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.
  4 Comments
Star Strider
Star Strider on 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 on 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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Accepted Answer

Star Strider
Star Strider on 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 Comments
Star Strider
Star Strider on 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.

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