## Preliminary Automatics Course

**Chapter 20. How does feedback work?
**

**Chapter 20.1 Introduction**

**Fig. 20-1**

**Fig. 20-1a
**The output signal

**y(t)**is fed back to input. But what is a cause and effect? What was first egg or hen? I will not ask this existential questions. Let’s treat this configurattion as a curio.

**Fig. 20-1b**

There is a need to contact with the outside world. It’s realised as error

**e(t)=x(t)-y(t)**calculation in the so-called

**subtractor**. It’s the most important configuration in the control theory. Its working must be for you pure and simple, as a biking. This is main goal of all this course!!!

**Chapter 20.2 How does positive feedback work?**

**Chapter. 20.2.1 Introduction**

**Fig. 20-2
**There is

**adder**instead of

**subtractor.**It realizes sum of signal

**e(t)=x(t)+y(t)**instead of difference as in

**Fig. 20-1**. It’s something wrong rather. Linesman mistakes the input “+” for “-” for example. The positive feedback is easier to understand and that’s the reason of this subchapter.

**Chapter 20.2.2 Big positive feedback.**

Why “big”? You will know by the “small” positive feedback.

Call Desktop/PID/06_sprzezenie_zwrotne/01_sprzezenie_zwrotne_+_duze.zcos

**Fig. 20-3**

Click “start”

**Fig. 20-4
**The single “needle” pulse

**is passing through inertial**

**x(t)****and is returning as**

**G(s)****to the**

**y(t)****input again**

**G(s)****.**Please note that

**is active during a small time! The**

**x(t)****then. The returnig**

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

**y(t)****K=1.5**and returning once more… etc. The classical avalanche effect! Please note that

**K=1.5>1!**

We suppose that every positive feedback means instability-avalache effect. Really? Let’s go to the next chapter.

**Chapter 20.2.3 Small positive feedback.**

Why “small”? The previous **K=1.5>1**. Let’s try **K=0.8<1**

Call Desktop/PID/06_sprzezenie_zwrotne/02_sprzezenie_zwrotne_+_male.zcos

**Fig. 20-5**

The input **x(t) **is step type instead of “needle” and **K=0.8**.

Click “start”

**Fig. 20-6
**The response

**y(t)**is stable and inertial type.

The parameters

**Kz=4**and

**Tz=15 sek**show that:

-the

**gain**increased from

**0.8**up to

**4**

– time constant increased from

**T=3 sec**up to

**Tz=15 sec**. The unit is more “lazy”.

The main conclusion is that dynamical object

**G(s)**with

**positive feedback**may be sometimes stable!

**Chapter 20.2.4 Transfer function Gz(s) with the positive feedback.**

**Fig****. 20-7
**

**Fig. 20-7a**

The formula for transfer function

**Gz(s)**with the

**The formula for transfer function**

**negative feedback****Fig. 20-7b****Gz(s)**with the

**. See**

**positive feedback****Fig.19-17d**and put

**E(s)=X(s)+Y(S)**instead of

**E(s)=X(s)-Y(S)**. The effect is this formula. We will check this formula for the

**G(s)**of the

**Fig. 20-3**and

**Fig. 20-5**

**Fig**

**. 20-8**

**Fig. 20-8a**– Transfer function

**Gz(s)**when

**positive feedback**

**Rys. 20-8b**– an example when

**G(s)**is

**unstable**as

**Fig. 20-3**. Note that

**K=1.5>1**and the nominator root

**s=+1/6**is positive.

**Rys. 20-8b**– an example when

**G(s) is stable**as

**Fig. 20-8**. Note that

**K=0.8<1**and the nominator root

**s=-1/15**is negative.

The nominators were the polynomians degree

**n=1**here. But we can generalize for all polynomians degree

**n**.

## The system is stable when the G(s) takes all negative roots in the nominator.

Do you know so-called **complex numbers**?* The statement will be more precisely now.

## The system is stable when the G(s) takes all roots with the negative real part in the nominator.

*We will discuss it in the **Chapter 23.**

**Chapter 20.2.5 Positive feedback conclusions**

**1-** It’s something wrong rather.

**2-** The cause of unstability mostly-but not always.

**3-** Any **G(s) **note.

Positive nominator root = unstability

More precisely

roots with the positive real part = unstability

**Chapter 20.3 How does negative feedback work?**

**Chapter 20.3.1 Introduction
**The main

**negative feedback**advantage is that the output signal

**y(t)**tries to tollow the input signal, so-called set point

**x(t)**.

**Chapter 20.3.2 How does the inertial unit come to the steady state?**

Call Desktop/PID/06_sprzezenie_zwrotne/10_otwarty_inercyjny.zcos

**
Fig. 20-9
**Inertial unit as a open loop configuration. (no feedback)

The oscilloscope observs the signals:

-output

**y(t)**

-intput x

**(t**

**)**

– goal

**K*x(t) (K=5)**<– What is this? You will know at the moment.

Click “Start”

**Fig. 20-10**

The steady state

**y=5*x(t)=5**is a goal of the output signal

**y(t),**when

**x(t)=1**is a step type. The difference-

**gren line**

**U=5*x(t)-y(t)**is a

**“motive force” of the output signal**

**y(t)**.

The “motive force”

**U=5*x(t)-0=5**is maximal at the start time=1sec and the

**y(t)**speed signal is maximal now.

The

**“motive force”**

**U=5*x(t)-y(t)**is decreasing then, because

**y(t)**is increasing.

The “motive force”

**U=5*x(t)-5=0**after time=24 sec and the

**y(t)**speed signal is

**null**. The signal

**y(t)**is steady now.

**The main conclusion**

The output y(t) increases when “motive force”(gren line)is positive.

This conclusion is so obvious, that reader wonders-what’s the matter?

**Chapter 20.3.3 How does inertial unit with the negative feedback come to the steady**** state?**

The open loop unit (as previous) operation is easier to understatnd, than with the negative feedback. But you will be be convinced that the mode of action is such as inertial in open loop in previous chapter. This is the answer for the question also–> see above. **
**Call desktop/PID/06_sprzezenie_zwrotne/03_zamkn_inercyjny

**Fig. 20-11**

Inertial unit with negative feedback.

The oscilloscope observs the signals:

-output

**y(t)**

-intput x

**(t**

**)**

– goal

**K*e(t) (K=5)**<– It’s similar to previous goal

**K*x(t)**–>

**Fig.20-10**!

Click “Start”

**Fig. 20-12**

The

**y(t)**aims to goal as previous, but the goal

**5*e(t)**is time variable now! The “motive force”

**U=5*e(t),**which wants to coapt the goal

**5*e(t)**and

**y(t)**is a difference

**U=5*e(t)-y(t)**–>

**green line**. The bigger is a “motive force”–> the

**blue**and

**red lines**wish stronger to cuddle up together!

The output y(t) increases when “motive force”(green lineU)increases.

And when will **y(t) **come to a stop? Other words. When will be balance-steady state? Signal **y(t)** will come to a stop when the “motive force”-**green line **will disappear. **I. e. y(t)=K*e(t)**. This experiment confirms something important in **control theory**.

There is always y(t)=K*e(t) in the stable system with the negative feedback

This is fullfiled for all **static dynamic ****dynamic** **units. **Not only **inertial**.

It must be as clear as biking for you. Please note that it involves stable systems only.

Do you remember **Fig. 19-17e ****Chapter 19? **You weren’t convinced yet that **y(t)=K*e(t)**. You know now and the the proof is experiment **Fig. 20-12**. The lines **5*e(t)** and** y(t)** crosspassed in the balance point **5*e(t)****=****y(t)**. You can easy derive the formula for steady gain **Kz** when negative feedback

**
Fig. 20-13
**The

**y=0.83**in the

**Fig. 20-12**confirms this formula.

**Chapter 20.3.4 How does inertial unit with the big negative feedback come to the steady**** state?**

We will observe the additional signal **error e(t)**.

Call Desktop/PID/06_sprzezenie_zwrotne/04_zamkn_inercyjny_K100.zcos

**Fig. 20-14**

Click “Start”

**Fig. 20-15
**You observe all “variable goal”

**100*e(t)**. It strikes eye the big size of this signal next to others

**x(t),y(t)**and

**e(t)**.

You will know that is

**Proportional**or

**P**type control then. There is a typical big control signal at the beginning. The

**x(t),y(t)**and

**e(t)**signals aren’t good visible here. Let’s reprise this experiment with changed oscilloscope parameters.

Call desktop/PID/06_sprzezenie_zwrotne/05_zamkn_inercyjny_K100_inny_widok.zcos

The scheme is exactly as **Fig. 20-14** but there are different oscilloscope parameters now.

Click”Start

**Fig. 20-16
**The main conclusion is the short setting time when

**K**is big. The time constatnt

**T≈0.03sec**instead

**T=3sec**when open loop. The negative feedback is good for dynamics!

**Chapter 20.3.5 How does double inertial unit with the big negative feedback come to the steady**** state?**

The previous **inertial unit **came to steady state “underarm”–> It was **y(t)<5*e(t) **always. But the control process is with the oscillations very often.

Call Desktop/PID/06_sprzezenie_zwrotne/06_zamkn_dwuinercyjny.zcos

**Fig. 20-17**

Click “Start”

**Fig. 20-18
**The

**y(t)**achieved the state

**5e(t)=y(t)**quickly in the

**6 sec**. It’s supposedly “steady state” but the

**y(t)**goes like the clappers on. It isn’t real steady state because the

**y(t)**derivatives are’t null now. It will be real steady state

**5e(t)=y(t)**in

**18 sec**. and after. Please note that the

**green**“motive” force was

**postive**up to

**6 sec.**It’s

**negative**then! (6…11sec). It means that

**green**“motive” force changed to “braking” force! The steady state was

**y(t)=5*e(t)**achieved after some oscillations here.

**Chapter 20.3.6 Static and astatic units**

There are to methods to calculate steady gain **K**.

**1- **When **time run** used as **Fig. 20-18 **–>**K=0.83/1=0.83.
2-** as

**G(s=0)**see under

**Fig. 20-19**

**Fig. 20-19a**inertial unit

**Fig. 20-19b**double inertial unit

**Fig. 20-19c**real differential unit. Note that steady

**y(t)=0!**

The above gains

**K**are obvious. Their step respons are

**finit**and tey are

**static units**.

The problem is with the

**Fig. 20-19d**and

**Fig. 20-19e**How to divide by

**0**. It’s forbidden! But on the other side

**the K=infinity**. Imagine for a moment that the nominator is very small instead of

**0**for example

**0.000001**–>

**K=1000000**very big , infinity almost! There are

**astatic units**examples.

**Fig. 20-19d**integrating unit

**Rys. 20-19e**integrating unit with inertia

What are the

**astatic units**steady states?

**Chapter 20.3.7 How does astatic unit come to the steady**** state?
**The title is provocative a little. You will know why.

Call Desktop/PID/06_sprzezenie_zwrotne/07_otwarty_calkujący

**Fig. 20-20**

The special

**x(t)**signal type will be given to see typical astatic units behaviour.

Click “Start”

**Fig. 20-21**

It maybe an

**ideal**direct current motor with the counter on its axis. Have you problem with the digital

**y(t)**output istead of analog as usually? So imagine that there is

**digital/analog**converter installed. The output is analog now.

**phase 0…3 sec**input voltage

**x(t)=0V**–> motor stays–>

**y(t)=0**

phase 0…10 secinput voltage

phase 0…10 sec

**x(t)=+1V–>**motor rotates right –>

**y(t)**increases linearly

**phase 10…30 sec**input voltage

**x(t)=0V–>**motor stays–>

**y(t)=7 right revolutions**

**phase 30…40 sek**napięcie

**x(t)=-1V**silnik ze stałą prędkością kręci w lewo–>

**y(t)**maleje liniowo.

**etap 40…60 sek**napięcie

**x(t)=0V**–> motor stays–>

**y(t)=-3 left revolutions**

You know the provocation in the title? There is no

**astatic unit**steady state when

**x(t)**is step type. When time

**t=infinity**–>

**y(t)=infinity**–>

**K=infinity**.

**Chapter 20.3.8 How does Integrating unit with the negative feedback come to the steady**** state?
**The

**Integrating unit**is an example of the

**astatic unit**

**Fig. 20-22**

**Fig. 20-22a**Steady state gain

**Kz**for the static units

**Rys. 20-22b**Steady state gain

**for the astatic units**

**Kz**It means that there is

**error e=0**for steady state. It’s every automatics engneer goal.

Call Desktop/PID/06_sprzezenie_zwrotne/08_zamkn_calkujący.zcos

**Fig. 20-23**

Click “Start”

**Fig. 20-24
**The

**y(t)**goal is steady and

**y(t)=x(t)**when steady. The error

**e(t)=0**in steady state. The “motive” force-

**green line**is

**e(t)**! When

**e(t)**is no null then this signal

**decreases**or

**increases**the

**y(t)**up to

**y(t)=x(t)**stable state. The

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

**e(t)**value. The

**y(t)**speed is

**null**when

**e(t)=0**and achieves the stable state

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

**Chapter 20.3.9 How does Integrating unit with inertia and with the negative feedback come to the steady**** state?**

There was an ideal motor with the feedback in the **Fig. 20-23**. Let’s analyse the real motor with the feedback. It’s an ** Integrating unit with inertia **example.

Call Desktop/PID/06_sprzezenie_zwrotne/09_zamkn_calkujący_rzeczywisty

**Fig. 20-25**

Click “Start”

**Fig. 20-26**

The **+e(t)** “motive force” drives **y(t)** to balance state **x(t)=y(t) **and** -e(t)** breaks into this state too. This state **x(t)=y(t)=1 ** was achieved after some oscillations.

**Chapter 20.7 Conclusions**

**Chapter. 20.7.1 Closed loop Gz(s) transfer function
Fig. 20-27**

**Chapter 20.7.2 Static units with the negative feedback-steady state
**

**Fig. 20-28**

Our dream is that output

**y(t)**follows set point

**x(t)**. This is fulfilled when

**K**is big. There is

**Kz=1 Ke=0**“almost”.

**Rozdz. 20.7.3 Układy astatyczne z ujemnym sprzężeniem zwrotnym
**

**Fig. 20-29**

**Chapter 20.7.4 What a important subject was absent in this chapter?**

**1-** **Unstability** –> The next chapter

**2-** Open loop **Fig. 20-9** assures **x(t)=y(t) **in steady state when **K=1**. It means that **e(t)=0! **The question is now. Why do complicate the life to use **negative feedback** to assure **x(t)=y(t) **steady state. The answer is simple.

**Negative feedback **assures:

– resistance for disturbances

– better dynamiccs- You don’t need exactly mathematical object **G(s) **description. There is a need of the good sensor only.

** **

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