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I want to find the 3D contour of the equation1:

C1 = 10^(-10);
C2 = 0.1*C1;
R = 50;
Tb = 0.1;
Geb = 5*10^-15;
Z0 = 50;
L[Te_] := 10^-9 + 10^-9*(Te - 0.1);
Zlcr[Te_, w_] := (1/R + 1/(I*L[Te]*w) + I*C1*w)^-1;
Zload[Te_, w_] := -I*w*C2 + Zlcr[Te, w];
\[CapitalGamma][Te_, w_] := (Zload[Te, w] - Z0)/(Zload[Te, w] + Z0);
y[Te_, w_] := (Abs[\[CapitalGamma][Te, w]])^2;
eqn1[w_, Te_, Pprobe_] := (1 -   y[Te, w]) Pprobe  ==  (Te - Tb) Geb
ContourPlot3D[
 eqn1[w, Te, Pprobe], {w, 0, 5*10^9}, {Te, 0, 1}, {Pprobe, 0, 10^-14}]
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  • $\begingroup$ I don't understand this part: == (Te - Tb) Geb! Why do use == inside definition of a function? $\endgroup$
    – Mahdi
    Jul 15, 2014 at 10:05
  • $\begingroup$ (Te-Tb)Geb is simply equal to (Te - 0.1)* (5*10^(-15)) $\endgroup$
    – user44840
    Jul 15, 2014 at 10:07
  • $\begingroup$ Yes that is correct but I think this definition (1 - y[Te, w]) Pprobe == (Te - Tb) Geb is wrong. Are (1 - y[Te, w]) Pprobe and (Te - Tb) Geb equal? $\endgroup$
    – Mahdi
    Jul 15, 2014 at 10:09
  • $\begingroup$ Use Evaluate[eqn1[w, Te, Pprobe]] inside ContourPlot3D. $\endgroup$
    – Stephen Luttrell
    Jul 15, 2014 at 12:22

1 Answer 1

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If you change your last function just to expression and evaluate in plot:

exp = (1 - y[Te, w]) Pprobe == (Te - Tb) Geb;
ContourPlot3D[
 Evaluate[exp], {w, 0, 5*10^9}, {Te, 0, 1}, {Pprobe, 0, 10^-14}, 
 Mesh -> False]

enter image description here

Improve this answer
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1
  • $\begingroup$ Brilliant, you are a master at graphics! $\endgroup$
    – user44840
    Jul 15, 2014 at 12:14

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