EE522 NOTES: APERTURE ANTENNAS

The equivalent Principle

• Volume region: V, surface of V is S, the outward normal

direction of S is n .

• Original problem: Source in V, Field generated by thesource is E and H.

• General equivalent problem: Field internal to V is anyfield 1 1,E H , Equivalent magnetic current M and electriccurrent J on S. If the equivalent currents satisfy

( ) ( )1 1 1 1ˆ ˆ,S SJ n H H M E E n= × − = − ×

Then the field generated by the equivalent sourceexternal to V are the same as the field in the originalconfiguration.

• There is no restriction of the internal field 1 1,E H in theabove equivalent problem. Hence one choice is to make

1 10, 0E H= = . For this case, the equivalent currents are:

ˆ ˆ, ( )S SJ n H M E n on S= × = ×

Since the field inside S is zero in this case, we are free tointroduce any material to S (1. Same as external, 2. PEC, 3.PMC). For planar surface S, the problem can be simplifiedusing PEC or PMC material inside S.

Field radiated by Js and Ms: is equal to the summation ofthe field radiated by Js and that by Ms:

( ) ˆ 'ˆ ˆ , ( ') '4

j rj r r

A SS

eE j A A A J r e dS

r

ββ

θ φω θ φ µπ

−⋅= − + = ∫

( ) ˆ 'ˆ ˆ , ( ') '4

j rj r r

F SS

eH j F F F M r e dS

r

ββ

θ φω θ φ επ

−⋅= − + = ∫

Since E and H are related by

( )( )

ˆ ˆˆ ˆ ˆ

ˆ ˆ

F FE H r j F r F r

j F F

θ φ

θ φ

η ωη θ φ

ωη φ θ

= × = − × + ×

= − − +

We have the total field:ˆ ˆ( ) ( )E j A F A Fθ φ φ θω η θ η φ = − + + −

Let Js and Ms exist on Sa, and define

ˆ ˆ' '( ') ', ( ') 'a a

j r r j r ra aS S

P E r e dS Q H r e dSβ β⋅ ⋅= =∫ ∫

(1) Aperture in ground plane:

( )

( )

cos sin2

cos sin cos2

j r

x y

j r

x y

j eE P P

r

j eE P P

r

β

θ

β

φ

β φ φπ

β θ φ φπ

−

−

= +

= − +

(2) Aperture in free-space:

( )

( )

1 coscos sin

4 2

1 cossin cos

4 2

j r

x y

j r

x y

j eE P P

r

j eE P P

r

β

θ

β

φ

β θ φ φπ

β θ φ φπ

−

−

+= +

+= − +

Example: Uniform rectangular aperture

Aperture field: 0ˆ , | | , | |2 2

yxa

LLE y E x y= ≤ ≤

To calculate Py:( )ˆ ˆ ˆ ˆ' ' ' ' 'r r r xx yy ux vy⋅ = ⋅ + = +

' sin( )'

a j ux

a

uae dx a

uaβ β

β−=∫

Hence :

( ) ( )

/ 2 / 2' '0 / 2 / 2

0

' '

sin / 2sin / 2

/ 2 / 2

x y

x y

L Lj ux j vyy L L

yxx y

x y

P E e dx e dy

vLuLE L L

uL vL

β β

βββ β

− −= ×

=

∫ ∫

The radiation field (for aperture in free-space):

[ ]0

sin ( / 2)sin ( / 2)sin

2 ( / 2) ( / 2)

j ryx

x yx y

L vL uj eE E L L

r L u L v

β

θ

βββ φπ β β

− =

[ ]0

sin ( / 2)sin ( / 2)cos cos

2 ( / 2) ( / 2)

j ryx

x yx y

L vL uj eE E L L

r L u L v

β

φ

βββ θ φπ β β

− =

E-Plane ( 90oφ = ): cos 0, 0, 0, sinE u vφφ θ= = = =

sin ( / 2)sin

( )( / 2)sin

y

Ey

LF

L

β θθ

β θ =

H-Plane ( 0oφ = ): sin 0, 0, sin , 0E u vθφ θ= = = =

[ ]sin ( / 2)sin

( ) cos( / 2)sin

xH

x

LF

L

β θθ θ

β θ=

Half-Power beamwidth ( ,x yL L λ>> ):

Since sin(1.39) /1.39 0.707= ,From ( )1( / 2)sin 1.39, sin 2 1.39 / 2 0.443 /x H H x xL L Lβ θ θ λ π λ−= = × ≈

0.886 ( ), 0.886 ( )E Hy x

HP Rad HP RadL L

λ λ= =

Note: the HPs are inversely proportional to the size in thecorresponding cutting plane.The approximate HP expressions give good estimationseven when Lx, Ly as small as 1.5 wavelength.

Directivity: 2

4u x yD L L

πλ

= . (100% aperture efficiency)

Example: Determine the half-power beam width for E andH plane and the directivity of a rectangular aperture(10 20λ λ× ) with uniform aperture field.Solution:

0.886 0.0443 ( ) 2.48 ,

20

0.886 0.0886 ( ) 4.96 .10

oE

oH

HP Rad

HP Rad

λλ

λλ

= = =

= = =

2 2

4 410 20 2513.27 (34 )u x yD L L dB

π π λ λλ λ

= = × =

0.5 1 1.5 2 2.5 310

20

30

40

50

60

70

80

90

Ly / λ

HP

y

E−plane

sin[(βa/2)sinθ]/[(βa/2)sinθ]

Exact50.76λ / Ly

0.5 1 1.5 2 2.5 310

20

30

40

50

60

70

80

90

Lx / λ

HP

x

H−plane

cosθ sin[(βa/2)sinθ]/[(βa/2)sinθ]

Exact50.76λ / Lx

Rectangular aperture with tapered aperture fieldConsider constant direction of E:

ˆ ( ) ( ), ( , )a x y aE e E x E y x y S= ∈Then the aperture integration is

/ 2 / 2' '

/ 2 / 2( ') ' ( ') '

x y

x y

L Lj ux j vye x yL L

P E x e dx E y e dyβ β

− −= ×∫ ∫

Example: open-ended rectangular waveguide:(aperture on ground plane):

The aperture field is constant in y-direction and cosinevariation in x-direction.

[ ]/ 2 '

/ 2

sin ( / 2)'

( / 2)

b j vy

b

b ve dy b

b vβ β

β−=∫

[ ] [ ]

[ ]

/ 2 / 2' ( / ) ' ( / ) '

/ 2 / 2

2

' 1cos ' '

2

2 sin ( ) / 2 2 sin ( ) / 21 1

2 ( ) / 2 2 ( ) / 2

cos( sin cos / 2)

1 ( / 2)( sin cos / 2)

a aj ux j u a x j u a x

a a

xe dx e e dx

a

j ua j ua

j ua j ua

a

a

β β π β ππ

β π β πβ π β πβ θ θ

π β θ φ

+ −

− −

= + + −

= ++ −

=−

∫ ∫

For E-plane ( 90oφ = ): sin( sin / 2)

( )sin / 2E

bF

b

β θθβ θ

=

For H-plane ( 0oφ = ): [ ]2

sin( sin / 2)( ) cos

1 ( sin ) /H

aF

a

β θθ θβ θ π

=−

Directivity calculationAssumption: (1) Pattern peak near broadside

(2) Aperture relatively large to λ .(3) Aperture field close to plane wave.

2

2

'4

'a

a

aS

aS

E dSD

E dS

πλ

=∫∫

Example: Using the above formula to calculate thedirectivity for the open-ended waveguide.

2 2/ 2 / 2 2

/ 2 / 2

/ 2 / 22

/ 2 / 2

' 2cos( ) ' '

'cos ( ) ' '

2

a b

a b

a b

a b

x adx dy b

a

x abdx dy

a

ππ

π

− −

− −

× =

× =

∫ ∫

∫ ∫

2 2

2 2 2 2

4 (2 / ) 4 8 4(0.81)

/ 2

a b abD ab

ab

π π π πλ λ π λ

= = = In the above, the directivity is reduced by a factor of 0.81compared to uniform aperture of the same size. Thisreduction is caused by the tapered aperture, hence the 0.81here is called aperture taper efficiency that is defined by

/t t uD Dε =where

tD is the directivity of the tapered aperture field.

uD is the directivity of a uniform field aperture of thesame size.

Example: calculate the aperture taper efficiency for arectangular aperture whose aperture is given by

| ' |ˆ 1 , ( ', ')

/ 2a a

xE y x y S

a = − ∈

Solution: the x-direction tapering is called triangular taping.

2/ 2 / 2

/ 2 / 2

| ' | | ' |1 ' , 1 '

/ 2 2 / 2 3

a a

a a

x a x adx dx

a a− −

− = − = ∫ ∫

Aperture taper efficiency 21 ( / 2) 3

/ 3 4t

ab

ab abε = = .

Example: An aperture operating at 10GHz has physicalaperture area of 0.785 2m , a gain of 38dB, a directivity of39dB. Compute (1) Effective aperture, (2) Maximumeffective aperture, (3) aperture efficiency, (4) radiationefficiency, (5) aperture taper efficiency.

Solution: At 10GHz, 8 93 10 /(10 10 ) 0.03mλ = × × =

2 2

38 6309.6

39 7943.3

4 / 4 (0.785) /(0.03 ) 10961 (40.4 )u p

G dB G

D dB D

D A dBπ λ π

= ⇒ == ⇒ == = =

(1) Effective aperture 2 2 2/(4 ) 6309.6 0.03 /(4 ) 0.452eA G mλ π π= = × =(2) Maximum effective aperture:

2 2 2/(4 ) 7943.3 0.03 /(4 ) 0.569emA D mλ π π= = × =(3) Aperture efficiency: / 0.452 / 0.785 0.576ap e pA Aε = = =

(4) Radiation efficiency: / 6309.6 / 7943.4 0.794re G D= = =(5) Aperture taper efficiency: / 7943.4 /10961 0.725t uD Dε = = = .

H-plane sectoral Horn Antenna

Aperture field distribution:

21( / 2 )

0ˆ cos j R xa

xE yE e

Aβπ − =

21

/ 2 / 2' ( / 2 )( ') '0 / 2 / 2

cos( '/ ) ' 'A bj ux j R x j vy

y A bP E x A e dx e dyβ β βπ −

− −= ×∫ ∫

Steps to evaluate the integral

2

1/ 2 ' ( / 2 )( ')

/ 2cos( '/ ) '

A j ux j R x

AI x A e dxβ βπ −

−= ∫

(1) Convert cosine function into exponentials:'/ '/cos( '/ ) (1/ 2) j x A j x Ax A e eπ ππ − = +

(2) Write exponential into 2( ' )p x q c+ +

2 21 1

12 2

21

1 1 12

1 11 1

/( / ) ' ( / 2 )( ') ( / 2 ) ( ') '

/(2 )

1 2 / 2 2 / 2 2 / 2( / 2 ) ( ') '

2 /(2 ) /(2 ) /(2 )

2 / 2 2 / 2( / 2 ) ' ( / 2 )

/(2 ) /(2

A uA u x R x R x x

R

A u A u A uR x x

R R R

A u A uR x R

R R

π βπ β β ββ

π β π β π βββ β β

π β π ββ ββ β

++ + = +

+ + + = + + − + += + −

2

2

2

1 11 1

)( ' )

2 / 2 2 / 2( / 2 ), , ( / 2 )

/(2 ) /(2 )

p x q cWhere

A u A up R q c R

R R

π β π ββ ββ β

= + +

+ += = =−

(3) Evaluate the integral into Fresnel integrals:

Let 2 2( ' ) ( / 2)p x q π τ+ =

[ ]

2 22

1

/ 2 ( ' ) ( / 2)

/ 2

2 1 2 1

'2

( ) ( ) ( ) ( )2

A tjp x q j

A te dx e d

p

C t C t jS t jS tp

π τπ τ

π

+

−=

= − + −

∫ ∫

Where

1 2( / 2), ( / 2)2 2

t q A t q Ap p

π π= − = +

And the Fresnel integrals are defined by:2 2

0 0( ) cos( / 2) , ( ) sin( / 2)

x xC x d S x dπτ τ πτ τ= =∫ ∫

The total radiation field is

( )

( )

1 cos sin4

1 cos cos4

j r

y

j r

y

eE j P

r

eE j P

r

β

θ

β

φ

β θ φπ

β θ φπ

−

−

= +

= +

Directivity: For a given axial length R1, there is anoptimum value of A/Wavelength for which the directivityis maximum. The optimum H-plane horn: 13A Rλ= .

Explanation: For given R1, when A increases frombeginning from A=a, the aperture size increases, hence thedirectivity will increase. However when A is too large, thephase error will be so large as to cancel the radiation in themain beam direction, resulting in reduction in directivity.

Computer assignment

Write a computer program to calculate the radiation patternof H-sectoral horn antenna (H-plane, use equation 7-119).

The input parameters to your program are: , , , .a b A λAnd the output is the radiation pattern as a function of θfrom 0 to 80 degrees with increment of 1 degree. Yourprogram may have the following structures:

Start of programInput parametersCalculate derived parameters (such as 1, , .R etcβ ).Determine the number of integration points N and dx.Open a file for output.Loop for theta=0 to 80 Perform the numerical integration to get ( )f θ :

Sum=0 Loop over integration grids j=1,2,…N Calculate x' Calculate integrand h Sum=sum + W * h End loop j. F=abs(sum) Write (theta, F) to output file.End loop theta.Close file.End of program.

Usually 0.05λ as the initial grid size. Hence the number ofgrid point is [ ]int 0.5 /(0.05 )N A λ= + , and the actual grid sizeis determined by /( 1)x A N∆ = − . A simple method toevaluate the numerical integration is to use the trapezoidalrule. The following FORTRAN code shows the applicationof the trapezoidal rule to calculate

/ 2 'sin

/ 2( ) '

A j x

AF e dxβ θθ

−= ∫

IMPLICIT NONEINTEGER I, J, NREAL WL, A, DX, W, THETA, BETA, PICOMPLEX F, CXPARAMETER(PI=3.1415926)WRITE(*,*) 'Input A, WL=?'READ(*,*) A,WLOPEN(1,FILE='sample.ptn',status='unknown')BETA = 2.0 * PI/WLN = 0.5 + A/ (0.05 * WL)IF (N < 2 ) N=2DX = A / FLOAT(N-1)DO I=0, 80 THETA = FLOAT(I)*PI/180.0 F=0.0 DO J=1,N X= - 05.* A + DX * FLOAT(J-1) W=DX IF((J .EQ.1).OR.(J.EQ.N) ) W=0.5 * DX CX=(0.0,1.0) * BETA * X * SIN(THETA) F=F+W * EXP(CX) END DO WRITE(1,*) THETA, ABS(F)END DOCLOSE(1)STOPEND

Circular apertures:Aperture is on x-y plane (z=0).Aperture radius=a.Will consider: uniform and tapered distribution.

1. Uniform aperture field

0ˆ , ( , )a aE xE x y S= ∈2 ˆ '

0 0 0ˆ ' ' '

a j r rP xE e d dπ β ρ ρ φ⋅= ∫ ∫

Again,ˆ ˆ ˆ ˆsin cos sin sin cosr x y zθ φ θ φ θ= + +

ˆ ˆ' 'cos ' 'sin 'r x yρ φ ρ φ= +ˆ ' 'sin cos cos ' 'sin sin sin '

'sin cos( ')

r r ρ θ φ φ ρ θ φ φρ θ φ φ

⋅ = += −

2 'sin cos( ')0 0 0

0 00

20 1

ˆ ' ' '

ˆ 2 ( 'sin ) ' '

ˆ ( ) 2 ( sin ) /( sin )

a j

a

P xE e d d

xE J d

xE a J a a

π βρ θ φ φ φ ρ ρ

π βρ θ ρ ρ

π β θ β θ

− =

=

=

∫ ∫

∫

The radiation field:

2 1

0

2 ( sin )ˆ ˆ( cos cos sin ) ( )4 sin

j rj e J aE E a

r a

ββ β θθ φ φ θ φ ππ β θ

−

= −

E-plane pattern ( 0φ = ): 12 ( sin )( )

sin

J aF

aθ

β θθβ θ

=

H-plane pattern ( 90φ = ): 12 ( sin )( ) cos

sin

J aF

aφβ θθ θ

β θ=

For large apertures, the HP values for E and H plane arealmost the same, they equal to

1.02 /(2 ) ( )HP a radλ=Side lobe level ( a λ>> ): -17.6dB

Directivity: ( )22

4uD a

π πλ

= .

Example: An aperture antenna with uniform aperture fieldhas HP=2 degrees. Find the size and the directivity of theantenna.Solution: From 1.02 /(2 ), 0.61 / 17.48HP a HPλ α λ λ= = =

2 2 2( ) / 4 17.48 12061, (40.8 )uD a dBπ π λ π π= 4 = × × =

Tapered aperture field:Aperture taper efficiency will reduceSLL will reduceHP will increase.

Parabolic taper:2

22

' 8 ( sin )1 , ( ) , 0.75, 24.6 , 1.27

( sin ) 2a t

J aE f SLL dB HP

a a a

ρ β θ λθ εβ θ

= − = = = − = Parabolic squared taper:

22

33

48 ( sin )'1 , ( ) , 0.55, 30.6 , 1.47

( sin ) 2a t

J aE f SLL dB HP

a a a

β θρ λθ εβ θ

= − = = = − =

Parabolic reflector antennas

Parameters: D---Aperture diameterF/D---Focal length-to-diameter ratio

Points on surface can be specified in two ways:2

2

(1) ( ', ) : ( ') 4 ( )

(2) ( , ) : 2 /(1 cos ) sec ( / 2)

Re ' :

2 sin' sin 2 tan

1 cos 2

f f

f f f f f

f

f ff f

f

by z F F z

by r r F F

lation of and r

Fr F

ρ ρ

θ θ θρ

θ θρ θ

θ

= −

= + =

= = =+

Normal direction: ˆˆ cos( / 2) sin( / 2)f f f fn r θ θ θ= − +

Properties:(1) Rays from the focal point "O" will be reflected by the

parabolic surface

(2) All reflected rays are parallel to the axis fz .(3) The ray path from "O" to reflecting points to focal

plane have the same distance (that is equal to 2F).

Feeder pattern: Generalˆ ˆ( , ) ( , ) ( , )f f f f f f f fF F Fθ φθ φ θ θ φ φ θ φ= +

Feeder pattern: Linear:ˆ( , ) ( , )f f i f f fF u Fθ φ θ φ=

Methods to obtain the radiation pattern of the reflectorantenna:

(1) GO/Aperture distribution methodFind the aperture field distribution and integrateover the aperture (flat, focal plane) to get theradiation field.

(2) PO/Surface current methodUse PO approximation to calculate the currentdistribution over the reflecting surface (curved)and integrate over the parabolic surface to get theradiation field.

PO/Aperture Distribution Method

Ray tube: formed by rays originated from "O" in a smallangular region " dΩ ". This ray tube has a footprint on thefocal plane of "dA".

Conservation of power: Powers flowing through any crosssection in the ray tube are the same.

Using the power conservation concept, we can show thatthe electric field magnitude over the focal plane is inverselyproportional to the distance from "O" to the reflectingsurface:

1( )a f

f

Er

θ ∝

The reducing in field magnitude given in the aboveequation is called spherical spreading loss (because thereflected rays are parallel, the power does not spread overthe reflecting part of the ray tube). Based on this analysis,the electric field distribution on the focal plane can beshown to be

2

0

( , ')ˆ( , ') f fj F

a f rf

FE V e u

rβ θ φ

θ φ −= (for linear feed)

Where ˆru is the same as the reflected electric field and canbe found using the Snell's law:

ˆ ˆ2( ) ,

ˆ ˆ ˆ ˆ ˆ2( )r i i

r i i

E n E n E

u n u n u

= ⋅ −= ⋅ −

( )

( )

1 coscos sin

4 2

1 cossin cos

4 2

j r

x y

j r

x y

j eE P P

r

j eE P P

r

β

θ

β

φ

β θ φ φπ

β θ φ φπ

−

−

+= +

+ = − +

ˆ '

2 'sin cos( ')0 0 0

( ') '

( , ')ˆ ' ' '

a

j r raS

a f f jr

f

P E r e dS

FV u e d d

r

β

π βρ θ φ φθ φρ ρ φ

⋅

−

=

=

∫

∫ ∫In the above equation,

( )2 2

, Angle of observation

', ' Integration variables

, Variables related to ', ':

4 ' /(4 ), 2 tan( '/ 2 )

f f

f f

r

r F F F

θ φρ φ

θ ρ φ

ρ θ ρ

− − − −− − − −− − −

= + =

Example:Let the feed be a short dipole the is parallel to ˆ fx -axis. Then the primary pattern is given by

2 2

2

( , ') 1 cos sin '

ˆ ˆ ˆ ˆ( )ˆ

ˆ ˆ1 ( )

f f f

f f f fi

f f

F

r r x xu

r x

θ φ θ φ= −

⋅ −=

− ⋅

For axial symmetric feed pattern (primary pattern),( )f fF θ is independent of variable 'φ , then

0 00

( )ˆ( , ) 2 ( 'sin ) ' '

a f fr

f

FP V u J d

r

θθ φ π βρ θ ρ ρ= ∫