# slab_waveguide by liwenting

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```									Chapter 1

Dielectric Slab Waveguide

We will start off examining the waveguide properties of a slab of dielectric shown in Fig. 1.1.

x
n1
z
d               n2

n1

Figure 1.1: Cross-sectional view of a slab waveguide.

n2 , |x| < d/2
n(x) =                                                            (1.1)
n1 ,    else

1.1       Propagating Ray

We will initial look at the light traveling in the slab as a propagating ray. Even though this is not technically
accurate, it provides some intuitive feel for what is going on. Figure 1.2 shows that if the propagation angle
is greater than the critical angle then the ray will bounce off of the surface and will be con ned to the core
region. Therefore, the propagation is con ned to be

n2
θ1 > θc = sin−1          .                                       (1.2)
n1

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   θa .
sin θa = n2 sin (90 − θ1 )                                       (1.3)

= n2 cos (θ1 )                                           (1.4)

ECEn 562                                            1                                         January 17, 2007

θ1>θc

θ1<θc        θ1=θc                                       core      n2

Figure 1.2: Cross-sectional view of a slab waveguide.

Since   θ1 > θc

sin θa < n2 cos θc                                                   (1.5)

< n2       1 − sin2 θc                                     (1.6)

2
n1
< n2       1−                                              (1.7)
n2
n2 − n2
2    1
< n2                                                       (1.8)
n2
2

sin θa <          n2 − n2 ≡ N A
2    1                                            (1.9)

n=1.0
n1

θ1=θc
θa          90−θc                               n2

n1

Figure 1.3: Numerical aperture of an slab waveguide.

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. 1.4. These allowable propagation
angles are called the modes of the waveguide.

In this ray optics analysis the The particular modes of a waveguide can be characterized by their propagation
angle. The mode can be thought of as a plane wave that is either traveling upwards or downwards in the
waveguide. The resulting plane waves are given by

E(x, z) = E o e−jko n(± cos θ1 x+sin θ1 z) .                             (1.10)

The mode is essentially a standing wave pattern in the    x-direction and a traveling wave in the z -direction as
given by

E(x, z) = E m (x) exp (j (ωt − βz)) ,                                    (1.11)

ECEn 562                                             2                                         January 17, 2007
n1

λ

n2 d
θθ

n1

Figure 1.4: The rays must remain in phase after multiple re ections.

where   β   is called the propagation constant and is given by

β = ko n2 sin θ1 .                                       (1.12)

Since the propagation angle is in the range given by

θc < θ1 < 90◦ ,                                         (1.13)

the propagation angle is in the range given by

ko n2 sin θc < β < ko n2 sin (90◦ )                              (1.14)
n1
ko n2     < β < ko n2                              (1.15)
n2
ko n1 < β < ko n2                              (1.16)

If you divide the propagation angle by the free-space wavevector you get the effective index of the mode as
given by

β
nef f ≡      .                                        (1.17)
ko

n1 < nef f < n2                                          (1.18)

1.2      Wave Equation

Now that we have a qualitative understanding of waveguide modes, we want to calculate the exact values
of the supported mode, which we will characterize by the propagation constant       βm and the transverse mode
eld   Em (x).

× E = −jωB = −jωµH                                      ·D =     · E = ρv          (1.19)

× H = jωD + J = jω E + J                                ·B =     · µH = 0          (1.20)

ECEn 562                                              3                                       January 17, 2007
First, we rewrite Ampere's Law for the case of no sources resulting in

× H = jω E                                             (1.21)

Likewise, if we have no free charges     ρv = 0 and thus          ·D =0
If we take the curl of Faraday's law:

×     × E = −jωµ               ×H                                        (1.22)
2
= −jωµ(jω E) = ω µ E                                      (1.23)

There is a vector identity
2
×       ×E =      (     · E) −        E                            (1.24)

so that
2
(   · E) −        E = ω2µ E                                    (1.25)

From Gauss' law we get        · D = 0 since ρv = 0.       Since    D= E     we get     ·   E = 0. If is independent
of position then we can pull it outside of the spatial derivatives resulting in            · E = 0 and thus

· E = 0.                                              (1.26)

Plugging Eq. 1.26 into Eq. 1.25 and rearranging results in the Homogeneous Wave Equation given by

2
E + ω2µ E = 0                                          (1.27)

1.3       Dielectric Slab Waveguide

Since the waveguide is homogeneous along the z axis, solutions to the wave equation can be taken as

E(x, t) = E m (x) exp (j (ωt − βz))                                       (1.28)

H(x, t) = H m (x) exp (j (ωt − βz)) .                                      (1.29)

In time harmonic form the      eld equations become

E(x, t) = E m (x) exp (−jβz)                                         (1.30)

H(x, t) = H m (x) exp (−jβz) .                                          (1.31)

Plugging the general     eld solutions into the wave equation (Eq. 1.27) results in

∂2       ∂2      2
E + 2 E + ko n2 E = 0
i                                                  (1.32)
∂x2     ∂z
∂2
E + (−jβ)2 + ko n2 E = 0
2
i                                                  (1.33)
∂x2
∂2         2
E + ko n2 − β 2 E = 0
i                                                        (1.34)
∂x2
where     ni is either n1 or n2 depending on which region we are de         ning the   eld in.

ECEn 562                                                 4                                          January 17, 2007
The portion in parenthesis is a constant in terms of          x.   The differential equation is a constant coef cient
equation.

For the   elds in the core region (|x|   < d/2) ni = n2 and the solution is given by

Em = Aejhx + Be−jhx ,                                                   (1.35)

or

Em = A sin(hx) + Bcos(hx),                                                (1.36)

where

h=       ko n2 − β 2
2
2                                                     (1.37)

For the   elds in the cladding region (|x|     > d/2) ni = n1 and the solution is given by

Em = Aejgx + Be−jgx ,                                                   (1.38)

where

g=      ko n2 − β 2 .
2
1                                                      (1.39)

However, since   β > ko n1 the argument of the square root is actually negative resulting in

Em = Aeqx + Be−qx ,                                                   (1.40)

where

q=      β 2 − ko n 2 .
2
1                                               (1.41)

The total electric   eld of the mode is given by

                           d
 A sin hx + B cos hx |x| < 2
d
Em (x) =       C exp(−qx)       x> 2                                                 (1.42)

D exp(qx)      x < −d2

The unknowns are     A, B , C , D, q ,   and   h.   The solution of the unknows requires applying the boundary
conditions.   Since the boundary conditions depend on the vector quantities, we will break up the mode
into two orthogonal polarization cases. The directions of both the electric and magnetic                 elds need to be
perpendicular to the rays shown in Fig. 1.4.

One possible solution is to have the electric         eld in the   ˆ
y -direction.   In this case the electric   eld is per-
pedicular to the direction of power        ow (z -direction). This case is called Transverse Electric (TE). For
TE-polarization the magnetic     eld has both       x and z components.
The other case is when the magnetic            eld is in the   ˆ
y -direction.   In this case the magnetic       eld is per-
pedicular to the direction of power       ow (z -direction). This case is called Transverse Magnetic (TM). For
TM-polarization the magnetic      eld has both      x and z

ECEn 562                                                 5                                             January 17, 2007
1.3.1     TE Modes

The electric     eld for TE polarization is in the y-direction as given by
                                   d
 (A sin hx + B cos hx) e−jβz |x| < 2
Ey (x) =        C exp(−qx − jβz)        x> d .
2                                        (1.43)

D exp(qx − jβz)        x < −d 2

The magnetic       eld is

×E
H=                                                             (1.44)
−jωµ
resulting in

j ∂Ey
Hz (x) =             .                                            (1.45)
ωµ ∂x

The boundary conditions are that the tangential components of both           E   and   H   are equal across a boundary.

The tangential component of the electric        eld at   x = d/2 is given by
1                     1               1
A sin      hd + B cos            hd   = C exp − qd                                 (1.46)
2                     2               2
and at   x = −d/2 it is given by
1                     1               1
−A sin      hd + B cos            hd   = D exp − qd                                 (1.47)
2                     2               2

The continuity of the tangential components of the magnetic magnetic              eld essentially becomes continuity
of the derivative of the electric     eld across the boundary resulting in

1                      1                 1
hA cos      hd − hB sin            hd   = −qC exp − qd                               (1.48)
2                      2                 2
at   x = d/2 and
1                      1                1
hA cos      hd + hB sin            hd   = qD exp − qd                                (1.49)
2                      2                2
at   x = −d/2.
These four equations can be combined to produce

1                  1
2A sin      = (C − D) exp − qd
hd                                                              (1.50)
2                  2
1                        1
2hA cos     hd = −q (C − D) exp − qd                                              (1.51)
2                        2
1                    1
2B cos      hd = (C + D) exp − qd                                             (1.52)
2                    2
1                     1
2hB sin      hd = q (C + D) exp − qd                                            (1.53)
2                     2

ECEn 562                                                 6                                            January 17, 2007
The solutions of the TE modes may be divided into two classes:

(a) Symmetric (A      = 0 and C = D):
1
h tan          hd       =q                                     (1.54)
2

(b) Antisymmetric (B       = 0 and C = −D):
1
h cot         hd    = −q                                       (1.55)
2

There are now four unknowns (A or               B , C , h, and q ). The rst term (A or B ) can be thought of as the
amplitude of the mode. Let call this term           Eo . The last two terms (h and q ) are both related to β so they are
actually only one unknown. Let's combine these two together as given by

h2 + q 2 =             2                   2
ko n2 − β 2 + β 2 − ko n2 − β 2
2                   1                                  (1.56)

=       2
ko n2
2    −   ko n2
2
1                                          (1.57)

and   C   is just the continutity of the electric       eld at the boundary. Putting all of this together we get

                                                    d
 E1 e−qx−jβz
                                               x>   2


       sin hx
E0             e−jβz                          |x| ≤ d
Ey =         cos hx                                                                (1.58)



    −
         E1 e+qx−jβz                           x < −d
2
+
where

qd                         sin hd
2
E1 exp −                  = Eo                                                   (1.59)
2                         cos hd
2
qd      sin hd
2
E1 = Eo exp                                 .                  (1.60)
2      cos hd
2

So now the only unknown is        β.   We determine         β   by solving these two equations

2
h2 + q 2 = ko n2 − n2
2    1                                    (1.61)

hd                          hd
h tan               = q OR − h cot         =q                                    (1.62)
2                           2
We can solve these nonlinear transcendental equations using a nonlinear solver on a computer or calculator.
However, they can also be solved graphically to calculate the number of modes and estimate the approximate
solutions.

Since the argument of the      tan and cot is in terms of hd/2 we will plot the term qd/2 along the x-axis and
hd/2 along the y -axis.     The   rst equations becomes

2               2
hd              qd                 1
+                   =        (ko n2 d)2 − (ko n1 d)2                   (1.63)
2               2                 22
π 2 2
=       d     n2 − n2 ≡ V 2
1                             (1.64)
λ

ECEn 562                                                        7                                      January 17, 2007
This is the equation of a circle with a radius of V as given by     x2 + y 2 = V 2 .
The boundary condition equation for the symmetric modes is

hd
h tan         =q                                          (1.65)
2
hd           hd     qd
tan            =                                            (1.66)
2            2      2
which becomes

x tan (x) = y.                                             (1.67)

and for the antisymmetric modes it is

hd
−h cot            =q                                       (1.68)
2
hd           hd     qd
−      cot            =                                         (1.69)
2            2      2
which becomes

−x cot (x) = y.                                             (1.70)

In summary the equations are

2
h2 + q 2 = ko n2 − n2
2   1             ⇒ x2 + y 2 = V 2                          (1.71)

hd
h tan       =q             ⇒ x tan (x) = y                           (1.72)
2
hd
−h cot        =q             ⇒ −x cot (x) = y                          (1.73)
2

The zero crossing of the   tan are 0, π, ...mπ and the zeros of the cot are π , π , 3π , ... π (1 + 2m).
2 2 2            2

1.3.2   TM Modes

We can repeat the whole process for TM modes. In this case, we have

Hy (x, z, t) = hm (x) exp (j (ωt − βz))                                  (1.74)

j ∂
Ex (x, z, t) =       Hy                                  (1.75)
ωµ ∂z
j ∂
Ez (x, z, t) = −       Hy                                  (1.76)
ωµ ∂x
and
                           d
 A sin hx + B cos hx |x| < 2
Hm (x) =       C exp(−qx)       x> d2                                        (1.77)

D exp(qx)      x < −d 2

ECEn 562                                                8                                        January 17, 2007
The eigen equations become

1         n2
2
h tan          hd       = q                                             (1.78)
2         n2
1
1              n2
h cot      hd        = − 2q                                              (1.79)
2              n2
1

1.3.3       Parameter Meanings

What are the physical meanings of            h, q , and β ? If we look back at the ray optics treatment, then β is the
z -component      of the wave,   h   is   the x-component, and q speci es the rate at which the eld decays with
distance away from the core.

β ≡ kz                                                          (1.80)

h ≡ kx                                                          (1.81)

q≡α                                                            (1.82)

Dielectric Waveguide Example

How many modes exist in a dielectric waveguide that has the following parameters? index of refraction of
the coren1 = 1.6, index of refraction of the cladding n2 = 1.5,                           wavelength   λ = 1.0µm,   waveguide core
thickness 2d = 10µm.

The equations are

αd = ky d tan (ky d)                                           (1.83)

αd = −ky d cot (ky d)                                           (1.84)
2            2
(ky d) + (αd) = (ko d)                 n2
1   −   n2
2                                (1.85)

Using      ky d = x and αd = y these equations become

y = x tan x                                         (1.86)

y = −x cot x                                           (1.87)
2      2                 2
x + y = (ko d)                n2
1   −   n2
2                                     (1.88)

For this example the radius of the circle is given by

2π 10
r =        n2 − n2
1    2                                                      (1.89)
1.0 2
r = 17.5µm                                                              (1.90)

π 3π 5π
The equation     x tan x is equal to zero when x = 0π , 2π , 3π , ... mπ and is equal to ∞ when x =                      2, 2 , 2 ,
π
...
2   + mπ .
π 3π 5π         π
The equation     −x cot x is equal to zero when x =              2 , 2 , 2 , ... 2        + mπ and is equal to ∞ when x = π , 2π ,
3π , ...   mπ . And when x = 0 −x cot x = −1.

ECEn 562                                                         9                                               January 17, 2007
The radius of the circle for this problem is r = 17.5 = 5.56π .    There are 6 even modes (0,   π , 2π , 3π , 4π , 5π
) and 6 odd modes (0.5π , 1.5π , 2.5π , 3.5π , 4.5π , 5.5π ).

What is the waveguide thickness for single mode operation? We need

r < 0.5π                                          (1.91)
2π                     π
d    1.62 − 1.52 <                                               (1.92)
1.0                    2
d < 0.449                                         (1.93)

1.4     Asymmtric Slab Waveguides

In practice most slab waveguides are asymmetric. An asymmetric slab waveguide is given by

 n1 ,   x<0
n(x) =   n2 , −t < x < 0                                                (1.94)

n3 ,  x < −t
Sometimes rather than using   n1 , n2 , and n3 these indices are labeled as cover index nc , waveguide index nw ,
and substrate index ns . If we assume that n1 < n3 < n2 then the range for β is given by ko n3 < β < ko n2 .
The process used to calculate the mode     eld pro le is similar to the process describe above except that the
boundary conditions will be different at the top and bottom boundary.

For a TE mode the electric    eld is given by

Ey (x, z, t) = Em (x)ej(ωt−βz) ,                                        (1.95)

where the mode pro le is given by

 C exp −qx                                x>0
q
Em (x) =   C cos(hx) − h sin(hx)               −t < x < 0 ,                                 (1.96)
             q
C cos(ht) + h sin(ht) exp[p(x + t)]     x < −t
where

h=     2
k2 − β 2                                               (1.97)

q=           2
β 2 − k1                                               (1.98)

p=           2
β 2 − k3 .                                             (1.99)

The mode condition equation is given by

q
h sin(ht) − q cos(ht) = p cos(ht) +          sin(ht)                             (1.100)
h

For a TM mode the     elds are given by

Hy (x, z, t) = Hm (x)ej(ωt−βz)                                 (1.101)

i ∂Hy       β
Ex (x, z, t) =          =      hm (x)ej(ωt−βz)                                (1.102)
ωµ ∂z        ωµ
j ∂Hy
Ez (x, z, t) = −                                         (1.103)
ω ∂x

ECEn 562                                             10                                         January 17, 2007
where the mode pro le is given by

 −C h exp(−qx)
    q
¯                                      x>0

h
Hm (x) =   C − q cos(hx) + sin(hx)
¯                               −t < x < 0 ,                                     (1.104)


 −C h cos(ht) + sin(ht) exp[p(x + t)]     x < −t
¯
q

where

n2
2
¯
q≡   q                                                         (1.105)
n2
1
n2
2
p ≡ 2p
¯                                                               (1.106)
n3
The mode condition equation is given by

p ¯
h(¯ + q )
tan(ht) =       2 − pq
(1.107)
h    ¯¯

1.5       Effective Index Theory

A slab waveguide only con nes light in one dimension. In practive it is necessary to con ne light in both di-
rections. Exact analytic treatment of rectangular dielectric waveguides is not possible for arbitrary structures.
These type of waveguides can be analyzed using numerical techniques. There are also several approximate
analytical approaches. One of the simplest approaches is the effective index theory.

Figure 1.5 shows a ridge waveguide. The three regions of the ridge waveguide (I, II, I) are treated as slab
waveguides resulting in three different effective indices (nef f,I ,    nef f,II , and nef f,I ).   Referring to Fig. 1.5
nef f,I   is calculated by solving for the mode of a slab waveguide with a thickness of             d and for nef f,II   the
waveguide thickness is    t.   The ridge waveguide effective index is then calculated by treating the effective
indices as the cover, waveguide, and substrate indices with the waveguide thickness being the ridge width                 a.

n1
t
n2                                              d

a
n3
I                  II               I
y=-a/2               y=a/2

Figure 1.5: Rectangular waveguide.

Example: Consider a ridge waveguide made of GaAs (n        = 3.5) waveguiding layer on an AlGaAs (n = 3.2)
substrate. The thicknesses are    t = 0.4λ, d = 0.25λ, and a = 0.5λ.

ECEn 562                                             11                                                 January 17, 2007

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