Operation with units: f is not defined - Mensajes
#1 Publicado: 24/6/2016 19:13:55
Hello,
I am getting the following:
Basile
I am getting the following:
Basile
#2 Publicado: 24/6/2016 21:39:49
Hi Basile.
That's is the expected way to work with smath.
To get the value of X.C you must to call the function after defining X.C with X.C(f)=
Best regards.
Alvaro.
That's is the expected way to work with smath.
To get the value of X.C you must to call the function after defining X.C with X.C(f)=
Best regards.
Alvaro.
#3 Publicado: 24/6/2016 22:34:38
Alvaro,
Look to the left the of X.C , the X.L
Should behave the same?
Basile
Look to the left the of X.C , the X.L
Should behave the same?
Basile
#4 Publicado: 24/6/2016 23:02:38
Alvaro,
Look to the left the of X.C , the X.L
Should behave the same?
Basile
Depending which version, it could be. I guess that must to be an election from author, this is, if eval functions in that way. The "not evaluation" option, as the attached picture, it's ok for me.
Best regards.
Alvaro.
#6 Publicado: 25/6/2016 0:46:19
Same problem as before: you have no circuit, then no LC project.
The only thing you can do is calculate the resonance frequency: 'fo'
Then plot it as sin(fo)
The only thing you can do is calculate the resonance frequency: 'fo'
Then plot it as sin(fo)
#7 Publicado: 25/6/2016 9:41:27
Jean,
I am calculating the reactance of the inductor and the capacitor at a frequency f.
Basile
I am calculating the reactance of the inductor and the capacitor at a frequency f.
Basile
#8 Publicado: 25/6/2016 13:53:44
In my times, pure L, C were strictly imaginary. Just by themselves they don't have much use
whereas there is no such thing as pure L , pure C. They both have physical resistiviy.
whereas there is no such thing as pure L , pure C. They both have physical resistiviy.
#9 Publicado: 25/6/2016 18:30:56
Jean,
You are right that:
X.L=jωL
and
X.C=-j(1/ωC)
and the magnitude is that I calculate.
You can simply consider ideal elements without the loss resistance.
Basile
You are right that:
X.L=jωL
and
X.C=-j(1/ωC)
and the magnitude is that I calculate.
You can simply consider ideal elements without the loss resistance.
Basile
#10 Publicado: 25/6/2016 21:51:28
Hi Basile. That's isn't.
Correct XL and XC (reactances) are without j, as in your worksheets.
ZL and ZC (impedances) are with j.
Actually the total impedance is Z = R + jX, where X = XL + XC is the total reactance.
Best regards.
Alvaro.
Correct XL and XC (reactances) are without j, as in your worksheets.
ZL and ZC (impedances) are with j.
Actually the total impedance is Z = R + jX, where X = XL + XC is the total reactance.
Best regards.
Alvaro.
#11 Publicado: 26/6/2016 0:35:47
Alvaro,
You are confusing things.
What you say are correct but have to do nothing to do with the question.
X.L, X.C are reactance (with j or without j)
Z is impedance of R and X.
Basile
You are confusing things.
What you say are correct but have to do nothing to do with the question.
X.L, X.C are reactance (with j or without j)
Z is impedance of R and X.
Basile
#12 Publicado: 26/6/2016 1:03:00
Alvaro,
You are confusing things.
What you say are correct but have to do nothing to do with the question.
X.L, X.C are reactance (with j or without j)
Z is impedance of R and X.
Basile
Hi Basile. I don't confusing. Reactances ( https://en.wikipedia.org/wiki/Electrical_reactance ) are reals and impedances ( https://en.wikipedia.org/wiki/Electrical_impedance ) are complexes. So X.L, X.C can't have j.
About the numerical evaluation for a function, check this old post: http://en.smath.info/forum/yaf_postst640_Function-as-a-program.aspx
In X.L, X.C definitions as functions, f must to be a local variable. X.L(f) -> (symb evaluation) must to set the answer with "f", not with "100 Hz". And the numerical evaluation X.L(f) = must to be in red, as f as local variable isn't defined.
As a choice of the program writer, and what you can see in the last post of the above referenced, you can set all the local variables as globals with f(x) = 1*(line ... ). Actually I'm not comfortable with that behavoir, but is the choice of Andrei.
Best regards.
Alvaro.
#13 Publicado: 26/6/2016 2:01:29
Just like this Basile, for a family of "C's"
at a fixed frequency 8 the magenta line.
On the plot, 'f' => 'x'.
at a fixed frequency 8 the magenta line.
On the plot, 'f' => 'x'.
#14 Publicado: 26/6/2016 14:59:10
Jean,
Yes, I can plot X.C(f) and X.L(f) but if I try to plot the difference or take the aritmetic difference does not work.
Please see figure and figure in post #1 on top, if it helps.
Basile
Yes, I can plot X.C(f) and X.L(f) but if I try to plot the difference or take the aritmetic difference does not work.
Please see figure and figure in post #1 on top, if it helps.
Basile
#15 Publicado: 26/6/2016 19:42:21
Basile, let's make it clear on three points:
1. You have no work sheet, then there is no project [no help].
2. Read this link again https://en.wikipedia.org/wiki/LC_circuit
to help you start onstruct a project.
3. Those gyzmas: they do nothing circuit wise.
Jean
1. You have no work sheet, then there is no project [no help].
2. Read this link again https://en.wikipedia.org/wiki/LC_circuit
to help you start onstruct a project.
3. Those gyzmas: they do nothing circuit wise.
Jean
#16 Publicado: 27/6/2016 0:00:18
#17 Publicado: 27/6/2016 10:15:28
#18 Publicado: 27/6/2016 10:17:47
Note that I have removed the unit system because
"Units don't match" error mesage ... so: zap !
"Units don't match" error mesage ... so: zap !
#19 Publicado: 27/6/2016 11:03:08
Jean,
"why units do not match? Are they both Ω?"
Basile
PS: Thank you for the discussion.
"why units do not match? Are they both Ω?"
Basile
PS: Thank you for the discussion.
#20 Publicado: 27/6/2016 13:54:29
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