0
$\begingroup$

I have this expression (see below for context):

6.79018*10^-9/f0 - (2.16138*10^-9 ArcTan[Sqrt[Tan[4.62667*10^8 d f0]^2]])/f0

and want to solve it for f0.

$d$, $l$, and $f0$ are real positive numbers.

Why can't Mathematica solve it by:

Solve[
  l == 6.79018*10^-9/f0 - (2.16138*10^-9 ArcTan[Sqrt[Tan[4.62667*10^8 d f0]^2]])/f0,
  f0
]

Context

The following two equations need to be solved for f0 (dependent on d and l) because I need a contour plot of x axis d, y axis l, and contour f0.

 d = 2.16138*10^-9/f0 * ArcTan[Sqrt[ZL/50]]

 l = 6.79018*10^-9/f0 - (2.16138*10^-9 ArcTan[ZL/(Sqrt[50ZL])])/f0

My idea is:

  1. solve d for ZL (worked)
  2. replace ZL in l with the solution of 1. (worked)
  3. solve new l for f0. (Error)

d and l are actual lengths (real, positive) and f0 is a frequency of about 6-6.3Ghz when d=0.0053-0.0055 and l = 0.0049-0.0051

enter image description here

I made the assumption that ZL - Z0 = ZL. As ZL >> Z0. If Mathematica can manage that without the assumption it would be ever better.

The expressions above show original formula. With

Z0 = 50 (standart characteristic impedance of measurement devices).

$$ \beta = 2\pi \sqrt{4.88}f_0 / c $$ $$ \beta = 2\pi / \lambda $$

A reference would be this dissertation: (p.48) but beta is defined by some own measurements.

$\endgroup$
7
  • 2
    $\begingroup$ In general, a transcendental equation like yours does not admit a closed form solution. You might want to try FindRoot[] with a good initial guess instead. $\endgroup$ Commented Jun 17, 2016 at 13:29
  • $\begingroup$ Thanks for the fast reply! Unfortunately I cant figure it out. I edited my post and describe the original problem. It seems to be a simple task to solve two equations but it wont work. Thanks! $\endgroup$
    – mggiable
    Commented Jun 17, 2016 at 14:32
  • $\begingroup$ I have no time to answer your question right now however I recommend to take a look at this post: Solve symbolically a transcendental trigonometric equation and plot its solutions. If you read it carefully you'll understand your actual problem and find an appropriate solution. $\endgroup$
    – Artes
    Commented Jun 17, 2016 at 15:09
  • $\begingroup$ @mggiable Cab you tell me what this is specifically calculating and any references to the source of the equations? That will help me refine my answer. $\endgroup$
    – Young
    Commented Jun 17, 2016 at 15:47
  • $\begingroup$ I took the approach of eliminating f0 and trying to solve for ZL -- you can readily show that there are no solutions. $\endgroup$
    – george2079
    Commented Jun 17, 2016 at 15:59

2 Answers 2

3
$\begingroup$

Updated based on the following formulas provided as additional clarification:

enter image description here

Beta = 2 Pi / Lambda = 2 Pi Sqrt[4.88] f / c
Z0 = 50 and ZL>>Z0

Solved:

 Solve[{dvar == 1/((2 Pi)/wL) ArcTan[Sqrt[ZL/50]], 
  lvar == wL/2 - 1/((2 Pi)/wL) ArcTan[(ZL)/Sqrt[ZL 50]]}, {wL, ZL}]

{{wL -> 2*(dvar + lvar), ZL -> 50 Tan[(dvar Pi)/(dvar + lvar)]^2}}

v = 299792458/Sqrt[4.88];
ContourPlot[
 f0[dvar_, lvar_] = v/(2 (dvar + lvar)), {dvar, 0, 
  0.06}, {lvar, 0, 0.06}, PlotLegends -> Automatic]
ContourPlot[
 Z[dvar_, lvar_] = 50 Tan[(dvar Pi)/(dvar + lvar)]^2, {dvar, 0, 
  0.06}, {lvar, 0, 0.06}, PlotLegends -> Automatic]

enter image description here

$\endgroup$
1
  • $\begingroup$ Thank you sooooo soooo much!!! Thanks excatly what i needed! $\endgroup$
    – mggiable
    Commented Jun 19, 2016 at 16:55
0
$\begingroup$

For such complicated functions you can use FindRoot over a range of your parameters to get an idea.

data = Flatten[Table[{10^8 l, 10^8 d,
 f0 /. FindRoot[l == 6.79018*10^-9/f0 
        - (2.16138*10^-9 ArcTan[Sqrt[Tan[4.62667*10^8 d f0]^2]])/f0, {f0, 0.5}]}
, {l, 2. 10^-8, 50. 10^-8, 10^-8}, {d, 2. 10^-8, 50. 10^-8, 10^-8}], 1];

ListPlot3D[data, AxesLabel -> {"l 10^-8", "d 10^-8", "f0"}]

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.