How to solve circle diagram problem by Mathcad ?
Symbolic maximum problem.
Don't know. I see no error (no red) in what you post.
- You forgot to hit the F9 key, to have it calculate?
- You might want to inspect what it finds as symbolic results for the Zm and Zr functions, by adding a symbolic evaluation in their definition:
This is in red in your worksheet i opened after hitting the F9 button:
it says: "keine Lösung gefunden"
This is the symbolic evaluation i got:
I guess Mupad doesn't know how to deal with solving an equation that contains the signum() function.
Before giving up, you might try some symbolic manipulations like simplify or expand. maybe that can do away with the absolute in the magnitude which I suspect is the cause for the signum to appear in the derivative.
Guess you experience the difference between Mathcad 11 with Maple which is used by Luc and Mathcad 15 with muPad.
If you add "vereinfachen" ("simplify") before the solve command you get a pretty simply result
As an alternative you might add "simplify" when you symbolically evaluate the derivative.
Thats surprising and it looks that its not the signum function (alone) which causes "solve" to fail.
I know you solve electric circuit very carefully.
And you are the professional of Mathcad usage.
I only want to mention about your above responcse,
Zr => 1/jwC ... This is the answer when R=0.
Werner Exinger 2017/03/11 2:05
Anyway, I thank all answers to my help message. I am still under learning how to use the Mathcad.
The answer I want to get by Mathcad is in my first figure of the Help-2.
> Zr => 1/jwC ... This is the answer when R=0.
I was just showing a way to make Luc's sheet work in Versions of Mathcad with MuPad as symbolic engine (rather than Maple in Luc's version 11).
If you follow Lucs comments in his sheet you see, that r = 1/jwC simply is the value of r, where the magnitude of of Z is extreme. According to Mathcads result that value is independent of R (which is not true). Not sure what muPad is doing wrong here!??
Maybe a bug???
At least here is a working sheet using a user defined function for the magnitude rather than the built-in absolute value:
Hi Tetsuroh Tokoro,
we could represent the transfer function of the given quadrupole, on a particular graph as well, it is represented by the blue arc, namely:
Hi Tetsuroh Tokoro,
Since I like to do things well done, clear and understandable, and given that in the worksheets that I have presented, there is only a "sea" of formulas, I have written something to be clearer. It seems, too, that I have answered your questions. Follows a laborious collage shown here: