Step index lectures

Step index lectures - THE FIBER FORUM Fiber Optic...

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Dr. Joseph C. Palais step-index fiber 1 THE FIBER FORUM Fiber Optic Communications Dr. JOSEPH C. PALAIS PRESENTED BY
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Dr. Joseph C. Palais step-index fiber 2 Step-Index Optical Fiber
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Dr. Joseph C. Palais step-index fiber 3 r a y x φ Core Cladding Step-Index Fiber
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Dr. Joseph C. Palais step-index fiber 4 n 1 y z Assume: (129 = (n 1 - n 2 ) /n 1 << 1 Cladding is infinite in extent. Electric field is polarized in y direction. z n 1 n 2
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Dr. Joseph C. Palais step-index fiber 5 E y   is the instantaneous   electric field. E y satisfies the wave equation. (2) d 2 E y = μ 0 ε ( c 2 E y / c t 2 ) / ε 0 ) = n 1 2 , r < a / ε 0 ) = n 2 2 , r > a (3) d 2 E y = ( c 2 E y / c x 2 )+ ( c 2 E y / c y 2 ) + ( c 2 E y / c z 2 ) = d 2 t E y + ( c 2 E y / c z 2 )
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Dr. Joseph C. Palais step-index fiber 6 Assume the wave travels in z-direction. Then we can write: (4) E y = E y (x,y) e i( ϖ t- β z) Combining equations (2) - (4): (5) d 2 E y = d 2 t E y +(- i β ) 2 E y e i( ϖ t- β z) = μ 0 ε (i ϖ ) 2 E y e i( ϖ t- β z) i= j= -1
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Dr. Joseph C. Palais step-index fiber 7 (6) d 2 t [ E y e i( ϖ t- β z) ] - β 2 E y e i( ϖ t- β z) = - ϖ 2 μ 0 ε E y e i( ϖ t- β z) Then (7) d 2 t E y (x,y) + ( ϖ 2 μ 0 ε - β 2 )E y (x,y) = 0 (8) ϖ 2 μ 0 ε = ϖ 2 μ 0 ε 0 ( ε / ε 0 ) = k 0 2 n 2 (9) k 0 = ϖ ( μ 0 ε 0 ) 1/2 = ϖ / c
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Dr. Joseph C. Palais step-index fiber 8 Now writing d 2 t in cylindrical coordinates: (10) ( c 2 E y / c r 2 ) + (1/r) ( c E y / c r) + (1/r 2 ) ( c 2 E y / c φ 2 ) + (n 2 k 0 2 - β 2 )E y = 0 Use the method of separation of variables to solve this differential equation. Assume that the solution can be written in the form of a product of a function of r and a function of φ . That is: (11) E(r, φ ) = R(r) Φ ( φ )
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Dr. Joseph C. Palais step-index fiber 9 Now (10) can be written as: (12) Φ ( c 2 R / c r 2 ) + ( Φ /r)( c R / c r) + (R/r 2 )( c 2 Φ / c φ 2 ) + (n 2 k 0 2 - β 2 )R Φ = 0 (13) (1/R) ( c 2 R / c r 2 ) + (1/Rr)( c R / c r) + (1/ Φ r 2 )( c 2 Φ / c φ 2 ) + (n 2 k 0 2 - β 2 ) = 0 Placing the terms involving r and φ on opposite sides of the equation yields: (14a) (1/R)[r 2 ( c 2 R/ c r 2 ) + r( c R/ c r) + r 2 (n 2 k 0 2 - β 2 )R] = - (1/ Φ )( c 2 Φ / c φ 2 )
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Dr. Joseph C. Palais step-index fiber 10 The left hand side depends upon radial position r and the right hand side depends upon angular position φ . The only way for the equality to hold is if both sides are constant (independent of the values of r and φ ). Thus we conclude that: (14b) (1/R)[r 2 ( c 2 R/ c r 2 ) + r( c R/ c r) + r 2 (n 2 k 0 2 - β 2 )R] = - (1/ Φ )( c 2 Φ / c φ 2 ) = constant = l 2 Thus, we can write that: (15) - (1/ Φ )( c 2 Φ / c φ 2 ) = l 2
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Dr. Joseph C. Palais step-index fiber 11 The solution to this second order differential equation is: (16) Φ ( φ ) = A cos( l φ ) + B sin ( l φ ) Because E y ( φ + 2 π ) = E y ( φ ), must be an integer. Thus: = 0,1,2,3…
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Dr. Joseph C. Palais step-index fiber 12 From (14b), we now have: (18) r 2 ( c 2 R/ c r 2 ) + r( c R/ c r) + [ r 2 (n 2 k 0 2 - β 2 ) - l 2 ]R = 0 where n = n 1 , r < a n = n 2 , r > a We recognize this as a Bessel equation (equations 1, 1’, and 9 in the section on Bessel functions).
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Dr. Joseph C. Palais step-index fiber 13 Consider r > a (in the cladding). We expect fields which decay in the radial direction. The only Bessel function that does this is the modified Bessel function of the second kind: K l (x) From Bessel notes (20), as x approaches infinity: ( 29 - 1 2 x l K x e x π =
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Dr. Joseph C. Palais step-index fiber 14
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This note was uploaded on 05/11/2010 for the course EEE EEE-546 taught by Professor Palais during the Spring '10 term at ASU.

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Step index lectures - THE FIBER FORUM Fiber Optic...

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