Optical Fiber. Chapter 1. n 1 n 2 n 2. index. index

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1 Chapter 1 Optical Fiber An optical ber consists of cylindrical dielectric material surrounded by another cylindrical dielectric material with a lower index of refraction. Figure 1.1 shows that the transistion between the core and the cladding region can either be a discrete transistion (Step Index or a gradual transistion (Graded Index. step index graded index a a index n 2 n 2 index Figure 1.1: End cross-section of an optical ber. First let's look and the step index optical ber, where the core has an index of refaction of and the cladding has an index of n 2, where > n 2. We will initial look at the light traveling in the optical ber as a propagating ray. Even those this is not technically accurate, it provides some intuitive feel for what is going on. Figure 1.2 shows a cross-sectional view of a step index optical ber. If the propagation angle is greater than the critical angle then the ray will bounce of the surface and will be conned to the core region. ECEn January 24, 27

2 Therefore, the propagation is conned to be ( θ 1 > θ c = sin 1 n2. (1.1 cladding θ 1 >θ c θ1 <θ c θ 1 =θ c core n 2 cladding Figure 1.2: Cross-sectional view of a step index optical ber. In order to maintain that the propagation angle is greater than the critical angle, the entrance angle into the optical ber must be less than θ a. sin θ a = sin (9 θ c (1.2 sin θ a = sin θ in < = cos (θ c (1.3 = 1 sin 2 θ c (1.4 ( 2 n2 = 1 (1.5 = n 2 1 n2 2 n 2 1 (1.6 n 2 1 n2 2 NA (1.7 n 2 1 n2 2 (1.8 n=1. θa 9 θ c θ 1 =θ c n 2 Figure 1.3: Numerical aperture of an optical ber. ECEn January 24, 27

3 In addition to requiring the propagation angle to be greater than the critical angle, there are also only a descrete set of propagaton angles that remain in phase as illustrated in Fig These allowable propagation angles are called the modes of the waveguide. λ θ θ n 2 d Figure 1.4: The rays must remain in phase after multiple reections. 1.1 Field Relationships The optical ber is cylindrical in dimension with no variation in the z-direction. Therefore, we assume a z-dependence of exp [j (ωt βz]. Therefore, E E z = jβe and t = jωe. Similar relations exist for H. Since the waveguide shape is cylindrical we will use a cylindrical coordiant system. There are six eld quantities E z, E φ, E ρ, H z, H φ, and H ρ. Rather than solve the wave equation for all six equations we can relate all of the equations in terms of E z and H z. To demonstrate this relationship for one component start with the φ-component of Faraday's law as given by ( E = jωµh φ, (1.9 which in cylindrical coordinates becomes Solve for H φ to yield jβe r E z r = jωµh φ. (1.1 H φ = β ωµ E r + 1 E z jωµ r (1.11 Now use the r-component of Ampere's law as given by ( H = jωɛe which becomes r, ( H z r φ + jβh φ = jωɛe r (1.13 ECEn January 24, 27

4 Now substitute Eq into Eq to yield Combine the two E r terms to yield 1 H z r φ + jβ2 ωµ E r + β E z ωµ r = jωɛe r (1.14 Use the substitution κ 2 = ω 2 µɛ β 2 to yield j ( ω 2 µɛ β 2 E r = ωµ H z r φ + β E z r E r = j ( ωµ H z κ 2 r φ + β E z r The other four equations are solved in a similar manner as given by E φ = j ( β E z κ 2 r φ ωµ H z r H r = j κ 2 ( β H z r H φ = j κ 2 ( β r ωɛ r E z φ H z φ + ωɛ E z r (1.15 (1.16 (1.17 (1.18 (1.19 We will, therefore, only have to solve for E z and H z and we can relate all other eld component in terms of these two. 1.2 Determining the General Field Form We assume that the different variable can be separated (this is actually a very good assumption as given by E z (ρ, φ, z = R(ρF (φz(z (1.2 This is known as separation of variables. The waveguide does not vary in the z-direction and we know that the eld is a wave that is traveling only in the z-direction. Therefore, the z-dependent term is given by Z(z = e jβz (1.21 The φ term relates to the rotation and must be a periodic function of the angle with a period of 2π resulting in where m is an integer. F (φ = e jmφ, (1.22 The last term R(ρ is determined by using the wave equation that is given by 2 E z + (nk o 2 E z =. (1.23 ECEn January 24, 27

5 Remember that the wave equation is derived directly from Maxwell's equations. Since we have cylindrical geometry, we expand out the wave equation in cylindrical coordinates as given by 2 E z ρ E z ρ ρ E z ρ 2 φ E z z 2 + n2 koe 2 z =, (1.24 where the refractive index is given by { n1 ρ a n = ρ > a n 2 (1.25 The eld (Eq. 1.2 is then plugged in the wave equation (Eq to give ( 2 R ρ 2 (F (Z + ( 1 ρ ( R 1 (F (Z + (R ρ (R(F ρ 2 2 F φ 2 (Z + (1.26 ( 2 Z z 2 + ( n 2 ko 2 (R(F (Z = (1.27 Divide through by (R(F (Z to yield ( 1 R ( 2 ( R 1 ρ 2 + R ( 1 ρ R ρ + ( ( ( F 1 2 Z F ρ 2 φ 2 + Z Z 2 + n2 ko 2 =. (1.28 Since F = e jmφ and Z = e jβz the second derivatives of these functions yield and 2 F φ 2 = m2 F ( Z φ 2 = β2 Z (1.3 The wave equation becomes ( 1 R ( 2 ( R 1 ρ 2 + R ( 1 ρ R ρ m2 ρ 2 β2 + n 2 k 2 o =. (1.31 (1.32 Multipling through by R results in 2 R ρ R ρ ρ + (n 2 k 2o β 2 m2 ρ 2 R =. (1.33 Remember that n changes depending on whether we are looking in the core or cladding region. The range of β we can determine based on the ray trace appraoch. We know that the propagation angle is in the range ( sin 1 n2 < θ < π ( ECEn January 24, 27

6 and that the propagation constant is given by β = k o sin θ (1.35 resulting in n 2 k o < β < k o. (1.36 Let's dene two new variables p 2 ( k o 2 β 2 (1.37 and q 2 β 2 (n 2 k o 2. (1.38 These two variables can be combined to yield p 2 + q 2 = ( n 2 1 n 2 2 ko = NAk o (1.39 The ρ dependent equation (Eq then becomes ( 2 R + 1 R ρ 2 ρ ρ + p 2 m2 R = ρ < a ( ρ 2 2 R + 1 R ρ 2 ρ ρ q 2 m2 R = ρ > a ρ 2 (1.4 Equatio.4 is a differential equation satised by the Bessel functions. Its general solution in the core and cladding regions are given by R(ρ = { AJm (pρ + BY m (pρ ρ a CK m (qρ + DI m (qρ ρ > a (1.41 Since lim Y m(x =, (1.42 x B =. Similarly, since lim I m(x =, (1.43 x D =, resulting in { AJm (pρ ρ a R(ρ = CK m (qρ ρ > a (1.44 A similar process can also be followed for determining H z resulting in E z = { AJm (pρ exp (jmφ exp (jβz ρ a CK m (qρ exp (jmφ exp (jβz ρ > a (1.45 ECEn January 24, 27

7 Im(x Km(x Ym(x Jm(x x Figure 1.5: The bessel functions. and H z = { BJm (pρ exp (jmφ exp (jβz ρ a DK m (qρ exp (jmφ exp (jβz ρ > a (1.46 Note that the unknown constants were changed. We now recognize that we are left with 4 unknown amplitudes in our equations. We must compute B, C, and D in terms of A. Also, we must determine β which can then be used to compute p and q from our dispersion relations. If we enforce continuity of E z, H z, E φ, and H φ at the dielectric boundary, we will have our required four equations. The terms E φ, E ρ, H φ, and H ρ can be derived from E z and H z using Maxwell's equataions. In the core ECEn January 24, 27

8 region, these relationsips become E ρ = j ( p 2 β E z ρ + µ ω o ρ H z φ E φ = j ( β E z p 2 ρ φ µ oω H z ρ H ρ = j ( p 2 β H z ρ ɛ on 2 ω E z ρ φ H φ = j p 2 ( β ρ H z φ + ɛn2 ω E z ρ These equations can be used in the cladding region after replacing p 2 with q 2. (1.47 (1.48 (1.49. (1.5 Now we just need to solve for the unknowns, which are A, B, C, D, and β. These unknowns are determined by applying the boundary conditions. The boundary conditions are tangential E at the core-cladding interface and tangential H at the core-cladding interface. E z (ρ = a = E z (ρ = a + (1.51 E φ (ρ = a = E φ (ρ = a + (1.52 H z (ρ = a = H z (ρ = a + (1.53 H φ (ρ = a = H φ (ρ = a + (1.54 (1.55 One method to determine these unknowns is to set these equations up into a matrix equation as given by J m (pa k m (qa A J m (pa K m (qa jmβ J ap 2 m (pa µω p J m(pa jmβ K aa 2 m (qa µω q K B m(pa C = (1.56 ɛ on 2 1 ω p J m(pa jβm ɛ J ap 2 m (pa on 2 2 ω q K m(pa jβm K aq 2 m (pa D [M] 4x4 A B C D = (1.57 These equations have a nontrivial solution only if the determinant of the coefcient matrix ( M = is equal to zero. After some algebra, this condition leads to the eigenvalue equation [ ] [ J m (pa pj m (pa + K m(qa J m (pa qk m (qa pj m (pa + n2 2 K ] ( m(qa n 2 = m2 1 1 qk m (qa a 2 p ( 1 q 2 p 2 + n2 2 1 n 2 1 q 2, (1.58 where the prime indicates differentiation with respect to the argument. In general this eigenvalue equation can have multiple solutions for each integer value m. It is customary to enumerate these solutions in descending numerical order and denote them as β mn. In general, both E z and H z are nonzero (except for m =. Fiber mode, are therefore referred to as hybrid modes an are denoted as HE mn or EH mn depending on whether H z or E z demoninates. In the special case m =, HE n and EH n are also denoted as T E n and T M n. ECEn January 24, 27

9 The p and q terms are related by the dispersion relation and result in the following equation (pa 2 + (qa 2 = (k o a 2 ( n 2 1 n 2 2. (1.59 The left hand side of the equation is called the normalized frequency V V = k o a n 2 1 n2 2 (1.6 = k o a NA (1.61 A few other useful parameters are The effective index n = β k o The normalized propagation constant b = β/k o n2 n1 n2 = n n2 n1 n Weakly Guiding Optical Fibers We can also look at the eigenvalue equation for weakly guiding optical bers ( 1, where = n core n clad n core. (1.62 This is a very accurate approximation for practical optical ber because the index contrast is made low in order to limit the maximum dispersion. In this case the EH and HE modes are degenerate. (They have the same β. Therefore, we call them LP (linear polarized modes instead. The resulting mode equation becomes J m±1 (pa J m (pa = ± q K m±1 (qa p K m (qa. (1.63 The modes are labeled as LP mn, where m referes to the m in the equation and n is the root number. 1.4 Single Mode Operation A mode is cut-off when q =. The lowest order mode is when m =, so the cut-off of the lowest order mode is when xj 1 (x J (x = lim qa qak 1(qa K (qa = (1.64 xj 1 (x = (1.65 x = (1.66 So there is no cut-off, similar to a symmetric waveguide. What is the cut-off of the second mode? First let's look for the LP 2 mode. This is the second zero of xj 1 (x = which is x = We also have to look ECEn January 24, 27

10 at the LP 11 mode, which will have a cut off of xj 2 (x J 1 (x = lim qa qak 2(qa K 1 (qa = 2 (1.67 xj 2 (x = 2 J 1 (x (1.68 x = 2.45 (1.69 The cut-off for the LP 11 is lower than the LP 2 mode. So the single mode operating range for a step index optical ber is < V < Approximations The normalized propagation constant can be approximated in the range 1.5 < V < 2.5 as This approximation is valid to with.2% over this range. Since b(v ( /V 2. (1.7 the effective index is b(v = n n 2 n 2, (1.71 n = n 2 + b( n 2 ( Mode Field Prole For a weakly guiding ber the z-components of the eld are very small. Hence, the dominant mode HE 11 is approimately linearly polarized and is also denoted as LP 1. The eld prole can be derived to be and H y = n 2 ɛo /µ o E x. E x = { J (pρ J (pa exp(jβz ρ < a K (qρ K (qa exp(jβz ρ > a (1.73 The fundamental mode is often approimated by a Gaussian mode as given by E x = E o exp ( ρ2 w 2 exp (jβz. (1.74 The Gaussian beam width w is determined by tting the Gaussian beam to the actual eld prole. An approximation for this t (withi% for 1.2 < V < 2.4 is given by w a V 3/ V 6. (1.75 With a Gaussian mode we can more easily match the ber mode to an incident laser beam. We can also calculate the fraction of the power that is within the core region of the ber as given by Γ = P a core = E x 2 ρdρ P ( total E x 2 ρdρ = 1 exp 2a2 w 2 (1.76 ECEn January 24, 27