solution 2 antenna

Solutions 2 – Fundamentals Antennas and Propagation, Frühjahrssemester 2010

Problem 2.1 : 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: 1mD = : 900 MHz 0.33 mcff

λ= ⇒ = =

Far-field distance is given by: ( )222 2 1 60.333f

Dd mλ

= = =

That is, far-field starts at 6 m distance and extends to infinity.


Problem 2.2 : 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 = 7.112mm, b = 3.556mm⇒ D = 7.951× 10-3m Reactive near-field region:

R < R1, where

R1 = 0.62D3

λ= 0.62

7.951× 10−3( )3

λ=

4.396 × 10−4

λ m( )

Radiating near-field (Fresnel) region

R1 ≤ R < R2 , where R2 =

2D2

λ=

1.264 × 10−4

λ m( )

Far-field region

R ≥ R2

a

b D


Problem 2.3 : A 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

Ar

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):

0.3arctan 26.60.6

α = = ° , 0.3cos 1 cos arctan 0.110.6

h r r r rα ⎡ ⎤⎛ ⎞= − = − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ , 2

2 0.66rh srrπΩ = =

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 πr 2 = 3.14 × 10−4


Problem 2.4 : The power radiated by a lossless antenna is 10 Watts. The directional characteristics of the antenna are represented by the radiation intensity of

30 cos (W/sr) 0 /2, 0 2U B θ θ π φ π= ≤ ≤ ≤ ≤

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 ⋅ dφθ =0

π2

∫φ =0

2π

∫ = B0 ⋅ cos3θ ⋅ sinθ ⋅ dθ ⋅ dφθ =0

π2

∫φ =0

2π

∫

Prad = 2π B0

cos4 θ4

⎛

⎝⎜⎞

⎠⎟0

π2

=π2

⋅ B0 = 10 ⇒ B0 =20π

= 6.3662

3 30( ) cos 6.3662cosU Bθ θ θ= =

W =Ur 2 =

6.3662r 2 cos3θ =

6.366210002 cos3θ = 6.3662 × 10−6 ⋅ cos3θ

Wmax = Wcosθ =1

= 6.3662 × 10−6 Watts/m2 @ θ = 0

b)

max0

4 4 6.3662 8 9.0310rad

UD dBP

π π ⋅= = = =

c)

The radiation half-power occurs when : 1

3 1 31

1 1cos cos 37.5 0.212 2θ θ

θ θ π−

=

⎛ ⎞= ⇒ = = ° =⎜ ⎟

⎝ ⎠

The half power beamwidth is given as : 1 12 0.42 74.9d θ πΘ = = = ° . 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

01 2

41,253 41,253 7.35 8.6674.9 74.9d d

D dB= = = =Θ Θ ⋅

. This compares quite well with the result of (b).

d)

cos3θ = 0

θ =θ2

⇒ θ2 =π2

= 90o ⇒ FNBW = 180°


Problem 2.5 : The normalized radiation intensity of a given antenna is given by:

1. 1 sin sinU θ φ= ⋅



The intensity exists only in 0 180 , 0 180θ φ≤ ≤ ≤ ≤ 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)

max1,2,3

,1,2,3

4

rad

UDPπ= 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 22radP U d d d d

π π π π

φ θ φ θ

πθ θ φ φ φ θ θ π= = = =

= = ⋅ = ⋅ =∫ ∫ ∫ ∫ , 14 (1) 4 6.02D dBπ

π= = =

2 2,2

0 0 0 0

sin sin sin2 2radP U d d d d

π π π π

φ θ φ θ

π πθ θ φ φ φ θ θ= = = =

= = ⋅ = ⋅∫ ∫ ∫ ∫ , 216 5.09 7.07D dBπ

= = =

3,3

0 0 0 0

4sin sin sin 23radP U d d d d

π π π π

φ θ φ θ

θ θ φ φ φ θ θ= = = =

= = ⋅ = ⋅∫ ∫ ∫ ∫ , 33 4.71 6.732

D dBπ= = =

using b) The half-power beamwidths are equal to

( )0 1 01, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0

1, 2 90 sin (1/ 2) 120elevationHPBW −= ⋅ − =

( )0 1 02, 2 90 sin (1/ 2) 90azimuthHPBW −= ⋅ − = ( )0 1 0


( )0 1 03, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0


c)

Directivity using approx. formulas ( 1,2,341253

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


Problem 2.6 : 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

= 101.75 = 56.234


Problem 2.7 : The normalized far-zone field pattern of an antenna is given by

( )1/22 3sin cos 0 , 0 and 22 2

0 elsewhereE

π πθ φ θ π φ φ π⎧⎪ ⋅ ≤ ≤ ≤ ≤ ≤ ≤⎪⎪= ⎨⎪⎪⎪⎩

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 120η π= Ω ,

2

21 sin cos2 2

EU θ φ

η η= = max

1 for and 02 2

U πθ φη

= = =

a)

Prad = 2 U sinθ ⋅ dθ ⋅ dφθ =0

π

∫φ =0

π / 2

∫ = 21

2ηsin2 θ ⋅cos2 φ ⋅ dθ ⋅ dφ

θ =0

π

∫φ =0

π / 2

∫ =1η

⋅π4

⋅π2

=π 2

8η

D0 =4πUmax

Prad

=4π 1

2ηπ 2

8η

=16π

= 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 θθη

= , max

1sin2

UU

θ= = o, 30HPBW Eθ = 0 0 0, 2(90 30 ) 120HPBW E = − =

H-plane HPBW: θ = π 2 :

2cos( )2

U φφη

= , 2

max

1cos2

UU

φ= = o, 45HPBW Hθ = 0 0, 2 45 90HPBW H = ⋅ =

solution 2 antenna

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