Chapter 20.1 Introduction
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.
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
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.
The single “needle” pulse x(t) is passing through inertial G(s) and is returning as y(t) to the G(s) input again. Please note that x(t) is active during a small time! The x(t)=0 then. The returnig y(t) is amplified 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
The input x(t) is step type instead of “needle” and K=0.8.
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.
The formula for transfer function Gz(s) with the negative feedback
The formula for transfer function Gz(s) with the positive feedback. See 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. 20-8a – 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
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?
Inertial unit as a open loop configuration. (no feedback)
The oscilloscope observs the signals:
– goal K*x(t) (K=5)<– What is this? You will know at the moment.
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 more difficult 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.
Inertial unit with negative feedback.
The oscilloscope observs the signals:
– goal K*e(t) (K=5)<– It’s similar to previous K*e(t)!
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 line U) 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 remeber 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
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).
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.
The scheme is exactly as Fig. 20-14 but there are different oscilloscope parameters now.
The main conclusion is the short setting time when K is big. The time constatnt T≈0.03sec instaead 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.
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. 20-19a 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.
The special x(t) signal type will be given to see typical astatic units behaviour.
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 sec input voltage 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. 20-22a Steady state gain Kz for the static units
Rys. 20-22b Steady state gain Kz for the astatic units
It means that there is error e=0 for steady state. It’s every automatics engneer goal.
Click “StartFig. 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.
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
Chapter 20.7.2 Static units with the negative feedback-steady state
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
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.