Re: Completion of Controller Synhesis


<pnachtwey@xxxxxxxxx> schrieb im Newsbeitrag
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On Jun 28, 12:20 pm, "JCH" <j...@xxxxxxxxxxxxxxxxxxx> wrote:
<pnacht...@xxxxxxxxx> schrieb im
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On Jun 28, 9:42 am, "JCH" <j...@xxxxxxxxxxxxxxxxxxx> wrote:
<pnacht...@xxxxxxxxx> schrieb im
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On Jun 28, 4:51 am, "JCH" <j...@xxxxxxxxxxxxxxxxxxx> wrote:
<pnacht...@xxxxxxxxx> schrieb im
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On Jun 27, 8:44 am, "JCH" <j...@xxxxxxxxxxxxxxxxxxx> wrote:
<pnacht...@xxxxxxxxx> schrieb im
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You can SIMULATE/STUDY the time behavior of cranes on a model!
If
that
works
on a model then it will also work on real cranes under the same
physical
conditions.

Your model depends on being able to move the crane instantly from
one
point to another. Everyone but you knows this isn't possible.

<AGAIN: Citation [17. Juni 2007 14:26]>
I didn't specify the time scale. It can be milliseconds,
seconds,
minutes,
houres, etc. It can also be 3 seconds, 1/1.1223 minutes.
The distance could be inches, meters, kilometers, etc. Can you
find
such
specifications?

You don't use units or show you work so people can't pin you down
on
your calculations.

Look
athttp://home.arcor.de/janch/janch/_control/20070613-pd2(pid)z1z2/
Tell me where you find a dimension (unit). I wouldn't mind if
using
1
pet
for 3.5564 seconds and 1 tim for 12.4325 inches. Then read
distance
in
tims,
velocity 1 tim/pet and acceleration in tims/pet^2.

Why don't you want to put units on your work? Do you have
something
to hide?

I don't know either speed nor process transfer function
of:http://www.mast.queensu.ca/~dtyner/ControlLab/single-pendulum.mpg

Read: On all conditions
ofhttp://home.arcor.de/janch/janch/_control/20070613-pd2(pid)z1z2/
then anything is ok.

The process transfer function (Page 4) works with ALL dimensions!
It
also
works with your example if you have a process transfer function
(F1(s)
=
v1/v2).

Notes:
The controller output signal v2 (e.g. mA) can change very fast!
Regard
v2
instantaneous. v2 is NOT the position of the cart.

You said above that everything was OK and now you change your
diagram! So if v2 is not a control signal, where does it move the
cart? You need another transfer function to move the cart!
As your equations stand now you have the load directly moved by a
control signal. See equation 1 on page 1 on your site. look at
equation 1. If the control signal v2 is turned off your equation 1
says the load will move back to the 0 position. That doesn't make
sense. This is why you need another transfer function simular to G/
(s*(s+a)) like I used to convert the control signal to a cart
position. In my example the cart will just stop and stay where it
is
because position systems like this are typically type 1 systems.
If
you look at my example you can see how the load swings under the
crane
and when the control output goes to 0 at the end of travel the cart
and load stay where they are. The load lags at first and then leads
as the crane slows the load down. If you don't have a crane or cart
then how can you show the swinging of the load? The trick is to
make
the load swing once during the whole move. I don't see this in your
examples. I see the load swinging in your movies before the load
is
balanced.

v2 was always and is the controller output value (e.g. mA).
F1(s)=v1/v2
is
as before. I din't change anything physically. If doing so I would
tell
you.

All is correct
inhttp://home.arcor.de/janch/janch/_control/20070613-pd2(pid)z1z2/

Give us your

G=?
a=?
See the .pdf file I posted a link to in the 27 june thread. I show
G=0.04 m/s per % control output and alpha ( a ) which is the corner
frequency of the crane or cart is 1 Hz or 6.28 radians per second.
Notice that my crane does not move instantly. There is lag between
the control output and the motion of the crane. In fact the crane
bandwidth is quite low. I even show the formulas for calculating
the gains so the resulting closed loop system will have three poles at
-lambda at -3.14 radians/sec. The resulting closed loop bandwidth is
at about 0.509*lambda. I am just trying to show how control is
possible without breaking the laws of physics.
I am assuming there is only feedback on the cart or crane and the load
swings freely below. The trick is to move the load to the
destination or set point and have it stable when it arrives. You can
do this by hand with a little practice. Your hand and the cart must
vary the speed along the way at specific points or time relative to
the natural frequency of the load for it to be stable when it arrives
at the set point or destination.

G=0.04
a=6.28

0.04 1
F1(s) = --------- = ------------------------
s(s+6.28) 25*s^2 + 2*delta*5*s + ?

T = 5
T^2 = 25
delta = 15.7 (highly damped!)

Don't try to make a second order system out of the controller. The
crane response to the controller is

crane(s)/control(s) =G/(s*((s/alpha)+1)) or 0.04/(s*((s/(2*PI))
+1))
crane(s)/control(s) =G*alpha/(s*((s+alpha)) or 0.04*2*PI/(s*(s+2*PI))

!!!Note this is a correction!!!!

The load response is as follows

load(s)/crane(s)=1/((s/omega)^2+2*zeta*(s/omega)+1)

G=0.04 This the gain in meters per second per percent
control output
At 100% ouput the crane can move 4 meters per
second.
alpha = 2*PI This is the corner frequency or band width of the
crane motor with its loads.
omega = 1 This is the natural frequency of the load. At 9.8 m
of cable the sqrt(g/l)=1
zeta = 0.001 This is from your example. Actually it has been
determine by crane research.

The total transfer function is the combination of the two transfer
functions

load(s)/control(s)=G/(((s*((s/alpha+)1))*((s/omega)^2+2*zeta*(s/omega)
+1))


I used delta = 0.001 (Page
3)http://home.arcor.de/janch/janch/_control/20070613-pd2(pid)z1z2/

What about the missing '?-value'?

There is no missing ? value. The crane only has ONE corner
frequency, pole, bandwidth or time constant. The time constant is 1/
(2*PI). It takes the motor 5 time constants, 5/(2*PI) seconds
, to reach the within 1 percent of final stead state speed in response
to a control output change. The resulting crane velocity must be
integrated to provide crane positions as a function of control
output.

If interpretation is wrong give me the correction.

Notice that the system is now two transfer functions.


Process input value v2 (e.g. 4...20mA)
Process output value v1 (e.g. 0...10meter)

Time behavior measured for total system F1(s)=v1/v2!
See red box in Page 3.

One is for the
crane position as a function of the control output. The other is the
one you have always had which is the load position as a function of
the crane position. Combine the two and simulate. It will be
messy. I have it done the simulation both ways. The easiest is to
just control the crane motor and use the crane position for
feedback. The hardest way is using load position feedback. One must
calculate many dervatives ( 3 ) if the load is to be controlled. This
means one would need an accelerometer on the load as well as the
position feedback.



Evaluating a step process transfer function F1(s) and using sytem
identification methods you get approximated T, e.g.

1
F1(s) = -------------------------
T^2*s^2 + 2*delta*T*s + 1

Time domain:

T^2*v1'' + 2*delta*T*v1' + v1 = K1*v2

Feedback via math model

v1 = load position
v1' = load speed
v1'' = load acceleration (no accelerometer necessary!)

are known and used: See Page 1, Fig.1
http://home.arcor.de/janch/janch/_control/20070613-pd2(pid)z1z2/

Other compensations are necessary for controlling disurbances z1 and z2. See
formulae in Page 1.

Just using PID wouldn't work. See Page 6 (NEW)

Note:
F1(s) can be any plausible process transfer function of any order. It must
be known and match the system behavior as good as possible! One can't do
more and should be happy if at least it works.


--
Regards/Grüße http://home.arcor.de/janch/janch/menue.htm
Jan C. Hoffmann eMail aktuell: janch@xxxxxxxxxxxxxxxxxxx
Microsoft-kompatibel/optimiert für IE7+OE7



.

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