Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
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.
s = pa 2 = 25pl2 G0r = G 0t = 16.3dB G 0r = G 0t = 101.63 = 42.66
f = 10 GHz l = 3 cm æ l ö÷2 G0rG0t Pr ÷ = s çç çè 4pR R ÷÷ø 4p Pt 1 2
a)
R1 = R2 = 200 λ = 6 m :
Pr = 9 nW
b)
R1 = R2 = 500 λ = 15 m :
Pr = 0.23 nW
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
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 kl e - jkr I sin y 4p 0 r Ey kl e - jkr Hc » j I0 sin y = r 4p h Ey » j h
z χ
ψ
y
x 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 cos a = a ⋅ b . Therefore it is convenient to perform the following conversion sin y =
1 - cos2 y =
1 - aˆx ⋅ aˆr
2
where the vector aˆr in spherical coordinates can be expressed as aˆr = aˆx sin q cos f + aˆy sin q sin f + aˆz cos q
and therefore sin y =
1 - ( sin q cos f )2
The far field of the dipole oriented in y -direction can be written as Ey » j h
kl e - jkr I 4p 0 r
1 - ( sin q cos f )2
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
z
y x
When the radiation pattern stays the same as for an z -oriented dipole, also the
b)
directivity has to be the same: U = U 0 ( 1 - sin2 q cos2 f ) 2p p
2p p é ù 2 2 3 2 ê ú 1 sin q cos f sin q d q d f U = ) 0 ò ê ò ( sin q - sin q cos f )d q úd f = ò( ê úû 0 0 0 ë 0 W 2p é 2p ù 4 4 ù 8 é = U 0 êê ò 2d f - ò cos2 fd f úú = U 0 ê 4p - p ú = U 0 p 3 3 3 ë û êë 0 úû 0
Prad =
Ud W = U 0 ò òò
and D0 =
4pU 0 3 = = 1.5 8 2 U0 p 3
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
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) b)
( (
2 ì ï I0 1 + z ' ï ï l I z (z ') = ï í ï 2 ï I0 1 - z ' ï ï l î
I z (z ') = I 0 cos
a)
( (
( pl z ' )
2 ìï ïï I 0 1 + z ' l I z (z ') = ïí ïï 2 ïï I 0 1 - l z ' î 2 = I0 1 - z ' l
(
) )
-l / 2 £ z ' £ 0 0 £ z ' £ l /2
,
-l /2 £ z ' £ l /2
) ) )
.
-l / 2 £ z ' £ 0 0 £ z ' £ l /2 for
-l /2 £ z ' £ l /2
The magnetic vector potential can be found as m e - jkr ⋅ ⋅ A(r ) » az ⋅ 4p r
l
2
ò
I z ( z ' )e jkz '⋅cos qdz '
-l2
m I 0e- jkr » az ⋅ ⋅ ⋅ 4p r
l
2
ò
-l
2
(1 - 2l z ' )e (
)
jkz '⋅cos q
dz '
l ïìï sin kl cos q ïüï 2 ïï m I 0e z ' jkz '⋅cos q ïï 2 » az ⋅ ⋅ ⋅ íl -2ò e dz ' ý kl ïï ïï 4p r l l q cos - 2 ïîï ïþï 2 The integral in the brackets can be solved as - jkr
Solutions 4 – Fundamentals & Wire antennas l
2
ò
-l2
Antennas and Propagation, Frühjahrssemester 2011
z ' jkz '⋅cos q e dz ' l l
=
2
z ' jkz '⋅cos q e dz 'l
ò 0
l
=
0
ò
-l2
l
2
z ' jkz '⋅cos q e dz ' l
2
z ' jkz '⋅cos q z' e dz '+ ò e - jkz '⋅cos qdz ' l l
ò 0
0
l
2
= 2ò 0
1
z' 2 l é kl ù cos [ kz '⋅ cos q ] ⋅ dz ' = x = z ' = ò x ⋅ cos ê x ⋅ cos q ú ⋅ d x l l 2 ë2 û 0
(
)
( (
) )
ì üï kl kl ï ï ïï q q 1 sin cos cos cos lï 2 2 ï ïý = í + 2 kl ïï 2ï kl ï cos q cos q ï ïï ï 2 ï 2 î þï Thus:
(
)
kl cos q - jkr 1 - cos ml I 0e 2 A(r ) » az ⋅ ⋅ ⋅ 2 4p r kl cos q 2 The r , q , and f components of the vector potential can be found as: kl cos q - jkr 1 - cos ml I 0e 2 cos q Ar = Az cos q = ⋅ ⋅ 2 4p r kl cos q 2 kl 1 - cos cos q ml I 0e - jkr 2 ⋅ ⋅ sin q Aq = -Az sin q = 2 4p r kl cos q 2 Af = 0 In far-field zone: Er » 0; Ef » 0
(
)
(
)
(
)
(
(
E q » -j wAq = j w
H r » 0;
H q » 0;
- jkr
ml I 0e ⋅ 4p r
Hf =
)
)
( kl2 cos q ) sin q ( kl2 cos q )
1 - cos ⋅
2
Eq h
b) By using a formula from the script, the far-zone field of a finite length dipole can be found as: l /2
Eq =
ò
-l / 2
k ⋅ e - jkr sin q ⋅ dE q = j h 4pr
l /2
ò
-l / 2
Ie ( x ', y ', z ' ) ⋅ e jkz ' cos qdz '
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
In our case, we have: k ⋅ e - jkr Eq = j h sin q ⋅ I 0 4pr
l /2
ò
cos
-l / 2
( plz ' ) ⋅ e
jkz ' cos q
dz '
By letting a = jk cos q and b = p / l , and using the following integral formula
ò
cos (bz ) ⋅ eazdz =
eaz ( a cos (bz ) + b sin (bz ) ) a 2 + b2
we may write
Eq = j h
- jkr
k ⋅e 4pr
ìï üïl / 2 ïï jkz ' cos q e pz ' p pz ' ù ïïï é sin q ⋅ I 0 ïí jk cos q cos sin ⋅ ⋅ + ê úý ïï p 2 l l l û ïï 2 ë ïîï l - ( k cos q ) ïþï-l / 2
( )
jkl cos q jkl cos q é ù ê ú 2 2 k ⋅ e - jkr e p e p ú sin q ⋅ I 0 êê = jh + 2 ú p 2 p l l 4pr ê ( )2 - ( k cos q )2 úú êë l - k cos q l û kl cos q 2 cos I 0k ⋅ e - jkr p 2 sin q ⋅ = jh 2 l p 4pr - ( k cos q )2 l p I 0 ⋅ e - jkr cos 2 cos q = jh ⋅ 2pr sin q kl cos q 2 cos - jkr I k ⋅e p 2 sin q ⋅ Hf = j 0 4pr l p 2 - ( k cos q )2 l p I 0 ⋅ e - jkr cos 2 cos q = j ⋅ 2pr sin q
( )
( )
(
)
(
)
( ) ( )
( ) ( )
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
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 50 W . Find the input VSWR when a)
l = l/4,
b)
l = l /2 ,
c)
l = 3l / 4 ,
d)
l = l.
VSWR =
a)
1+ G , 1- G
l = l 4,
kl
G=
Rin - Z 0 , Rin + Z 0
p 2 = 4,
kl =
p , 2
Rin =
Rrad , kl sin2 2
Z 0 = 50W
( )
2kl = p
{ ( p2 ) - C ( p2 ) + 21 sin ( p2 ) ⋅ éêë S p - 2S ( p2 ) ùúû } 1 = 60 { 0.5772 + 0.45158 - 0.470 + [ 1.85 - 2 ⋅ 1.3698 ]} 2
Rrad = 60 C + ln
i
i
(
= 6.8388 W The input impedance is calculated as Rin =
Rrad 6.8388 = = 13.6776 W 2 p 2 kl sin sin 4 2
( )
( )
The reflection coefficient
G=
b)
13.6776 - 50 = -0.5704 VSWR = 3.6555 13.6776 + 50
l = l 2,
kl
p 2 = 2,
kl = p,
2kl = 2p
)
i
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
{ {
}
1 p + C i ( 2p ) - 2C i ( p ) ùú Rrad = 60 C + ln ( p ) - C i ( p ) + cos ( p ) ⋅ éêC + ln 2 2 ë û 1 = 60 0.5772 + 1.14473 - 0.059 - [ 0.5772 + 0.45158 - 0.0227 - 2 ⋅ 0.059 ] 2 = 73.13 W Rin =
G=
c)
( )
}
Rrad 73.13 = p = 73.13 W kl sin2 sin2 2 2
( )
( )
73.13 - 50 = 0.18785 73.13 + 50
l = 3l 4 ,
kl
3p 2= 4 ,
kl =
VSWR = 1.4626
3p , 2
2kl = 3p
{ ( 34p ) - C ( 32p ) + 21 sin ( 32p ) ⋅ éêë S 3p - 2S ( 32p )ùúû } 1 = 60 { 0.5772 + 1.5502 - (-0.19839) - 1.67473 - 2 ⋅ 1.611 } 2
Rrad = 60 C + ln
i
i
(
[
)
i
]
= 185.965 W Rin =
G=
d)
Rrad 185.965 = = 371.93 W 2 3p 2 kl sin sin 4 2
( )
( )
371.93 - 50 = 0.7630 371.93 + 50
l = l,
kl
2 = p,
kl = 2p,
VSWR = 7.4386
2kl = 4p
{
1 Rrad = 60 C + ln ( 2p ) - C i ( 2p ) + cos ( 2p ) ⋅ [C + ln ( p ) + C i ( 4p ) - 2C i ( 2p ) ] 2 = 60 { 0.5772 + 1.8378 - (-0.0227) 1 + [ 0.5772 + 1.14473 - 0.006 - 2 ⋅ (-0.0227) ] 2 = 199.099 W Rin =
G=
Rrad 199.099 = =¥ kl sin2 ( p ) sin2 2
( )
1 - 50 ¥ ¥ - 50 = =1 ¥ + 50 1 + 50 ¥
VSWR = ¥
}
}
Solutions 4 – Fundamentals & Wire antennas
Antennas and Propagation, Frühjahrssemester 2011
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 ?
l =
l , 2
Zc = 50 W
Z in = ( 73 + j 42.5 ) W
Yin =
1 1 = = ( 0.01023 - j 0.0059563 ) S = Gin - jBin Z in 73 + j 42.5
The capacitance that has to be put across the antenna terminals is
Bin = wC in = 2p fC in and thus
Bin 5.9563 ⋅ 10-3 = = 0.94797 ⋅ 10-12 F 2p f 2p ⋅ 109 = 0.94797 pF
C in =
The resonant antenna will only have
Rin =
Gin = 10.23 ⋅ 10-3 S
1 = 97.75 W Gin
The reflection coefficient at the input terminals is
Rin - Zc = 0.3232 Rin + Zc 1 + Gin = 1.955 VSWR = 1 - Gin
Gin =
. , i.e.
