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.

m=(2d/lambda)*(sqrt(n1^2-n2^2))
I am looking for the relation as given in the attachment.

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ok don't access the pdf i'll send you the calculated values and the equations to plot shortly... any help is hugely appreciated.
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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

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:

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Hiee Star Strider..I have tried a lott on this question.
If you could help me..it would be of greatt help.
----> Implement an error detection mechanism using the standard FCS algorithm Write two programs, call them "Generator" and "Verifier".
---> The Generator program reads from a standard input a "k-bit" message as a string of "0s and 1's" as a line of ASCII text. The generator program also reads from a second line a "(n+1)-bit" generator pattern also in ASCII. The Generator program outputs to a standard output a line of ASCII text with "k+n" 0's and 1's representing the message to be transmitted (including the original k-bit message + the n-bit FCS bits). It also outputs the polynomial (just as it reads it in).
---> The Verifier program reads in the output of the Generator program and outputs a message indicating whether it is correct or not.
---> Finally, write a program "Alter" that inverts one bit on the first line (assume that that the leftmost bit position is #1) but copies the rest of the two lines correctly. The output of this program should indicate whether the error is detected or not.
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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on 30 Oct 2014

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