# Automatics

**Chapter 25 I Control**

**Chapter 25.1 Introduction**

You have already met the simplest type **P** controller consisting only of a **comparator** element and an amplifier **Kp**. It roughly ensured **approaching** the setpoint **x(t)** and suppressing disturbances **z(t)**. Then came the **PD** controller which did the same thing, only much **faster** and with smaller **oscillations**. Still, however, the **steady-state** closed-loop gain **Kz** and the **error** gain **Ku** was as follows:**Fig. 25-1**

The formulas show that **Kz** is always slightly smaller than **1**. The closer to **1**, the larger **K**. Similarly, **Ke** is always slightly larger than **0**. The closer to **0**, the larger **K**. Let me remind you that **K** is the steady gain of the entire open control system. That is, taking into account the **Kp** of the controller and the **Ko** of the **object**. It so happens that it is mostly **Ko=1**, then **K=Kp**. It is convenient when selecting the controller settings–> **chapter 31**.**And now** the most important!

For type **I** control, the formulas for **Kz** and** Ke** are as follows:**Fig. 25-2**

**Cheaper**, sorry, it couldn’t be

**easier**! And what does that mean? If we enter

**x(t)**in the form of a

**unit step**, then after some time it will be

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

**e(t)=0**! This is the main goal of every

**automatics engineer**. The control time may be

**long**, and there may also be fading

**oscillations**. But when certain additional conditions are met (Hurwitz!, Nyquist!…) this will always happen. And when they are not

**fulfilled**? Then the system is

**unstable.**

**Note**

The formulas in

**Fig.**

**25-2**are actually a special case of

**Fig. 25-1**. Because what is the gain

**K**of the object with the

**integral unit**in steady state.

**y(t)=infinity**when

**t=infinity**! So

**K=infinity**! Substitute this value into

**Fig. 25-1**and you will get

**Fig. 25-2**.

The

**I**controller consists of a

**comparison**element and an

**I**

**integrating**element with

**adjustable**integration time

**Ti**. It is rarely used as a standalone controller. The integral unit

**I,**however, is present in the

**PI**or

**PID**controllers and its purpose is to bring the error

**e(t)**to

**zero**.

You’ve met

**I**unit him earlier in

**chapter 4**, but I suggest a

**quick replay**!

**Chapter 25.2 Integral unitChapter 25.2.1 Introduction**We will examine

**3**units

**I**with different integration

**rates**

**-Slow**

**-So so**

**-Fast**

They are not yet

**I**controllers, because they do not have a

**comparator**that calculates the error

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

**Chapter 25.2.2 Slow I unit Ti=2sec**

**Fig. 25-3**

**Ti=2sec**

The

**integral unit**with the integration time

**Ti=2sec**. The signal increases at a constant rate.

**Ti=2sec**is the time after which

**y(t)**equals the step

**x(t)**.

**Chapter 25.2.3 “So so” I unit Ti=1 sec**

**Fig. 25-4****Ti=1 sek**After time

**Ti=1 sec, y(t)**equaled the step

**x(t)**. That is

**2**times

**faster**than before.

**Chapter 25.2.4 Fast I unit Ti=0.5sec**

**Fig. 25-5****Ti=0.5 sec**.

After the time **Ti=0.5 sec** **y(t)** equaled the step **x(t)**. Again **2** times** faster** than before.

**Chapter 25.3 Servo controller that reduces e(t) error to 0 as an example of I control****Chap. 25.3.1 Introduction**This is a good example that

**type I control**reduces the

**e(t) error**to

**0**

Fig. 25-6

Fig. 25-6

The input is the voltage

**x(t)**across potentiometer

**A**

**Fig. 25-6a**. The

**DC**motor is controlled by the voltage difference

**e(t)=x(t)-y(t)**. The motor shaft is mechanically coupled through a gear with the

**slider**of the potentiometer

**B**. The motor is ideal, i.e. when there is a voltage on it even as

**small**as

**+1 µV**, the slider will slowly move up, i.e. the voltage

**y(t)**will slowly

**increase**. And when

**+2 µV**is

**+2 times**faster. Similarly

**-1 µV**the slider will go down. This is the classical

**integrating**unit, i.e. the

**I**unit.

Slider

**B**will only stop in one situation. When

**x(t)=y(t)**, i.e. when the voltage on the motor is

**zero**.

**At first**, both sliders are down. So

**x(t)=y(t)=0V**and the voltage on the motor is also

**zero**, because

**Kp*[x(t)-y(t)]= Kp*(0-0)V=0V**and the motor stops. Suddenly you will set potentiometer

**A**

**+5V**. As if you gave a unit step

**x(t)=+5V**Then the motor will go up with a

**high**initial

**speed**. The

**speed**will

**decrease**, because at the input – of the amplifier there will be a voltage “subtracting” from the

**slider. b**

In this way, slider

**B**will reach

**+5V**after a theoretically infinite time. The voltage on the motor will be

**zero**.

And what if the gear motor had

**inertia**and the slider would exceed

**+5V**. For example, it would be

**+5.1V**. Then a

**negative voltage**will appear and the slider

**B**will move back towards

**+5V**, until after some time it will be back to

**+5V**.

That’s how it is in an ideal world. In real life, there are some

**hysteresis**, dead

**zones**and other clutter, so that the slider would be set to, for example,

**+4.999V**. Then the

**very small**voltage on the motor would not be able to move it.

So you see that under ideal assumptions the error will be

**zero**, which is typical for type

**I**control.

**Fig. 25-6b**is a block version of the diagram in

**Fig. 25-6a**. Here the

**Kp**of the amplifier is obvious. The motor’s

**Km**is greater the higher the speed of the motor at a given voltage. The

**Kg**parameter of the gear is like a

**derailleur**in a bicycle.

In total, it is an

**integral**unit with some integration time

**Ti**covered by a

**negative**feedback loop–>

**Fig. 25-6c**.

**Chapter 25.3.2 Slow servo model**

The servo control boils down to the very simple

**Fig. 25-6c**.

**Fig. 25-7**

The input **x(t)** is the voltage **x(t)** across potentiometer** A Fig. 25-6a**. The **DC** motor is controlled by the voltage difference **e(t)=x(t)-y(t)**. The motor shaft is mechanically coupled through a **gear** with the **slider** of the potentiometer **B**. The motor is ideal, i.e. when there is a voltage on it even as small as **+1 µV**, the slider will slowly move up, i.e. the voltage **y(t)** will slowly increase . And when +**2 µV** is **+2** times faster. similarly **-1 µV** the slider will go down. This is the classical integrating term, i.e. the **I** unit.

Slider **B** will only stop in one situation. When** x(t)=y(t)**, i.e. when the voltage on the motor is **zero**.

At first, both sliders are down. So **x(t)=y(t)=0V** and the voltage on the motor is also **zero**, because **Kp*[x(t)-y(t)]= Kp*(0-0)V=0V** and the motor stops. Suddenly you will set potentiometer **A +5V**. As if you gave a unit step **x(t)=+5V** Then the motor will go up with a high initial speed. In a moment the speed will decrease, because at the input – of the amplifier there will be a voltage “subtracting” from the slider. **b**

In this way, slider **B** will reach **+5V** after a theoretically **infinite** time. The voltage on the motor will be **zero**.

And what if the gear motor had **inertia** and the slider would exceed **+5V**. For example, it would be **+5.1V**. Then a **negative** voltage will appear and the slider** B** will move back towards **+5V**, until after some time it will be back to **+5V**.

That’s how it is in an ideal world. In real life, there are some **hysteresis**, dead zones and other clutter, so that the slider would be set to, for example, **+4.999V**. Then the very small voltage on the motor would not be able to move it.

So you see that under ideal assumptions the error will be **zero**, which is typical for type** I** control.**Fig. 25-6b** is a block version of the diagram in **Fig. 25-6a**. Here the **Kp** of the amplifier is obvious. The motor’s **Km** is greater the higher the speed of the motor at a given voltage. The **Kg** parameter of the gear is like a derailleur in a bicycle.

In total, it is an **integral**unit with some integration time **Ti** covered by a **negative feedback** loop–>**Fig. 25-6c**.**Chapter 25.3.3 Slow servo control**

The servo control boils down to the very simple diagram in **Fig. 25-6c**.

**Fig. 25-8**Servo model when

**Ti=1 sec**

The system reacted

**4**times

**faster**. It reached the steady state

**y(t)=7.5 V**after about

**7.5 seconds**.

**Chapter 25.3.4 Conclusions from the servo test**

Integration

**I**appeared in the control system, because the

**DC**motor is an

**integrating**unit. This brought the

**error e(t)**to

**zero**. This is the most important advantage of integration. In previous chapters, the

**proportional**component

**P**was unable to reduce the

**error**to

**0**. Here, the

**I**component did it. Because no matter how small the error is, even a very small voltage will always spin the motor so that

**y(t)=x(t)**. Of course, theoretically. In practice, however, the

**deadband**and

**hysteresis**of the motor will cause some

**error**.

**Chapter 25.4 I control with an one-inertial object****Chapter 25.4.1 Introduction**We will examine a system in which the

**I**controller controls an

**inertial**object with a time constant

**T=10 sec**. We will start with the most conservative setting

**Ti=36 sec**, then

**Ti=16 sec**and finish with

**Ti=8 sec**. Patience will be required as the experience time is

**2**minutes. We will start, as usual, with the study of the

**inertial**object itself.

**Fig. 25-9**

An inertial object with a time constant **T=10 sec**.**Chapter 25.4.2 I control Ti=36 sec**

**Fig. 25-10**The only setting of the controller is

**Ti=36 sec**. We start with such a slow

**integration**, because we are afraid of

**oscillations**and even

**instability**. I guess the fears were exaggerated. The system very

**slowly**came without

**oscillations**(in other words –

**aperiodically**) to the state of equilibrium, where

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

**error e(t)=0**. In fact, after

**120 seconds**, the state

**y(t)=1**is not yet there. It will happen later. But I guess you believe it will. As long as the

**error e(t)**is different from

**0**(here more than

**0**), the

**I**controller will push the signal to the state where

**y(t)=x(t)=1**. In theory, it’s

**infinitely**long, in practice after about

**130 seconds**.

**Chapter 25.4.3 Control I when Ti=16 sec**

The previous run was

**slow**. Faster integration

**Ti=16**should improve the situation.

**Fig. 25-11****I** control **Ti=16 sec**Great joy at first. As

**s(t)**grows faster,

**y(t)**does the same. Unfortunately, there was an

**oscillation**. Nevertheless, after about

**120 seconds**, the output signal

**y(t)**reached the state of

**equilibrium**, in which, as the

**steady-state**error is

**zero**. Is the response much better than the previous one? Suppose it does despite the

**oscillation**.

**Chapter 25.4.4 I control Ti=8 sec**

Let’s shorten thecontrol time even more by giving

**Ti=8**. Maybe it will be better?

**Fig. 25-12**I control

**Ti=8 sec**

Slightly shorter control

**time**but larger

**oscillations**. We will therefore assume that the setting

**Ti=16 sec**is

**optimal**for the

**inertial**unit with the time constant

**T=10 sec**.

**Chapter 25.5 I control with an Two-inertial object**

**The**

**Chapter**25.5.1 Introduction**I**control controls the

**two-inertial**object. We will start with the study of the object itself. Already the

**two-inertial**object turned out to be difficult to control,

**long**control time and

**oscillations**. It will probably be even worse here.

**Fig. 25-13****Two-inertial** object with time constants **T1=3** sec and **T2=5 sec**

You can see the inflection point characteristic of **multi-inertial** objects. It is difficult to read the parameters **T1=3sec** and **T2=5sec** from the time chart.

There are ways to do it, but let’s just give it a rest.**Chapter 25.5.2 Control I when Ti=25 sec**

**Fig. 25-14****I** control **Ti=25sec**

Surprise. The **time chart** is’nt **better** or worse than “easier” **I** control for the **one-inertial** unit from **Fig. 25-11**! I was prepared for the **worse** answer **y(t)**. Can someone explain it?**Chapter 25.5.3 Control I when Ti=15 sec**

**Fig. 25-15**

If we don’t care about oscillations, it’s better. Shorter adjustment time.**Chapter 25.5.4 Control I when Ti=6 sec**

**Rys. 25-16****I** control **Ti=6 sek**

However, we overdid it. Not only is it a long control time, but dangling is unacceptable. We will therefore assume that the setting **Ti=15 sec** is **optimal** for the **two-inertial** unit with time constants **T1=3 sec** and **T2=5 sec**.

**Chapter 25.6 I control with an three-inertial object****Chapter 25.6.1 Introduction**First, the object itself.

**Fig. 25-17**A

**three-inertia**l object with time constants

**T1=0.5 sec, T2=3**sec and

**T3=5 sec**.

The response is similar for the

**two-inertial**one in

**Fig. 25-13**, although we know very well that the

**three-inertial**one is more difficult to control.

**Chapter 25.6.2 I control Ti=30 sec**

We will start with careful control, i.e. slow integration

**Ti=30 sec**.

**Fig. 25-18**

Control **I** **Ti=30 sec**This shows careful control. The control signal

**s(t)**is slightly larger than

**y(t)**.

Just like a teacher who is undemanding to the student. Therefore, the response is slow and without oscillation.

**Chapter 25.6.3 I control Ti=10 sec**

**Fig. 25-19**I Control

**Ti=10 sec**

Control time shorter although oscillations appeared.

**Chapter 25.6.4 I control Ti=5 sec**

**Fig. 25-20**

I control **I** **Ti=5 sek**Terrible oscillations. We will therefore assume that the setting

**Ti=10 sec**is optimal for this

**three-inertial**unit

**Chapter 25.6.5 When we overdo the integration i.e. Ti=1.5sec**

So with the speed of integration, e.g.

**Ti = 1.5 sec**

**Fig. 25-21**

I control **Ti=1.5 sek**

You can’t see it, but there is a different oscilloscope scale. Can you guess why?

Beautiful instability. The oscillation amplitude increases to** +/-infinity**.

**Chapter 25.7 I control with disturbances****Chapter 25.7.1 Introduction**

The same **3** objects will be controlled. Their inputs will be affected by large disturbances **z(t)=+0.5** or **z(t)=-0.5**. The resulting error **e(t)** will “annoy” the **integrating** component** I** so much that it will always bring it to** e(t)=0**. You just need to be **patient**, because due to the slow time charts typical of type **I** control, the experiments will last up to **4** minutes!**Chapter 25.7.2 Positive disturbance with a one-inertial object z(t)=+0.5, Ti=16sec**

**Fig. 25-22**

The disturbance **z(t)=+0.5** will appear in **130** seconds.

Up to **130** seconds, i.e. until the appearance of a **disturbance**, the waveform is the same as in **Fig. 25-11**, taking into account, of course, a different time scale on oscilloscopes. Initially, the disturbance **z(t)=+0.5** caused an increase in the **y(t)** signal, but then the **I** component “forced” **y(t)** to return to its previous value, i.e. to **y(t)=1**. This corresponds to the appropriate reduction of the error **e(t)** to **0**. The positive disturbance-“heating”, the control** s(t)** responded with a reduction in power.**Chapter 25.7.3 Negative disturbance with a one-inertial object z(t)=-0.5, Ti=16sec**

**Fig. 25-23**

The disturbance **z(t)=-0.5** will occur at **130 seconds**.

On the **negative** disturbance – “cooling”, the control **s(t)** reacted by increasing the power. The error **e(t)** was reduced to **0**.**Chapter 25.7.4 Positive disturbance with a two-inertial object z(t)=+0.5, Ti=15sec**.

**Fig. 25-24**

The disturbance **z(t)=+0.5** will occur at **130 seconds**.

For a** positive** disturbance **z(t)=+0.5** – “heating”, control **s(t)** reacted with power r**eduction**.**Chapter 25.7.5 Negative disturbance with a two-inertial object z(t)=-0.5, Ti=15sec**

**Fig. 25-25**

The disturbance **z(t)=-0.5** will occur at **130 seconds**.

On the **negative** disturbance – “cooling”, the control **s(t)** reacted by **increasing** the **power**.**Chapter 25.7.6 Positive disturbance with a three-inertial object z(t)=+0.5, Ti=10sec**

**Fig. 25-26**The disturbance

**z(t)=+0.5**will occur at

**130 seconds**.

On a

**positive**positive disturbance

**z(t)**– “heating”, the control

**s(t)**reacted with a

**reduction**in power.

**Chapter 25.7.7 Negative disturbance with a three-inertial object z(t)=-0.5, Ti=10sec**

**Fig. 25-27**The disturbance

**z(t)=-0.5**will occur at

**130 seconds**.

On the

**negative**disturbance

**z(t)**– “cooling”, the control

**s(t)**reacted by

**increasing**the power.

**Chap. 25.8 Comparing Type I, P and PD Regulations.****Chap. 25.8.1 Introduction**

**PD, P**and

**I**type regulators will control the

**two-inertial**object. Their inputs will receive unit steps

**x(t)**in

**3**

**seconds**, and

**positive**disturbance

**z(t)=+0.5**or negative disturbance

**z(t)=-0.5**in

**120 seconds**.

We will compare the output signals

**y1(t), y2(t)**and

**y3(t)**, especially their

**steady states**and

**control times**. We will only observe a unit step

**x(t)**, a disturbance

**z(t)**and

**3**output signals

**y1(t), y2(t)**and

**y3(t)**. Missed

**e(t)**and

**s(t)**signals have already been observed in previous experiments.

**Chapter 25.8.2 Positive disturbance z(t)=+0.5.**

**PD, P**and

**I**type regulators will control the

**two-inertial**object. Their inputs will receive unit steps

**x(t)**in

**3 seconds**, and positive disturbance

**z(t)=+0.5**or negative disturbance

**z(t)=-0.5**in

**120 seconds**.

We will compare the output signals

**y1(t), y2(t)**and

**y3(t)**, especially their steady states and control times. We will only observe a unit steps

**x(t)**, a disurbance

**z(t)**and

**3**output signals

**y1(t), y2(t)**and

**y3(t)**. Missed

**e(t)**and

**s(t)**signals have already been observed in

**previous**experiments.

**Chap. 25.8.3 Positive disturbance z(t)=+0.5**

Fig. 25-28

Fig. 25-28

The controller settings

**Kp=10**of the

**PD**and

**P**controllers according to the formula for

**Ke**in

**Fig.25-1**will provide a steady state

**e(t)=9%**. In addition, setting

**Td=5 sec**of the

**PD**controller will give us the shortest possible

**control time**with a small

**overshoot**.

The

**I**controller ensures

**zero**steady state

**error e(t)**regardless of the

**Ti**setting.

**Fig. 25-29****PD**–**green** control is the **fastest**, although it gives a **non-zero** steady error** e(t)**.**P**–**blue** control is **slower** and with **overshoots** and also gives the same **non-zero** steady error** e(t)**.**I**–**red** control is the **slowest**, but provides **zero** steady error** e(t)**. And this is the basic **advantage** of type **I**, i.e. **integral**, control.**Chap. 25.8.4 Negative disturbance z(t)=-0.5**The scheme is the same as in

**Fig. 25-28**, only

**negative disturbance**, i.e.

**z(t)=-0.5**.

**Fig. 28-30**

I leave the analysis to the reader.

**Chapter 25.9 Conclusions**

Why is type **I** control so** slow**? Look at the control signals **s(t)** in earlier experiments. For **P** and even more so for** PD** they are large at the beginning of **x(t)** and **z(t)**. This is what determines the **quick control time**. In addition, in the case of **PD**, the **derivative** component gives a **braking** effect that prevents oscillations. Control **I** does not have this initial kick** s(t)**.

What should be done so that the system reaches a** steady state** relatively** quickly**, but with a **zero steady-state error**? The method imposes itself.

Combination of the **P** controller with the **I** controller resulting in the **PI** controller–>**chapter 26**.

Or even better – a combination of **PD** with an** I** controller giving a **PID** controller–>**chapter 27**.