Whites, EE 692CEM
Lecture xx
Page 1 of 7
Lecture xx: MM Solution for a Microstrip
Using Pulse Expansion-Point Match
The second example we will consider is the quasi-static moment
method solution of a microstrip. For simplicity, we will assume
that the entire upper half space is filled with the same material.
y
ε
w
d
x
Conducting plane
We’ll imagine that a time-harmonic voltage source has been
connected between the strip and the ground plane. This will
cause a non-uniform charge distribution on the strip as well as
the ground plane.
y
+
ρl [C/m]
+ + + + ++
ε
V
-
-
- - - - - - - - - - -
-
-
-
x
Conducting plane
Provided the height and width of the microstrip are electrically
very small, we can expect that a quasi-static solution will be
quite accurate.
© 2004 Keith W. Whites
Whites, EE 481/692
Lecture xx
Page 2 of 7
Quasi-static means that the time harmonic fields are simply
those of the static solution, but are oscillating sinusoidally in
time. In other words, there are no radiation fields present so all
electric field lines begin and end on charges.
Integral Equation
Next, we’ll employ the image theory method to create an
equivalent problem for the upper half space ( y ≥ 0 ).
y
ρl [C/m]
ε
+ + + + ++
+
V
-
d
(x,y) O.b. point
r′
r
x
-
ε
-V
+
d
Φe = 0 satisfied naturally on
this plane.
-- - -
--
−ρl [C/m]
You’ve learned that the potential at a point r in a homogeneous
space produced by a line charge density ρl ( r ′ ) is given by
⎛ 1 ⎞
1
ρl ( r ′ ) ln ⎜
Φe ( r ) =
⎟ dl ′
∫
′
2πε C ′
⎝ r −r ⎠
(1)
1
2
2
′
′
′
ρ
r
x
x
y
y
dl ′
ln
=−
−
+
−
(
)
(
)
(
)
l
2πε C∫′
Note that we must integrate along all contours C ′ that have
surface charge. Hence C ′ includes both the strip and its image.
(
)
Whites, EE 481/692
Lecture xx
Page 3 of 7
Now, we will apply the boundary condition that on the top strip
Φe ( r ) = V
∀r ∈ {upper strip}
(2)
Using (2) in (1) and accounting for both the + ρl and − ρl charge
distributions on the strips then
1 ⎧⎪
2
2
′
′
−
+
−
ln
V =−
ρ
r
x
x
d
d
dl ′ +
(
)
(
)
(
)
⎨∫ l
2πε ⎪⎩ top
(
∫
− ρl ( r ′ ) ln
bottom
)
(
or
w
)
⎫
2
2
′
′
( x − x ) + ( d + d ) dl ⎬
⎭
⎡ ln x − x′ − ln
′
ρ
r
(
)
)
∫ l ⎢ (
(
)
+ 4d 2 ⎤⎥ dx′ (3)
2πε 0
⎣
⎦
This is the integral equation for the line charge density on the
strip.
V =−
1
( x − x′ )
2
(Another approach to deriving this integral equation for the
charge density would be to first find the Green’s function and
then convolve it with the forcing function.)
Pulse Expansion-Point Match
For a simple MM solution, we will use a pulse expansion for the
charge density
N
ρl ( r ′ ) ≈ ∑ α n Pn ( x′; xn −1 , xn )
n =1
(4)
Whites, EE 481/692
Lecture xx
Page 4 of 7
and point match the discretized integral equation.
Substituting (4) into (3) gives
w
1 ⎡N
⎤
′
′
′
;
,
V =−
α
P
x
x
x
∑
n n(
n −1
n ) ⎥ G ( x − x ) dx
⎢
∫
2πε 0 ⎣ n =1
⎦
where
G ( x − x′ ) = ln ( x − x′ ) − ln
(
( x − x′ )
2
+ 4d 2
)
(5)
(6)
In (5), we can interchange the order of integration and
summation, since these are linear operators, except perhaps
when x = x′ . Assuming this isn’t the case, then (5) becomes
xn
1 N
V =−
α n ∫ G ( x − x′ ) dx′
(7)
∑
2πε n =1 xn−1
We’ll generate N linearly independent, constant coefficient
equations by point matching (7) at the centroid of every segment
in the discretized strip model. We’ll denote these points as
ˆ m + yd
ˆ .
rm = xx
Point matching (7) at these points rm gives
xn
1 N
V =−
α n ∫ G ( xm − x′ ) dx′
∑
2πε n =1 xn−1
which is a matrix equation of the form
[Vm ] = [ Z mn ] ⋅ [α n ]
N ×1
where
N ×N
Vm = V
αn = αn
m = 1,…, N
(8)
(9)
N ×1
(10a)
(10b)
Whites, EE 481/692
and
Lecture xx
Z mn = −
xn
1
2πε
∫ G( x
m
− x′ ) dx′
Page 5 of 7
m≠n
(10c)
xn −1
As before, the process for obtaining a numerical solution from
(9) is to “fill” [V ] and [ Z ] , then solve this system of equations
for the line charge density coefficients [α ] . In particular,
• For [V ] – choose V = 1 V, for example.
• For [ Z ] – compute (10c) analytically, if possible, or use
numerical integration.
In this particular problem, we are able to evaluate (10c)
analytically since a simple anti-derivative of the integrand is
available.
Analytical Evaluation of Zmn
To accomplish this analytical evaluation of (10c), we will begin
with a segment of width ∆ located at the origin that is supporting
a uniform line charge density ρl :
y
ρl
(x,y) O.b. point
r
x
-∆/2
∆/2
The electrostatic potential at point r produced by this “pulse” of
line charge density is
Whites, EE 481/692
Lecture xx
∆ 2
ρ
Φ e ( r ) = − l ∫ ln ⎡
2πε −∆ 2 ⎢⎣
Page 6 of 7
( x − x′ )
2
+ y 2 ⎤ dx′
⎥⎦
(11)
When r does not lie anywhere on this strip, the potential is
Φe ( r ) = −
ρl
2πε
{( x + ∆ 2) ln ⎡⎣( x + ∆ 2) + y ⎤⎦
2
2
− ( x − ∆ 2 ) ln ⎡( x − ∆ 2 ) + y 2 ⎤ − 2∆
⎣
⎦
2
(12)
⎡ −1 ⎛ x + ∆ 2 ⎞
⎤⎫
−1 ⎛ x − ∆ 2 ⎞ ⎪
+2 y ⎢ tan ⎜
− tan ⎜
⎟
⎟⎥ ⎬
y
y
⎝
⎠
⎝
⎠ ⎦ ⎪⎭
⎣
Using (12) in (10c), it can be shown that for m ≠ n
Z mn
⎡ ( ∆ mn + ∆ 2 )2 ⎤
1 ⎪⎧
=−
⎥
⎨( ∆ mn + ∆ 2 ) ln ⎢
2
2
4πε ⎪
⎢⎣ ( ∆ mn + ∆ 2 ) + 4d ⎥⎦
⎩
⎡ ( ∆ mn − ∆ 2 )2 ⎤
− ( ∆ mn − ∆ 2 ) ln ⎢
⎥
2
2
⎢⎣ ( ∆ mn − ∆ 2 ) + 4d ⎥⎦
(13)
⎡
⎛ ∆ +∆ 2⎞
−1 ⎛ ∆ mn − ∆ 2 ⎞ ⎤ ⎫
−4d ⎢ tan −1 ⎜ mn
−
tan
⎟
⎜
⎟⎥ ⎬
2d
2d
⎝
⎠
⎝
⎠⎦ ⎭
⎣
where
∆ mn ≡ xm − xn
(14)
Whites, EE 481/692
Lecture xx
Page 7 of 7
Evaluation of Zmm
A more nettlesome situation occurs if the observation point lies
on the segment. This will happen every time m = n while the
matrix [Z] is being filled.
For these “self-cell” evaluations, the observation point will be
located at the center of the segment in the chosen pointmatching scheme
Referring to the figure above, if x = y = 0 (at the center of the
strip) it can be shown that
Φe ( 0) =
ρl
⎡⎣1 − ln ( ∆ 2 ) ⎤⎦
2πε
(15)
Consequently, from this result it can be shown that for m = n
Z mm =
∆
⎡⎣1 − ln ( ∆ 2 ) ⎤⎦
2πε
1 ⎧
2
2⎤
−1 ⎛ ∆ ⎞ ⎫
⎡
+
∆
∆
+
−
∆
+
ln
2
4
2
8
tan
d
d
(
)
⎨
⎜
⎟⎬
⎣
⎦
4πε ⎩
⎝ 4d ⎠ ⎭
(16)