# Automatics

**Chapter 31 Influence of Non-linearity on Control**

**Chapter 31.1 Introduction **The most influential

**non-linearity**on the control process is

**saturation**in the

**power amplifier**(another name –

**actuator**) in front of the controller. Therefore, the title of the chapter could be more explicit – “Impact of actuator

**saturation**on the quality of control”. You’ll soon know what it’s all about. Even one

**non-linear**element “spoils” the whole scheme. So you can’t uncritically apply the

**theory**you’ve learned so far, in which all elements were

**linear**.

**Transmittance reminder**.

When you use

**transmittance**it means you are in

**linear**objects. In these objects, the output signal

**y(t)**in

**steady**state is proportional to the input signal

**y(t).**

**Characteristics of a linear object in a steady state.**

Fig. 31-1

Fig. 31-1

And in the

**unsteady**(or transient)

**state**,

**linear**objects are governed by

**linear differential equations**, in which all

**coefficients**at

**y(t), y'(t), y”(t), … x(t), x'(t), …**are constant. such as:

Fig. 31-2

Fig. 31-2

**Differential**equation of a

**linear**object

Fig. 31-3

Fig. 31-3

**Examples**of differential equations that are

**not linear**

The

**red**ones are bugs that make the

**differential**equation not linear,

**read**– difficult to solve. Let’s return to the

**linear differential equation**in

**Fig. 31-2**, which corresponds to the transmittance

**G(s)**.

Fig. 31-4

Fig. 31-4

The transmittance

**G(s)**which is described by the

**differential**equation in

**Fig.**

**31-2**

For the exercise, associate parameters

**1, 2**and

**3**of the numerator and 1, 2 of the denominator of the above transmittance

**G(s)**with the

**linear differential**equation in

**Fig. 31-2**. If you have problems, look at

**Fig. 15-7**in

**chapter 15**. So far, we have dealt with

**linear objects**all the time, except for the chapter on

**on-off**control. And only for them the concept of transmittance

**G(s)**makes sense.

**G(s)**transmittances correspond to

**linear differential**equations, and these have long been figured out by

**mathematicians**. Therefore, the theory of the behavior of

**linear dynamic**objects is very elegant. From the transmittance

**G(s)**, it is relatively easy to predict how the system will behave after closing it with a feedback loop. Will it be

**stable**? What will be the

**steady**state? How to select the parameters of the

**PID**controller? If at least

**1**

**non-linear**element appears in the block diagram, then (almost

**100%**) the entire object will become

**non-linear**. That is, one that can no longer be described by a

**linear differential**equation.

**Chapter 31.2 Non-linear elements****Chapter 31.2.1 Introduction**

Examples of **non-linear** elements.

Fig. 31-5**A typica**l feature of each **real object** (and not an **ideal one!**) is that sooner or later it reaches a state of **saturation**. For example, for an amplifier, **saturations** are **+/-15V** supply **voltages**. This can be seen in the characteristics above. Well, maybe apart from **Fig. 34-5d**.

Typical **non-linear** characteristics**Fig. 31-5a** – “increasingly faster with saturation” – A strange but accurate name.**Fig. 31-5b** – linear with saturation**Fig. 31-5c** – “increasingly slower with saturation”.**Fig. 31-5d** – hysteresis. Note that the value of **y(t)** depends on whether **x(t)** is **increasing** or** decreasing**. Here you can’t see **+/- saturation**. They would appear with larger amplitudes of the input signal **x(t)**.**Fig. 31-5e** – linear with dead zone and saturation

In the next **2** subsections for element inputs:**-linear** with saturation (**b**)**-linear** with dead zone and saturation (**e**)

we will give the** black** sine wave **x(t)** and examine the **red** response** y(t)**.

**Chapter 31.2.2 Non-linear element with saturation**

**Non-linear element with**

Fig. 31-6

Fig. 31-6

**saturation**and sine input

**x(t)**

**Fig. 31-7**

**Saturation**occurs at

**y(t)=+/-0.5**.

Before saturation

**x(t)=y(t)**

After saturation

for

**x(t)>0.5 y(t)=+0.5**

for

**x(t)<0.5 y(t)=-0.5**

If the saturation was

**+/-1**, it would be

**x(t)=y(t)**all the time and you would see

**red**

**sine wave**

**y(t)**.

**Chap. 31.2.3 Non-linear element with dead zone**

Fig. 31-8

Fig. 31-8

**Non-linear**element with

**dead zone**for sinusoidal

**x(t)**

**Fig. 31-9**The

**dead**zone exists for

**-0.5….+0.5**.

If the

**dead**zone was

**zero**(otherwise-it doesn’t exist), then it would be

**x(t)=y(t)**all the time and you would see a

**red**sine

**y(t)**.

If the

**dead**zone was in the range

**-1…+<1**, it would still be

**y(t)=0**and you would see a flat

**red y(t)=0**and a

**black**sine

**x(t).**

**Chapter 31.3 How does an ideal control system work?**

A typical control system with a** PID** controller and a **two-inertial** object. You have already met him in **Fig. 27-18** **chapter 27**. It can be, for example, a temperature controller with a **1 kW** heater output. The heater is immersed in a small **tank** with liquid. Let’s assume that the liquid is perfect, i.e. it neither **freezes** nor **evaporates**. We expect that in the **steady** state there will be **zero** error **e(t)**, i.e. **y(t)=x(t)=+1**.

**Fig. 31-10**

Let **+1** be **+100°C**.

And we rightly expect **e=0**, because there is an integral component** I**. The output temperature reached **+100°C** after only **1.5 seconds**! Then there was a slight overshoot of **+10 °C** and after **10 seconds** a steady state **y(t)=x(t)=+100°C** occurred. On the other hand, the **sPID(t)** control signal has large **overshoots**. “Positive” power – **heating** is sometimes greater than **+2**, “negative” power – **cooling** lower than **-1**. Therefore, the temperature** y(t)** reached the steady state very quickly, after only** 1.5 seconds**, although the object has **2** inertias of **3** seconds and **5** seconds!

This is the good job of **negative** feedback, more generally – **automatics**. If it was an **open** system, read **without feedback****,** it would take **half a minute**, not just **1.5 second**s. You are also using** negative feedback** by yourself when cooking **soup**. You turn the gas up to **full** and then turn it **down** when the temperature reaches the** boiling** point. I talked a bit, so let’s go back to the **time charts**. The **sPID(t)** control signal is clearly visible, cut off by the oscilloscope at the level** -1** and **+2**, i.e.** -100°C … +200°C**. To see how large these overshoots of the **sPID(t)** control signal are, let’s change the range of the oscilloscope from** -1…+2** to **-20…+160**. So at** -2000°C…+16000°C**! I remind you that in our course there are liquid temperatures lower than **-2000** **° C** and higher than **+16000°C**.

**Fig. 31-11**

Although the signals** x(t), y(t)** of approx. **+100°C** are barely visible now, you can see the whole** sPID(t)** control signal . At the beginning, the** sPID(t)** varies between **-5…+160**. So for a short moment the **positive** (heating) power “wants” to **heat** the liquid to **+16000°C** and in a moment the **negative** (cooling) power “wants” to cool down to **-500°C**! In the** steady** state, the control **heater** draws **1kW**, but during the **peak**, as much as **160kW**! After all, no constructor will give such a huge **heater**! Maximum **3 kW**. Negative power-**cooling** is also rarely used. Therefore, the narrower range **0…3** instead of **-5…+160** will be the limitation. Even narrower. After all, when the power is **turned off**, the temperature drops not to **0°C**, but to the ambient temperature, e.g. **+20°C**. This is where **theory** differs from **practice***. The** linear systems** we have considered so far are **elegant, easy** (relatively, relatively!) and answers many **general** questions. **Stability, Hurwitz**, etc. On the other hand, **real systems** –** non-linear** will **differ** from **linear** ones.

How far to **deviate**? How much **worse** they are? How reliable is their linear approximation? This will be discussed in a moment.*** Once** in a noble automatic company I saw a poster with its motto:

–** Practitioner** – Everything works, but doesn’t know why.

– **Theorist** – Nothing works, but he knows why.

We combine **theory** with **practice**. Nothing works and we don’t know why.

I liked them right away.

**Chapter 31.4 How does a real automation system work?****Chapter 31.4.1 Introduction**

There are no ideals in life. The transmittance **G(s)** is only an approximation of the **real **object. The question arises. How far is it from** reality**. The first **approximation** of a real system is something like **this**.**Fig. 31-12**In the following points I will try to justify that the

**non-linearity**is mainly introduced by the

**Power Amplifier**. Therefore, we will consider the

**Power Amplifier, PID Controller**and

**Object**in this respect

**Chapter 31.4.2 Power Amplifier with Saturation**

The diagram differs from

**Fig. 31-10**only with the

**Power Amplifier**in the middle of the diagram. So far, we have assumed that the signal from the

**sPID(t)**controller always has sufficient power to control the object. It is known that this is not so. Each

**amplifier**has its own saturation. Even the big power plant i will not generate more than

**4,000 MW**. It is also a gigantic power amplifier. The input is the power knob (it doesn’t matter if it’s computerized) in the Central Dispatch Room, and the output is the power given to the

**energy network**.

The

**Power Amplifier**in the diagram is nothing but a

**linear element**with

**saturation**from

**Fig. 31-5b**. It does not have to be in the electric version. It can be, for example, a

**valve**with an

**actuator**. The signal from the

**controller**controls the degree of

**valve**opening, thus indirectly the

**flow**, e.g. the

**supply of steam**to the

**turbine**. The flow

**F**will not be arbitrary, only in the range of

**0 … Fmax**So again we have a

**Power Amplifier.**The

**electric power amplifier**or

**valve**or something else is collectively called the

**actuator**. All right, we have the

**actuator**, which as a

**non-linear**element

**spoils**the

**whole**scheme. It

**spoils**, i.e. t

**he transmittance**of the entire

**opened**and

**closed**system cannot be calculated. What about

**PID**and

**object**?

**Chapter 31.4.3 PID controller**

This is an almost perfect linear system with transmittance

**The**

Fig. 31-13

Fig. 31-13

**controller**settings can be e.g. as

**follows**:

**Kp=10**

**Ti=7 sec**

**Td=2 sec**(derivative, as in any decent controller, is real, i.e. with

**T**inertia)

**One more**comment on “almost perfect linear system”.

Modern

**PID**controllers are usually implemented in

**microprocessor**technology. So the response of the controller is

**not continuous**but

**quantized**by the

**clock pulses**. However, with

**frequent**quantization, the

**controller**response can be treated as

**continuous**. The output of the

**controller**is a

**digital/analog**converter. So a

**continuous**signal enters the

**power amplifier**. By assumption, the

**power amplifier**will enter

**saturation**earlier than the

**PID**. Therefore, with a clear conscience, we can treat

**PID**as the transmittance

**Gpid(s).**

Chapter 31.4.4 Object G(s)

Chapter 31.4.4 Object G(s)

The object can be a

**furnace, a rectification column**or

**a rocket**. Of course, each of them has different dynamic properties. In

**Fig. 31-12**, an example object is a

**two-inertial**unit. So we assumed that it is a

**linear**unit with transmittance

**G(s)**.

There are

**weaker**linearity arguments for the

**object**than for the

**PID**. The static

**characteristic**of most typical objects is “increasingly slower with saturation” –>

**Fig. 31-5c**.

What is the conclusion? The fact that the

**denominator**, i.e. the

**slope**of the characteristic, is equal to

**unity**for the

**zero**operating

**point**. and then it

**decreases**. Similarly, the

**time**constants can vary within certain

**limits**. So the “true” transmittance

**parameters**may depend on the operating point and the true value is kind of

**fuzzy**within the

**limits**.

Fig. 31-14

Fig. 31-14

**Fig. 31-15a**-based on the

**observation**of the

**object**, or on the basis of the mathematical model of the object, we assume that it is a

**two-inertial**object with gain

**k=1**, time constants

**T1=10 sec**and

**T2=5 sec**.

**Fig. 31-15b**-The true parameters can be, depending on the operating point and measurement error, in 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**ones, can be approximated as

**inertia with delay**. Why? An

**exact**model is better than an approximate one. It’s better to be

**healthy and rich**than

**poor and sick**.

**First**, it is easier to determine its parameters

**K, T**and

**To**by observing the response to a

**unit step**.

**Secondly**, for the transmittance determined in this way, the parameters of the

**PID**controller settings are already ready. Then the

**response**of the

**PID**-controlled system is optimal under

**some criterion**. For example, the shortest

**control time**with possibly

**small overshoots**.

The

**T**inertia with

**To**delay and

**K**gain have already been mentioned much earlier. As a reminder, I will show the response of

**y(t)**to the

**x(t)**unit step.

Fig. 31-15

Fig. 31-15

Inertia with delay.

**Chapter 31.4.5 Conclusions**

Referring to

**Fig. 31-12**, we found that:

**– The**

**PID**controller is a

**linear**object with transmittance

**Gpid(s)–>Fig. 31-13**

**– With**some forbearance, the object

**G(s)**is also a

**linear**object. The

**closer**to

**linearity**the

**smaller**the

**input**signal.

Often

**objects**are approximated by

**inertia**with a

**delay**.

**– The**

**actuator**is a

**linear**element with

**saturation**, i.e. it is a

**non-linear**element. Mainly, it “spoils” the whole system, depriving it of

**linearity**. So it is impossible to calculate the transmittance from

**Fig. 31-12**.

**– All the**theory taught so far –>

**stability**studies,

**selection**of controller

**settings**, etc…, concerns

**linear systems**.

**Considering**that in real control systems there is always an

**actuator**, we can conclude that the whole

**control theory**is just a beautiful mathematical trinket. There are no

**material benefits**in it, only

**spiritual ones**. Fortunately, it’s not quite like that.

**Chap. 31.5 Comparison of an ideal and a real system****Chap. 31.5.1 Introduction**

We will study the response to **x(t)=+1** unit step, and **z(t)= +0.4** and **z(t)=-0.4** **disturbances** at **actuator** saturations of **0…+1.5** and **0…+5.** Common sense expects that the **later** saturation **occurs**, the more **ideal** the system is. Also, reducing the **x(t)** step amplitude has the same effect. What about the **z(t)** disturbance? It will turn out that the **responses** to a small **disturbance** are **similar** for **real** and **ideal** systems. The more the less disturbance.

In the **next** experiments, the **PID** controller with the settings **Kp=10, Ti=7 sec, Td=1.5 sec** will control a **two-inertial** object with the parameters **K=1, T1=10 sec** and **T2=5 sec**. as optimal for this object. Assuming, of course, that the object was ideal **linear**.**Chap. 31.5.2 Saturation 0…+1.5, x(t)=+1, z(t)=+0.4 (heating)**

We have **2** identical **tanks** with **identical liquid** next to each other. The **parameters** of the** controllers** are also identical. The difference is only in the **power** of the **heaters**.

The **first-ideal** system has an **unlimited** power. So there are **no saturation**. It can even give a **heating** power of **+1 000 000kW** and a **cooling** power of **-1 000 000kW **when needed.

The **second real-**system already has a **real actuator**, i.e. a **heater** with a **limited** power of **+1.5 kW**. In addition, a **negative** signal from the controller does not **cool**, but **turns off** the **heater**. The **temperature** then drops to the environment, e.g. **to 0°C**. So the controller range is **0…+1.5 kW**.

The **disturbance** is the activation of the **+0.4 kW** heater.** x(t)** setpoint and **z(t)+0.4** disturbance act in **parallel** on** 2** systems. For sure **Ideal** will be better than **Real**. But how much better? Will it be **100:0** or just only **5:2** as in football match** score**?

Fig. 31-16

At the top there is an **ideal system**, i.e. with **an actuator** without **saturation**.

At the bottom there is a **real system** with **an actuator** with **saturation 0…+1.5**.

How to interpret it? The **maximum value** of the controller corresponds to the **heater power** of **+1.5 kW**, which in a steady state will give a **liquid** temperature of **+150°C**. Lower saturation** 0 kW** means the heater is** turned off** and the temperature then drops to the **ambient** temperature, e.g. **to 0ºC**.

**Fig. 31-17**

**Up**to

**60 sec**, we only have

**x(t)=1**, in

**60 sec**, there will be a

**disturbance +0.4-heating.**

**IDEAL SYSTEM**– actuator without saturation

The

**output**signal

**yi(t)**and the control signal

**sPIDi(t)**relatively quickly led to the state

**yi(t)=1**, in which the fixed error is

**zero**.

**Very**large fluctuations of the

**sPID(t)**control signal at the beginning of the

**x(t)**step –>

**Fig. 31-11**. In

**60 seconds**we have a disturbance

**z(t)=+0.4**–>

**activation**of an additional

**heater**with a power of

**0.4 kW**. The

**PID**controller correctly assessed the situation and also

**reduced**the power by

**0.4 kW**. There was, of course, a transitional state of several seconds, but everything ended as it should.

**REAL SYSTEM**actuator with saturation

Previously, the control signal

**sPIDi(t)**could roam within the limits of

**-infinity …+infinity**, now the actuator gives only a modest

**sPIDr(t)**within

**0…+1.5**. Therefore, instead of the

**160 pin,**there is a

**green flat time chart**with an amplitude of

**+1.5**from the beginning of the step to about

**18 sec**.

Good and that. During this time,

**yr(t)**tends to

**+1.5**instead of

**160**. Therefore,

**yr(t)**increases more slowly. In addition, there is a large overshoot to approx.

**+1.3**, because the braking from the derivative element

**D**is weaker than in the ideal case. Note that then the

**green**control signal

**sPIDr(t)**no longer enters

**saturation**and therefore

**yr(t)**comes to

**1**in steady state (zero error), as in any decent

**linear**system, i.e.

**ideal**. Despite the fact that

**overshoot**and control

**time**are greater in the

**real**system, they are not

**100 times**greater, as if they resulted from the maximum signals

**sPIDi(t)=+160**and

**sPIDr(t)=+1.5**.

In the

**60th**second, a disturbance

**z(t)=+0.4**occurred. The

**yr(t)**and

**yi(t)**time charts now perfectly coincide because no saturation worked. The

**real**system now thinks it’s

**ideal**.

**Conclusion**

**Non-linearity**, including mainly the

**saturation**of the

**actuator**, obviously deteriorates the quality of

**control**, but not so

**much**!

**that you may not read, because there has already been something about it**

A note

A note

What do

**Kp, Ti, Td**do?

**Kp=10-proportional**

It quickly brings the output signal

**yi(t)**to

**0.91**

**Fig. 31-18**

So the

**Kp**component

**does not**ensure

**zero**control error.

**Ti=7sec-integrating component**

Brings the

**error**to

**zero**. Question. If the

**integrating**component brings the error to

**zero**, then what the hell is

**Kp**for? The answer is simple.

**Kp**quickly brings

**yi(t)**to

**0.91**. The

**I**component also

**reduces**the

**error**to

**zero**from the very beginning, but it does so

**slowly**. In short and not very precisely –

**Kp**quickly makes

**yi(t)=0.91**, and

**Ti**is responsible for the rest, i.e. for bringing

**yi(t)**to

**1**.

**Td=1.5 sec-differetiating component**

Even more than

**Kp**, it speeds up the approach to

**yi(t)=1**. At the very beginning of the step,

**Kp**gives a kick in the

**sPID(t)**signal to the value of

**10**, and for the rest, i.e.

**150**, corresponds to the

**differentiating**component, i.e.

**Td**. After all, this differentiates the

**x(t)**step. And so it is fortunate that this is

**real**and not i

**deal**

**differentiation**. Then we’d have an

**infinitely**large pin, not a miserable

**sPID(t)=160**.

Let’s go back to that

**x(t)**kick. Thanks to it,

**yi(t)**grows very quickly. As if he wanted to go not to

**1**but to

**160**. In a moment

**x(t)**is flat–>

**sPID**drops quickly. But

**differentiation**from increasing

**yi(t)**works. And it works in the opposite direction of the initial

**160**kick! It even gives negative

**sPID(t)**values for a while. And what does that mean? That the

**Td**component now acts as a

**brake**or stabilizer. This prevents overshoots and even instability

**Chapter 31.5.3 Saturation 0…+1.5, x(t)=+1, z(t)=-0.4 (cooling)**

The diagram differs from

**Fig.31-16**only in the disturbance. Previously it was

**z(t)=+0.4**-heating, now it’s

**z(t)=-0.4**-cooling

**Fig. 31-19**

The time charts up to **60 sec** are identical to **Fig. 34-18**. To the disturbance **z(t)=-0.4**-cooling, the controller reacted correctly by **increasing** the heating power by** +0.4**. The control signal has not entered saturation, the controller “thinks” that the system is **linear**. Therefore, the **sPIDi(t)** and **sPIDr(t)** signals coincide.**Chapter 31.5.4 Saturation 0…+5, x(t)=+1, z(t)=+0.4 (heating)****This** and the **next** scheme differ from the previous ones only in saturation. It will now be greater **0…+5** instead of** 0…+1.5**. Bigger, i.e. the system is more “ideal”, i.e. we expect better control. We now have a more powerful actuator->read more **expensive**.

**Fig. 31-20**

In** Fig. 31-17**, you saw the effect of saturation **+1.5** on the real control signal **sPIDr(t**)– (after the power amplifier). There is a **higher** upper saturation here – **+5**, and that’s why you can’t see this **green** and flat** sPIDr(t)**. In any case, the higher maximum **sPIDr(t)** resulted in **faster** control and less **overshoot**. It can be said that the system has come close to **ideal**. The response to a** disturbance** has not improved, because it is still the same as for an **ideal** system. Why? Because the **sPIDr(t)** control signal did not enter** saturation**.**Chapter 31.5.5 Saturation 0…+5, x(t)=+1, z(t)=-0.4 (cooling)**

The scheme differs from the previous one only in the disturbance **z(t)=-0.4** (cooling).

**Fig. 31-21 **

The difference is only in the reaction to the disturbance **z(t)=-0.4**-cooling.**Chap. 31.5.6 Conclusions**

Compared to the **Ideal** system, the** Real** system comes to a steady state more **slowly** and with greater **overshoots**. This is of course a **downside**, but you can live with it. Especially that **zero** error is ensured and **disturbance** suppression is almost identical to the **ideal** one. Also take into account that the **automatics **mainly suppresses disturbances **z(t)** and changes of the setpoint **x(t)** are much **less** frequent. Although the **ideal** system gives more than **100** times more kick than the **real** system at the beginning of the step (**160 **to** 1.5**), the reaction is not** 100** times** worse** – read slower. Therefore (with some leniency) we can treat **real** systems as **linear**. But only when **x(t)** is between the saturation of the **actuator**–> in our case **x(t)=0…+1.5**.

This condition is not difficult to meet. And this means that for **real** systems we can use all the **rich** knowledge for **linear** systems.**Coming back** to football, the relationship between the **Ideal** and the **Real** system is more **Germany-Austria** than **Germany-Gibraltar**.

**Chap. 31.6 How does the system behave when the setpoint x(t) is greater than saturation?****Chap. 31.6.1 First attempt – only x(t), yi(t) and yr(t)**

The **maximum** power behind the controller is **+0.95 kW**, which is able to heat the liquid to **+95°C**. But I want the controller to heat the liquid to **+105 °C**. It is physically **impossible**! As if the **manager** demanded from the** employee** to perform a task above his competence.

Fig. 31-22

This time it will not be a single unit step of the setpoint** x(t)**, but a signal from the **step generator**. The real** PID** controller has limits **0…+0.95** on the output of the **actuator**.

There will be **3** periods for step **x(t)** where:**Period 1****x(t)=+0.3** (below saturation)**Period 2****x(t)=1.05** (above saturation!)**Period 3****x(t)=+0.5 (again below saturation)**

The reactions of ideal **yi(t)**, i.e. without saturation, are obvious. Negative feedback will work, which will bring the** zero** error. The signal **yi(t)** tries to imitate **x(t)**. The entire course so far has tried to convince you of this. And the reactions of the controller with **saturation**, i.e.** real**? Try to anticipate them. For now, we are not interested in the “internal” signals **sPID(t), sPIDir(t) and sPIDr(t)** from **Fig. 31-22**.

**Fig. 31-23**

The performance of the ideal **yi(t)** controller is always better than the real **yr(t)**. Therefore, let us examine only the real** yr(t)**.**Period 1. x(t)=+0.3**

Setpoint **x(t)** is less than **saturation=+0.95**. So there will be feedback. The signal **yr(t)** will go to **x(t)=yr(t)=+0.3**.**Period 2. x(t)=+1.05**

The setpoint **x(t)** makes the** real controller** reach the state **yr(t)=+1.05**, i.e. the output temperature** +105°C**. But how is the poor** PID** supposed to do that when the **actuator** has a saturation of **+0.95**? It’s not possible! Simply** yr(t)** will tend to **+0.95**. Negative feedback won’t work! As if it was an **open** system.

This can be seen in the** time chart**, where only at the end of period **2** the signal **yr(t)** reached a steady state of **+0.95**.**Period 3. x(t)=+0.5**

Setpoint **x(t)** is less than** saturation=+0.95**. So a negative feedback will work which will bring the error to** zero–>yr(t)=x(t)=+0.5**. Of course, the real** yr(t)** is **slower** (worse!) than the **ideal yi(t)**. But where did the extra **dead time** come from that wasn’t there in **period 1**? After all, this phenomenon did not exist then. **Dead time** is extra **delay** The one that is always bad for **control**!

**Fig. 31-24**

The signals of an **ideal** system are **obvious**. The stronger and shorter **sPIDi(t)** resulted in a fast** yi(t)** response as usual. However, in **period 2**, the influence of the actuator **saturation** on the **sPIDr(t)** cut of the** real** controller is **clear**. There is **no feedback** here anymore, and therefore the answer is as for an **open** system. But where did this period **3 dead** time come from? Then let’s observe** sPIDir(t)** before the **actuator**.

**Chapter 31.6.3 Third approach – even more thorough**

We observe **yellow ideal sPIDi(t)**, **black real sPIDr(t)** and **green sPIDir(t)** in front of the** actuator** in **Fig.31-22**

**Fig. 31-25****Ideal system**

The time chart of the **yellow sPIDi(t)** is obvious and requires no comment.**Real System**

The** green sPIDir(t)** signal in front of the **actuator** can have any numeric value because it is calculated by the microprocessor and does not contain power. The letter** i** in **sPIDir(t)** suggests that the signal is after the **ideal** part of the real **PID** controller.**Period 1 x(t)=+0.3**

Until **tz=6.5 sec** the **green sPIDir(t)** is greater than **+0.95** saturation. This will result in** cutting o**f the the black **sPIDr(t)**. Then **yr(t)** tends to **+0.95** as in an **open** system. **Feedback** will occur after **tz=6.5 sec** when **sPIDi(t)** drops below **+0.95**. Then it will reduce the output signal yr(t) to **yr(t)=x(t)=0.3**. **Fast** but of course **slower** than **yi(t)** in the **Ideal System**.**Period 2 x(t)=+0.95****Green sPIDir(t)** jumped up. And what happens to it next, is not visible because the **scope** of the oscilloscope is **too** small. Anyway, the signal after the **actuator** is **cutted** at **sPIDr(t)=+0.95** and **yr(t)** tends to **+0.95** as in **open** circuit (no feedback). You can’t see all the **green sPIDir(t)**! And interesting things happen there. You will find out in the next experiment.**Period-3 x(t)=+0.5**

I guess there’s something dawning with **dead time**. It can be seen that at this time **yir(t)** is greater than **+0.95**. Then **yr(t)** is roughly truncated at **+0.95** as well. When **yir(t)** becomes less than **+0.95** the feedback will bring **yr(t)** to **+0.5**.**Chapter 31.6.4 The same, only a larger oscilloscope range**

We will get visually the same scheme, but with a different oscilloscope range **-100…+140**, previously **-1…+2**. This will allow you to see what was not visible in **Fig. 31-25**.

**Fig. 31-26**

Signals **x(t), yi(t), yr(t)** are now tiny, tiny. However, the control signals** sPIDi(t)** and **sPIDir(t)** are visible. The** green sPIDir(t)** mostly **covers** the **yellow sPIDi(t)**.**Period 1 x(t)=+0.3**

Falling** green yir(t)** **pin** until** tz=6.5 sec**. You didn’t see it in **Fig. 31-25**.**Period 2 x(t)=+0.95**

Feedback does not work all the time because **sPIDir(t)>0.95**. Therefore, we have the classic **PID** response to a** positive** unit step in an **open** system!**Period 3 x(t)=+0.5**

The setpoint **x(t)** has decreased from **+1.05** to **+0.5**. So we have again the classic **PID** response to a **negative** unit** step x(t)**. It is now clear that the** dead time** is the **rise** of the integral component of the** PID** controller to a value where **sPIDir(t)** becomes **less** than saturation **+0.95**. Then the **feedback** will bring **yr(t)** to **yr(t)=x(t)=0.5**–>** zero** control error.**Chapter 34.6.5 Conclusions****1. If the real controller** “tells” the object to enter a state that is** possible** to meet* by the **actuator**, then the control works in a way similar to **an ideal system **(with feedback). The response** yr(t)** is of course **slower** (worse) than for an **ideal** one, but in the end **yr(t)=x(t)** so **zero** control error will be ensured.**2. If the real controller** “tells” the object to enter a state that is** impossible** to meet, i.e. when **x(t)>0.95**, then the output signal **yr(t)** slowly reaches the saturation state as in an** open** system -> period** 2** in **Fig. 31-25**.**3.** If the real controller goes out of saturation, as in period **3** in **Fig. 31-25**, a **dead time** will occur. The reason for this is the **integral** component of the controller->period** 3** in **Fig.31-26**. Another name for the **dead time** of the **PID** controller is the i**ntegral dead zone.**

How to fight the **integral dead zone**?

Dead time **To** introduces a **delay** into the system which is obviously a **disadvantage**. Because would you like to drive a car in which you turn the steering wheel to the** left** and the wheels react only after **To=1 sec**?

Currently, the **controller** are made in microprocessor technology. Therefore, when the controller finds that despite the control signal **increasing** from the **integral** component, the output signal does not increase, it **turns off** the **integral** component. So in the **saturation** state, the **PID** controller becomes the **PD** controller.

This is an example of a controller with a **variable structure**. This switching can occur, for example, when the **actuator** enters the saturation state, and it is easy to detect by the microprocessor controller.