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.
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 To–>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
Call Desktop PID/07_sprzężenie_zwrotne_a_stabilność/01_czlon_opoz.zcos
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.
Delay unit To=1sec K=0.75 with the feedback.
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
Please note that:
y(2) was calcculated beacause we know y(1),
y(3) was calcculated beacause we know y(2),
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.
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
Delay unit To=1sec K=1 with the feedback.
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
The single input pulse x(t) unbalanced the system but the oscillations are steady now.
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
Delay unit To=1sec K=1.25 with the feedback.
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
K>1 and the system is unstable
Chapter 21.2.6 Conclusions
The delay unit with the feedback is:
–on the border….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 delay 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
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
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.
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 Fig. 21-4. There is some analogy here, but the response is, hm how to say it, a liltle fuzzy. There is a maksimum in 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
Input x(t) is symilar to Dirac.
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
Do you predict the increased gain K influence?
The oscillations are bigger now but the system is stable yet.
Chapter 21.3.4 The triple inertial with the feedback K=10.035
Why does K=10.035 exactly?
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
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
The input signal x(t) is a step type. Do you predict the behaviour?
Chapter. 21.4.2 Stable state
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
The oscillations amplitude is growing up to +/-infinity. But y(t) output signal has a constant component y=0.923!