Figure out the waveguide width at cutoff (d) for each of TE0,TE1,TE2 and TE3 modes. Assume n1=1.48, n2=1.46 and wavelength is .82um. Sketch the transverse mode patterns (Ey(x)) at cutoff for TE0 and TE1.

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m=(2d/lambda)*(sqrt(n1^2-n2^2))
I am looking for the relation as given in the attachment.
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Star Strider
Star Strider 2014 年 10 月 30 日
Download and attach the PDF to your original Question. (Use the ‘Edit’ function.) That way we know exactly what the problem statement is.
garry
garry 2014 年 10 月 31 日
Okay... These are my two equations and i need to plot these in one graph...:-
Equation 1:- (22.2*10^3)*sin((161*10^4)*x) x < d/2
Equation2 :- (42.32*10^9)*exp(-2.896*10^6) x > d/2
here d/2 = 5*10^-6 m.

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Star Strider
Star Strider 2014 年 11 月 1 日
Here you go:
d2 = 5E-6;
fv = @(x) [(22.2E+3)*sin((161E+4).*x).*(x<d2) + (42.32E+9)*exp(-2.896E+6.*x).*(x>=d2)];
x = linspace(0,2*d2);
figure(2)
plot(x, fv(x))
grid
producing:
  5 件のコメント
Star Strider
Star Strider 2014 年 11 月 10 日
Communications engineering is far from my areas of expertise. I’m not at all familiar with the ‘FCS algorithm’. It might be best if you posted this as a new Question for others to see.
garry
garry 2014 年 11 月 11 日
This is the link where i have posted the exact question along with the flyer explaining the exact procedure..

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