Chapter 7.1 Introduction
It’s the simplest dynamical unit except Proportional.

Fig. 7-1
Inertial unit parameters:
K -steady state gain
T – time constant
We will do some experiments with different time constants.
We will do some experiments with different input signals x(t). The output signals y(t) will be observed on the bargraf or on the oscilloscope.

Chapter 7.2 T=5 sec, step x(t) from the virtual potentiometer, y(t) on the bargraf
Call Desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/01_inercyjny_suwak_bargraf.zcos

Remember that parameter K=1 is a gain in the steady state!

Push “Start” (as you did for proportional before)

You see virtual potentiometer window (TKsource), bargraf window (BARXY) ont the block diagram window. You don’t see Scilab window because it’s covered by the diagram window. Move the windows, windows that you see block diagarm  with the digital meters.
Move the slider from to 1 in a flash so you give the step signal x(t) and observe the bargraf! The response isn’t up to date as in proportional unit now! Wait when will be the steady state. When x(t)=y(t)=1.000. The digital meters will be usefull here!
The maximal velocity  y(t) signal is when started. The y(t) velocity decreases (velocity, not y(t) value!) then. The velocity is 0 when steady state.
“Swing” the slide several times. You observe the smoothened effect of the inertial unit. Click the red button “finished the experiment”. The button colour changed to grey again. The experiment is finished now.
It should be opened the 3 windows only before we start the new experiment:
Xcos with the block diagra  –>”01_proporcjonalny_bargraf…”
-“naked” Xcos  “Untitled”
Close all the other windows.
Do it always (as on the Fig. 7-4) and  the Scilab will work perfectly.

Fig. 7-4

What about new  time constant T=10 sec? You will teach to change  Go(s) parameters in Xcos by the way.

Chapter 7.3 T=10 sec, step x(t) from the virtual potentiometer, y(t) on the bargraf
We change the parameters in existing block diagram and don’t call a new one.
Move the mouse on transmision G(s) change parameters 1+5*s –>1+10*s.

Parameters change
Click “start” when accepted

The T=10sec is doubled. The object G(s) will be “doukled lazy” too. I hope that you see it
Close the Xcos but don’t save the parameters change!  The “old”  T=5sec parameter would be when you call  this G(s) again.

Chapter 7.4 T=5 sec, steps “+1 -1” x(t) from the virtual potentiometer,  y(t) on the oscilloscope.
We test the respons y(t) for to  consecutive “+1,-1” steps  input signal x(t)
Call Desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/02_inercyjny_suwak_oscyloskop.zcos

Block “0” draws y=0  axis

Click “start”

All is clear I hope.
Experiment ends in 60 seconds. Remember that 3 windows should be opened, before you you “start” new experiment in existing block diagram . There are SCILAB, “naked” untitled XCOS window and existing block diagram  XCOS window.
We are ready to new experiment in existing block diagram now.

Chapter 7.5 T=5 sec, random  x(t) from the virtual potentiometer,  y(t) on the oscilloscope.
Click “start” i and “swing” the potentiometer slider.

You observe the smoothened effect of the inertial unit.

Chapter 7.6 T=5 sec, x(t) from the step genearator,  y(t) on the oscilloscope.
Step generator enables more precisely G(s) parameter testing.
Call desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/03_inercyjny_skok_oscyloskop.zcos


Click “start”

Transient performance shows “who i who” in the transfer function G(s).
For example here:
1 in the G(s) numerator–> gain K=1 in steady state.
5 in the G(s) denominator–>time constant T=5 sec.
The y(t) velocity at the x(t) start is maximal. Imagine that this velocity is constant all the time. When does y(t) attain state y(t)=1? After 5 sec of course. This time is called contant time T=5sec. of the inertial unit.

Chapter 7.7 Two different inertial units comparison.
Call Desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/04_porownanie_2_inercyjnych.zcos


The step pulse x(t) given for two inputs simultaneously. Can you predict the y1(t) and y2(t) signals?

Have you predicted. I propose to repeat the chapter if the answer is “no”.

Chapter 7.8 T=1 sek,  Dirac pulse x(t),  y(t)  oscilloscope
Dirac pulse δ(t) is a funny  type.
– the puls
e δ(t)  time is infinitely short
-the pulse amplitude δ(t) is infinitely high
-but the pulse δ(t) energy (δ(t) area)  is =1–>normal value 

Its mechanical interpretation is a hammer hit.
The ideal Dirac pulse δ(t) doesn’t exist in the real world. There is always finitely short  δ(t) pulse time and big but finitely  δ(t) amplitude.

Call desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/05_inercyjny_dirac_oscyloskop.zcos

Dirac pulse area=1

Click “start”

The output y(t) is rising very fast during 3..3.01 sec. But not immediately! The final state y(t)=0. This final state  y(t)=0 for input x(t)=δ(t) dirac type,  is typical for all so-called dynamic static units.  We will discuss them later.

Chapter 7.9 T=1 sek, “better”  Dirac pulse x(t),  y(t)  oscilloscope
The previous Dirac pulse wasn’t ideal of course. But we try do it “better”. I propose pulse with amplitude 100 and time 0.01 sec.
Note that it’s energy-area =100*0.01=1!
Call Desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/06_inercyjny_idealniejszy_dirac_oscyloskop.zcos
The diagram is the same as Fig.7-16. The pulse parameters are different only:
– amplitude=100
– time=0.01 sec.
Click “start”

It looks like ideal Dirac pulse. But it’s oscilloscope limitation!

Chapter 7.10  Why do we need Dirac pulses?
Their execution is difficult.  It requires big power in the short time.
Let’ go to the Chapter 18 for a moment.
We give the  Dirac pulse δ(t) to the G(s) object.
The Laplace transform  of this y(t) is Y(s)=G(s)*δ(s)=G(s) because Laplace transform Dirac pulse  δ(s)=1 !

Chapter 7.11  What to do when G1(s) is as undermentioned?
Call Desktop/PID/01_podstawowe_człony_dynamiczne/02_człon_inercyjny/07_rozne_postacie_tych_samych_czlonow.zcos

We suppose that the G1(s) inertial type. But what are K and T? We don’t see it for a first look. So make the G1(s) denominator as 1+… (not as previous 7+21*s)–> divide denominator and numerator by 7. You have G2(s)=G1(s). You see the K=2 and T=3 sec. here. Check it.
Click “start”

You see that G1(s)=G2(s)
G2(s) form shows:
T=3 sek

Chapter 7.11  Typical inertial units
RC circuit

R = 100 kΩ
  and C =10 µF –>T = R*C = 100 000 Ω * 10*0,000001F = 1 sec.

The direct-current motor
Input x(t) – voltage step
Output y(t) – rotational speed 
The rotational speed at  start  is 0. It rises as in inertial units and attain maximum value. The K and T parameter depend on the  mechanical and electrical  qualities. This is  simplified  motor model. The real is more complicated of course.

The bathtub with the taken out plug
Input x(t) – you open immediately the valve –> Q(t) input flow step
Output h(t) – the water level.
The h(t) level speed is maximal at the start. The  h(t) level rises with maximal initial speed. This speed declines because the output flow  through the open hole rises when h(t) rise.  The maximum level h(t)  is when input flow=output flow.
This is smplified “bathtub with the taken out plug” model. More–> see Chapter 17.

Chapter. 7.12 Summary
The proportional unit is the first approximation of the every* dynamic units.
The second and more accurate  approximation approximation of the every* dynamic units is the  inertial unit.  When you give x(t) input step type signal, then the response y(t) will reach the steady level y after some steady-state time t.
We assume that it’s a inertial unit with parameters:
K=y  because x(t)=1
T is 4…5 times smaller than steady-state time t.
This G(s) approximation is very primitive of course, but much better than proportional unit approximation.
almost –> doesn’t concern integral unit for example. 


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