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.

2 vues (au cours des 30 derniers jours)
m=(2d/lambda)*(sqrt(n1^2-n2^2))
I am looking for the relation as given in the attachment.
  4 commentaires
Star Strider
Star Strider le 30 Oct 2014
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 le 31 Oct 2014
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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Réponse acceptée

Star Strider
Star Strider le 1 Nov 2014
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 commentaires
Star Strider
Star Strider le 10 Nov 2014
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 le 11 Nov 2014
This is the link where i have posted the exact question along with the flyer explaining the exact procedure..

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