### Automatics

**Chapter 4 Integrating Unit**

**Chapter 4.1 Introduction**

**Fig. 4-1 **Transmittance of the

**Integral unit**

It is recommended when the transmittance

**numerator**has a value of

**1**. Then the parameter – integration time

**Ti**has an easy interpretation ->

**p.4.4**. And when not? E.g.

**G(s)=3/9s**? Divide the numerator and denominator by

**3**and you get the same

**G(s)=1/3s**, just in normalized form. Here you can see that

**Ti=3sec**.

Note for those unfamiliar with integrals and derivatives.

Don’t worry for now. It is enough for you to associate the output signal

**y(t)**with the input signal

**x(t)**(usually a unit step) and the integration time

**Ti**.

**Chapters**

**11**and

**12**will be devoted to

**differentiation**and

**integration**

**Chap. 4.2 Scheme with slider and bar graph**

As usual, we start the description of the new **dynamic unit** with the **slider** and the **bar graph**. The response to the **slider** is more visual than **math.**

**Fig. 4-2**.

Observe the slider **x(t)** and bargraf **y(t)**

The **input** signal **x(t)** is a **slider** and a **digital meter**

The **output** signal **y(t)** is a **bar graph** and a **digital meter**

A **typical** feature of the **integral unit** is the **speed** of the output signal **y(t)** which is proportional to the input signal **x(t)**.**You** will find that:**– increased positive** signal **x(t)** causes a rapid **y(t)** increase**– increased negative** signal **y(t)** causes a rapid **y(t)** decrease

-zero signal **x(t)** stops **y(t)****Note**:

I had a problem setting **x(t)=0**, that’s why you see “**almost x(t)=0**” on the gauge.

That’s why it’s “almost stop” then

What does that remind you of? You have such a device at home. It’s a **TV** remote.

Once you enter the **“plus”** value ->the volume **increases**.

You will enter **“zero”** -> **constant** volume.

You enter **“minus”** -> the volume **decreases**.

**Chapter 4.3 Scheme with slider and oscilloscope**

And now** x(t)** input from the slider but **y(t)** output from the **oscilloscope**.

**Fig****. 4-3**I changed the

**x(t)**signal with the

**slider**. This can be seen in the

**x(t)**gauge and the

**x(t)**graph. The time plot clearly shows that the rate of change of the output

**y(t)**is proportional to the input

**x(t)**. For example, when

**x(t)**doubled around

**13**second, the signal

**speed y(t)**also

**doubled**. Also

**x(t)=0**kept

**y(t)**constant.

**Chapter 4.4 Scheme with unit step function x(t)=0.1 and oscilloscope. Chapter 4.4.1 Ti=1sek**

**F****ig. 4-4**The previous scheme, when the

**x(t**) input signal came from the

**slider**, allowed for the

**initial**analysis of the integrator. Therefore, we will replace the

**slider**with a more accurate

**unit step generator**. More precisely, it will be a step

**x(t)=0.1**.

Let’s repeat the experiment for

**Ti=2 se**c.

**Chapter. 4.4.2 Ti=2sek**

**F****ig. 4-5**As expected, the signal will grow

**2**times slower. The alignment took place after time

**t=Ti=2 sec**. Now we can define the integration time

**Ti**for the integral term from

**Fig. 1-1**. This is the time

**Ti,**after which the output

**y(t)**equals the step input

**x(t)**.

**Chapter 4.5 Scheme “with stairs” and an oscilloscope****Chapter 4.5.1 “2 stairs” and an oscilloscope**

**Fig. 4-6**Response to

**x(t)**“2 stairs”

When the

**input**signal

**increased 2**times, the speed of the

**output**signal also

**increased 2**times.

**Chapter 4.5.2 “4 stairs” and an oscilloscope**

**Fig. 4-7**Response to

**x(t) “4 steps”**

With each

**step**, the speed increases

**Chapter 4.6 Scheme with a x(t) ramp-up signal and an oscilloscope**

It is no coincidence that the previous input signals** x(t)** were **stepped**. Why? Because a linearly increasing signal can be treated as with an infinite number of small steps.

**Fi****g. 4-8**When the input signal increases linearly

**x(t)=t**then

**y(t)**increases as a

**square**function. Check by substituting different values for t. The

**integrating unit**, as the name suggests, integrates the input signal

**x(t)**giving the output a parabola

**y(t)**.

On the other hand,

**calculus**says that the

**definite integral**of the function

**x(t)=t**is also a

**parabola**

**Check that e.g. for**

Fig. 4-9

Fig. 4-9

**t=4**

**x(t)=5, y(t)=12.5**

Indeed, put

**t=5**and you will see. So theory matches practice. If you don’t know math, don’t worry. For now, consider that the definite integral from

**0**to

**t**of

**x(t)**is the signal

**y(t)**of the

**1/s**integrating

**unit**.

**Chapter 4.7 Scheme with positive and negative step and oscilloscope**

This is better version of the experiment in **Fig. 4-3**, in which the **x(t)** signal is supplied from the** generator**, which is a more **precise device** than **the slider**.

**Fig. 4-10**In a similar way, you control the

**volume**on the

**TV**with the

**remote**control, but here you can also control the

**speed**of the

**bar**. It can also be an

**actuator**as a

**DC motor**with a

**gear**. The

**input**is the

**voltage**and the output is the position of the

**actuator lever**.

**Chap. 4.8 Scheme with a Dirac pulse and an oscilloscope**

**Fig. 4-11**There is no perfect

**Dirac**impulse in nature. There is only an approximation of it. Here, a rectangular

**pulse**with an amplitude of

**100**and a duration of

**0.01 sec**.

**Chapter 4.9 Typical integrating units.****Capacitor.****Input** – capacitor charging current**Output** – voltage on the capacitor

When a capacitor is charged with a constant current, the voltage increases linearly, the faster the smaller the capacitance C of the capacitor

**The bathtub is clogged with a cork**.

I assume the** bathtub** is a rectangular** prism**.**Input** – water flow from the tap, indirectly the degree of opening of the tap**Exit** – water level

If the **flo**w is **constant**, the water** level** will increase **linearly**. The faster the smaller the **surface** of the **bathtub**.

**Electric actuator**.

I assume we have an ideal **dc gear motor**.**Input** – motor voltage**Output** – position angle of the actuator shaft

A** constant** voltage on the motor will cause the angle to **increase** linearly. The **rate** of increase is **proportional** to the **voltage**.