### Automatics

**Chapter 16 How does feedback work?**

**Chapter 16.1 Introduction**

Feedback consists in the fact that the

**output**signal

**y(t)**

**returns back**to the

**input**.

Fig.16-1

Fig.16-1

**Fig. 16-1a**

The

**y(t)**signal from the output returns directly to the input. But what is the cause and what is the effect? Did the egg come first or the chicken? Not to mention more existential questions. So let’s treat the scheme as a curiosity.

**Fig.16-1b**

So you need some contact with the outside world. It is implemented as a calculation of the difference

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

**subtractor node**. This is the

**most important**scheme in automatics because

**y(t)**tries to imitate

**x(t)**. It must be as obvious and instinctive to you as riding a bike! Understanding the mechanism of this imitation is the main goal of this chapter.

**Chapter 16.2 How does positive feedback work?****Chapter 16.2.1 Introduction**

Fig. 16-2**Positive feedback.**

That is, in which the** sum** of signals **e(t)=x(t)+y(t)** is fed to the input, instead of the **difference** as

in **Fig. 16-1b**. Rather, it is something harmful. Accident at work. For example, when the fitter gets the cables wrong, and instead of the “-” input of the regulator, he will go to the “+” input. It is easier to understand than negative feedback, and that is why we are analyzing this case. It’s a bit like a guy looking for a key under a lamppost, because you can see better than in the darkness. Though he might have lost the key out of reach of the lamp.**Chapter 16.2.2 Positive Feedback “Large”**

Why “big”? You’ll find out when “small” is mentioned.

This is the version of **Fig. 16-2** when **G(s)** is the inertial term with the parameters **K=1.5, T=3 sec**. The signal from the output returns to the input and is added to the input Dirac signal **x(t)** – **“pins”** or in other words** “flick”**. Note that **K=1.5>1**!

**Fig. 16-3**The sum of the signals

**s(t)=x(t)+y(t)**enters the input of the

**inertial bloc**k. Initially,

**x(t)=y(t)=0**. So

**s(t)=0+0=0**and nothing happens. But in the

**fifth**second, a

**Dirac=”pin”**appears. Oh, such a little prick. This signal will pass through the inertial unit as a “damped spike” with a lower amplitude and

**increased by**

**1.5**times. At first, this “repressed pin” is so small that it cannot be seen on the

**time chart**, but it will immediately return to the input and again be increased by

**1.5**times. It will come back again and will be increased by

**1.5 times…**etc. Classic

**avalanche**effect. The system was unstable from the beginning, although it was not visible

**until the fifth**second. But all it took was a small “flick” for a disaster to ensue.

In automatics,

**positive feedbacks**are

**avoided**. However, in

**social life**,

**economy**… it will not always work out. A common example is the

**stock market**. The company appears to be stable. All it takes is a rumor (the equivalent of a “pin”) that its stock will start to rise. As they grow, people buy –> the more they grow –> the more they buy…etc.

Returning to our example, we could make such a hypothesis. A system with

**positive feedback**is always unstable. Any unbalance of it causes the signal to

**grow**to

**+/- infinity**. Is it always like this? Let’s check the next example.

**Chapter 16.2.3 Positive feedback “Small”**

Why “small”? Previously, the transmittance counter was

**1.5>1**. And when the numerator is

**less**than

**1**, e.g.

**0.8**? Check it out. I will anticipate the facts. Despite the positive feedback, the system will be stable. Therefore, we can afford a

**step-type**input

**x(t)**.

**Fig. 16-4**Here the

**numerator**of the transmittance

**G(s)**is

**0.8**. Or in other words –

**steady-state gain**of an open system (i.e. before closing the feedback loop)

**K=0.8 and T=3sec**. The response of a closed system resembles an inertial unit. And so it is! In this case, let us determine the gain

**Kz**of this system from the

**time chart**and its time constant

**Tz**see–>

**Fig. 3-4**

**Chapter 3**.

The result

**Kz=4**and

**Tz=15 sec**tells us that

**positive feedback**increases:

**– amplification**from

**K=0.8**to

**Kz=4**

–

**time constant**from

**T=3**sec to

**Tz=15 sec**– read

**more inertia**

A long time ago, when there was no

**electronics**yet, only

**radio engineering**, and

**lamps**or

**transistors**were

**very expensive**,

**positive feedback**was used to

**increase**the

**gain**. Unfortunately, at the expense of

**greater inertia**, i.e. at the cost of limiting the

**frequency response**. In the next section, we will prove that it is indeed an

**inertial unit**. Now we only assume that it is. And the conclusion is that a

**positive feedback system**can be

**stable**!

**Chap. 16.2.4 Transmittance Gz(s) of a positive feedback system**

Derive the formula for

**Gz(s)**for a system with

**positive feedback**.

**Fig. 15-13**,

**chapter 15**will be helpful in this. Instead of

**E(s)=X(s)-Y(S)**put

**E(s)=X(s)+Y(s)**.

**Fig. 16-5**

For comparison purposes,

**Gz(s)**of the system with

**negative feedback**are shown ->

**Fig. 16-5a**.

If you managed to derive the formula for

**positive feedback**, you will get–>

**Fig. 16-5b**.

Let’s check this formula, e.g. for the transmittance in

**Fig. 16-3**and

**Fig. 16-4**

Fig. 16-6**Fig. 16-6a –** transmittance of the system with **positive feedback****Fig. 16-6b –** an example when **K=1.5>1**, as in **Fig. 16-3**. The system is unstable. The **root** of the **denominator** is **positive**.**Fig. 16-6c** – example when **K=0.8<1**, as in **Fig. 16-4**. The system is stable. The **root** of the denominator is **negative**.

Here the denominator was a **polynomial** of **degree n=1**. Our observation can be generalized to **any polynomial** of degree **n**.

By the way, we learned one of the basic theorems of automatics.**The system is stable if the transmittance denominator contains only negative roots**.**More precisely**, for those who know the so-called **complex numbers**.

They will be discussed in **Chapter. 19**.**The system is stable if the transmittance denominator contains only roots with a negative real part.**

So we will replace the word **negative root** with the word** negative real part** of the **root**.**Chap. 16.2.5 Positive Feedback Conclusions****1-** We rather try to avoid it**2-** Usually, but not always, **positive feedback** is the cause **of instability**.**3-** General remark regarding any transmittances.

The **positive** root of the **denominator** of the transmittance **G(s)** is **instability**!

**Chap. 16.3 How does negative feedback work?****Chap. 16.3.1 Introduction**

In automatics, **negative feedback** is used, the most important feature of which is that the output signal **y(t)** “tries” to **imitate** the input signal **x(t)**. That’s why the rockets hit the target, because they track something there, the furnace tries to maintain the temperature of the liquid **+75ºC** when the set value **x(t)** is also **+75ºC**, etc…**Chap. 16.3.2 How does the inertial unit in an open system come to a steady state?**

Before we consider how the **negative-feedback inertial unit** comes to **steady state**, let’s answer the easier question above.

**Fig. 16-7**

**Inertial unit**in an open system.

We are observing signals:

–

**input signal**-step

**x(t)=1(t)**

**– output signal y(t)**of the inertial unit

**K=5 and T=3sec**

**– goal K*x(t) (K=5)**<– What is a goal? You’ll know in a moment.

The steady state

**y=K*x(t)=5**is the goal

**y(t)**is aiming at. The “driving force” that causes

**y(t)**to increase is the

**difference**–the

**vertical green**

**bar**

**U=K*x(t)-y(t)**. At the beginning of the step

**x(t)**in

**1 sec**the signal

**y(t)=0**. Therefore, the “driving force” – the

**vertical green**

**bar**is the greatest here and

**y(t)**starts at maximum speed. Then

**U**gradually decreases (e.g. in 5 s

**U≈1.3**) and the

**growth rate**also

**decreases**. After

**25 s**, the driving force

**U**has dropped to

**zero**. The goal was achieved

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

The most important conclusion

**Output y(t) tends to equilibrium**

**y(t)=K*x(t)**

**in which the “driving force U” disappears – a vertical green line**

The conclusion is so obvious that the reader wonders what the author means? And this is just an introduction to the next chapter.

**Chap. 16.3.3 How an inertial unit with negative feedback comes to a steady state.**

**F****ig. 16-8**

**Inertial**unit with

**negative**feedback.

**The**signals are:

– output

**y(t**)

– goal

**K*e(t)**(K=5)

Here

**y(t)**also pursues the goal

**K*e(t)**, which, however, unlike the open system, is

**time-varying**. The “driving force” that tries to bring the

**goal K*e(t)**and

**y(t)**together is also the difference

**U=K*e(t)-y(t)**in the form of a

**vertical green**line. The more the difference

**U**– “driving force” is

**greater**, the stronger the goal lines

**K*e(t)**and

**y(t)**“stick” to each other.

**The output y(t) tends to the state of equilibrium y(t)=K*e(t) in which the “driving force U” disappears – the vertical green line**

And when will

**y(t)**stop growing? Or in other words, when will it be

**steady state**. Then, when the

**goal is achieved**, when the

**driving force**(vertical green U line)

**disappears**! That is, when

**y(t)=K*e(t)**.

This is the case not only for the

**inertial unit**, but for each

**dynamic units**with

**negative feedback**, which has a gain

**K**in steady state. The experiment confirms something very important in automatics.

Just like a

**pendulum out of balance**, it will return to its

**steady state**, which is

**the lowest position**

**Thus, a stable system with negative feedback will return to the state y(t)=K*e(t)**.

This only applies to stable systems. And that won’t always work.

Do you remember

**Fig. 15-13e**in

**Chapter 15**? In it, you weren’t completely convinced yet that

**steady state**with

**negative feedback**is

**y=K*e**. Now you know that this is the case and the proof of this is the time charts

**5*e(t)**and

**y(t)**that met at

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

**The lower formula for the**

Fig. 16-9

Fig. 16-9

**Kz**gain of a

**closed**system results directly from

**y(t)=K*e(t)**.

Also

**y=0.83**in

**Fig. 16-8**confirms our calculations.

**Rozdz.**

**16.3.4 How does an inertial unit with large negative feedback come to a steady state?**

We will also add the observation of the control error

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

**Fig. 16-10**The oscilloscope settings are such that you can see the entire “variable target”

**100*e(t)**where

**e(t)=x(t)-y(t)**. The very large initial

**100*e(t)**compared to the signals

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

**e(t)**are

**conspicuous**. This is the main

**conclusion**of this experiment. Later you will learn that it is a

**P-type**control system, in which

**s(t)=100*e(t)**is the

**control signal**. The “pin”

**100*e(t)**at the beginning of the stroke is characteristic here.

On the one hand, a good “pin” causes almost instantaneous arrival to a steady state compared to an open system. This causes a technical problem. If it was a

**steady-state furnace**, e.g.

**10kW**, the “pin” wants

**100 times more**– >

**1MW**! We’ll come back to the topic.

With these oscilloscope settings, there is a very inaccurate view of what is most important, i.e. the

**y(t)**,

**x(t)**and

**e(t)**signals. It can be roughly seen that in steady state there is almost

**x(t)=y(t)=1**and

**e(t)=0**. So let’s change the settings of the oscilloscope to see these signals.

You will get a

**time char**t exactly the same as

**Fig. 16-11**. The difference is only in what you can’t see – in the oscilloscope settings.

**Fig. 16-11**You see the same time chart, only in the range

**y=0…1.2**and not

**y=1…100**as before. Although the “pin” has been cut off, the remaining signals –

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

**e(t)**are clearly visible. Thanks to the strong feedback (the

**y(t)**signal returns

**100 times stronger**!) and

**y(t)**almost immediately it reaches a steady state

**y(t)=0.99**. This is a confirmation of the steady-state

**Kz**amplification formula from

**Fig. 16-9**.

**Chap. 16.3.5 How does a two-inertial unit with negative feedback come to a steady state?**

The previous

**inertial unit**was coming to a steady state “from below”. i.e.

**y(t)**was always

**smaller**than

**5*e(t)**. However, it is known that control systems often do this with oscillations. An example will be the

**two-inertial**

**unit**with feedback.

**Fig. 16-12**The signal

**y(t)**reached the value in which

**y(t)=K*e(t)**in just

**6**seconds. It seems to be an

**equilibrium state**, but

**y(t)**keeps increasing. Why is it like that?

**– First**, it is not in

**equilibrium**, because

**y(t)**is “moving”. It is true that

**y(t)=K*e(t)**but the derivatives of

**y(t)**(e.g. velocity

**y(t)**) are not

**zero**!

**– Secondly**, this is a

**two-inertial unit**. The

**y(t)**signal can increase for a while when the input signal is

**zero**or even

**negative**!

Go back to

**Fig. 16-8,**that is to

**inertial unit**, for a moment. Here

**K*e(t)**is always greater than

**y(t)**. Or in other words

**y(t)**arrives at

**K*e(t)**from

**below**. Because the “driving force”

**U**– the

**vertical green line**was

**positive**all the time.

And what about

**Fig. 16-12**?

Here,

**up to 6**seconds, the “driving force” in the form of a green line is

**positive**. Further, the accumulated energy of the

**two-inertial**term makes it exceed the apparent equilibrium state

**y(t)=K*e(t)**and the “driving force” becomes

**negative**. So the “

**driving**force” becomes the “

**braking**force”. The

**braking**effect is clearly visible.

**Red**

**y(t)**decreases the growth rate until it becomes

**negative**. Now

**y(t)**“turns around” again and after a

**few oscillations**it reaches a

**real**(and not

**“fake”**as in

**6 sec**)

**equilibrium state**, in which

**y(t)=5*e(t)=0.83**.

**Chapter 16.3.6 Static and astatic units**

Let me remind you that for

**G(s)**steady-state gain

**K**we can calculate

**2 methods**:

**1-**From the

**time charts**–>

**K=y(t)/x(t)**in steady state, e.g. in

**Fig.16-12**

**2-**Directly from the

**transmittance**by calculating

**K=G(s=0)**as below

Fig. 16-13

Fig. 16-13

**Fig. 16-13a**inertial unit – no comment

**Fig. 16-13b**two-inertial unit – no comment

**Fig. 16-13c**the real differentiating unit. In steady state

**y(t)=0**!

The above gains

**K**in the

**steady state**are probably obvious. Maybe a small question mark is

**Fig. 16-13c**but it will pass somehow. These were the so-called

**static units**which give a specific finite response in a steady state per unit

**step x(t)=1(t)**.

However, the problem arises with the ideal

**integrating unit**in

**Fig. 16-13d**, where we have to break the holy rule “Remember, damn,

**never divide**by

**zero**“. However, if we break it, i.e., divide it by a number

**infinitely**close to

**0**, we get

**K=infinity**. This

**unit**and others (e.g.

**Fig. 16-13e**) that have at least one

**zero root**in the denominator are

**astatic units**.

**Fig. 16-13d**

**integrating unit**

**Fig. 16-13e**the

**real integrating unit**, i.e. with i

**nertia**

And what are

**steady states**for the

**astatic units**?

**Chapter 16.3.7 Going to a steady state of the astatic unit in an open system**

The title is provocative. You’ll find out why in a moment.

In order to show the typical features of the

**integrating unit**as

**astatic**, the input will be given a signal

**x(t)**which:

– At the beginning it is a

**positive unit step**

– then it is

**0**

– then it is a

**negative unit step**

– ends as

**0**

**Fig. 16-14**It can be, for example, an i

**deal DC motor**.

**Ideal**, because it does not take into account the

**mechanical**(mainly rotor mass) and

**electrical**(winding inductance)

**inertia**. The

**output**signal

**y(t)**is the

**position**as positive or negative

**angle**of

**rotation**. The

**input**signal is the

**supply voltage**

**x(t)**which changes in

**4 stages**:

**stage 0…3 sec**voltage

**x(t)=0V**–-> the motor is

**stationary**, i.e.

**y(t)=0**—

**>0V**

**stage 3…10 sec**. voltage

**x(t)=+1V**the motor turns

**right**

at a constant speed–->

**y(t) increases**linearly –>

**0…+7V**.

**stage 10…30 sec**voltage

**x(t)=0V**–> the motor

**is stationary**,

i.e. at the level

**y(t)=7**–>

**+7V**

**stage 30…40 se**c, the voltage

**x(t)=-1V**, the motor turns

**left**

at a constant speed–->

**y(t)**

**decreases**linearly–>

**+7V…-3V**.

**stage 40…60 se**c voltage

**x(t)=0V**–> the motor is

**stationary**, i.e.

**y(t)=-3**–>

**-3V**

**Conclusions**

**1- The**output signal

**y(t) increases**at a constant rate when the input signal

**x(t)**is constant

**positive**

**2- The**output signal

**y(t) decreases**at a constant rate when the input signal

**x(t)**is constantly

**negative**

**3- The**rate of

**rise**or

**fall**is proportional to the

**input**signal

**x(t)**

**4- Outpu**t signal

**y(t**) will stop when

**input**signal

**x(t)=0**.

Now you know why the title was

**provocative**. With a unit step

**y(t)**will never reach a

**steady state**, i.e. one in which it is

**stationary**.

**More precisely**, in the

**steady**state

**y(t)**is

**infinity**, i.e. the gain

**K**is formally

**infinity**.

**Chapter 16.3.8 How does the negative feedback integral unit come to a steady state?**

The

**integral unit**is an example of an

**astatic unit**

Fig. 16-15

Fig. 16-15

**Fig. 16-15a**is a tirelessly repeated formula for gain

**Kz**of a

**closed**system for

**static units**.

**Fig. 16-15b**are analogous formulas for

**astatic units**. They strike with their

**simplicity**and this is due to the fact that for

**astatic gain K=infinity**.

In

**steady state y(t)=x(t)**. So the output

**y(t**) ideally reproduces the input

**x(t)**. That is, it provides

**zero**fixed error

**e(t)**. This is the goal of

**every automatics engineer**. So says the theory. What about intuition?

**Fig.16-16**You can see how the

**integrating unit**comes to a steady state

**y(t)=x(t)=1**. Compare with

**Fig. 16-8**with the

**inertial unit**which is

**static**. In it, the “driving force”

**u=5*e(t)-y(t)**that strives for the state of

**equilibrium**defined by the formula in

**Fig. 16-8**was

**y=0.83**when

**u=5*e(t)-y(t)=0**.

Here, however, the

**driving force**is the error

**e(t)**. Simplicity is beautiful!

When

**x(t)>y(t)**(or

**e(t)>0**) then

**y(t)**increases as in any decent

**integrating unit**. I will add that it is

**growing**at a

**slower**and

**slower**

**speed**because

**e(t)**is constantly

**decreasing**. In a

**steady state**, this

**speed**is, of course,

**zero**. That is

**e(t)=0**and

**y(t)=x(t)=1**.

Let’s check how a slightly more complicated astatic unit, i.e.

**integrating with inertia**, comes to

**equilibrium**.

**Chapter 16.3.9 How does the integral unit with inertia and negative feedback come to a steady state?**

As in the previous point, we studied the

**ideal integral**

**unit**in the feedback, so now we will consider the

**real**

**integral unit**. It will still be a

**DC motor**, but taking into account the

**electromechanical inertia**. This time we will not study

**open system**. It will give similar results as in

**Fig. 16-14**, only the time charts will be smoother, as in

**Fig. 8-5**,

**chapter 8**.

**Fig. 16-17**You can see how positive

**e(t)**(like a

**green line**) drives

**y(t)**towards equilibrium, and negative

**e(t)**slows it down. Note that although, for example, in

**6 sec**. there is a state

**x(t)=y(t)=1**but only temporary. The motor has a certain energy that “makes” it continue to spin above the steady state

**x(t)=y(t)=1**. Then a negative

**e(t)**makes him brake. Eventually, after several oscillations, the steady state

**x(t)=y(t)=1**will occur.

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

**Chapter 16.4 Conclusions****Chap. 16.4.1 Transmittances of a closed system**

**The formulas apply to all**

Fig. 16-18

Fig. 16-18

**G(s)**–

**static**and

**astatic**units

**Chapter 16.4.2 Static systems with negative feedback****Fig. 16-19**In automatics,

**y(t)**to tries to imitate

**x(t)**. This is true (“almost”) when the open-loop gain K is large. Then it is “almost”

**Kz=1 Ke=0.**

**Chapter 16.4.3 Astatic systems with negative feedback**

**Fig. 16-20**

The gain

**Kz**is equal to

**1**and the error

**Ke**equals

**0**. It follows that

**astatic**systems provide

**zero**fixed error

**e(t)**.

That is in steady state

**y(t)=x(t)**. They are easier to understand than

**static**ones. Later you will learn that the

**zero**fixed error

**e(t)**is provided by the integral component

**I**of the

**PID**controller.

**Chapter 16.4.4 What was not in this chapter**

**1- Instability**in

**negative feedback**–> We will deal with this topic in the next chapter.

**2- In a static**, e.g.

**inertial**, open system when

**K=1**, the steady state provides

**y(t)=x(t)**. Then why complicate life with some feedback? The answer is simple

**Negative feedback**provides:

**– resistance**to external disturbances

**– improves**the dynamics, i.e. the speed of reaction.

** **