Cartesian Coordinates
Daniel S. Weile
Department of Electrical and Computer Engineering
University of Delaware
ELEG 648—Cartesian Coordinates
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
Separation of Variables
Away from sources, in Cartesian coordinates, we have
The Helmholtz Equation
∂2ψ ∂2ψ ∂2ψ
+
+
+ k 2ψ = 0
∂x 2
∂y 2
∂z 2
To apply separation of variables, we assume
ψ(x, y , z) = X (x)Y (y )Z (z)
and plug in:
∂2X
∂2Y
∂2Z
YZ
+
X
Z
+
XY
+ k 2 XYZ = 0
∂x 2
∂y 2
∂z 2
D. S. Weile
Cartesian Coordinates
Separation of Variables
We now divide by XYZ :
1 ∂2Y
1 ∂2Z
1 ∂2X
+
+
+ k2 = 0
X ∂x 2
Y ∂y 2
Z ∂z 2
We now make the following key observation:
D. S. Weile
Cartesian Coordinates
Separation of Variables
We now divide by XYZ :
1 ∂2Y
1 ∂2Z
1 ∂2X
+
+
+ k2 = 0
X ∂x 2
Y ∂y 2
Z ∂z 2
We now make the following key observation:
Justification of Separation of Variables
Each term in the equation depends on one variable and not the
others. Since only one term depends on x, it cannot vary with x
since then the other terms would need to vary with x to keep
the equation true. Since this is a contradiction, every term in
the equation must be constant.
D. S. Weile
Cartesian Coordinates
Separation of Variables
We thus have
1 ∂2X
X ∂x 2
1 ∂2Y
Y ∂y 2
1 ∂2Z
Z ∂z 2
= −kx2
= −ky2
= −kz2
subject to the separation equation
kx2 + ky2 + kz2 = k 2
We have also seen that solutions to these equations are of the
form
h (kx x) ∼ cos (kx x) , sin (kx x) , e−jkx x , e+jkx x
D. S. Weile
Cartesian Coordinates
Separation of Variables
A function of the form
ψkx ky kz (x, y , z) = hx (kx x) hy (ky y ) hz (kz z)
(subject to the separation equation) is called an elementary
wave function. In problems with boundaries, the allowed values
of ki are determined by the boundary conditions and are called
eigenvalues. The functions themselves are called
eigenfunctions.
Superpositions of these functions also work:
XX
ψ =
Akx ky hx (kx x) hy (ky y ) hz (kz z)
kx
ky
Z Z
f (kx , ky ) hx (kx x) hy (ky y ) hz (kz z) dkx dky
ψ =
kx ky
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
The General Form
A general Cartesian elementary wave function is of the form
ψ(x, y , z) = e−jkx x e−jky y e−jkz z
A more telling version of this expression defines
k = kx ux + ky uy + kz uz
r = xux + y uy + zuz
so that
General Plane Wave Expression
ψ(r) = e−jk·r
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Cartesian Coordinates
Interpretation of the General Form
In the time domain, the wave becomes
ψ(r) = cos(ωt − k · r)
Thus, the wave is traveling in the k direction with phase velocity
vp =
ω
|k|
This is of course the smallest phase velocity that might be
assigned, and it is along the direction perpendicular to the
equiphase planes.
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Cartesian Coordinates
Complex Propagation Constant
Suppose
k = β − jα.
The strange notation allows us to define
γ = jk = α + jβ.
If the propagation material itself is lossless, k 2 = ω 2 µ must be
real. This implies
k 2 = k · k = β 2 − α2 − j2α · β
is real, and so either α = 0 or
α · β = 0.
What does this mean about the equiphase planes and the
equiamplitude planes?
D. S. Weile
Cartesian Coordinates
Complex Propagation Constant
Suppose
k = β − jα.
The strange notation allows us to define
γ = jk = α + jβ.
If the propagation material itself is lossless, k 2 = ω 2 µ must be
real. This implies
k 2 = k · k = β 2 − α2 − j2α · β
is real, and so either α = 0 or
α · β = 0.
What does this mean about the equiphase planes and the
equiamplitude planes?
What type of wave has nonzero α?
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
The Rectangular Waveguide
Let us consider TM modes in an a × b waveguide along the
z-direction. (We assume a > b, and a is along the x-direction.)
The Setup
A = µψuz
ψ = hx (kx x)hy (ky y )e−jkz z
Now, from our earlier work,
Ez =
1 2
k − kz2 ψ
ŷ
The most important thing to note is that Ez ∝ ψ.
D. S. Weile
Cartesian Coordinates
TM Modes
This immediately tells us that we need
Boundary Conditions
ψ(x = 0, y , z) = ψ(x = a, y , z) = 0
ψ(x, y = 0, z) = ψ(x, y = b, z) = 0
Now
hx (kx x)
=
Ax cos(kx x) + Bx sin(kx x)
hx (0) = 0 ⇒ Ax = 0
mπ
hx (kx a) = 0 ⇒ kx =
for m = 1, 2, . . .
a
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Cartesian Coordinates
The TM solution
Therefore, we have the
TM Wave Potential
ψ(x, y , z) = sin
mπx a
sin
nπy b
e−jkz z
subject to
mπ 2
a
+
nπ 2
b
+ kz2 = k 2
and
m = 1, 2, . . .
n = 1, 2, . . .
From here, we can find all of the fields from our TM formulas.
D. S. Weile
Cartesian Coordinates
TE Modes
Now, let us consider TE modes.
The Setup
F = ψuz
ψ = hx (kx x)hy (ky y )e−jkz z
Now, from our earlier work,
∂ψ
∂y
∂ψ
∂x
Ex
= −
Ey
=
D. S. Weile
Cartesian Coordinates
TE Modes
This immediately tells us that we need
Boundary Conditions
∂ψ(x = 0, y , z)
∂ψ(x = a, y , z)
=
∂x
∂x
∂ψ(x, y = 0, z)
∂ψ(x, y = b, z)
=
∂y
∂y
= 0
= 0
Now
hx (kx x) = Ax cos(kx x) + Bx sin(kx x)
∂hx (kx x)
= −kx Ax sin(kx x) + kx Bx cos(kx x) = kx hx0 (kx x)
∂x
hx0 (0) = 0 ⇒ Bx = 0
mπ
for m = 0, 1, 2, . . .
hx0 (kx a) = 0 ⇒ kx =
a
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Cartesian Coordinates
The TE solution
Therefore, we have the
TE Wave Potential
ψ(x, y , z) = cos
mπx a
cos
nπy b
e−jkz z
subject to
mπ 2
a
+
nπ 2
b
+ kz2 = k 2
and
m = 0, 1, 2, . . .
n = 0, 1, 2, . . .
m = n = 0 excluded
From here, we can find all of the fields from our TE formulas.
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Cartesian Coordinates
Cutoff
From the formula for kz , it is obvious that cutoff wavenumber for
the TEmn or TMmn mode is
The Cutoff Wavenumber
kcmn =
r
mπ 2
a
+
nπ 2
b
In terms of this wave number we have the propagation constant
(
p
jβ = jp k 2 − kc2
for k > kc
γmn = jkz =
2
2
α =
kc − k
for k < kc
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Cartesian Coordinates
Cutoff Frequency and Wavelength
√
From the cutoff k and the formula k = 2πf µ = 2π/λ, we find
the
Cutoff Frequencies
fcmn
1
= √
2 µ
r
m 2
a
+
n 2
b
and
Cutoff Wavelengths
λcmn = q
D. S. Weile
2
m 2
+
a
n 2
b
Cartesian Coordinates
Alternate Expression for γ
Since
fc
kc
=
k
f
we can always write the
Propagation Constant as a Function of Cutoff Frequency

r
2


 jβ = jk 1 − ffc
for f > fc
r
γmn = jkz =
2


 α = kc 1 − ff
for f < fc
c
D. S. Weile
Cartesian Coordinates
TE Impedance
Recall from earlier that for TE modes
Ex = − ∂ψ
∂y
Ey =
Hx =
∂ψ
∂x
Hy =
1 ∂2ψ
jωµ ∂x∂z
1 ∂2ψ
jωµ ∂y ∂z
=
=
−jkz
jωµ
−jkz
jωµ
From this we have
jωµHx
= −jkz ∂ψ
∂x
jωµHy
−jkz ∂ψ
∂y
=
= −jkz Ey
= jkz Ey
This gives the
TE Modal Impedance
TE
Z0mn
=
Ey
Ex
ωµ
=−
=
Hy
Hx
kz
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Cartesian Coordinates
∂ψ
∂x
∂ψ
∂y
TM Impedance
By precisely the same logic, we find the
TM Modal Impedance
TM
Z0mn
=
Ey
kz
Ex
=−
=
Hy
Hx
ω
In the application of these formulas, we must remember that kz
is real above cutoff and imaginary below.
Indeed, these formulas are not particular to rectangular cross
section, as we will see.
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
TM Modes
Let us assume we form a cavity by placing metal walls on our
waveguide at z = 0 and z = c. Because we need to satisfy the
BCs on the other four walls, we can start with our expression
for TM waveguide modes.
nπy mπx sin
[Az cos(kz z) + Bz sin(kz z)]
ψ(x, y , z) = sin
a
b
We need the x- and y -directed electric fields to vanish at
z = 0 and z = c.
Both of these fields are proportional to
D. S. Weile
∂ψ
∂z ,
Cartesian Coordinates
so
TM Modes
Let us assume we form a cavity by placing metal walls on our
waveguide at z = 0 and z = c. Because we need to satisfy the
BCs on the other four walls, we can start with our expression
for TM waveguide modes.
nπy mπx sin
[Az cos(kz z) + Bz sin(kz z)]
ψ(x, y , z) = sin
a
b
We need the x- and y -directed electric fields to vanish at
z = 0 and z = c.
Both of these fields are proportional to
∂ψ
∂z ,
We need hz0 (0) = hz0 (kz c) = 0.
D. S. Weile
Cartesian Coordinates
so
TM Modes
hz0 (0) = 0 ⇒ Bz = 0
pπ
hz0 (kz c) = 0 ⇒ kz =
c
Thus, we have the
Elementary Wave Functions for TM Cavity Modes
Az = µψ
mπx nπy pπz ψ(x, y , z) = sin
sin
cos
a
b
c
subject to
m, n = 1, 2, . . .
p = 0, 1, 2 . . .
D. S. Weile
Cartesian Coordinates
Resonant Frequencies and TE Elementary Wave
Functions
The Resonant Frequencies
1
frmnp = √
2 µ
r
m 2 n 2 p 2
+
+
a
b
c
By similar effort
Elementary Wave Functions for TE Cavity Modes
Fz = ψ
mπx nπy pπz ψ(x, y , z) = cos
cos
sin
a
b
c
subject to
m, n
p
= 0, 1, 2, . . .
= 1, 2 . . .
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
Alternate Mode Sets
The most important way of finding waveguide modes is by
assuming them TE or TM to z.
This characterization holds for non-Cartesian coordinates
as well.
In Cartesian coordinates, other choices are possible and
sometimes useful.
Thus, we can create a set of TMx modes by choosing
A = µψux
or a set of TEx modes with
F = ψux
D. S. Weile
Cartesian Coordinates
TMx Modes
Using the standard formulas for computing fields from
potentials we can write
2
∂
2 ψ
Hx = 0
Ex = ŷ1 ∂x
2 + k
Ey =
Ez =
1 ∂2ψ
ŷ ∂x∂y
1 ∂2ψ
ŷ ∂x∂z
Hy =
Hz =
∂ψ
∂z
− ∂ψ
∂y
From the formula for Ez , it is easy to find the
Wave Potential for TMx Waves in a Rectangular Guide
mπx nπy sin
e−jkz z
ψ(x, y , z) = cos
a
b
D. S. Weile
Cartesian Coordinates
TEx Modes
Similar methods give rise to TEx mode formulas:
2
∂
2 ψ
Ex = 0
Hx = ẑ1 ∂x
2 + k
Hy =
Hz =
1 ∂2ψ
ẑ ∂x∂y
1 ∂2ψ
ẑ ∂x∂z
Ey = − ∂ψ
∂z
Ez =
∂ψ
∂y
From the formula for Ez , it is easy to find the
Wave Potential for TEx Waves in a Rectangular Guide
mπx nπy ψ(x, y , z) = sin
cos
e−jkz z
a
b
D. S. Weile
Cartesian Coordinates
Mode Impedance
Despite the novel mode formulation, we can still compute the
impedance of a mode relative to z.
For TMx modes, we have
mπ 2 1 ∂2
2
2
Ex =
+ k ψ ⇒ jωEx = k −
ψ
ŷ ∂x 2
a
Similarly,
Hy =
∂ψ
= −jkz ψ
∂z
We thus have the
TEx Mode impedance
TE
Z0mn
k 2 − mπ
Ex
a
=
=
Hy
ωkz
2
We can find modal impedance for TMx waves similarly.
D. S. Weile
Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
Partially Filled Waveguide
As an example of the need for alternative modes, consider the
partially filled guide
y
d
b � ,µ
1
1
z
�2 , µ2
a
x
To find the modes in this, we will start with TMx modes.
D. S. Weile
Cartesian Coordinates
TMx Modes in Partially Filled Waveguide
We will expand Ax differently in the two regions, taking into
account the step change in µ. We thus write
nπy −jkz z
e
b
nπy −jkz z
= C2 cos [kx2 (a − x)] sin
e
b
ψ1 = C1 cos kx1 x sin
ψ2
Note:
1
The form of the solution is determined by previous slides.
2
The solutions in the y -direction must be the same since the
boundary conditions have not changed.
3
The solutions in the z-direction must have identical kz to
ensure continuity of fields.
4
The behavior of the solution at x = 0 and x = a is
determined by the metal waveguide wall.
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Cartesian Coordinates
TMx Modes in Partially Filled Waveguide
For these to be legitimate solutions, they must satisfy the usual
separation conditions:
2
kx1
+
nπ 2
+ kz2 = k12 = ω 2 µ1 1
b
nπ 2
2
kx2
+
+ kz2 = k22 = ω 2 µ2 2
b
Now we must enforce continuity at the boundary. This implies
1
Ey 1 (x = d) = Ey 2 (x = d)
2
Ez1 (x = d) = Ez2 (x = d)
3
Hy 1 (x = d) = Hy 2 (x = d)
4
Hz1 (x = d) = Hz2 (x = d)
How many of these conditions do we actually need?
D. S. Weile
Cartesian Coordinates
TMx Modes in Partially Filled Waveguide
For these to be legitimate solutions, they must satisfy the usual
separation conditions:
2
kx1
+
nπ 2
+ kz2 = k12 = ω 2 µ1 1
b
nπ 2
2
kx2
+
+ kz2 = k22 = ω 2 µ2 2
b
Now we must enforce continuity at the boundary. This implies
1
Ey 1 (x = d) = Ey 2 (x = d)
2
Ez1 (x = d) = Ez2 (x = d)
3
Hy 1 (x = d) = Hy 2 (x = d)
4
Hz1 (x = d) = Hz2 (x = d)
How many of these conditions do we actually need? Two. Why?
D. S. Weile
Cartesian Coordinates
Finding the Eigenvalues
From our work with alternative mode sets, we find
nπ
nπy −jkz z
1
C1 kx1
sin kx1 x cos
e
jω1
b
b
1
nπ
nπy −jkz z
C2 kx2
sin [kx1 (a − x)] cos
e
jω2
b
b
Ey 1 = −
Ey 2 =
These must be equal at x = d so we have the
E-Field Continuity Condition
1
1
C1 kx1 sin kx1 d = − C2 kx2 sin [kx2 (a − d)]
1
2
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Cartesian Coordinates
Finding the Eigenvalues
A similar condition can be found from the continuity of Hz :
Hz1 =
Hz2 =
nπ
nπy −jkz z
C1 cos kx1 x cos
e
b
b
nπ
nπy −jkz z
C2 cos [kx2 (a − x)] cos
e
b
b
These must be equal at x = d so we have the
H-Field Continuity Condition
C1 cos kx1 d = C2 cos [kx2 (a − d)]
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Finding the Eigenvalues
Dividing these two equations yields a transcendental equation
for kz :
TMx Eigenvalue Equation
kx2
kx1
tan kx1 d = −
tan [kx2 (a − d)]
1
2
Once kz is found (and hence kx1 and kx2 , either one of the
previous equations may be used to find C1 /C2 . TEx modes can
be found similarly, and lead the the
TEx Eigenvalue Equation
kx1
kx2
cot kx1 d = −
cot [kx2 (a − d)]
µ1
µ2
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Cartesian Coordinates
Partially Filled Waveguide Observations
The resulting kz can be shown to be between those
obtained for homogeneous guides filled with material 1 or
2.
The TEx and TMx modes are no longer degenerate.
Computing the cutoff frequency can be done without
solving transcendental equations by setting kz = 0 and
finding kx1 and kx2 .
Knowledge of the cutoff frequency is no longer sufficient
using the usual formulas to find kz ; a transcendental
equation must be solved at each frequency.
The dominant mode is affected by the orientation of the
inhomogeneity as well as the shape of the guide.
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Cartesian Coordinates
Outline
D. S. Weile
Cartesian Coordinates
The Slab Waveguide
Waves can be guided without metal. (This should be known to
everyone here, especially those studying optics. The simplest
example of a dielectric waveguide is the slab waveguide:
�0 , µ0
a
x
z
�d , µd
This problem is simple primarily because it is two dimensional.
We assume no variation with y .
D. S. Weile
Cartesian Coordinates
Slab Guide Setup
It turns out that we can examine modes either
TEx and TMx , or
TEz and TMz .
We choose the latter. Given the lack of y variation, the TMz
equations are:
Ex
Ez
Hy
kz ∂ψ
ω ∂x
1 2
=
k − kz2 ψ
jω
∂ψ
= −
∂x
= −
Also, because of the symmetry of the problem, the solutions
must either be odd or even functions of x.
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Cartesian Coordinates
Form of the Solution
In the dielectric region we choose
Odd Dielectric Mode
ψdo = A sin ux e−jkz z
for |x| <
a
2
In the air we choose
Odd Air Mode
ψao
= Be−vx e−jkz z
ψao = −Bevx e−jkz z
for x >
for x < − a2
Why are these expression of different forms?
D. S. Weile
a
2
Cartesian Coordinates
Separation Equations
As usual, all of this is subject to the separation equations,
which say the length of the wavevector is the wavenumber.
The guide wavenumber kz must be the same in both
media.
Separation Equations
u 2 + kz2 = kd2 = ω 2 µd 0
−v 2 + kz2 = k02 = ω 2 µa 0
Of course, Ez and Hy must be continuous.
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Cartesian Coordinates
Continuity of Ez
From our equations of TM propagation, we find that
Ez =
Ez =
Ez =
A
2
−jkz z
jωd u sin ux e
−B 2 −vx −jkz z
e
jω0 v e
B
2 vx −jkz z
jω0 v e e
Enforcing continuity at x =
a
2
for |x| <
for x >
a
2
for x < − a2
gives the
E-Field Continuity Condition
va
A 2
ua
B
u sin
= − v 2 e− 2
d
2
0
D. S. Weile
a
2
Cartesian Coordinates
Continuity of Hy
By the same token
Hy = −Au cos ux e−jkz z
Hy =
Bve−v |x| e−jkz z
Enforcing continuity at x =
a
2
for |x| <
for |x| >
gives the
H-Field Continuity Condition
va
ua
= −Bve− 2
Au cos
2
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Cartesian Coordinates
a
2
a
2
The Odd TM Characteristic Equation
Dividing the E-field continuity condition by the H-field condition
gives
The Characteristic Equation
ua
d va
ua
tan
=
2
2
0 2
Coupling this with the
Separation Equations
u 2 + kz2 = kd2 = ω 2 µd 0
−v 2 + kz2 = k02 = ω 2 µa 0
we can find the kz ’s and cutoffs of the odd TM modes.
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Cartesian Coordinates
Even TM Modes
Doing precisely the same for even modes, we can write
ψde = A cos ux e−jkz z
for |x| <
ψae
for |x| >
= Be−v |x| e−jkz z
Enforcing continuity of Ez and Hy gives the
Even TM Characteristic Equation
ua
d va
ua
cot
=
−
2
2
0 2
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Cartesian Coordinates
a
2
a
2
TE Modes
TE modes are completely dual to TM modes. The characteristic
equations are
Characteristic Equations
ua
ua
tan
2
2
ua
ua
−
cot
2
2
=
=
d va
0 2
d va
0 2
(odd)
(even)
The fields are given by
Hx
Hz
Ey
kz ∂ψ
ωµ ∂x
1
=
k 2 − kz2 ψ
jωµ
∂ψ
= −
∂x
= −
D. S. Weile
Cartesian Coordinates
Cutoff in Dielectric Waveguides
Above cutoff, the wave propagates in the z-direction
unattenuated.
In free space, kx is imaginary.
In the dielectric, kx is real.
Below cutoff, kz is complex due to radiation out of the core.
The condition for guidance is
Guidance Condition
k0 < kz < kd
1
After all, if kz < k0 , there will be propagation away from the
guide.
2
If kz > kd , there evanescence everywhere (and the
equations cannot be satisfied).
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Cartesian Coordinates
Cutoff Computation
1
By the previous argument, cutoff occurs when v = 0.
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Cartesian Coordinates
Cutoff Computation
1
By the previous argument, cutoff occurs when v = 0.
2
This implies that kz = k0 .
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Cartesian Coordinates
Cutoff Computation
1
By the previous argument, cutoff occurs when v = 0.
2
This implies that kz = k0 .
3
This in turn implies that u =
D. S. Weile
q
kd2 − k02 .
Cartesian Coordinates
Cutoff Computation
1
By the previous argument, cutoff occurs when v = 0.
2
This implies that kz = k0 .
3
This in turn implies that u =
q
kd2 − k02 .
We thus have the equations
q
a
2
2
kd − k0
tan
= 0
2
q
a
cot
kd2 − k02
= 0
2
(odd)
(even)
for both TE and TM modes.
D. S. Weile
Cartesian Coordinates
Cutoff Computation
The cutoff equations are solved when
Cutoff Wavenumbers
q
a
nπ
kd2 − k02 =
2
2
n = 0, 1, 2, . . .
Consistent with Murphy’s law, odd mode numbers go with even
modes!
Cutoff Frequencies
fc =
n
2a d µd − 0 µ0
√
D. S. Weile
Cartesian Coordinates
Observations
1
The TE0 and TM0 modes have no cutoff regardless of a.
2
If the slab is thin, the guidance is weak in that v → 0.
3
The mode number n is roughly the number of half
wavelengths in the dielectric; this characterization is more
accurate for large refractive indices.
4
If the slab is weakly guiding, the cutoff of higher order
modes will be at very high frequency.
D. S. Weile
Cartesian Coordinates
Graphical Method for Characteristic Equation Solution
Eliminating kz from the separation equations, we have
u 2 + v 2 = kd2 − k02 = ω 2 (d µd − 0 µ0 ) .
Given this, the TE characteristic equations become
r
ua 2
ωa 2
µ0 ua
ua
tan
=
(d µd − 0 µ0 ) −
µd 2
2
2
2
r
ua 2
µd ua
ua
ωa 2
(d µd − 0 µ0 ) −
−
cot
=
µ0 2
2
2
2
We now assume µd = µ0 and let
x=
ua
2
The right hand side is thus a circle of radius
ωa √
r=
d µd − 0 µ0
2
D. S. Weile
Cartesian Coordinates
Graphical Method for Characteristic Equation Solution
4
3.5
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Why bother with graphical methods when we have computers?
D. S. Weile
Cartesian Coordinates