# Automatics

**Chapter 26. PI Control**

**Chapter 26.1 Introduction**

From **chapter 25.8** of the previous chapter, we learned that:**– P Control**, and even more so** PD**, reacts quickly to the **setpoint x(t)**, but does not provide **steady error e(t)=0**.

– With type **I** **Control** it is the other way around. The system** slowly** approaches the **setpoint x(t)**, but it provides a steady error **e(t)=0**.

Therefore, the idea of a **PI** controller with a proportional **P** unit and an integrating **I** unit suggests itself. We expect its response to a setpoint **x(t)** and a disturbance **z(t)** to be faster than that of the **I** controller (although not as fast as **P** or **PD**!), but in steady state the **error e(t)** will be **zero**. We will start by studying the structure of **PI**. This is not yet a **PI** controller but a **PI** unit because the input is only a single signal **x(t)** and not the error **e(t)=x(t)-y(t)**.

**Chapter 26.2 PI unit****Chapter 26.2.1 PI unit Kp=1 Ti=1 sec**

**Fig. 26-1**

**Kp=1 Ti=1 sek**

The input signal is the unit step

**x(t)**. The proportional component

**yp(t)**and the integral component

**yi(t)**are clearly visible. The component

**yp(t)**is for

**Kp=1**a repetition of the jump

**x(t)**. The component

**yi(t)**is an increasing “ramp” for which the value after time

**Ti=1sec**is also

**yi(t)=1**. The controller signal

**y(t)**is then twice the value of the initial signal

**y(t)**. Hence another name for the integration time

**Ti**is the doubling time

**Ti**. With the same

**Ti**and different

**Kp**of the

**PI**controller, the

**slew**rates of the output signal

**y(t)**are different, but their parameters

**Ti**will of course be the same.

**Chapter 26.2.2 PI unit Kp=1 Ti=10 sec**

**Fig.26-2**

** Kp=1 Ti=10 sec.**It integrates

**10**times

**slower**than before, but after

**Ti=10 sec**the output signal is also

**doubled**.

**Chapter 26.2.3 PI unit Kp=3 Ti=10 sec**

**F****ig. 26-3**It integrates

**3**times faster than before, but after

**Ti=10 sec**the output signal is also

**3 times bigger**. But the

**Kp**setting has no effect on

**Ti**.

**Chap. 26.3 Control system with Didactic PI controller****Chap. 26.3.1 Didactic PI controller****The didactic PI controller** that I tinkered with will show that the component:**– proportional** **P** for **faster** operation**– integrating I** ensures **zero** error**– at unit** step **x(t)** starts as **proportional** **P** and ends as integral **I.** This transformation will occur at some point. In a real **PI** controller, the transformation **P–>I** occurs continuously. You’ll find out later.

Fig. 26-4

The structure of the controller depends on the position of the contact:**-Fig. 26-4a**-contact “top”**P-type** structure. Note that at this time at **contact b**, thanks to the strong **negative** feedback involving the integrating unit, the voltage at** contact b** follows the voltage at **contact a**. If you don’t believe it, put** G(s)=100/sTi** to the most important formula in automation –> **Fig. 15-3d** **chapter 15**.**-Fig.26-4b**-“down”contact**-type I** structure. Thanks to the above-mentioned tracking, after switching from contact **a** to** b**, the system will behave as in the animation in **Fig. 26-5**

**Fig. 26-5**

Up to** 25** seconds, the controller is** P**-type and** y(t)=Kp*e(t)=3*1=3**. In the **25th** second, the contacts drop (you can see it in the animation) and the controller becomes type **I**. Thanks to “tracking”–>**Fig. 26-4a**, the integration starts from **+3V** and not from **0V**. After the so-called at the** doubling** time **Ti=25 sec**, the **I** component equaled the** P** component. If, for example, in **40 sec**. if the error **e(t)** increased a little bit, then **s(t)** would start to grow like a dashed line. There will be no signal spike s(t) because only the integral component **I** acts.**Chapter 26.3.2 Control system with didactic PI controller**

We will show that the component:

-proportional **P** is responsible for a quick approach of** y(t)** to a state close to the setpoint **x(t)**

-integrating **I** is responsible for reaching the state in which **y(t)=x(t)**. So it will provide steady error** e(t)=0**.

After that, it will be easier for you to understand the operation of a real** PI** controller.

**Fig. 26-6**Didactic

**PI**controller with settings

**Kp=10**and

**Ti=50sec**was used.

Before

**12 sec**, only the proportional component

**P**works, and after

**12sec**, the integral component

**I**.

Before you press the video button, take a good look at the signals:

–control

**blue s(t)**

–output

**red y(t)**

– green

**error e(t)**

In

**12**seconds, the structure of the controller changed from

**P**to

**I**, i.e. the contacts from

**a**to

**b**were dropped.

Fig. 26-7

Fig. 26-7

**Fig. 26-7a**shows all the time charts. The control signals

**sP(t), sI(t)**and

**sPI(t)**overlap and therefore

**Fig. 26-7b**was created, where these signals are shown separately. Here it is clear that in

**13 sec**the

**proportional**component

**sP(t)**disappears and the

**integrating**component

**sI(t)**appears. The total signal

**sPI(t)**is the

**sum**of the components

**sP(t)**and

**sI(t)**. Due to the different scales, the time charts in

**Fig. 26-12a**and

**Fig. 26-12b**are not identical.

In

**3 seconds**, a unit step

**x(t)**occurred. Then the control signal is only the

**proportional**component

**sP(t)**and the output signal

**y(t)**will very quickly reach the steady

**state y=0.91**. Certainly faster than if only the

**I**controller was working. In

**12 sec.**the controller changes its character from

**P**to

**I**and since then only the

**integrating**component

**sI(t)**is active. We know very well that it can reduce the error

**e(t)**to

**0**, which it does willingly. Therefore, after about

**90 seconds x(t)=y(t)=1 or e(t)=0**.

**Short and to the point**.

In

**PI**control, the

**P**component quickly brings the output signal

**y(t)**to a signal close to the setpoint

**x(t)**(but smaller) here to

**y(t)=0.91**.

**P**-control is unable to bring the steady-state signal

**y(t)**to

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

The didactic controller knows this handicap and therefore switches off

**sP(t)**and switches on

**sI(t)**in

**12 seconds**. And the

**integrating**component

**sI(t)**will lead the signal

**y(t)**to the value

**y(t)=x(t)=1**after some time, i.e. it will ensure

**zero**control

**error**.

**Chap. 26.3.3 Comparison of didactic PI with P-control**

See

**Fig. 26-7a**. The

**P**-type controller is a special case of the

**didactic controller**, in which there will never be a switch from

**P**to

**I**. That is, after

**12 sec**. the same state

**y=0.91**will remain. Compared to the didactic

**PI**controller, the error of the

**P**-type control is not

**zero**.

**Chap. 26.3.4 Comparison of didactic PI with Type I control**

The

**I-**type integral controller is optimal for a

**single-inertial**object with a time constant

**T=10 sec**, when its setting

**Ti=16 sec**. Let’s compare this control with the

**didactic PI**.

**Fig. 26-8**

For **2** control systems with an **one-inertial object** controlled by the:

– didactic** PI **controller

– classical** I** controller (integrating)

the same unit step **x(t)** is given. **Both** controllers should reduce the **steady** error **e(t)** to **0**. Which will do it better? i.e. **faster** and with** less** oscillation. **5:0** for didactic **PI**! It comes to a steady state **y(t)=1** **faster** and with less **oscillation** than type **I** control. You will soon see that real (not didactical)** PI** control is even better!

**Chapter 26.4. PI controller with one-inertial object****Chapter ****26.4.1 Introduction**

Time for a “real” **PI** controller, which is simply a **PI** controller.

This topic in **chapter 26.4, 26.5, 26.6 **are **PI** controllers for objects:**–single-inertial,****-two-inertial,****-three-inertial.**

We will study **control systems** with different **Kp** and **Ti** settings. We will always start with the **integration** action turned **off**, and then the **PI** controller will become a **P** controller. Why? Because you will see the main advantage of the **PI** controller. Reducing the control **error** to **e(t)=0** while operating much faster than the **I** controller.**Chapter 26.4.2 One-inertial object**

**Fig. 26-9****One-inertial** object with time constant **T=10 sec K=1**

By the way, you will remember how to read the parameters of an **one-inertial** object **K=1** and **T=10 sec** from the time chart.**Chap. 26.4.3 PI control Kp=3 integration disabled**

So we have a typical **P-**type control. **Kp=3** disables integration (**0** in the numerator of the integrating unit **I** disables integration).

**Fig. 26-10**

Up to approx. **19 sec**., the control works as **PI** didactical control up **to 12 sec**. in **Fig. 26-6**. So it is a **P**-type control. It is true that **Fig. 26-6** shows a different **Kp**, but it does not prevent us from analyzing. The output signal reached after **19** sec. steady state **y(t)=0.75** and this is consistent with the **time chart** and the theory for **P**-type control. So there was a **non-zero** error** e(t)=0.25**. In order to eliminate the **non-zeroness** of the error **e(t)**, let us introduce an **integrating** component **I**. We will start with a careful (“slow”) integration of **Ti=12 sec.**

**Chapter 26.4.4 PI control Kp=3 Ti=12 sec**

If you want to remember how the

**PI**unit reacts to a step

**x(t)**in an open system, go back to

**chapter 26.2**. After adding the

**I**component, the

**P**controller became a

**PI**controller. Will the error

**e(t)**be set to

**zero**?

**Fig. 26-11**Yes, yes, yes!!! After about

**50**seconds), the

**red**

**y(t)**coincided with the

**black x(t)**. So the error

**e(t)=0**. Such a quite good time chart was obtained with quite careful settings, low

**Kp**and high

**Ti**. We expect that with more “aggressive” settings, the system will reach the steady state

**y(t)=1**faster. Perhaps at the expense of

**oscillation**, because nothing in life is

**free**.

**Chapter 26.4.5**

**More detailed analysis of the PI Regulation Kp=3 Ti=12 sec**

Let’s try to find an analogy with the already well-developed

**PI**didactic controller, i.e. with the time chart in

**Fig. 26-6**. Same control system as before. But now we study not only the control signal of the

**sPI(t)**controller, but also its

**proportional**component

**sP(t)**and

**integrating**component

**sI(t)**.

**Fig. 26-12**The didactic

**PI**controller in

**Fig. 26-6**starts the action as (only) proportional

**P**. This allows it to reach a steady state

**y(t)**very quickly. Let me remind you that in the

**P**control at the beginning of the the control signal

**sPI(t)**is very large and this is what forces a time chart. Then, when the steady state is reached (which is always less than

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

**I**integral action is switched. Now the

**I**integral controller will do what it does best. It will complete the action, i.e. it will bring the control error to

**e(t)=0**. The

**red**

**y(t)**will coincide with the

**black x(t)**.

And how does a “real”

**PI**controller in

**Fig. 26-12**do this?

Similarly. At the beginning of the step

**x(t)**only the

**proportional**component

**sP(3)=3**is active, because the integrating

**sI(t)**“has not yet grown up” (

**sI(3)=0**). As in

**didactical**unit.

And in a

**steady state**, e.g. in

**55**seconds?

Here it is the opposite, only the

**integrating**component

**sI(t)**works because the

**proportional**one has disappeared –

**sP()=0**. Just like in

**didactical**too.

And between

**t=3 sec**and

**t=55 sec**?

Then

**sP(t)**disappears and sI(t) increases. So, to use elegant words, the

**P**controller changes into the

**I**controller. In a continuous manner, not stepwise as the didactic

**PI**controller! That’s it. Are there questions? I do not see. So let’s do a more aggressive (or faster)

**integration**. Hopefully this will speed up the time chart without exaggerating the oscillation.

**Chapter. 26.4.6 PI control Kp=3 Ti=4 sec**

More aggressive integration will certainly speed up the time chart, but at the expense of large oscillations?

**Fig. 26-13**The scope of the oscilloscope has since changed. It was

**0…3**, it’s

**-1…2**. The scale remained the same. Change due to appearance of negative

**e(t)**.

Faster runs compared to

**Fig. 26-12**when

**Ti=12sec**. The steady state is already after approx.

**25 sec**. at the expense of some control time. Some may like the previous

**Ti=4 sec**better. But let’s assume that our client’s priority was the

**time**to reach a

**steady state**. Even at the expense of the appearance of

**oscillations**. How about an even

**faster**integration?

**Chapter. 26.4.7 PI control Kp=3 Ti=2 sec**

**Fig. 26-14**

I think we overdid it. Large oscillation and control time similar to the previous one. So far we have studied the system by changing **Ti** when **Kp=3**. We will continue to do the same, but with **Kp=10**. We expect **faster** time charts and **larger** oscillations.**Chapter 26.4.8 PI control Kp=10 without integration**

We’ll start with a non-integral control. that is, from the type** P **control.

**Fig. 26-15**

From the time chart and theory, for the proportional control when **Kp=10**, the output signal **y(t)** in the steady state is** y=0.91** and the error **e=0.09**. In **Fig. 26-16** is the same time chart, but on a different scale. You’ll see the entire “unclipped” control signal **s(t)**.

**Fig. 26-16**This control signal

**s(t)**is large.

In order for the steady error to be

**zero**, the

**integration**must be turned on.

**Chapter 26.4.9 PI control Kp=10 Ti=15 sec**

**Fig. 26-17**Better than

**Kp=3**and

**Ti=4sec**in

**Fig. 26-13**. So let’s increase the intensity of the integration even more.

**Chapter 26.4.10 PI control Kp=10 Ti=5 sec**

**Fig. 26-18**I guess it’s better. There is a slight overshoot, but the steady state came much faster. We will increase the integration again.

**Chapter 26.4.11 PI control Kp=10**

**Ti=1.75 sec****Fig. 26-19**If we care about the control time and the overcontrol of about

**20%**is not a problem, then the settings

**Kp=10 Ti=1.75sec**are the best for our

**one-inertial**object. Otherwise,

**Kp=10 Ti=5sec**are optimal. Of course there could be even better settings. Only the responses with other combinations of

**Kp**and

**Ti**need to be investigated.

Now let’s move on to looking for the optimal setting for a slightly more difficult to control object –

**Two-inertia**l.

**Chapter 26.5 PI controller with two-inertial object**

**Chapter 26.5.1 Two-inertial object**

**Fig. 26-20**

**Two-inertial**object with time constants

**T1=3 sec, T1=5 sec, K=1**. You see a point of inflection typical for

**multi-inertial**units.

**Chapter 26.5.2 PI control Kp=3 integration disabled**

So we start with the P control.

**Fig. 26-21**The output signal reached after

**35 sec**. steady state

**y(t)=0.75**, and this is consistent with the

**time chart**and the theory for

**P**-type control. There was a non-zero

**error e(t)=0.25**. To make th

**e(t)=0**, let us introduce an integrating component

**I**. We start carefully with

**Ti=32 sec**.

**Chapter 26.5.3 PI control Kp=3**

**Ti=32 sec**

**Fig. 26-22**The error signal

**e(t)**reaches the

**zero**state. The chart doesn’t show it,because the time of the experiment was too short.

**Chapter 26.5.4 PI control Kp=3**

**Ti=8sec****Fig. 26-23**

Here we have already reached the steady state when **e(t)=0**. It got better. Let’s move on**Chapter 26.5.5 PI control Kp=3 Ti=5 sec**

**Fig. 26-24**

We went a bit overboard with the intensity of the **integration**. It was better before.**Chapter 26.5.6 PI control Kp=10 integration disabled**

**Fig. 26-25**

Steady state consistent with the theory for** P** control, i.e. **e(t)=0**.**Chapter 26.5.7 PI control Kp=10 Ti=20 sec**

**Fig. 26-26**

At **60** seconds, the error is not yet **zero**. Let’s reduce **Ti**. Note the oscillation amplitude of the control signal **sPI(t)** compared to the output signal **y(t)**. It is generally larger.**Chapter 26.5.8 PI control Kp=10 Ti=10 sec**

**Fig. 26-27**Not bad. But the setting

**Kp=3**

**Ti=8 se**c from

**Fig. 26-23**is better for a controlled

**two-inertial**object. And we’re used to the fact that a large

**Kp**is mostly good. Let’s reduce

**Ti**further. Will it help?

**Chapter 26.5.9 PI control Kp=10 Ti=5 sec**

**Fig. 26-28**

Did not help. Too much oscillation.

Now let’s move on to the most difficult object – the **three-nertial** one.

**Chapter 26.6 PI controller with three-inertial object**

**Chapter 26.6.1 Three-inertial object**

**Fig. 26-29**

Three-inertial object –** K=1** and time constants **T1=0.5 sec T2=3 sec, T3=5 sec**.

At first glance, it is difficult to distinguish from a **two-inertial** one.

The more inertia (here **T1, T2** and **T3**) the more clearly the **delay** parameter is marked – here **To=1.3 sec**.**Chapter 26.6.2 PI control Kp=10 integration disabled**

**Fig. 26-30**Steady state,

**i.e e(t)=0**, consistent with the theory for P-regulation.

**Chapter 26.6.3 PI control Kp=10**

**Ti=16 sek**

**Fig. 26-31**

The output signal slowly reaches a steady state **y(t)=1**. So let’s increase the to **Ti=10 sec**.**Chapter 26.6.4 PI control Kp=10 Ti=10 sek**

**Fig. 26-32**Increasing the

**integration**rate had the desired effect. After

**45 sec**. the condition

**x(t)=y(t)**occurred, i.e.

**e(t)=0**. Let’s reduce

**Ti**further.

**Chapter 26.6.5 PI control Kp=3**

**Ti=4 sec**

**Fig. 26-33**It is worse. And what will happen when we increase

**Kp**?

**Chapter 26.6.6 PI control Kp=10 integration disabled**

I,e

**P**control.

**Fig. 26-34**According to the theory, for a steady state of

**P**-type control, after

**60 seconds**, a steady state will occur

**y(t)=0.91**and

**e(t)=0.09**. In order to bring the control error

**e(t)**to 0, we will start with a cautious (i.e.

**slow**, i.e. large

**Ti**) integration.

**Chapter 26.6.7 PI control Kp=10**

**Ti=16 sec**

**Fig. 26-35**The

**y(t)**oscillations occur around

**x(t)=1**. Finally in agony after

**60 sec**. control error will be

**zero**. So let’s reduce

**Ti**to make it

**faster**.

**Chapter 26.6.8 PI control Kp=10**

**Ti=8 sec**

**Fig. 26-36**It is worse. So for this

**three-inertial**object, the optimal

**PI**control settings are

**Kp=3**and

**Ti=10 sec**from Fig. 29-60. We got used to Kp=10 giving better results than Kp=3. So it’s not worth getting into habits, especially in automation. And what will happen when we increase the integration intensity to Ti=2.5 sec?

Chapter 26.6.9 PI control Kp=10

Chapter 26.6.9 PI control Kp=10

**Ti=2.5 sec**

**Fig. 26-37**

Notice the unusual scope of the oscilloscope. Although it is an **unstable** system, up to **3** seconds nothing happens at the output because nothing happens at the input. In** 3** seconds we “knocked” the system with **Dirac pulse**, bringing it out of equilibrium in this way.

**Chapter 26.7 PI control with disturbances ****Chapter 26.7.1 Introduction**The same

**one-, two- and three-inertial**objects will be controlled. Their inputs will be affected by input signals

**x(t)**(as before) and

**disturbances**signals

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

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

**Disturbances**will cause an

**error e(t)**which will “annoy” the

**PI**controller so much that it will always bring it to

**0**. More precisely. The

**proportional**component

**P**will bring the error

**e(t)**to a value close to

**0**and the

**integrating**component

**I**will finish the job bringing it to

**0**. The experiment will last

**2**minutes. The settings that gave the “prettiest” response in previous experiments will be used. The “prettiest” i.e. relatively

**fast**and with possibly

**small**

**oscillations**. Let’s call these settings

**optimal**, although they can be even better. After all, we studied only a few combinations of

**Kp, Ti**.

**Chapter 26.7.2 Positive disturbance with a one-inertial object z(t)=+0.5 Kp=10 Td=5sec**

The disturbance

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

**70 seconds**.

**Fig. 26-38**

Up to **70 seconds**, i.e. until the appearance of the disturbance, the time chart is the same as in **Fig. 26-18**, taking into account the different time scale on the oscilloscopes. Initially, the disturbance **z(t)=+0.5** caused the signal **y(t)** to **increase**, but then the** I** component “forced” **y(t)** to return to its previous value, i.e. to **y(t)=1**. This corresponds to bringing the error **e(t)** to **0**. On a **positive** “heating” disurbance, the control **s(t)** responded by reducing **power**.**Chapter 26.7.3 Negative disturbance with a one-inertial object z(t)=-0.5 Kp=10 Td=5sec**The disturbance

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

**70 seconds**.

**Fig. 26-39**On the

**negative**“cooling” disurbance, the

**sPI(t)**control responded by increasing the

**power**.

**Chapter 26.7.4 Positive disturbance with a two-inertial object z(t)=+0.5 Kp=3 Ti=8sec**

The disturbance

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

**70 seconds**.

**Fig. 26-40**The object is “more difficult” to control, hence the longer control times.

**Chapter 26.7.5 Negative disturbance with a two-inertial object z(t)=-0.5 Kp=3 Ti=8sec**

The disturbance

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

**70**seconds.

**Fig. 26-41**The disturbance

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

**70**seconds.

**Chapter 26.7.6 Positive disturbance with a three-inertial object z(t)=+0.5 Kp=3 Ti=10sec**

The disturbance

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

**70 seconds**.

**Fig. 26-42**Even “more difficult” to control and longer control times

**Chapter 26.7.7 Negative disturbance with a three-inertial object z(t)=-0.5 Kp=3 Ti=10sec**.

The disturbance

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

**70 seconds**.

**Fig. 26-43**On the negative “cooling” disturbance, the control

**s(t)**reacted with an increase in power. The error

**e(t)**was reduced to

**0**.