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

1

Figure 1: E-Plane Horn Antenna. Taken from course Text, Antenna Theory by C. Balanis.

2

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,

4

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,

8

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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