## Preliminary Automatics Course

**Chapter 21. Instability i.e. how does the oscillations arise?
**

**Chapter 21.1 Introduction**

You start the differential solution process when you click “start” icon in the

**Xcos**window.

The step type response

**y(t)**may be:

– without osccillations

– with decaying oscillations

and ends with the steady state

**y**.

**Fig. 21-1
**The stable systems steady state.

These are the

**stable**systems examples. But it isn’t so nice always. There are steady or increasing oscillations sometimes.

These systems are

**instable**.

What is the source of the

**instability**. The answer is very simply. This is a differential solution result. No discussion, fullstop. But such a treatment left some shortage! Let’s try to explain this problem more humanly.

Imagine that you drive the motor boat to the lighthouse. The control algorithm is very simple. The lighthouse is a little left–> your steering wheel reaction is a little right. Your route is a little wiggly line now. Imagine that there is a

**delay unit**between steering wheel and the rudder. How is it made? Not important. What will be your route now? There is wind blow hit now. Your steering wheel reaction is immediate but the boat reaction is delayed now! What will be a route? A little wiggly line? No- It will a big boa snake route. The bigger is delay

**T**

**o**–>the bigger are oscillations. The main conclusion is that

**dealys**and

**inertias**of the object are a source of the

**instability**.

**Chapter 21.2 The delay unit with the feedback**

**Chapter** **21.2.1 Introduction**

It’s possible a analysis without differential equations. Using the fingers almost! The conclusion will be similar to the **Nyquist **benchmark. Nyquist will be discussed later otherwise.

We close the **delayed unit **–> dealy unit with the feedback

**The **system will be** **:

**-stable** when **K<1**

**-instable** when **K>1**

**-on the border** when **K=1**

We will test **delay unit** in open loop first. We did it once but reportedly **repetitio est mater studiorum**.

**Chapter 21.2.2 Delay unit in open loop**

**Fig. 21-2**

Delay unit

Call Desktop PID/07_sprzężenie_zwrotne_a_stabilność/01_czlon_opoz.zcos

**Fig. 21-3**

Click “Start”

**Fig. 21-4**

The response **y(t)** is a absolute copy **x(t)** delayed **To=1 sec.**

**Chapter 21.2.3 Delay unit in closed loop K=0.75**

Let’s close **Delay unit**.

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/02_czlon_opoz_-sprzez_0.75.zcos

**Fig. 21-5**

Delay unit **To=1sec K=0.75 **with the feedback.

Clck “Start”

**Fig. 21-6
**Input pulse

**x(t=5sec)=1**is given in

**5 sec**. Please note that

**y(t=5sec)=0**in this time, so there is only a single pulse

**x(t=5sec)*0.75=+0.75**on the

**delay unit**input.

**For t=6 sec**

**y(1) = 0.75*x(0) = +0.75**

The

**subtractor**inverts only the signal after

**t=6 sec**because

**x(n)=0**oraz

**e(n)=x(n)-y(n)=-y(n)**.

So the formula for

**y(n+1)**is:

**y(n+1)= – 0.75*y(n)**

Further

**y(n)**pulses are:

**y(2) = – 0.75*y(1)= – 0.75*0.75= – 0.5625**

**y(3) = – 0.75*y(2)= + 0.422**

**y(4) = – 0.75*y(3)= – 0.316**

**y(5) = – 0.75*y(4)= + 0.237**

**y(6) = – 0.75*y(5)= – 0.180**

**y(7) = – 0.75*y(6)= + 0.133**

**y(8) = – 0.75*y(7)= – 0.100**

**y(9) = – 0.75*y(8)= + 0.075**

**etc…**

Please note that:

**y(2)**was calcculated beacause we know

**y(1)**,

**y(3)**

**was calcculated beacause we know**

**y(2)**,

etc…

An example of the so-called recurrence formula.

Further

**y(n)**are smaller and are aiming to

**0**. The single input pulse

**x(t)**unbalanced the system but the oscillations are decaying up to

**0**again.

**Conclusion:**

**K=<1**and the system is stable.

**Chapter 21.2.4 Delay unit in closed loop K=1**

Call Desktop/ PID/07_sprzężenie_zwrotne_a_stabilność/03_czlon_opoz_-sprzez_1.0.zcos

**Fig. 21-7**

Delay unit **To=1sec K=1 **with the feedback.

Click “Start”

**Fig. 21-8
**We calculate

**y(n+1)**as previously but

**K=1**now.

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

y(2) = -1*y(1) = – 1

y(3) = -1*y(2) = +1

y(4) = -1*y(3) = – 1

y(5) = -1*y(4) = +1

y(6) = -1*y(5) = – 1

etc…

The single input pulse

y(2) = -1*y(1) = – 1

y(3) = -1*y(2) = +1

y(4) = -1*y(3) = – 1

y(5) = -1*y(4) = +1

y(6) = -1*y(5) = – 1

etc…

**x(t)**unbalanced the system but the oscillations are

**steady now.**

**Conclusion:**

**K=1**and the system is on the stability border

**Chapter 21.2.5 Delay unit in closed loop K=1.25**

Call Desktop/ PID/07_sprzężenie_zwrotne_a_stabilność/04_czlon_opoz_-sprzez_1.25.zcos

**Fig. 21-9**

Click “Start”

**Fig. 21-10
**Delay unit

**with the feedback.**

**To=1sec K=1.25**The oscillations are increasing up to

**+/-infinity**

**y(1) = 1.25*x(t) = + 1.250**

y(2) = -1.25*y(1) = – 1.562

y(3) = -1.25*y(2) = + 1.953

y(4) = -1.25*y(3) = – 2.441

y(5) = -1.25*y(4) = + 3.051

y(6) = -1.25*y(5) = – 3.815

y(7) = -1.25*y(6) = + 4.768

y(8) = -1.25*y(7) = – 5.960

y(9) = -1.25*y(8) = + 7.451

etc…

y(2) = -1.25*y(1) = – 1.562

y(3) = -1.25*y(2) = + 1.953

y(4) = -1.25*y(3) = – 2.441

y(5) = -1.25*y(4) = + 3.051

y(6) = -1.25*y(5) = – 3.815

y(7) = -1.25*y(6) = + 4.768

y(8) = -1.25*y(7) = – 5.960

y(9) = -1.25*y(8) = + 7.451

etc…

**Conclusion:**

**K>1 **and the system is **unstable**

**Chapter 21.2.6 Conclusions**

**Fig. 21-11
**The delay unit with the feedback is:

–

**stable**…………….when

**k**

**<1**

–

**on the border**….when

**k=**

**1**

–

**unstable**…………when

**k>**

**1**

Please note that the conclusion about closed loop stability/instability was based on open loop parameter

**K**. It’s similar to the

**Nuquist**benchmark–>

**Chapter 22**.

**Chapter 21.3 The triple inertial unit with the feedback instability**

**Chapter 21.3.1 Introduction
**We tested

**delay unit**with the feedback. The main conclusion was that delay time

**To**and gain

**K**are a source of the

**instabilty**. This unit and its analyzis is simple. The differential calculus wasn’t necessary here. The pure

**de**

**lay**isn’t popular in the real objects. There are their

**multiinertia units**approximations rather. By the way. The

**multiinertia units**may be approximated as

**equivalent transfer function**

**Fig. 21-12**

**Equivalent transfer function**

Let’s arrange that

**triple inertial**transfer function will be the representant of all objects.

**Chapter 21.3.2 The triple inertial unit in open loop**

Call/Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/05_otwarty_3_inercyjny_dirac.zcos

**Fig. 21-13
**The object will be tested by the short rectagle pulse -symilar to Dirac pulse. The time

**t=0.02sec**and amplitude

**50**. So the pulse area-“energy”=

**50*0.02=1**.

Click “Start”

**Fig. 21-14**

The input

**x(t)**is acting during short time

**3…3.02 sec**. The pulse energy is stored in the

**G(s)**and is offloaded during

**3.02…11 sec**then. Please compare with the pure

**delay unit**. There is some analogy here, but the response is, hm how to say it, a liltle fuzzy. There is a maksimum in

**Fig. 21-4****t=4.8 sec**. (

**1.8 sec**after

**x(t)**). It may be treated as a some kind of

**To**delay time. The analogy to the ideal

**delay unit**isn’t absolutely full, but the action mechanisms are symilar. The

**inertia**and

**gain K**casuses instability for example.

**Chapter 21.3.3 The triple inertial with the feedback K=3**

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/06_sprzezenie_zwrotne_male_3_inercyjny_dirac.zcos

**Fig. 21-15
**Input

**x(t)**is symilar to Dirac.

Click “Start”

**Fig. 21-16
**The beginning is similar to the open loop–>

**Fig. 21-14**because

**y(t)**output signal is small yet. But the negative feedback from

**y(t)**causes its drop and even oscillations! Please note that maksimum is in the

**t=4.35**sec, earlier than in open loop

**Fig. 21-14**. Do you observe the analogy to

**Fig. 21-6**with the

**delay unit**?

**Chapter 21.3.4 The triple inertial with the feedback K=7**

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/07_sprzezenie_zwrotne_srednie_3_inercyjny_dirac.zcos

**Fig. 21-17
**Do you predict the increased gain K influence?

Call “Start”

**Fig. 21-18**

The oscillations are bigger now but the system is stable yet.

**Chapter 21.3.4 The triple inertial with the feedback K=10.035**

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/08_sprzezenie_zwrotne_graniczne_3_inercyjny_dirac.zcos

**Fig. 21-19**

Why does **K=10.035 **exactly?

**Fig. 21-20
**The system is on

**the instability border**. See

**Fig. 21-8**. There is

**delay unit**on

**the instability border**but

**K=1**other than

**K=10.035**. What is the reason? My answer will be dusty rather. This continuous

**triple inertial**object is more complicated than a simple

**delay unit**. But I will try to explain it in the next

**Chapter 22. Nyquist stability benchmark**

**subchapter 22.6.**

When

**K=10.035**–> the next amplitude is the same

When

**K<10.035**–> the next amplitude is lower so there are decaying oscillations.

Are you Einstein? What does cause the

**K=12>10.035**?

**Chapter 21.3.4 The triple inertial with the feedback K=12**

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/09_sprzezenie_zwrotne_duze_3_inercyjny_dirac.zcos

**Fig. 21-21**

Click “Start”

**Fig. 21-22
**The next amplitude is bigger so there are arising oscillations up to

**+/-infinity**. The system is instable.

**Chapter 21.4 The triple inertial with the feedback -input signal x(t) is a step type**

**21.4.1 Introduction**

The input signal x(t) is a step type. Do you predict the behaviour?

**Chapter. 21.4.2 Stable state **

Call Desktop/PID/07_sprzężenie_zwrotne_a_stabilność/10_sprzezenie_zwrotne_male_3_inercyjny_skok.zcos

**Fig. 21-23**

Click “Start”

**Fig. 21-24
**The steady

**state y(t)=0.75**state is compatible with with the classical formula.

**Chapter. 21.4.3 Instable state**

Some people say that steady gain **Kz** is nonsense for **instable systems**. I agree but there is something rational

Wywołaj PID/07_sprzężenie_zwrotne_a_stabilność/11_sprzezenie_zwrotne_duze_3_inercyjny_skok.zcos

**Fig. 21-25**

Click”Start”

**Fig. 21-26
**The oscillations amplitude is growing up to

**+/-infinity**. But

**y(t)**output signal has a constant component

**y=0.923**!