# Lecture Notes series On Antennas L20 MICROSTRIP ANTENNAS – PART II

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```					LECTURE 21: MICROSTRIP ANTENNAS – PART II
(Transmission-line model. Design procedure for a rectangular patch. Cavity
model for a rectangular patch.)

1. Transmission line model – the rectangular patch
The TL model is the simplest of all, representing the rectangular patch as
a parallel-plate transmission line connecting two radiating slots (apertures),
each of width W and height h:

In the figure above, z is the direction of propagation of the transmission line.
The TL model is not accurate and lacks versatility. However, it gives a
relatively good physical insight into the nature of the patch antenna, and the
field distribution for all TM00 n modes.
The slots represent very high-impedance terminations from both sides of
the transmission line (almost an open circuit). Thus, we expect this structure
to have highly resonant characteristics depending crucially on its length along
z, L. The resonant length of the patch, however, is not exactly equal to the
physical length due to the fringing effect. The fringing effect makes the
effective electrical length of the patch longer than its physical length,
Leff > L . Thus, the resonance condition β ( n ) Leff = n⋅π / 2 , n=1, 2,…, depends

1
on Leff , not L. A sketch of the E-field distribution for the first (dominant)
resonant mode, n =1, is shown in the figure below.
Leff > L

x                            L

patch (side view)

h                              E001                                       z
ground

A. Computing the effective patch length

∆L         (ε   reff   )   ⎛W        ⎞
+ 0.3 ⎜ + 0.264 ⎟
⎝ h       ⎠.
= 0.412                                         (21.1)
h
(ε   reff       ) ⎛W      ⎞
− 0.258 ⎜ + 0.8 ⎟
⎝ h     ⎠
For the computation of ε reff , see Lecture #20.
Leff = L + 2∆L                         (21.2)

∆L                  L          ∆L

2
B. Resonant frequency of the dominant TM001 mode

λ0          v                 c                          c
Leff =        =               =                 ⇒ f r(001) =                   (21.3)
2        2 f (001)       2 ε r f (001)              2 Leff ε r

The resonant frequency of a patch depends strongly on L, therefore, the
exact calculation of Leff is necessary to predict the antenna resonance:
c
f r(001) =                                   (21.4)
2 ε reff ( L + 2∆L )
The field of the TM001 mode does not depend on the x and y coordinates but it
strongly depends on the z coordinate, along which a standing wave is formed.
The figure below shows the field distribution along z when the patch is in
resonance.

⎛ π ⎞
Ex ∼ cos ⎜
⎜ Leff ⎟
z
⎟
⎝       ⎠

z

z = Leff / 2                         z = Leff

3
C. The patch width W
1           2     c       2
W=                         =                           (21.5)
2 f r µ 0ε 0 ε r + 1 2 fr ε r + 1
Expression (21.5) makes the width W equal to half-wavelength resonant
dimensions. Thus, the patch can be viewed as a continuous planar source
consisting of infinite number of infinitesimally thin half-wavelength dipoles.

D. Equivalent circuit of the patch
The dominant TM001 mode has a uniform field distribution along the y-
axis at the slots formed at the front and end edges of the patch. The
equivalent conductance and susceptance can be obtained from the theory of
uniform apertures (see chapter “Aperture Antennas” in any antenna
textbook).
W ⎡        1 ⎛ 2π h ⎞ ⎤
2
h   1
G=           ⎢1 − ⎜        ⎟   ⎥ , for      < ,        (21.6)
120λ0 ⎢ 24 ⎝ λ0 ⎠ ⎥                 λ0 10
⎣                 ⎦
W ⎡                ⎛ 2π h ⎞ ⎤
2
h   1
B=           ⎢1 − 0.636ln ⎜       ⎟   ⎥ , for   < .       (21.7)
120λ0 ⎢              ⎝  λ0 ⎠ ⎥           λ0 10
⎣                        ⎦
The limitation (h / λ0 ) < 0.1 is necessary since a uniform field distribution
along the x-axis is also assumed. The patch has two radiating slots (see figure
below).

L
x
1
t#
2
t#

z    slo
slo

y                                                     W
E (001)

4
The equivalent circuit of a slot is constructed as a parallel R-C circuit, using
the values computed by (21.6) and (21.7):

B              G

G =1/ R represents the radiation losses, while B = jωC is the equivalent
susceptance, which represents the capacitive nature of the slot.
More accurate values for the conductance G can be obtained through the
cavity model:
I
G=          ,                       (21.8)
120π 2

where
2
⎡ ⎛ k0W           ⎞⎤
π⎢  sin ⎜     cos θ ⎟ ⎥
I = ∫⎢ ⎝                ⎠ ⎥ sin 3 θ dθ = −2 + cos X + X ⋅ S ( X ) + sin X , (21.9)
2
cosθ
i
0 ⎢                   ⎥                                           X
⎢
⎣                   ⎥
⎦
x
sin y
and X = k0W, k0 =ω µ0ε 0 . Si denotes the sine integral, Si ( x) = ∫           dy .
0
y
The equivalent circuit representing the whole patch in the TM001 mode
includes the two radiating slots as parallel R-C circuits and the patch
connecting them as a transmission line whose characteristics are computed in
the same way as those of a microstrip transmission line.

B            G                              G     B
Z c , β g =ω µ0ε 0ε reff
Leff ≈ λg / 2
Here, Z c is the characteristic impedance of the line, and β g is its phase
constant. When losses are not neglected, we must include also the attenuation

5
constant α (see Lecture 20). For each slot, G represents the radiation loss
and B = ωC represents the capacitance associated with the fringe effect.

E. Resonant input resistance
When the patch is resonant, the susceptances of both slots cancel out at
the feed point regardless of the position of the feed along the patch. Thus, the
input admittance is always purely real. This real value, however, strongly
depends on the feed position along z. This is easily shown through the Smith
chart for the admittance transformation through a transmission line.
At the feed point, the impedance of each slot is transformed by the
respective transmission line representing a portion of the patch:
Leff ≈ λg / 2

B              G                        G         B
L1           L2

Yin        Y1 Y2

Yin = Y1 + Y2                           (21.10)
The admittance transformation is given by
YL + jY0 tan( β g L)
Yin =Yc                      =YL β L=π , Yc = Z c−1         (21.11)
Yc + jYL tan( β g L)      g

if the line is loss-free. Below, the Smith charts illustrate the slot-impedance
transformations and their addition, which produces a real normalized
admittance, in three cases: (1) the patch is fed at one edge ( L1 = 0 , L2 = L ),
(2) the patch is fed at the center ( L1 = L2 = L / 2 ), and (3) the patch is fed at a
distance (feed inset) z0 = 0.165λ .

6
L2 = L ≈ 0.46λ

Y2′         slot #1 Y
1

slot #2
Y2
line to slot #2

L ≈ 0.46λ

Yin =Y1 +Y2′ = 2×0.09 = 0.18

feed-point at the edge of cavity

7
slot #1 Y1

slot #2 Y
2

L ≈ 0.46λ

L1 = L2 = L / 2 ≈ 0.23λ
Yin =Y1′+Y2′ = 2×15 = 30

Y2′   Y1′

feed-point at the middle of cavity

8
slot #1 Y1

slot #2 Y
2

L ≈ 0.46λ

Yin =Y1′+Y2′ = 2×0.5 =1

line to slot #2                              line to slot #1
Y2′
Y1′

L2 = L − z0 ≈ (0.46 − 0.165)λ = 0.315λ                                                L1 = z0 ≈ 0.165λ
feed-point at inset z 0

9
The edge feed and the inset feed are illustrated below.

slot #1                     slot #2

z0

L
Yin                                             Yin
The two slots are separated by an electrical distance of 180 . However,
because of the fringe effect the physical length L of slightly less than λ/2.
The reduction of the length is not much. Typically, it is 0.48λ ≤ L ≤ 0.49λ.
Ideally, the resonant input impedance of the patch for the dominant TM001
mode is entirely resistive and equal to half the transformed resistance of each
slot:
1    1
Zin =    =      = Rin .                  (21.12)
Yin 2G1 ′

In reality, there is some mutual influence between the two slots, described by
a mutual conductance and it should be included for more accurate
calculations:
1
Rin =              ,                   (21.13)
′
2(G1 ± G12 )
where the “+” sign relates to the odd modes, while the “–“ sign relates to the
even modes. Normally, G12 G1 .    ′
For most patch antennas fed at the edge, Rin is greater than the
characteristic impedance Zc of the microstrip feed line (typically Zc = 50 to
75 Ω). That is why, the inset-feed technique is widely used to achieve
impedance match.
10
The figure below illustrates the normalized input impedance of a 1-D (along
the y axis) loss-free open-ended transmission-line, whose behavior is very
close to that of the dominant mode of the patch.

Fig 14.14, pp 735, C. Balanis

Using modal expansion, the input resistance for the inset-feed at z = z0 is
given approximately by
1        ⎡ 2 ⎛ π ⎞ G12 + B12              ⎛π ⎞ B               ⎛ 2π ⎞ ⎤
Rin =                ⎢ cos ⎜ z0 ⎟ +             sin 2 ⎜ z0 ⎟ − 1 sin ⎜         z0 ⎟ ⎥ . (21.14)
2 ( G1 ± G12 ) ⎣       ⎝L ⎠         Yc 2
⎝ L ⎠ Yc             ⎝ L ⎠⎦
Here, G1 and B1 are calculated using (21.6) and (21.7). For most feeding
microstrips, G1 / Yc 1 and B1 / Yc 1. Then,
1             ⎛π ⎞                        ⎛π ⎞
Rin =                  cos 2 ⎜ y0 ⎟ = Rin( z = 0 ) cos 2 ⎜ y0 ⎟ .       (21.15)
2 ( G1 ± G12 )       ⎝ L ⎠                       ⎝ L ⎠
Notice that the inset feeding technique for impedance match of the microstrip
antennas is essentially identical to the off-center or asymmetrical feeding
techniques for dipoles. In both cases, a position is sought along a resonant
structure, where the current magnitude has the desired value.

11
2. Designing a rectangular patch using the transmission line model
Input data: εr of substrate, h, fr
1) Calculate W using (21.5).
2) Calculate εreff using (21.5) and equation (6) from Lecture 20.
3) Calculate the extension ∆L due to fringing effect using (21.1).
4) Calculate actual (physical) length of the patch using
λ                          1
L = 0 − 2∆L or L =                       − 2∆L .      (21.16)
2                 2 f r ε reff µ0ε 0
6) Calculate resonant input resistance at patch edge using (21.12) or
′
(21.13) with G1 = G from (21.6).
7) If Rin calculated in step 6 is too large, calculate the inset distance z0
using (21.14) or (21.15).

3. Cavity model for the rectangular patch
The TL model is very limited in its description of the real processes
x
taking place when a patch is excited. It takes into account only the TM 00 n
modes where the energy propagates only in the longitudinal z direction. The
field distribution along the x and y axes is assumed uniform. It is true that the
x
dominant TM 001 is prevalent but the performance of the patch is very much
affected by higher-order modes, too.
The cavity model is a more general model of the patch which imposes
open-end conditions at the side edges of the patch. It represents the patch as a
- electrical walls (above and below), and
- magnetic walls (around the perimeter of the patch.
The magnetic wall is a wall at which
n × H = 0 (the H − field is purely normal)
ˆ
n ⋅ E = 0 (the E − field is purely tangential)
ˆ
It is analogous to the open end termination in the theory of transmission lines.
If we treat the microstrip antenna only as a cavity, we could not represent
radiation because an ideal loss-free cavity does not radiate and its input
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impedance is purely reactive. To account for the radiation, a loss mechanism
has to be introduced. This is done by introducing an effective loss tangent,
δeff.
The thickness of the substrate is very small. The waves generated and
propagating beneath the patch undergo considerable reflection at the edges of
the patch. Only a very small fraction of them is being radiated. Thus, the
antenna is quite inefficient. The cavity model assumes that the E field is
purely tangential to the slots formed between the ground plane and the patch
edges (magnetic walls). Moreover, it considers only TM x modes, i.e., modes
with no Hx component. These assumptions are, basically, very much true.

The TMx modes are fully described by a single scalar function Ax – the x-
component of the magnetic vector potential:
A = Ax x .
ˆ                        (21.17)
In a homogeneous source-free medium, Ax satisfies the wave equation:
∇ 2 Ax + k 2 Ax = 0 .                   (21.18)
For regular shapes (like the rectangular cavity), it is advantageous to use the
separation of variables:
∂ 2 Ax ∂ 2 Ax ∂ 2 Ax
+       +         + k 2 Ax = 0          (21.19)
∂x 2
∂y 2
∂z 2

Ax = X ( x)Y ( y ) Z ( z )              (21.20)
∂2 X   ∂ 2Y   ∂2Z
YZ 2 + XZ 2 + XY 2 = −k 2 XYZ
∂x     ∂x     ∂x
13
1 ∂ 2 X 1 ∂ 2Y 1 ∂ 2 Z
+          +            = −k 2                      (21.21)
X ∂x   2
Y ∂y  2
Z ∂z 2

δ 2X                δ 2Y                  δ 2Z
+ k x X = 0,
2
+ k y Y = 0,
2
+ kz Z = 0
2
(21.22)
δx 2
δy  2
δz 2

The characteristic equation is
kx + k y + kz = k 2 .
2      2      2
(21.23)
The solutions of (21.22) are harmonic functions:
X ( x)=∑ An cos(k xn x)+ An sin(k xn x)
c                 s

n
Y ( y )=∑ Bn cos(k yn y )+ Bn sin(k yn y )
c                s
(21.24)
n
Z ( z )=∑ Cn cos(k zn z )+Cn sin(k zn z )
c               s

n
When the functions in (21.24) are substituted in (21.20), they give the
general solution of (21.18). The particular solution of (21.18) depends on the
boundary conditions.
In our case, there are electric walls at x = 0 and x = h . There, the
tangential E-field components must vanish, i.e., E y = Ez = 0      . Having in
x = 0, h
mind that
1 ⎛ ∂ 2 Ax         ⎞            1 ⎛ ∂ 2 Ax ⎞           1 ⎛ ∂ 2 Az ⎞
Ex =        ⎜      + k Ax ⎟ , E y =
2
⎜       ⎟ , Ez =      ⎜       ⎟ ,(21.25)
jωµε ⎝ ∂x 2          ⎠          jωµε ⎝ ∂x∂y ⎠          jωµε ⎝ ∂x∂z ⎠
we set Ax at the top and bottom walls as
∂Ax
= 0.                          (21.26)
∂x x =0,h
At all side walls, we set a vanishing normal derivative for Ax :
∂Ax                     ∂Ax
= 0,                   = 0.              (21.27)
∂z   z = 0, L           ∂y   y = 0,W
This ensures vanishing H x and H y at z = 0 and z = L , as well as vanishing
H x and H z at y = 0 and y = W (magnetic walls), as follows from the
relation between the H-field and Ax :

14
1 ⎛ ∂Ax ⎞       1 ⎛ ∂A ⎞
H x = 0, H y =   ⎜     ⎟ , Hx = ⎜ x ⎟.            (21.28)
µ ⎝ ∂z ⎠         µ ⎝ ∂y ⎠
It is obvious now that the solution must appear in terms of the functions
π
X ( x) = ∑ An cos(k xn x), k xn = n
c

n                           h
π
Y ( y ) = ∑ Bn cos(k yn y ), k yn = n
c
(21.29)
n                           W
π
Z ( z ) = ∑ Cn cos(k zn z ), k zn = n
c

n               L
The spectrum of the eigenmodes in the cavity is discrete. The frequencies of
those modes (the resonant frequencies) can be calculated from (21.23) as
2       2         2
⎛ mπ ⎞ ⎛ nπ ⎞ ⎛ pπ ⎞
(          )
2
⎜    ⎟   +⎜  ⎟   +⎜    ⎟   = ωr( mnp ) µε ,       (21.30)
⎝ h ⎠ ⎝W ⎠ ⎝ L ⎠
2         2          2
1       ⎛ mπ ⎞ ⎛ nπ ⎞ ⎛ pπ ⎞
f r( mnp )   =             ⎜    ⎟ +⎜   ⎟ +⎜   ⎟ .          (21.31)
2π µε     ⎝ h ⎠ ⎝W ⎠ ⎝ L ⎠
The mode with the lowest resonant frequency is the dominant mode. Since
x
usually L > W, the lowest-frequency mode is the TM 001 mode, for which
1   π     c
f r(001) =         =        .             (21.32)
2π µε L 2 L ε r
x
The dominant TM 001 mode is exactly the mode considered by the
transmission-line model (see previous sections). The field distribution of
some low-order modes is given in the following figure:

15
Fig. 14.13, pp. 741, C.B.
(
The general solution for the Axmnp ) (see (21.20) and (21.24)) is
⎡ c      ⎛ π ⎞⎤ ⎡ c         ⎛ π ⎞⎤ ⎡ c               ⎛ π ⎞⎤
Axmnp ) = ⎢ Ax cos ⎜ m x ⎟ ⎥ ⎢ Bx cos ⎜ n y ⎟ ⎥ ⎢C x cos ⎜ p z ⎟ ⎥ ,
(
(21.33)
⎣        ⎝ h ⎠⎦ ⎣           ⎝ W ⎠⎦ ⎣                 ⎝ L ⎠⎦
or
⎛ π ⎞          ⎛ π ⎞              ⎛ π ⎞
Axmnp ) = Amnp ⋅ cos ⎜ m x ⎟ ⋅ cos ⎜ n y ⎟ ⋅ cos ⎜ p z ⎟ .
(
(21.34)
⎝ h ⎠          ⎝ L ⎠              ⎝ W ⎠
The respective solution for field vectors of the (m,n,p) mode is

Ex = − j
(
k 2 − kx
2
)
Amnp ⋅ cos(k x x) ⋅ cos( k y y ) ⋅ cos(k z z ) ,              (21.35)
ωµε
kx k y
Ey = − j             Amnp ⋅ sin(k x x) ⋅ sin(k y y ) ⋅ cos(k z z ) ,         (21.36)
ωµε
kx kz
Ez = − j            Amnp ⋅ sin(k x x) ⋅ cos( k y y ) ⋅ sin(k z z ) ,         (21.37)
ωµε
Hx = 0,                                       (21.38)
kz
Hy = −           Amnp ⋅ cos(k x x) ⋅ cos(k y y ) ⋅ sin(k z z ) ,          (21.39)
µ
ky
Hz =          Amnp ⋅ cos(k x x) ⋅ sin(k y y ) ⋅ cos(k z z ) .            (21.40)
µ
x
For the dominant TM 001 mode,
Ex = ⎡ − j (k 2 − π 2 / h 2 ) /(ωµε ) ⎤ A001 cos(π z / L), E y = Ez = 0, (21.41)
⎣                                ⎦
H y = −(π / µ L) A001 sin(π z / L), H x = H z = 0.               (21.42)
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4. Cavity model for the radiated field of a rectangular patch
The microstrip patch is represented by the cavity model reasonably well
assuming that the material of the substrate is truncated and does not extend
beyond the edges of the patch. The four side walls (the magnetic walls)
represent four narrow apertures (slots) through which radiation takes place.
To calculate the radiation fields, the equivalence principle is used. The
field inside the cavity is assumed equal to zero, and its influence on the field
in the infinite region outside is represented by the equivalent surface currents
on the surface of the cavity.

Js , M s       Js = n × H
ˆ
M s = −n × E
ˆ

Js , M s

Because of the very small height (h) of the substrate, the field is concentrated
beneath the patch. There is some actual electrical current at the top metallic
plate, however, its contribution to radiation is negligible. That is because (1)
it is backed by a conductor, and (2) it is very weak compared to the
equivalent currents at the slots. The actual electrical current density of the top
patch is maximum at the edges of the patch, but still its values are negligible
in comparison with the radiation effect from the slots.
In the cavity model, the side walls employ magnetic-wall boundary
condition, which sets the tangential H components at the slots equal to zero.
Therefore,
Js = n × H = 0.
ˆ                                  (21.43)
Only the equivalent magnetic current density
M s = −n × E
ˆ                             (21.44)
has substantial contribution to the radiated field.

17
Ms      Js = 0
M s = −2n × E
ˆ

Ms

The influence of the infinite ground plane is accounted for by the image
theory, according to which the currents M s in the presence of the infinite
plane radiate as if magnetic currents of double strength radiate in free space:
M s = −2n × E .
ˆ                              (21.45)
Note that an Ex field at the slots corresponds to M s density vector, which is
tangential to the ground plane. Thus, its image is of the same direction. The
x
equivalent magnetic current densities for the dominant TM 001 mode are
sketched below.

L

x                        #1
t                                        #2
slo                                          t
z                Ms                        slo
y                                                                         Ms
E (001)

At slots #1 and #2, the equivalent M s currents are co-directed and with
equal amplitudes. They are constant along x and y.

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Radiation from a slot with constant current density

P ( x, y , z )

z                   rPQ
r
Q ( x′, z ′)
θ
h                    r′
Ms             φ                                y

x         W
P′

The radiation from an (x-y) slot of constant M s currents is found using
the electric vector potential F . Since M s has only a y component, so does
F : F = Fy y .
ˆ
ε
h/2 W /2
My          − jkrPQ
Fy (r ,θ ,φ ) =
4π        ∫ ∫          r
e             dx′dz ′ .        (21.46)
− h / 2 −W / 2 PQ
Here, M y = −2 E0 , E0 being the phasor of the E-field at the radiating slot,
and rPQ = r − r ′ ⋅ r = r − x′ sin θ cos φ − y′ sin θ sin φ .
ˆ
e− jkr
h/2                                     W /2
Fy =−2ε E0
4π r        ∫      exp( jkx′ sin θ cos φ )dx′⋅      ∫        exp( jky′ sin θ sin φ )dy′ (21.47)
−h / 2                                 −W / 2
ε E0Wh − jkr sin X sin Y
⇒ Fy = −                   ⋅e   ⋅     ⋅                                     (21.48)
2π r         X     Y
where
kh
X=     sin θ cos φ ,
2
kW
Y=      sin θ sin φ .
2

19
According to the relation between the far-zone E-field and the vector
potential,
Er 0, Eφ = jωη Fθ , Eθ = − jωη Fφ ,          (21.49)
where η = µ / ε , Fθ = Fy cosθ sin φ , and Fφ = Fy cos φ .
WhE0 − jkr                  sin X sin Y
⇒ Eφ = jωηε           e      cosθ sin φ                           (21.50)
2π r                         X     Y
WhE0 − jkr           sin X sin Y
⇒ Eθ = − jωηε             e     cos φ                            (21.51)
2π r                   X     Y
Since ωηε = k ,
V         ⎛              sin X sin Y ⎞
Eφ = jkW 0 e − jkr ⎜ cosθ sin φ                   ⎟,             (21.52)
2πr       ⎝                X       Y ⎠
V         ⎛       sin X sin Y ⎞
Eθ = − jkW 0 e− jkr ⎜ cos φ                    ⎟.               (21.53)
2π r       ⎝          X       Y ⎠
Here V0 = hE0 is the voltage between the patch edge and the ground plane.
Slots #1 and #2 form an array of two elements with excitation of equal
phase and magnitude, separated by a distance L. Their AF is
⎛ kLeff         ⎞
AF12 = 2cos ⎜         cosθ ⎟ .                        (21.54)
⎝    2          ⎠
Here Leff = L + 2∆L is the effective patch length. Thus, the total radiation
field is
kWV0 − jkr ⎛            sin X sin Y ⎞ ⎡ ⎛ kLeff               ⎞⎤
Eφ = j
t
e     ⎜ cosθ sin φ               ⎟ × ⎢ cos ⎜       cos θ ⎟ ⎥ , (21.55)
πr        ⎝               X      Y ⎠ ⎣ ⎝ 2                   ⎠⎦
kWV0 − jkr ⎛          sin X sin Y ⎞ ⎡ ⎛ kLeff          ⎞⎤
Eθ = − j
t
e    ⎜ cos φ             ⎟ × ⎢ cos ⎜     cosθ ⎟ ⎥ .   (21.56)
πr          ⎝         X     Y ⎠ ⎣ ⎝ 2                ⎠⎦
kLeff
Introducing Z =           cosθ , the pattern of the patch is obtained as
2
sin X sin Y
f (θ ,φ ) = Eφ2 + Eθ2 = 1 − sin 2 φ ⋅ sin 2 θ ⋅              cos Z .   (21.57)
X    Y

20
E-plane pattern (x-z plane, φ = 0°, 0°     ≤   θ     ≤   180°)
⎛ kh     ⎞
sin ⎜ sin θ ⎟
f E (θ ) =     ⎝ 2      ⎠ ⋅ cos ⎛ kLeff cosθ ⎞ .
kh               ⎜            ⎟          (21.58)
sin θ         ⎝ 2          ⎠
2
90
1
120                         x                60

0.8

0.6
150                                                                   30

0.4

0.2

ε r = 2.2                            z
180                                                                                   0
r = (cos(.674 π cos(t)/2)) (sin(0.05 sin(t))/(0.05 sin(t)))
90
1
120                                        60

0.8

0.6
150                                                                 30

0.4

0.2

εr = 4
180                                                                              0
r = (cos(.5 π cos(t)/2)) (sin(0.05 sin(t))/(0.05 sin(t)))
21
H-plane pattern (x-y plane, θ = 90°, 0° ≤ φ ≤ 90° and 270° ≤ φ                            ≤   360°)
⎛ kh     ⎞    ⎛ kW        ⎞
sin ⎜ cos φ ⎟ sin ⎜     sin φ ⎟
f H (θ ) = cos φ ⋅ ⎝           ⎠⋅ ⎝ 2           ⎠
2
(21.59)
kh            kW
cos φ          sin φ
2              2

0

330
30

x

300
60

0.8
1

0.6

0.4

0.2

y

270
90

r = cos(t) (sin(0.01 cos(t))/(0.01 cos(t))) (sin(0.6 π sin(t))/(0.6 π sin(t)))
0

330
30

300
60

0.8
1

0.6

0.4

0.2

270
90

r = cos(t) (sin(0.01 cos(t))/(0.01 cos(t))) (sin(1.1 π sin(t))/(1.1 π sin(t)))

22
Fig.14.16, p. 744, Balanis

Fig. 14.17, p. 746, Balanis

23
Fig. 14.18, p. 747, Balanis

Non-radiating slots: It can be shown that the slots at y = −W / 2 and
y = W / 2 do no radiate in the principle E- and H-planes. In general, these
two slots do radiate away from the principle planes, but their field intensity is
everywhere small compared to that radiated by slots #1 and #2.
24

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