Find the far-field distance for an antenna with maximum dimension of 1 m and operating frequency of 900 MHz.
Largest dimension of the antenna: D =1m : f 900 MHz c 0.33 m f
l
= = =
Far-field distance is given by: 2 2 2 1( )2 6 0.333
f D
d m
l
= = =
An aperture antenna is radiating in free space. The dimensions of the aperture are a = 7.112 mm and b = 3.556 mm.
a) What are the three field regions of this antenna?
b) Determine the boundaries of these regions.
a = 7.112 mm, b = 3.556 mm D = 7.951× 10-3m Reactive near-field region:
R < R1, where R1= 0.62 D 3 λ = 0.62 7.951× 10−3
(
)
3 λ = 4.396× 10−4 λ( )
m Radiating near-field (Fresnel) regionR1≤ R < R2, where R2 = 2D 2 λ = 1.264× 10−4 λ
( )
m Far-field region R ≥ R2 a b DA circular disc is lying in the plane y = 0.6, whereas its center lies on the y-axis. The radius of the disc is r = 0.3.
a) Calculate the solid angle covered by this disc seen from the origin of the coordinate system.
b) Calculate the solid angle covered by this disc, if it is lying in the plane y = 30.
a)
The solid angle Ω which is covered by a certain object seen from the origin is given by:
2
A r
Ω = , where A stands for the surface of the sphere area “shadowed” by the object, and r denotes the distance to the origin.
α y z x 0.6 m 0.6 m h a r
The solid angle covered by the disc is calculated by using the formula to calculate the surface of a spherical calotte (Kugelhaube), which is given by A= 2πrh , where h is the height of the calotte (as shown in the figure above):
0.3
arctan 26.6
0.6
α = = ° , cos 1 cos arctan0.3 0.11
0.6 h= −r r α =r − = r , 2 2 0.66 rh sr r π Ω = =
Considering that “all directions” (full sphere) corresponds to 4π sr, this solid angle equals about 1/19 of a full sphere.
b)
The exact solution using the above formulas yields: α = 0.57°, h = 0.000’05, Ω = 0.000’314. An approximate formula can be used if the distance of the disc is large compared to its radius; then the surface angle is approximately given by: Ω = rdisc
2 π
r2 = 3.14 × 10
The power radiated by a lossless antenna is 10 Watts. The directional characteristics of the antenna are represented by the radiation intensity of
3
0cos (W/sr) 0 /2, 0 2
U =B q £ £q p £f £ p
a) Find the maximum power density (in watts per square meter) at a distance of 1000 m (assume far field distance). Specify the angle where this occurs.
b) Find the directivity of the antenna (dimensionless and in dB). c) Calculate the half-power beamwidth (HPBW).
d) Find the first-null beamwidth (FNBW).
a) Prad = U sinθ⋅ dθ⋅ dφ θ =0 π 2
φ =0 2π
= B0⋅ cos 3θ⋅sinθ⋅ dθ⋅ dφ θ =0 π 2
φ =0 2π
Prad = 2πB0 cos 4θ 4 0 π 2 =π 2⋅ B0 = 10 B0 = 20 π = 6.3662 3 3 0 ( ) cos 6.3662cos U θ =B θ = θ W =U r2 = 6.3662 r2 cos 3θ = 6.3662 10002 cos 3θ = 6.3662 × 10−6⋅cos3θ Wmax = W cosθ =1 = 6.3662 × 10 −6 Watts/m2 @ θ = 0 b) max 0 4 4 6.3662 8 9.03 10 rad U D dB P π π ⋅ = = = = c)The radiation half-power occurs when :
1 3 1 3 1 1 1 cos cos 37.5 0.21 2θ θ 2 θ θ − π = = = = ° =
The half power beamwidth is given as : Θ =1d 2θ1=0.42π =74.9°. The pattern is independent on φ, thus HPBW in a plane at right angle to the first one is Θ2d = Θ1d. The directivity can be estimated using an approximate formula (see lecture slides) as
0 1 2 41,253 41,253 7.35 8.66 74.9 74.9 d d D = = = = dB
Θ Θ ⋅ . This compares quite well with the result of (b). d) cos 3θ = 0 θ =θ2 θ2 =π 2 = 90 o FNBW = 180°
The normalized radiation intensity of a given antenna is given by: 1. U1 = sinq⋅sinf 2. 2 2 sin sin U = q⋅ f 3. 2 3 sin sin U = q⋅ f
The intensity exists only in 0 £ £q 180 , 0 £f £180 region, and is zero elsewhere.
a) Find the exact directivity (dimensionless and in dB).
b) Find the Azimuthal and elevation plane half-power beamwidths (in degrees). c) Find the directivity by using approximate formulas.
a) max 1,2,3 ,1,2,3 4 rad U D P π
= where Umax = 1, and it occurs when θ =φ =π / 2 (in all 3 cases 1,2,3)
2 ,1
0 0 0 0
sin sin sin 2
2 rad P U d d d d π π π π φ θ φ θ π θ θ φ φ φ θ θ π = = = = =
=
⋅
= ⋅ = , D1 4 (1)π 4 6.02dB π = = = 2 2 ,2 0 0 0 0sin sin sin
2 2 rad P U d d d d π π π π φ θ φ θ π π θ θ φ φ φ θ θ = = = = =
=
⋅
= ⋅ , D2 16 5.09 7.07dB π = = = 3 ,3 0 0 0 0 4sin sin sin 2
3 rad P U d d d d π π π π φ θ φ θ θ θ φ φ φ θ θ = = = = =
=
⋅
= ⋅ , 3 3 4.71 6.73 2 D = π = = dB using b)The half-power beamwidths are equal to
(
0 1)
0 1,azimuth 2 90 sin (1/ 2) 120 HPBW = ⋅ − − = HPBW1,elevation = ⋅2 90(
0−sin (1/ 2)−1)
=1200(
0 1)
0 2,azimuth 2 90 sin (1/ 2) 90 HPBW = ⋅ − − = HPBW2,elevation 2 90(
0 sin (1/ 2)1)
1200 − = ⋅ − =(
0 1)
0 3,azimuth 2 90 sin (1/ 2) 120 HPBW = ⋅ − − = HPBW3,elevation = ⋅2 90(
0−sin (1/ 2)−1)
=900 c)Directivity using approx. formulas ( 1,2,3 41253
azimuth elevation D ≈ Θ ⋅Θ , 1,2,3 2 2 72815 azimuth elevation D ≈ Θ + Θ ) :
This gives for case 1 (HPBWs of 120º,120º) : D ≈ 2.86 = 4.6 dB and D ≈ 2.53 = 4.0 dB It gives for cases 2,3 (HPBWs of 90º,120º) : D ≈ 3.82 = 5.8 dB and D ≈ 3.24 = 5.1 dB
The maximum of the radiation pattern of a horn antenna is +20 dB, while maximum of its first side-lobe is -15 dB. What is the difference between the two maxima
a) in dB,
b) as a ratio of the field intensities.
a) Difference = 20 dB – (-15) dB = 35 dB b) 20log10 Emax Esl = 35 log10 Emax Esl = 1.75 Emax Esl = 10 1.75 = 56.234
The normalized far-zone field pattern of an antenna is given by
(
2)
1/ 2 3sin cos 0 , 0 and 2
2 2 0 elsewhere E p p q f q p f f p ìï ⋅ £ £ £ £ £ £ ïï = í ïï ïî
where the E-field vector points in z-direction.
a) Find the directivity using the exact expression. b) Find HPBWs of E- and H-planes of the antenna.
Using free-space wave impedance η=120π Ω ,
2 2 1 sin cos 2 2 E U θ φ η η = =
max 1 for and 0
2 2 U θ π φ η = = = a) Prad = 2 U sinθ⋅ dθ⋅ dφ θ =0 π
φ =0 π / 2
= 2 1 2ηsin 2θ⋅cos2φ⋅ dθ⋅ dφ θ =0 π
φ =0 π / 2
= 1 η⋅ π 4 ⋅ π 2 = π2 8η D0 = 4πUmax Prad = 4π 1 2η π2 8η =16 π = 5.09 = 7.07 dB b)Maximum of the beam is directed along the x-axis, with E-field pointing in z-direction. Therefore, E-plane is elevation plane (φ = 0 ) and H-plane is azimuthal plane ( θ =π 2 ). E-plane HPBW: φ = 0 : sin ( ) 2 U θ θ η = , max 1 sin 2 U U = θ = o , 30 HPBW E θ = , 2(900 30 ) 1200 0 HPBW E= − = H-plane HPBW: θ =π 2 : 2 cos ( ) 2 U φ φ η = , 2 max 1 cos 2 U U = φ = o , 45 HPBW H θ = , 2 450 900 HPBW H = ⋅ =
