Radar cross section
Problem 4.1 :
A rectangular X-band horn, with aperture dimensions of 5.5 cm and 7.4 cm and a gain of 16.3 dB (over isotropic) at 10 GHz, is used to transmit and receive energy scattered from a perfectly conducting sphere of radius a = 5 λ. Find the maximum scattered power delivered to the load when the distance between the horn and the sphere is:
a) 200 λ b) 500 λ
Assume that the transmit power is 200mW, and the radar cross section is equal to the geometrical cross section.
2 25 2 a s = p = pl 1.63 0r 0t 16.3dB 0r 0t 10 42.66 G =G = G =G = = 10 GHz 3 cm f = l = 2 0 0 1 2 4 4 r r t t P G G P R R l s p p æ ö÷ ç ÷ = çç ÷÷ è ø a) R1 = R2 = 200 λ = 6 m : Pr = 9 nW b) R1 = R2 = 500 λ = 15 m : Pr = 0.23 nW
Coordinate transformation
Problem 4.2 :
A infinitesimal electric dipole of constant current I0 is placed symmetrically about the origin
and directed along the x-axis. Derive the a) far-zone fields radiated by the dipole b) directivity of the antenna.
a) The pattern of a dipole oriented in x -direction has to be the same as oriented along the z -axis. Two new coordinates y and c are introduced that are corresponding to q and f . Thus, the electric and magnetic far field of the dipole can be written as
0 0 sin 4 sin 4 jkr jkr kl e E j I r E kl e H j I r y y c h y p y p h -» » =
The new coordinate y now must be expressed in terms of the old coordinates q and f . Generally, an angle a enclosed by two vectors a and b can be calculated by
cosa = a b⋅ .
Therefore it is convenient to perform the following conversion
2
2 ˆ ˆ
siny = 1-cos y = 1- ax ⋅ar
where the vector ˆa in spherical coordinates can be expressed as r
ˆr ˆx sin cos ˆysin sin ˆz cos
a =a q f+a q f+a q
and therefore siny = 1-(sin cosq f)2
The far field of the dipole oriented in y -direction can be written as
( )2 0 1 sin cos 4 jkr kl e E j I r y h p q f -» -y x z
ψ
χb) When the radiation pattern stays the same as for an z -oriented dipole, also the directivity has to be the same:
(
2 2)
0 1 sin cos U =U - q f(
)
(
)
2 2 2 2 3 2 0 0 0 0 0 0 2 2 2 0 0 0 0 01 sin cos sin sin sin cos
4 4 8 2 cos 4 3 3 3 rad P Ud U d d U d d U d d U U p p p p p p q f q q f q q f q f f f f p p p W é ù ê ú = W = - = ê - ú = ê ú ë û é ù é ù ê ú = ê - ú = ê - ú = ë û ê ú ë û
òò
ò ò
ò ò
ò
ò
and 0 0 0 4 3 1.5 8 2 3 U D U p p = = =x
y
z
Wire antennas
Problem 4.3 :
A thin linear dipole of length l is placed symmetrically about the z-axis. Find the far-zone spherical electric and magnetic components radiated by the dipole whose current distribution can be approximated by a)
(
)
(
)
0 0 2 1 ' / 2 ' 0 ( ') 2 1 ' 0 ' / 2 z I z l z l I z I z z l l ìï + - £ £ ïï ï = í ïï - £ £ ïïî , b) I zz( ') I0cos( )
l z' l/2 z' l/ 2 p = - £ £ . a)(
)
(
)
(
)
0 0 0 2 / 2 ' 0 1 ' ( ') 2 1 ' 0 ' / 2 2 1 ' for / 2 ' / 2 z l z I z l I z I z z l l I z l z l l ì - £ £ ï + ïï ï = í ïï - £ £ ïïî = - - £ £The magnetic vector potential can be found as
( )
(
)
(
)
2 ' cos 2 2 0 ' cos 2 2 0 ' cos 2 ( ) ' ' 4 2 1 ' ' 4 sin cos ' 2 2 ' 4 cos 2 l jkr jkz z z l l jkr jkz z l l jkr jkz z l e A r a I z e dz r I e a z e dz r l kl I e z a l kl e dz r l q q q m p m p q m p q -⋅ -⋅ -⋅ -» ⋅ ⋅ ⋅ » ⋅ ⋅ ⋅ -ì ü ï ï ï ï ï ï ï ï » ⋅ ⋅ ⋅íï - ýï ï ï ï ï ï ï î þò
ò
ò
[ ]
(
)
(
)
2 ' cos 2 0 2 ' cos ' cos 0 2 2 2 ' cos ' cos 0 0 1 2 0 0 ' ' ' ' ' ' ' ' ' ' ' 22 cos ' cos ' ' cos cos
2 2
sin cos cos cos
2 2 2 cos 2 jkz l l jkz jkz l l l jkz jkz l z e dz l z z e dz e dz l l z z e dz e dz l l z l kl kz dz z d l l kl kl l kl q q q q q q x x x q x q q q ⋅ -⋅ ⋅ -⋅ - ⋅ = -= + é ù = ⋅ ⋅ = = = ⋅ ê ⋅ ú⋅ ë û -= +
ò
ò
ò
ò
ò
ò
ò
(
)
2 1 cos 2 kl q ì ü ï ï ï ï ï ï ï ï ï ï í ý ï ï ï ï ï ï ï ï ï ï î þ Thus:(
)
(
)
0 2 1 cos cos 2 ( ) 4 cos 2 jkr z kl l I e A r a r kl q m p q - -» ⋅ ⋅ ⋅ The r , q , and f components of the vector potential can be found as:
(
)
(
)
0 2 1 cos cos 2 cos cos 4 cos 2 jkr r z kl l I e A A r kl q m q q p q - -= = ⋅ ⋅(
)
(
)
0 2 1 cos cos 2 sin sin 4 cos 2 0 jkr z kl l I e A A r kl A q f q m q q p q - -= - = - ⋅ ⋅ = In far-field zone:(
)
(
)
0 2 0; 0 1 cos cos 2 sin 4 cos 2 0; 0; r jkr r E E kl l I e E j A j r kl E H H H f q q q q f q m w w q p q h -» » -» - = ⋅ ⋅ » » =b) By using a formula from the script, the far-zone field of a finite length dipole can be found as: ( ) / 2 / 2 ' cos / 2 / 2 sin ', ', ' ' 4 l jkr l jkz e l l k e E dE j I x y z e dz r q q q h p q -- -⋅ =
ò
= ⋅ò
⋅In our case, we have:
( )
/ 2 ' cos 0 / 2 ' sin cos ' 4 l jkr jkz l k e z E j I e dz r l q q h p q p -⋅ = ⋅ò
⋅By letting a = jkcos and q b = p/l, and using the following integral formula
( ) ( cos(2 ) 2 sin( )) cos bz e dzaz eaz a bz b bz a b + ⋅ = +
ò
we may write( )
( )( )
( )( )
( ) / 2 ' cos 0 2 2 / 2 cos cos 2 2 0 2 2 2 2 0 ' 'sin cos cos sin
4 cos sin 4 cos cos sin 4 l jkr jkz l jkl jkl jkr jkr k e e z z E j I jk r k l l l l k e e e j I r k l k l l l I k e j r q q q q p p p h q p q p q p p h q p p p q q p h q p -ì ü ï ï ï ï ⋅ ïï é ùïï = ⋅ íï ⋅ê ⋅ + úýï ë û ï - ï ï ï ï ï î þ é ù ê ú ⋅ ê ú = ⋅ ê + ú ê - - ú ê ú ë û ⋅ = ⋅
(
)
( )
( )(
)
2 2 0 cos 2 cos 2 cos cos cos 2 2 sin jkr kl l k l I e j r q p q p q h p q -⋅ = ⋅(
)
( )
( )(
)
0 2 2 0 cos 2 cos 2 sin 4 cos cos cos 2 2 sin jkr jkr kl I k e H j r l k l I e j r f q p q p p q p q p q -⋅ = ⋅ -⋅ = ⋅Wire antennas
Problem 4.4 :
A center-fed electric dipole of length l is attached to a balanced lossless transmission line whose characteristic impedance is 50W. Find the input VSWR when
a) l = l/ 4, b) l = l/ 2, c) l = 3 / 4l , d) l = l.
( )
in 0 rad in 0 2 in 0 1 VSWR , , , 50 1 sin 2 R Z R R kl Z R Z + G -= G = = = W - G + a) 4, 2 4, 2, 2 kl l = l = p kl = p kl = p( )
( )
( )
( )( )
{
}
[ ]{
}
rad 60 ln 2 2 12sin 2 2 2 1 60 0.5772 0.45158 0.470 1.85 2 1.3698 2 6.8388 i i i R = C + p -C p + p ⋅êéS p - S p ùú ë û = + - + - ⋅ = WThe input impedance is calculated as
( )
( )
rad in 2 2 6.8388 13.6776 sin sin 4 2 R R = kl = p = WThe reflection coefficient 13.6776 50 0.5704 VSWR 3.6555 13.6776 50 -G = = - = + b) 2, 2 2, , 2 2 kl l = l = p kl = p kl = p
( ) ( ) ( )
( )
( ) ( ){
}
[ ]{
}
rad 60 ln 12cos ln 2 2 2 1 60 0.5772 1.14473 0.059 0.5772 0.45158 0.0227 2 0.059 2 73.13 i i i R = C + p -C p + p ⋅êéC + p +C p - C p ùú ë û = + - - + - - ⋅ = W( )
( )
rad in 2 2 73.13 73.13 sin sin 2 2 R R kl p = = = W 73.13 50 0.18785 VSWR 1.4626 73.13 50 -G = = = + c) 3 3 3 , , , 2 3 4 kl2 4 2 l = l = p kl = p kl = p( ) ( )
( )
( )( )
{
}
[ ]{
}
rad 60 ln 34 32 12sin 32 3 2 32 1 60 0.5772 1.5502 ( 0.19839) 1.67473 2 1.611 2 185.965 i i i R = C + p -C p + p ⋅êéS p - S p ùú ë û = + - - - - ⋅ = W( )
( )
rad in 2 2 185.965 371.93 3 sin sin 4 2 R R = kl = p = W 371.93 50 0.7630 VSWR 7.4386 371.93 50 -G = = = + d) l = l, kl2 = p, kl = 2 ,p 2kl = 4p ( ) ( ) ( ) [ ( ) ( ) ( )]{
}
{ [ ]}
rad 60 ln 2 2 12cos 2 ln 4 2 2 60 0.5772 1.8378 ( 0.0227) 1 0.5772 1.14473 0.006 2 ( 0.0227) 2 199.099 i i i R = C + p -C p + p ⋅ C + p +C p - C p = + -+ + - ⋅ -= W( )
( ) rad in 2 2 199.099 sin sin 2 R R kl p = = = ¥ 50 50 1 1 VSWR 50 50 1 ¥ - - ¥ G = = = = ¥ ¥ + + ¥Wire antennas
Problem 4.5 :
A linear half-wavelength dipole is operating at a frequency of 1 GHz. Determine the capacitance or inductance that must be placed across the input terminals of the dipole so that the antenna becomes resonant (make the total input impedance real). What is then the VSWR of the resonant half-wavelength dipole when it is connected to a 50Ω line ?
, 50 2 c l = l Z = W ( ) in 73 42.5 Z = +j W ( ) in in in in 1 1 0.01023 0.0059563 S 73 42.5 Y j G jB Z j = = = - = -+
The capacitance that has to be put across the antenna terminals is
in in 2 in B = wC = pfC and thus 3 in 12 in 2 5.9563 109 0.94797 10 F 2 10 0.94797 pF B C f p p -⋅ = = = ⋅ ⋅ = .
The resonant antenna will only have
3 in 10.23 10 S G = ⋅ -, i.e. in in 1 97.75 R G = = W
The reflection coefficient at the input terminals is
in in in in in 0.3232 1 VSWR 1.955 1 c c R Z R Z -G = = + + G = = - G
