mathematically**and

**physically**i.e. we know that at an ambient temperature of

**0ºC**and

**s(t)=+3V**on the potentiometer, the temperature of the liquid should be set exactly at

**y(t)=+30ºC**.

**Thirdly**

We assumed the

**potentiometer**was perfect. i.e. that

**30 mm up**is exactly

**s(t)=+3V**

**Fourth, etc…**

In short, we assumed everything

**was perfect**. And these were just

**blissful wishes**. So we cannot be sure that

**30 mm up**on the potentiometer will heat the liquid to exactly

**+30ºC**. There can only be one

**conclusion**.

**The Client must constantly observe the output signal y(t) and correct it with the potentiometer slider.**

So the diagram should look like this.

**Man as a**

Fig. 22-13

Fig. 22-13

**Controller**

**Fig. 22-13a**

The

**Client**constantly observes the output signal

**y(t)**and, depending on the situation, corrects it with the voltage signal

**s(t)**from the potentiometer. So in the “head” he calculates the error

**e(t)=x(t)-y(t)**and combines it so that the error

**e(t)**in the steady state is as small as possible. The unachievable ideal is that

**e(t)=0**at any time. Then

**y(t)=x(t)**, i.e. the output signal

**y(t)=x(t)**. Otherwise–>

**y(t)**tries to follow the set value

**x(t)**.

**Fig. 22-13b**

The

**Client**used some algorithm to reduce the error

**e(t)**to

**zero**. It can also be performed by a

**microprocessor**.

Then the man is replaced by the

**controller**as above.

**Fig. 22-13c**

There are many such algorithms. One of them is surprisingly simple!

**step 1**calculate the current control error

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

**step 2**calculate the control signal

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

**step 3**return to

**step 1**

In this way, the

**algorithm**of the proportional controller

**P**will be implemented in which, with a

**step**change of the set value

**x(t)=1**, the output signal

**y(t)**will tend to a value almost equal to the set value, e.g.

**y(t)≈0.91**when

**Kp=10**. So the error will be “almost” zero->

**e(t)≈0.09**. The more precisely, the greater the

**gain**of the

**controller**and the

**Go(s)**object.

Go back to

**p. 16.3**from

**chapter 16**, then you’ll know why.

One more thing. The

**y(t)**signal in practice rarely goes beyond the range

**0…+5V**, i.e.

**0…+50ºC**. On the other hand, the control signal

**s(t)**enables the achievement of –

**10V…+10V**states! i.e.

**-100ºC…+100ºC**!* Why such distortion?

It enables a faster transition from one temperature to another. Generally from one state to another. Although the input

**x(t)**and output

**y(t)**signals change only in the range of

**0…+5V**, i.e.

**0…+50ºC**. Such overshoots are typical of any regulation, including

**PID**.

**Let’s check how the man-controller works**

The

**Go(s)**object itself is a tank with a

**heater**of such power that in about

**40 seconds**, the liquid reached a temperature of

**100ºC**. So I will control the

**Go(s)**object – an

**inertial member**with a gain

**K=1**and a time constant

**T=10 sec**. More precisely, I will try to make

**y(t)**, i.e. the temperature of the liquid, as similar as possible to the set value

**x(t)**in time, i.e. a

**step**from

**0**to

**+30ºC**. You will find out how hard the life of Mr.

**Controller**is. Your task will be to observe the setpoint value

**x(t)**and the output

**y(t)**, i.e. the

**temperature**of the water in the glass and controlling the potentiometer

**slider**by the

**Client**. The movements of the

**slider**will be visible in the

**upper left**corner.

**Fig 22-14**You can see how in the

**5th**second

**Client**gives the control signal

**s(t)**to the

**max**, i.e. he sets the potentiometer slider to

**+10V**, i.e.

**to +100°C**. Note that this is a larger signal than needed

**i.e. +30°C**. But thanks to this, the

**y(t)**signal will reach

**+30°C**faster. As you will see later, this is how the

**P, PI**and

**PID**controllers initially behave. Then, seeing how

**y(t)**increases,

**s(t)**decreases. A more experienced operator would even use intensive cooling to get to

**y(t)=+30°C**faster. With such control, even decreasing oscillations may occur. This can be seen even with “cautious” steering in

**Fig. 22-14**. After all,

**y(t)=s(t)**and is quite close to

**x(t)**. And why not exactly

**x(t)**? Well, the author is old and he treated a small error

**e(t)**as zero. And it is not.

**Chap. 22.6 Man-Regulator with disturbances****Chap. 22.6.1**

The **Go(s)** affected by** disturbances****z(t)=+20ºC****z(t)=-20ºC**

A disturbance may be, for example, a ** ** through which a hot liquid of **+20ºC** or a cooling liquid of** -20ºC** flows, as described in **Fig. 21-12**, **chapter 21**. In the diagrams, disturbances** z(t)** are simulated with a positive or negative voltage steps **+/-20ºC**.**Chapter 22.6.2 The Client as a controller not reacting to a disturbance z(t)=+20°C**

The Client will react only to **x(t)**, which is exactly as in **Fig. 22-14**. In **35 seconds**, a disturbance with **+20°C** will appear. You can think of it as putting an additional electric heater in the tank with such power that in steady state it will raise the temperature of the liquid by **+20°C**.

**Fig.22-15**

The** Client** does not respond to the **disturbance**. Then he made excuses explained that he was curious what the effect would be. And he saw. The temperature rose to** y(t)=+50°C**.**Chapter 22.6.3 The Client as a controller reacting to a disturbance z(t)=+20°C**

**Fig.22-16**Up to

**35 seconds**, the

**Client**behaves as in

**Fig. 22-14**and

**Fig. 22-15**. After

**35 seconds**, when

**z(t)=+20°C**(additional heater) appears, it worked properly. He reduced

**s(t)**until the temperature returned to approximately

**y(t)=+30°C**. Way to go. The green error

**e(t)**dropped to almost

**0**. And why “almost” and not “exactly” to

**0**. Well, that’s the

**Client’s**vision. When he was younger, he would have steered more precisely.

**Chapter 22.6.4 The Client as a controller not reacting to a disturbance z(t)=-20°C**

**Fig. 22-17**This is an experiment similar to

**Fig. 22-15**. The only difference. Instead of an additional

**heater**, a

**cooler**was inserted. Its power was selected so that in the steady state the temperature of the liquid

**y(t)**in the tank was lowered by

**-20°C**. It can be, for example, the previously mentioned so-called

**Peltier**element. The

**Client**reacts only to

**x(t)**and not to the disturbance

**z(t)=-20°C.**The liquid temperature y(t) actually dropped by

**-20°C**.

**Chapter 22.6.5 The Client as a controller reacting to a disturbance z(t)=-20°**

**Fig.22-18**

Control analogous to **Fig.22-7**

The problem here is **cooling** instead of **heating**. The **Client** reacted correctly to the **cooling** by increasing the **heating** with the **s(t)** potentiometer slider.

**Chapter 22.7 Conclusions****1**. In** continuous control**, the control signal **s(t)** assumes all values of **s(t)** in the **Min…Max range.****2. Automatics** is dealing with **2** types of **Go(s)** dynamic objects**-Static–>Fig.22-4****-Astatic–>Fig.22-5****3. The ranges** of the set value **x(t)** and the output value** y(t)** are the same. E.g. **0ºC…+50ºC** (0…+5V). And that’s how it is, or I’ll say it as a precaution, that’s how it should be in control systems with real **controllers**. So the steady-state gain between output and input is (always?) **K=1**.**4. The control** signal **s(t)** should have a larger range than the above** 0ºC…+50ºC**. In our examples it was** -100ºC…+100ºC** (-10V…+10V). This improves the dynamics! Often this range is even **greater**. And already a** smaller range** is unacceptable! Certain states would simply be **unattainable**. It would be rather difficult to** heat** a tub of water with a **“glass”** electric **heater**!**5. Although** you steered manually, it was** negative feedback**. It was closed by the **Client** as **Mr. Controller**. Therefore, the time charts are (slightly stretching reality) similar to systems with a “real”** P-type** **controller**, as you will see in the following chapters.**6. The Client-Controller** tries to properly control the **s(t)** potentiometer so that the output signal** y(t)** follows the setpoint value **x(t)**. For this glorious purpose, **disturbance** signals **z(t)** try to prevent it. When a **z(t)** heats, the **controller** tells it to **cool** and vice versa.