# Automatics

**Chapter 12 Integration**

**Chapter 12.1 Introduction**There are

**indefinite**and

**definite integrals**.

**Chapter 12.2 Indefinite integral F(t) from f(t)**

The **indefinite integral** from the function** f(t)** is the so-called **antiderivative** function **F(t)** and is often simply called an **integral**.

Fig. 12-1

The** derivative** of the **antiderivative** function **F(t)** is the function **f(t)** itself, and vice versa – the **indefinite integra**l of the function** f(t)** is the **antiderivative** function **F(t)**.

Fig. 12-2|**Integration** and **Differentiation** are mutually **inverse operations**. **Integration** as a** calculation** of the **antiderivative** function **F(t)** is a certain proficiency that **first-year** students of **technical universities** acquire. It is needed to calculate the **definite integrals**. They will be discussed in a moment. Usually, in higher **universities** years, only the** simplest integrals**, otherwise, **antiderivatives** are remembered. This is the right approach, because you can always go back to your lecture notes.

In short soldier words. The **antiderivative function** from the **derivative function** is this function (Fig. 12-2c).

**Chapter 12.3 The Definite Integral of x(t)****Chapter 12.3.1 Introduction**

In automatics, we are dealing more with a** definite integral** than an **indefinite** one. The latter as an **antiderivative** function is used for very easy calculation of the **definite integral** value.

**Chapter 12.3.2 The Definite Integral of x(t) as the area under x(t)**

Fig. 12-3

The **definite integral** from **t1** to **t2** is the area **S** under the **x(t)** function. So it is a specific number, e.g. **S=27.13**. For our purposes, we will narrow this definition down a bit. In automatics, something usually starts at time **t=0** and continues until time **t**. For example, the input signal **x(t)** can be a unit step or a **sawtooth**. It is therefore assumed that for **t<0** (negative!) the signal **x(t)=0**.–>**Fig. 12-3**.**Fig. 12-4**So this is the version from

**Fig. 12-3**where

**t1=0**and

**t2=t**. With this approach, the

**definite integra**l becomes a function

**y(t)**and not a specific number.

And now the most important. How we calculate the fields in

**Fig. 12-3**and

**Fig. 12-4**.

Fig. 12-5

Fig. 12-5

**F(t1)**and

**F(t2)**in

**Fig. 12-5a**are the values of the

**antiderivative functions**, i.e.

**indefinite integrals**for

**t1**and

**t2**.

We can also treat

**F(t1)**itself as a

**area**to

**the left**of

**t1**and, analogously,

**F(t2)**as a

**area**to the left of

**t2**. With this approach, the formula in

**Fig. 12-5a**as the areas

**difference**is obvious.

And in

**Fig.**

**12-5b**,

**F(t)**itself is simply an

**antiderivative**! In the next

**2 experiments,**we will check if this is the case. Does the theory match the practice?

**Chapter 12.3.3 The Definite integral of a Function that is a unit step x(t)=1(t)**

Why exactly a unit step **x(t)=1(t**)? Because there is no simpler function and it is easy to calculate the definite integral = area **S**.

Let’s treat the **definite integral** as the output **y(t)** of the integrating **unit** whose input is **x(t)=1(t)**. It turns out that **y(t)=t(t)<**–linear function

**Fig. 12-6**The input of the

**integral unit**is a unit

**step x(t)=1(t)**and its output is the integral

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

I remind you that:

**1(t)=0**for

**t<0**and

**1(t)=1**for

**t>0**

**t(t)=0**for

**t<0**and

**t(t)=t**for

**t>0**, i.e.

**t(t)**is the so-called

**saw**.

The

**integral**from 0 to t of any function is the

**are**a under the graph of this function.

Here

**x(t)=1(t)**is very simple and it is easy to calculate the

**area**under the function as the area of a rectangle with sides

**h=1**and

**t**. For any

**t>0**, the area

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

E.g. for

**t=5**

**y(t)=5**. So theory matches practice.

**So in short “the derivative of the definite integral in the range**

Fig. 12-7

Fig. 12-7

**0…t**is an integrand function.

**Fig. 12-7a**

**[t(t)]’=1(t)**, i.e. the

**derivative**of the

**integra**l is an

**integrand**function

**Fig. 12-7b**

Generalization of the previous one

Instead of the function

**t(t)**there is just a general

**f(t)**

**Fig. 12-7c**

Ultimately

The

**definite integral**of the

**derivative**, which is

**f'(t)**is a function of

**f(t)**. So

**differentiation**is the inverse of

**integration**.

**Chapter 12.3.4 The Definite Integral of a function that is a sawtooth x(t)=0.2*t**

For the saw, the **Definite Integral** can be computed as the **area** of a **triangle**. So let’s check if the output of the** integral unit** is a **definite integral**.**Fig. 12-8**The

**inpu**t of the

**integral unit**is a “saw”, otherwise a linearly increasing signal, here according to the formula

**x(t)=0.2*t**.

We will calculate the

**definite integral**of

**x(t**) as the

**area**of a

**triangle**with base

**t**and height

**x(t)=0.2*t**.

Here, for example, for t

**=9sec**, the saw

**x(t)=0.2*9=1.8**and the area of the triangle

**S=0.1t²=8.1**

Theory agrees with practice. The signal behind the

**integrating unit**is a

**parabola**with the given formula and for e.g.

**t=9**the output signal

**y(t)=8.1**.

**Chapter 12.3.5 The definite integral of the “slider swing” function. Otherwise type I manual control**

The **definite integral** is always the area under the function from time **t=0** to time **t**. Even when the function is more complicated like below. The **integral** can be **positive, zero, or negative.**

**Fig. 12-9**Observe the output signal

**y(t**) depending on the changing position of the slider

**x(t)**.

You will find that:

– The

**more**positive

**x(t)**, the

**faster y(t)**grows

– The

**more**negative

**x(t)**, the f

**aster y(t)**decreases

–

**x(t)=0**is stopping

**y(t)**at the last level.

This makes us believe that

**x(t)=y'(t)**. Imprecisely speaking, but accurate “

**x(t)**is the

**derivative**of the

**integra**l” or

**“**the circles in

**Fig. 12-8**are

**true”**.

In my experiment,

**52**seconds when

**y(t)=0**, then the

**positive**area

**x(t)**is equal to the

**negative**area (area counted up to

**t=52 sec**.)

**Chapter 12.4 Conclusion**

**Fig. 12-10**

**Integral units symbols**

**Fig. 12-10a**

Using the integral symbol

**Fig. 12-10b**

Using operational calculus