The Directivity of E-Plane, H-Plane and Pyramidal Horn Antennas
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Transcript of The Directivity of E-Plane, H-Plane and Pyramidal Horn Antennas
The Directivity of E-Plane, H-Plane and PyramidalHorn Antennas
November 23, 2001
1 Introduction
First introduced in the late nineteenth century, horn antennas were found to be most useful in
high frequency applications such as microwave transmission. The first recorded horn antenna
to appear in an experiment was the pyramidal horn used by J. Chunder Bose in 1897. In a
lecture at the London’s Royal Institution, Bose performed a demonstration using the horn,
which he referred to as a “collecting funnel” at an operating frequency of 60 GHz.
The most commonly used horns today are pyramidal and conical, which can be manufac-
tured in many shapes and sizes. For the purposes of this lecture, we will look at the pyramidal
horn and the E and H-plane horn of which it is comprised. These notes derive the formulas
for the e-plane horn but leave all other derivations up to the reader.
2 Aperture and radiated fields of the E-Plane Horn
An E-plane horn antenna is an aperture antenna that is flared in the direction of the E-field.
This results in radiated fields that have a high directivity in the E-plane of the antenna. A
detailed geometry for a E-plane antenna is given in Figure 1.
From Balanis the aperture fields for the E-Plane Horn are
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Figure 1: E-Plane Horn Antenna. Taken from course Text, Antenna Theory by C. Balanis.
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E ′z = E ′
x = H ′y = 0
E ′y(x
′, y′) ∼= E1 cos(π
ax′)e
−j[ ky′2(2ρ1)
]
H ′z(x
′, y′) ∼= jE1π
kaηsin(
π
ax′)e
−j[ ky′2(2ρ1)
]
H ′x(x
′, y′) ∼= −E1
ηcos(
π
ax′)e
−j[ ky′2(2ρ1)
]
(1)
where
ρ1 = ρe cos(ψe) .
We can use these equations and the equivalent fields approach to aperture find the surface
current densities,
~J ′s = −E1
ηcos(
π
ax′)e
−j[ ky′2(2ρ1)
]y (2)
and
~M ′s = E1 cos(
π
ax′)e
−j[ ky′2(2ρ1)
]x (3)
for the region −a2≤ x′ ≤ a
2and − b1
2≤ y′ ≤ b1
2. The current densities are zero elsewhere.
The E-field components for the radiated fields of an aperture antenna are given by
Eθ = −jke−jkr
4πr(Lφ + ηNθ) (4)
Eφ =jke−jkr
4πr(Lθ + ηNφ) (5)
where
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Nθ =
∫∫S
[Jx cos θ cosφ+ Jy cos θ sinφ− Jz sin θ]e+jkr′ cosψds′
Nφ =
∫∫S
[−Jx sinφ+ Jy cosφ]e+jkr′ cosψds′
Lθ =
∫∫S
[Mx cos θ cosφ+My cos θ sinφ−Mz sin θ]e+jkr′ cosψds′
Lφ =
∫∫S
[−Mx sinφ+My cosφ]e+jkr′ cosψds′ .
Looking at the equation for Nθ and (2) we have
noting from the geometry that,
r′ cosψ = ~r ′r = x′ sin θ cosφ+ y′ sin θ sinφ
we get,
Therefore,
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Nθ = E1πa
2
√πρ1
kej
[ky2ρ1
2k
] [cos θ sinφ
η
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
](6)
where,
kx = k sin θ cosφ
ky = k sin θ sinφ
t1 = − b1√2λρ1
t2 =b1√2λρ1
and
F (t1, t2) = [C(t2)− C(t1)]− j[S(t2)− S(t1)] .
Note: C(t) and S(t) are cosine and sine integrals, the solutions of which can be found in
Appendix III of the course text.
By similar analysis we can also find,
Nφ = E1πa
2
√πρ1
kej
[ky2ρ1
2k
] [cosφ
η
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
]
Lθ = E1πa
2
√πρ1
kej
[ky2ρ1
2k
] [− cos θ cosφ
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
]
Lφ = E1πa
2
√πρ1
kej
[ky2ρ1
2k
] [sinφ
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
].
(7)
Now, we can substitute the results of (6) and (7) into (4) and (5) to get the radiated
E-fields,
Eθ = −j a√πkρ1E1e
−jkr
8r
[ej
[k2yρ12k
]sinφ(1 + cos θ)
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
]. (8)
Eφ = −j a√πkρ1E1e
−jkr
8r
[ej
[k2yρ12k
]cosφ(1 + cos θ)
[cos(kxa
2)
(kxa2
)2 − (π2)2
]F (t1, t2)
](9)
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3 Directivity of an E-plane Horn
3.1 Maximum Radiation
To find the directivity of an E-plane horn we must first find the maximum radiation,
Umax =r2
2η|E|2max . (10)
For most horn antennas |E|max is directed mainly along the z-axis (ie. θ = 0). Thus,
|E|max =√|Eθ|2max + |Eφ|2max .
From (8) and (9) and noting kx = 0,
|Eθ|max =a√πkρ1
8r(2)
[1
(π2)2
]|E1 sinφF (t1, t2)| (11)
|Eφ|max =a√πkρ1
8r(2)
[1
(π2)2
]|E1 cosφF (t1, t2)| . (12)
Recalling that t1 = −t2 and using the fact that,
C(−t) = −C(t) (13)
and
S(−t) = −S(t) (14)
we get,
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Therefore we now have,
|Eθ|max =2a√πkρ1
8r|E1 sinφF (t)| (15)
|Eφ|max =2a√πkρ1
8r|E1 cosφF (t)| . (16)
Substituting these results into (10), gives the final result,
Umax =2a2kρ1
ηπ3|E1|2 |F (t)|2 (17)
where,
|F (t)|2 = [C2(b1√2λρ1
) + S2(b1√2λρ1
)]
3.2 Power Radiated
The total power radiated from a E-plane horn can be found by,
Prad =1
2
∫∫S
<e( ~E ′ × ~H ′∗) · d~s (18)
Using the ~E ′ and ~H ′-fields over the aperture of the horn given in (1),
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Now recall,
∫cos2 udu =
1
2(u+ sinu cosu) + C
Thus, we have,
Prad =ab14η
|E1|2 (19)
3.3 Putting it all together
Now that we have found the maximum radiation and the power radiated from the horn, we
can find the directivity using the formula,
DE =4πUmax
Prad
=64aρ1
πλb1|F (t)|2
(20)
4 Directivity of a H-plane Horn
Like the E-plane horn antenna, a H-plane horn is only flared in one direction. In this case it
is flared in the direction of the H-field, giving a better directivity in that plane. Figure 2 is a
diagram showing the geometry of the antenna.
Using the same approach as presented for the E-plane horn, and starting from,
E ′x = H ′
y = 0
E ′y(x
′) = E2 cos(π
a1
x′)e−jk
[x′22ρ2
]
H ′x(x
′) = −E2
ηcos(
π
a1
x′)e−jk
[x′22ρ2
]
ρ2 = ρh cosψh
it can be shown that,
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Figure 2: H-Plane Horn Antenna. Taken from course Text, Antenna Theory by C. Balanis.
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Umax = |E2|2b2ρ2
4ηλ
{[C(u)− C(v)]2 + [S(u)− S(v)]2
}(21)
where,
u =
√1
2(
√λρ2
a1
− a1√λρ2
)
v =
√1
2(
√λρ2
a1
+a1√λρ2
)
and that,
Prad = |E2|2ba1
4η.
Combining this gives us the H-plane horn directivity,
DH =4πbρ2
a1λ
{[C(u)− C(v)]2 + [S(u)− S(v)]2
}(22)
5 Directivity of an Pyramidal Horn
The final type of horn presented is the pyramidal horn antenna. This horn is a combination of
the E-plane and H-plane horns and as such is flared in both directions. A three dimensional
diagram of the pyramidal is given in Figure 3. For the E-plane and H-plane views, the
diagrams Figure 1 and Figure 2 can be used1.
The equations of interest for the pyramidal horn are,
1Note: p1 and p2 in figure 1 and figure 2 are referred to as pe and ph respectively in the pyramidal hornequations.
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Figure 3: Pyramidal Horn Antenna. Taken from course Text, Antenna Theory by C. Balanis.
Umax = |E0|2ρ1ρ2
2η
{[C(u)− C(v)]2 + [S(u)− S(v)]2
}×
{C2(
b1√2λρ1
) + S2(b1√2λρ1
)
}Prad = |E0|2
a1b14η
DH =8πρ1ρ2
a1b1
{[C(u)− C(v)]2 + [S(u)− S(v)]2
}×
{C2(
b1√2λρ1
) + S2(b1√2λρ1
)
}=
πλ2
32abDEDH
Also, for this type of horn antenna, the properties,
pe = (b1 − b)
[(ρeb1
)2 − 1
4
] 12
and
ph = (a1 − a)
[(ρha1
)2 − 1
4
] 12
are important. If these values are not equal, then the horn is not physically realizable.
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