Mode Theory For Circular Waveguide
Mode Theory For Circular Waveguide
Mode Theory For Circular Waveguide
3. TE νm TM νm : ν = 0, TE νm (E z = 0), TM νm (H z = 0) correspond to
meridional rays traveling within fiber
4. HE νm EH νm : hybrid modes, both Ez and Hz are nonzero
5. HE modes: E z dominates
6. EH modes: H z dominates
2.4 Mode theory for circular waveguide
Light propagates
only in core !
β/ k
Light propagates
in cladding !
2.4 Mode theory for circular waveguide
Approximate solution from scalar equations based on weakly guiding fiber
approximation:
For low-order modes the fields are tightly concentrated near the center of
slab (WG, optical fiber), with little penetration into cladding region
For high-order modes, the fields are distributed more toward the edges of the
guide and penetrate further into cladding region
2.4 Mode theory for circular waveguide
2.4.1 Over view of Modes
Cutoff condition : propagation angle for a given mode just equals the critical angle.
β 2π n1
θ β = k1 cos θ = cos θ
n1 λ
n2
Critical angle : cos θ =
k n1
h
n2 For guided modes, propagation constant in the range:
2π n 2 2π n1 k 2 ≤ β ≤ k1
≤β ≤ or
λ λ
Cutoff condition: β = k 2 = n2 k
Propagation constant
- Guided mode : bound mode guided inside of core β ≥ k 2 = n2 k
- Radiation modes : refracted mode by cladding β < k2
- Leaky mode : partially confined into core β < k2
2.4 Mode theory for circular waveguide
2.4.2 Summary of Key Modal Concept
Question : what parameter could determine if the fiber is Single-
mode fiber (SMF), or multi-mode fiber (MMF) ?
Parameters: Fiber : a, n1, n2 Source : λ
V number is an important parameter connected with cutoff condition,
determined how many modes a fiber could support.
2π a
V = n12 − n 22
λ
Lowest-order mode : HE11
Single mode condition : V < = 2.405
When V > 10 , the total number of modes : M ≈V2 /2
Pclad 4
Fraction of average optical power residing in cladding : ≈
P 3 M
2.4 Mode theory for circular waveguide
2.4.3 Maxwell’ Equations
∂
→ →
∂D
→ ∂2 E
→
∇× E = − B
→
∇× H = ∇ E − εμ
2
=0
∂t ∂t ∂t 2
→
→ →
∇⋅D = 0 ∇× B = 0
→ ∂ 2
H
∇ 2 H − εμ =0
∂t 2
¾ Light is electromagnetic wave
H = H ( r , φ ) e j (ω t − β z ) → ∂
0 ∇× H = D
∂t
B = μH , D = ε E
With help of ∂ → jω , ∂ → − j β , we can find following Eqs.
∂t ∂z
E H
1 ⎛ ∂H z ⎞ 1 ⎛ ∂Ez ⎞
r direction ⎜ + jr β H φ ⎟ = jωε Er , (2.33a ) ⎜ + jr β Eφ ⎟ = − jωμ H r (2.34a)
r ⎝ ∂φ ⎠ r ⎝ ∂φ ⎠
∂H z ∂E
φ direction j β Er + z = jωμ H φ (2.34b)
jβ H r + = − jωε Eφ (2.33b) ∂r
∂r
1⎛ ∂ ∂H r ⎞ 1⎛ ∂ ∂Er ⎞
⎜ − ⎟ = jωε Ez ⎜ (rEφ ) − ⎟ = − jωμ H z (2.34c)
z direction
r ⎝ ∂φ
( rH φ )
∂φ
(2.33c) r ⎝ ∂φ ∂φ ⎠
⎠
2.4.4 Waveguide Equations
1 ⎛ ∂Ez ⎞
⎜ + jr β Eφ ⎟ = − jωμ H r (2.33a) j ⎛ ∂Ez μω ∂H z ⎞
r ⎝ ∂φ ⎠ Er = − ⎜β + ⎟ (2.35a )
q 2 ⎝ ∂r r ∂φ ⎠
∂E
j β Er + z = jωμ H φ (2.33b)
H ∂r j ⎛ β ∂Ez ∂H z ⎞
Eφ = − ⎜ − μω ⎟ (2.35b)
1⎛ ∂ ∂Er ⎞ q ⎝ r ∂φ
2
∂r ⎠
⎜ ( rEφ ) − ⎟ = − jωμ H z (2.33c)
r ⎝ ∂φ ∂φ ⎠ j ⎛ β ∂H z μω ∂Ez ⎞
Hr = − ⎜ − ⎟ (2.35c)
q 2 ⎝ r ∂r r ∂φ ⎠
1 ⎛ ∂H z ⎞
⎜ + jr β H φ ⎟ = jωε Er , (2.34a ) j ⎛ β ∂H z ∂Ez ⎞
r ⎝ ∂φ ⎠ Hφ = − ⎜ + ωε ⎟ (2.35d )
q 2 ⎝ r ∂φ ∂r ⎠
∂H z
E jβ H r + = − jωε Eφ (2.34b)
∂r
1⎛ ∂ ∂H r ⎞ with q 2 = ω 2εμ − β 2 = k 2 − β 2
⎜ ( rH φ ) − ⎟ = jωε Ez (2.34c)
r ⎝ ∂φ ∂φ ⎠
Using equation 2.33a, 2.34b to find Hr and Eφ
Using equation 2.33b, 2.34a to find Er and Hφ
2.4.4 Waveguide Equations
Step 3: Find wave equations for E and H:
j ⎛ β ∂H z μω ∂Ez ⎞
Hr = − ⎜ − ⎟ (2.35c)
q 2 ⎝ r ∂r r ∂φ ⎠
j ⎛ β ∂H z ∂Ez ⎞
Hφ = − ⎜ + ωε ⎟ (2.35d ) ∂2 E ∂ ∂2
q 2 ⎝ r ∂φ ∂r ⎠ z + 1 Ez + 1 Ez + 2
∂r 2 r ∂r r 2 ∂φ 2
q Ez = 0 (2.36)
1⎛ ∂ ∂H r ⎞
⎜ ( rH φ ) − ⎟ = jωε Ez (2.34c)
r ⎝ ∂φ ∂φ ⎠
j ⎛ ∂Ez μω ∂H z ⎞
Er = − ⎜β + ⎟ (2.35a )
q 2 ⎝ ∂r r ∂φ ⎠
j ⎛ β ∂Ez ∂H z ⎞
Eφ = − ⎜ − μω ⎟ (2.35b) ∂2 H ∂ ∂2
z + 1 Hz + 1 Hz + 2
q ⎝ r ∂φ
2
∂r ⎠ q H z = 0 (2.37)
∂r 2 r ∂r r 2 ∂φ 2
1⎛ ∂ ∂Er ⎞
⎜ ( rEφ ) − ⎟ = − jωμ H z (2.33c)
r ⎝ ∂φ ∂φ ⎠
Ez1 (r < a ) = AJν (ur )e jνφ e j (ωt − β z ) (2.42) Ez 2 (r > a) = CKν ( wr )e jνφ e j (ωt − β z ) (2.44)
H z1 (r < a ) = BJν (ur )e jνφ e j (ωt − β z ) (2.43) H z 2 (r > a) = DKν ( wr )e jνφ e j (ωt − β z ) (2.45)
with u 2 = k12 − β 2 k1 = 2π n1 / λ with w2 = β 2 − k22 k2 = 2π n2 / λ
Solution β can be determined by Boundary conditions :
Tangential components Eφ , Ez , and Hφ , Hz at r = a must be continue
At r = a,
E z1 = E z 2 AJν (ua ) = CKν ( wa ) (2.47)
H z1 = H z 2 BJν (ua ) = DKν ( wa ) (2.51)
Eφ 1 = Eφ 2 j ⎛ β ∂Ez ∂H z ⎞ j ⎛ β ∂H z ∂Ez ⎞
Eφ = − 2 ⎜ − μω H
⎟ (2.35b) , φ = − 2 ⎜
+ ωε ⎟ (2.35d )
Hφ1 = Hφ 2 q ⎝ r ∂φ ∂r ⎠ q ⎝ r ∂φ ∂r ⎠
j ⎡ jνβ ⎤ j ⎡ jνβ ⎤
− A
⎢⎣ a Jν (ua ) − Bωμ uJν
'
(ua ) = C
⎥⎦ w ⎢⎣ a Kν ( wa ) − Dωμ wKν
'
( wa ) ⎥⎦ (2.50)
u2 2
j ⎡ jνβ ⎤ j ⎡ jνβ ⎤
− B
⎢⎣ a νJ (ua ) + Aωε 1uJν
'
(ua ) =
⎥⎦ w2 ⎢⎣ D Kν ( wa ) + Cωε 2 wKν
'
( wa ) ⎥⎦ (2.52)
u2 a
2.4 Mode theory for circular waveguide
2.4.6 Modal Equations
¾ A set of four Eqs, right side = 0, therefore, only if the determinant of the
coefficients is zero, there is a solution exists.
(2.47)
(2.50)
(2.51)
(2.52)
format long
clear
clc
z = (0:0.001:15)'; zero_j0 = 2.405; 5.520; 8.654; 11.792
zero_j1 = 3.832; 7.016; 10.173
j0=besselj(0,z); zero_j2 = 5.036; 8.317; 11.520
j1=besselj(1,z); zero_j3 = 6.180; 9.561;
j2=besselj(2,z);
j3=besselj(3,z);
figure(1)
plot(z,j0,'r',z,j1,'b',z,j2,'g',z,j3,'y');
ylabel('Jv(x)');xlabel('x');
axis([0 15 -0.5 1]);grid
title('Bessel functions of the 1st
kind');
legend('v=0','v=1','v=2','v=3');
zero_j0=z(find(abs(j0)<1.2e-4))'
z=z(100:end);
j1=j1(100:end);
j2=j2(200:end);
j3=j3(300:end);
zero_j1=z(find(abs(j1)<1.3e-4))'
zero_j2=z(find(abs(j2)<1.5e-4))'
zero_j3=z(find(abs(j3)<1e-4))'
2.4.7 Modes in Step-Index Fibers Equations
¾ Cutoff conditions: It means a mode is not longer bound to the fiber core
β = k 2 = n2 k w=0
¾ Normalized frequency V: V = ( u + w ) a = ⎛ 2π a ⎞ ( n 2 − n 2 ) = ⎛ 2π a ⎞ NA 2
2 2
2 2 2 2
⎜ ⎟ ⎜ ⎟
⎝ λ ⎠ ⎝ λ ⎠
1 2
¾ HE11 has no cut off, but we have single-mode condition which is: V < = 2 .4 0 5
¾ Question: how do we get this number ? Æ From 1st zero of J0
V2
M =
2
2.4 Mode theory for circular waveguide
2.4.7 Modes in Step-Index Fibers Equations