ENE 423
Lecture III
Frequency dependent losses in fiber
The total loss (dB) is the summation of fixed loss (owing to absorption and
scattering), La, and modulation-frequency dependent loss (owing to pulse spreading),
Lf.
For the Gaussian response, Lf can be expressed as
 0.693 f
f
L f (dB )  10 log10 e


3 dB



2




for f << f3-dB, Lf can be negligible.
Ex. The modulation-frequency dependent loss of a fiber is measured to be 6 dB at a
modulation frequency of 2 GHz. Find the 3-dB optical bandwidth of the fiber.
Soln
2


 f   
L f  10 log10 exp  0.693 
 
f

3

dB

  


2


 2   
6  10 log10 exp  0.693 
 
f

3

dB

  


( f3 dB )opt  1.416 GHz
Reflections at a plane boundary
Normal incidence
2
The reflection coefficient, , can be written as

n1  n2
n1  n2
where  = the ratio of the reflected electric field to the incident electric field
Reflectance or Reflectivity, R 
reflected power
. We know that P0  E2, this
incident power
yields
 n n 
R 1 2 
 n1  n2 
2
From the conservation of energy, R + T = 1 where T = transmittance.
Ex. Calculate transmittance, T, into fiber from air
air
Fiber; n = 1.5
T+R=1
 1  1.5 
R= = 
 = 0.04
 1  1.5 
2
2
T = 1 – R = 0.96 = 96%
Oblique incidence
s-polarization
p-polarization
3
If the electric field is polarized perpendicular to the incident plane, it is called
“s-polarization”. If the electric field is polarized parallel to the plane of incidence, this
is called “p-polarization”. The reflection coefficients for the p- and s- polarizations
known as Fresnel’s laws of reflection are
p 
s 
n22 cos i  n1 (n22  n12 sin 2 i )
n22 cos i  n1 (n22  n12 sin 2 i )
n1 cos i  (n22  n12 sin 2 i )
n1 cos i  (n22  n12 sin 2 i )
Note: R   2
Zero reflection (R=0) occurs only for the p-polarization at the angle called
“Brewster angle”.
 n2 

 n1 
 B  tan 1 
There is no incident angle that will make s = 0.
In case of  =1, which occurs at
n22  n12 sin 2 i  0 .
sin  i 
n2
n1
This occurs at i = c = critical angle
 n2 

 n1 
c  sin 1 
4
At critical angle, all the lights reflect back into the first medium. This is called
“Total reflection” and it happens only if n2 < n1.
In case of i > c,
n1 sin i  n2
then n22  n12 sin 2 i  0 .
Therefore,  can be written in the form of
A  jB
 

A  jB
A2  B 2 e j tan
A B e
2
2
1
( B / A)
j tan 1 ( B / A)
 e j 2
where  = tan-1(B/A)
Interference between the incident and reflected waves creates a standing wave
in the incident region. A field still exists in the 2nd medium, although all power is
reflected. This field is fading away and carrying no power called “evanescent”. This
evanescent E decays exponentially as e-Z, where  = attenuation factor.
  k0 n12 sin 2 i  n22
This  indicates how far the field extends into the second medium before returning to
the incident region.
5
Ex. A parallel-polarized ray is incident at an angle of 85° when traveling from a
medium of index 1.48 into a medium having index 1.465. The wavelength is 1.3 µm.
(a) Compute the reflection coefficient. (b) At what distance from the boundary in the
transmission medium does the evanescent electric field decay to 10% of its value at
the boundary?
Soln
(a)
p 

n22 cos i  n1 n22  n12 sin 2 i
n22 cos i  n1 n22  n12 sin 2 i
1.4652 cos85  1.48 1.4652  1.482 sin 2 85
1.4652 cos85  1.48 1.4652  1.482 sin 2 85
 174.48
(b)
  k0 n12 sin 2 i  n22 

2

2

n12 sin 2 i  n22
1.482 sin 2 85  1.4652  0.802 106 m 1
E  E0 e  Z
0.1E0  E0 e 0.80210
z  2.871  m
6
Z
6
Optical Waveguide
Optical waveguides are used for guiding lightwave.
n
1
They are regions of dielectric or semiconductor of
n2 < n1
n
3
<
n
n
1
1
n
2
<
known refractive indices.
n1
n3 < n1
Dielectric slab waveguide
Ray will bounce from 2 interfaces provided TIR occurs at each interface
without exiting the middle regiodn (slab). The lightwave travels in the middle layer
which has a refractive index n1 (n1 > n2 > n3). This neutral layer is sandwiched
between a top and bottom layers with indices n2 and n3, respectively.
n2 = n3 : “symmetric waveguide”
If
n2 ≠ n3 : “asymmetric waveguide”
7
Consider symmetric structure (resembles optic fiber)
n2
d/2

0
n1
y

-d/2
symmetric waveguide (optical fiber)
n2
n 
Light is guided for c < i < /2 where c  sin 1  2 
 n1 
k  k0 n1 
2
0
n1
Propagation factor,   k sin 
h  k cos 
Consider :  

where vg = guided velocity
vg
We define neff = effective refractive index
=
From
c c 


v g  k0
  k sin   k0 n1 sin   k0 neff
neff  n1 sin 
Fields in the waveguide can be expressed as
Inside:
E1 cos(hy )sin(t   z ) ...even mode
E ( y, z , t ) 
E1 sin(hy )sin(t   z )
...odd mode
Outside:
E2 e
 d
  y  
2

sin(t   z )
for y  d / 2
E ( y, z , t ) 


d
  y 
2
E2 e

sin(t   z )
for y  d / 2
where E2 is a magnitude of E at interface
z
8
For reflection from top and bottom interfaces to add up constructively, the
phase accumulation resulting from 2 bounces must be m(2), where m = integer. This
restricts  to discrete selected values and this can be described “modes”.
By applying boundary condition at interfaces (y =  d/2), it yields
n 2 sin   n22
 hd 
tan    1
n1 cos 
 2 
... for even solutions
n12 sin   n22
 hd  
tan 
 
... for odd solutions
n1 cos 
 2 2
where
hd (k0 n1 cos  )d 2 n1 cos  d d


.  . n1 cos 
2
2

2 2
Consider even solution case:
 d n1 cos 
tan 


n12 sin 2   n22



n1 cos 

For a given symmetric waveguide of n1 and n2
1. Choose values of  : c <  < π/2.
2. Calculate value of right-handed side of an equation above.
3. Find value of (d/2) that makes left-handed side of that equation equal to
the right-handed side for chosen .
This yields
- If (d/2) is fixed, there may be several allowed values of . Each  identifies
a mode. Each mode will have its own neff.
- If  is fixed, there may be several (d/) values that satisfies the equation.
Each (d/) identifies a mode and all modes have the same neff.
For TE polarization, modes are expressed as TE0, TE1, TE2, … where m = 0 is
called “fundamental mode”.
9
Symmetric waveguide
Asymmetric waveguide
For a fixed  value, (d/) for mth mode is related to (d/) mode by
m
d  d 
   
     0 2n1 cos
10
How to find number of modes
1. Using a mode chart
2. Finding cutoff modes by using  = c
m
d 
  
  m,c 2 n12  n22
If m is not an interger, it must be rounded down, for example, m = 2.45, then
we round it down to 2. Therefore, total number of modes, N = m+1.
Ex. Symmetric slab waveguide of n1 = 1.48, n2 = 1.46, d = 4 μm, and  = 0.82 μm
(a) Calculate highest allowed mode order, m.
(b) Total number of modes, N.
Soln
d 
m     2 n12  n22

 4 
2
2

  2 1.48  1.46  2.3657
 0.82 
Therefore, the highest mode is m = 2. Total number of modes = m+1 = 3
11
TM Mode
The solutions for TM mode are
n
 hd 
tan    2 1
n12 sin 2   n22
.......even mode
2
n
cos

 
2
n
 hd  
tan 
  2 1
n12 sin 2   n22 .......odd mode
2
2
n
cos



2
The cutoff values for TE and TM modes are the same. Therefore,
m
d 
can be applied to TM mode also.
  
  m,c 2 n12  n22
Ex. For TE mode in symmetric slab waveguide of n1 = 1.48, n2 = 1.46,
(a) find allowed (d/) and corresponding neff values at various s for the case m = 0,
(b) find (d/) for m = 1,2,3 for  = 82
Soln
 d n1 cos
tan 


c     / 2
 n2 
  80.57
 n1 
c  sin 1 
n12 sin 2   n22



n1 cos 

m
d 
d 
    
  m   0 2n1 cos
()
neff =
n1sin
R.H.S
(d/)0
(d/)1
(d/)2
(d/)3
80.57
1.46
0
0
2.0685
4.12
6.18
82
1.4656
0.6212
0.859
3.2865
5.71
8.14
…
…
…
…
…
…
…
88
1.479
4.857
8.575
18.308
27.7
37.4
90
1.48





12
Ex. For slab waveguide n1 = 1.48, n2 = 1.46. If d = 3.164 μm and  = 0.6328
μm. Find (a) number of allowed TE modes (b) neff and  for each allowed mode
Soln
(a) d/ = 3.164/0.628 = 5
From the chart, at (d/) = 5 and by drawing the vertical line, we have the
intercepts and modes. Therefore, total number of modes = 3.
(b) Projecting intercepts to LHS axis neff
Projecting intercepts to RHS axis  
m
neff

0
1.4773
87.8
1
1.4728
84.6
2
1.4632
81.5
13
Numerical Aperature (NA)
NA identifies the largest angle which light can be coupled to the waveguide,
so that rays will be guided as modes in the waveguide.
n0 sin  0  n1 sin 1
Snell’s law:
 n1 sin  90   
 n1 cos 
As we know no TIR for < c (cutoff at = c)
n0 sin  0  n1 cos 
 n1 1  sin 2  c
 n1 1 
n22
n12
n0 sin( 0 ) max  NA  n12  n22
If we define the fractional refractive index change as

If n1  n2, NA  n1 2
n1  n2
n1
14
Dispersion and distortion in the slab waveguide
We have learned that there was a distortion due to the dependence of n to .
There are other 2 additional distortions in waveguides: waveguide dispersion and
multimode distortion. We can conclude for the factors that lead to signal distortion as
1. Material dispersion
n  n  
   M . .l


l
where M 

   M .

 d 2n
c d 2
2. Geometrical confinement
2.1 Waveguide dispersion
The effective refractive index varies with wavelength for a fixed film
thickness. This is called “waveguide dispersion”. This variation in neff causes pulse
spreading just as the variation in n does.


l

   M g 

2
 d neff
where M g 
c d 2
2.2 Multimode dispersion
This can be called “Intermodal dispersion”. Waves with different modes are
propagating the slab with different net velocities. The lower the mode is, the faster the
wave velocity is. This causes distortion since the input energy is distributed among
several modes, each traveling at a different speed. This multimode distortion does not
depend on the  of source.


l
 n1  n1  n2 

cn2
