Optical Communication Components and Subsystems ECE 565 Spring 2005 Homework 2 Solutions Problem 2.1. We use Equation 2.1.5 in Agrawal, which states that the dispersion is given by 2 2 -4 ∆T / L L = (n1 /n2c)∆. But ∆= (n1- n2)/ n1 =(n1-1.45)/ n1. So, 10 ns/km = ( n1 /1.45 x 2.99x10 km/ns) (n1-1.45)/ n1 or 1.45 x 2.99x10 -3 = n12 – 1.45 n1. Solve this quadratic equation and obtain n1= 1.453. Thus, ∆ = 0.0021.
Approximate alternative solution: Assume n1≅ n2 = 1.45 and use 10 ns/km = ( n12/1.45 x 2.99x10-4 km/ns) (n1-1.45)/ n1 to find ∆ = 0.00207. Problem 2.2. (First show that the solution of 2.1.8 is 2.1.9)
Using α =2 in Equation 2.1.7, we obtain 2
n(ρ) = n1 [1 – ∆(ρ /a) ],
ρ≤a.
After differentiation, we obtain dn/d ρ = -2 n1 (∆/a2) ρ. Now since ∆<< 1 and ρ≤ a <1, ∆(ρ /a)2 << 1, and we can approximate n = n1 [1 – ∆(ρ /a)2] by n1. Thus, dn/d ρ ≅ -2 n (∆/a2) ρ and the ray equation becomes 2
2
2
d ρ/dz = -2 ( ∆/a ) ρ ≡ -k ρ. rz Propose a solution of the form ρ( z z) = e , plug into the above differential equation, and obtain the characteristic equation r 2+k =0. =0. Thus, r = = ± i(k )1/2 = ± i(2∆)1/2 /a ≡ ± ip. The general ipz -ipz solution is ρ( z z) = A e + B e . Now ρ(0) = ρ0 implies that A + B = ρ0, and ρ'(0) = ρ'0 implies * we can use iAp - iBp = ρ'0. Thus, A = (-iρ'0/ p p + ρ0)/2 and B= (iρ'0/ p p + ρ0)/2 = A . Therefore, we Euler’s identity and recast ρ( z z) as 2| A A |cos( pz pz + ∠ A) = 2| A A |[ cos( pz pz)cos(∠ A) - sin( pz pz)sin(∠ A). But A |A |cos(∠ A) = Re( A A) = ρ0/2 and A |A |sin(∠ A) = Im( A) = -ρ'0/2 p, which completes the derivation of 2.1.9 and we obtain ρ( z z) = ρ0cos( pz pz)cos(∠ A) + ρ'0/ p p sin( pz pz).
Special Problem: We first note that ρ( z periodic. Suppose that R 1 and R 2 are two z) is 2π /p- periodic. rays emerging from initial conditions ρ0 and σ0, respectively. Then it is true that at z = 2π /p, both rays return to their respective initial positions (viz., ρ0 and σ0). Now at z = 2π /p, both rays have traveled the same net distance, namely, 2π /p (justify this statement). By Fermat’s
principle, both rays must have traveled the same net distance at the minimum time required, so both times of flight must be the same! Thus, intermodal dispersion is completely absent in this type of fiber. An excellent derivation of Fermat’s principle can be found in the book by Born & Wolf (Principles of Optics, 7th edition, page 136). Problem 2.3.
Writing Eq.(2.2.1) in the frequency domain and by using B=µ0H, we obtain ∇ x E = iωµ0H.
Next, we write H in cylindrical coordinates: H= H ρ + H φ + z H z, E= E ρ + E φ + z E z and apply the curl in cylindrical coordinates: ∇ x E = (ρ−1∂ E z/∂φ − ∂ E φ/∂z ) - (∂ E z/∂ρ − ∂ E ρ/∂z) + z(ρ−1 E φ + ∂ E φ/∂ρ − ρ−1∂ E ρ/∂φ)
= iωµ0H. We now equate the three components from the two sides of the above equation and obtain: ρ−1∂ E z/∂φ − ∂ E φ/∂z = iωµ0 H ρ, ∂ E z/∂ρ − ∂ E ρ/∂z = - iωµ0 H φ, ρ−1 E φ + ∂ E φ/∂ρ − ρ−1∂ E ρ/∂φ = iωµ0 H z.
We can apply the same procedure to Eq.(2.2.2), namely, take the curl of H in cylindrical coordinates, ∇ x H = iωεE, and equate terms. This yields: ρ−1∂ H z/∂φ − ∂ H φ/∂z = - iωε E ρ, ∂ H z/∂ρ − ∂ H ρ/∂z = iωε E φ, ρ−1 H φ + ∂ H φ/∂ρ − ρ−1∂ H ρ/∂φ = − iωε E z.
We now replace all derivatives with respect to z by iβ since the z dependence is always of the form eiβ z. We then express E ρ, E φ, H ρ, and H φ in terms of E z and H z using the above six algebraic equations. For example, by substituting H ρ from the first equation in the fifth equation we obtain: E φ = (i/ωε)[-∂ H z/∂ρ + ( β/ωµ0)(ρ−1∂ E z/∂φ − iβ E φ)] or
(1 – [β2/ω2εµ0]) E φ = (i/ωε)[-∂ H z/∂ρ + ( β/ωµ0)(ρ−1∂ E z/∂φ)]. Use ε=ε0n2, 1/c2= ε0µ0, and k 0 = ω0/c we obtain 1 – [β2/ω2εµ0] = p2/ω2εµ0, where p2 = n2 k 0 - β2. Hence, we finally obtain E φ = (i/ p p2)[(β/ρ)∂ E z/∂φ-ωµ0∂ H z/∂ρ]. The other three equations are derived similarly.
Problem 2.5. 2 1/2 For single-mode fibers, the cut-off condition is given by V = 2k 0 a(n12 – n n2 ) = 2.405. Substitute 1 µm for λ, 1.445 for n2, and solve for a to obtain: a=3.1815 µm.
At λ = 1.3 µm, V =1.85. =1.85. From Eq.(2.2.45), w = 4.368 µm. From Eq.(2.2.46), Γ = 1-exp(-2a2/w2) = 0.6539.
