solution 3 antenna

Solutions 3 – Fundamentals Antennas and Propagation, Frühjahrssemester 2010

Problem 3.1 : A uniform plane wave is propagating in direction of the positive z-axis. Find the polarization (linear, circular, or elliptical), sense of rotation (CW or CCW), axial ratio (AR), and tilt angle τ (in degrees) for

a) x yE E= and 0φΔ = ,

b) x yE E≠ and 0φΔ = ,

c) x yE E= and /2φ πΔ = ,

d) x yE E= and /2φ πΔ = − ,

e) x yE E= and /4φ πΔ = ,

f) x yE E= and /4φ πΔ = − ,

g) 0.5x yE E= and /2φ πΔ = ,

h) 0.5x yE E= and /2φ πΔ = − .

In all cases justify the answer. a) Linear because 0φ = .

b) Linear because 0φ = .

c) Circular because x yE E= and 2φ π= ,

LHCP / CCW, AR 1= .

d) Circular because x yE E= and 2φ π= − ,

RHCP / CW, AR 1= .

e) Elliptical because φΔ is not multiples of 2π ,

CCW,

AR : 0x yE E E= = :

( )[ ]

( )[ ]

1/20 0

1/20 0

OA 0.5 1 1 2 1.30656

OB 0.5 1 1 2 0.541196

1.30656AR 2.4140.541196

E E

E E

⎫⎪= ⋅ + + = ⋅ ⎪⎪ ⇒⎬⎪= ⋅ + − = ⋅ ⎪⎪⎭

⇒ = =

( )( )11 2 1 cos 45 190 tan 90 90 45

2 1 1 2τ − ⋅ ⋅⎡ ⎤= − ⋅ = − ⋅ =⎢ ⎥−⎢ ⎥⎣ ⎦


f) Elliptical because φΔ is not multiples of 2π ,

CW,

AR OA/OB= : 0

0

OA 1.30656 1.30656AR 2.4140.541196OB 0.541196

E

E

= ⋅ ⎫⎪⎪ ⇒ = =⎬⎪= ⋅ ⎪⎭

( )190 90 452

τ = − ⋅ =

g) Elliptical because: x yE E≠ AND φΔ is not zero nor multiples of π .

CCW,

AR OA/OB= .

( )[ ]

( )[ ]

1/2

1/2

OA 0.5 0.25 1 0.75 1AR 20.5OB 0.5 0.25 1 0.75 0.5

y y

y y

E E

E E

⎫⎪= ⋅ + + = ⎪⎪ ⇒ = =⎬⎪= ⋅ + − = ⋅ ⎪⎪⎭

( )11 0 190 tan 90 0 902 0.75 2

τ − ⎡ ⎤= − ⋅ = − ⋅ =⎢ ⎥−⎣ ⎦

h) Elliptical because: x yE E≠ AND φΔ is not zero nor multiples of π .

CW,

AR OA/OB= . OA 1AR 2

0.5OB 0.5y

y

E

E

= ⎫⎪⎪ ⇒ = =⎬⎪= ⋅ ⎪⎭

( )190 0 902

τ = − ⋅ = .


Problem 3.2 : Write a general expression for the polarization loss factor (PLF) of two linearly polarized antennas if a) both lie in the same plane, b) both lie in different planes.

a) assuming that both polarization vectors are in the xy-plane:

cos sincos sin

w x w y w

a x a y a

a aa a

ρ φ φρ φ φ

= +

= +

( ) ( )

( )

22

22

PLF cos sin cos sin

PLF cos cos sin sin cos

TTw a x w y w x a y a

w a w a w a

a a a aρ ρ φ φ φ φ

φ φ φ φ φ φ

= ⋅ = + ⋅ +

= ⋅ + ⋅ = −

b) arbitrarily directed polarization vectors:

sin cos sin sin cossin cos sin sin cos

w x w w y w w z w

a x a a y a a z a

a a aa a a

ρ θ φ θ φ θρ θ φ θ φ θ

= + +

= + +

( )

( )

( )

2

2

2

2

PLF

sin cos sin sin cos

sin cos sin sin cos

sin cos sin cos sin sin sin sin cos cos

sin sin cos cos sin sin cos cos

sin sin

Tw a

Tx w w y w w z w

x a a y a a z a

w w a a w w a a w a

w a w a w a w a

w a

a a a

a a a

ρ ρ

θ φ θ φ θ

θ φ θ φ θ

θ φ θ φ θ φ θ φ θ θ

θ θ φ φ φ φ θ θ

θ θ

= ⋅

= + + ⋅

⋅ + +

= + +

= ⋅ + +

= ( ) 2cos cos cosw a w aφ φ θ θ⋅ − +

x y

z

φ

θ

x y

z

φ


Problem 3.3 : A linearly polarized wave traveling in the negative z-direction has a tilt angle τ of 45°. It is

incident upon an antenna whose polarization characteristics are given by 4

17x y

aa ja

ρ+

=

Find the polarization loss factor (PLF) dimensionless and in dB.

Polarization vector of the linearly polarized wave : 2

x yw

a aρ

+=

Polarization vector of the elliptically polarized wave : 4

17x y

a

a jaρ

+=

( )2

2 41 1 171 1 0.5 3dB342 17

Tw aPLF

jρ ρ ⎛ ⎞= ⋅ = ⋅ = = = −⎜ ⎟

⎝ ⎠

x y

z

elliptical

LH / CCW

τ


Problem 3.4 : Problem taken from a previous exam A right-hand circularly polarized antenna has normalized electric far-field pattern given as:

( )sin cos 0 180 , 90 90

,0 elsewhere

Eθ φ θ φ

θ φ⎧ ⋅ ≤ ≤ − ≤ ≤⎪⎪= ⎨⎪⎪⎩

a) Calculate the direction (θ,φ) of maximum radiation and the exact maximum directivity in dB.

b) Find the 3-dB beamwidths in azimuthal and elevation planes. c) A CW elliptically polarized plane wave propagates along –x-direction towards the

antenna. The major axis of the ellipse is positioned along the y-axis and is twice as large as the minor axis. Find the polarization loss factor (PLF). Note: It is assumed that the antenna is placed at the center of the coordinate system.

a) ( ) 21 ,2

U E θ φη

= max1 @ 90 & 02

U θ φη

= = =

max

0rad

4 UDPπ

=

2 22

rad0 0

2 2

22

rad0

2

rad

1sin sin cos2

1 1sin cos 22 2 2

2

P U d d d d

P d d

P

π ππ π

π πθ θφ φ

ππ

πθ φ

θ θ φ θ φ θ φη

πθ θ φ φ

η η

πη

= ==− =−

= =−

= ⋅ ⋅ = ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅

=

∫ ∫ ∫ ∫

∫ ∫

Thus

0

0

142 4

210 log 4 6.02dB

D

D

πη

πη

⋅= =

= ⋅ =

b) Azimuthal plane ( 90θ = ):

( )1 1

12 2 2

azimuthal

1 @ 0

1 1@ 0.5 cos cos 602 2

HPBW 2 60 120

U

U

φ

φ φ −

= =

= = ⇒ = =

= ⋅ =

Elevation plane ( 0φ = ):


( )( )

1 1

12 2 2

elevation

1 @ 90

1 1@ 0.5 sin sin 302 2

HPBW 2 90 30 120

U

U

θ

θ θ −

= =

= = ⇒ = =

= ⋅ − = c) the polarization vector of the RHCP receiving antenna at origin, looking in +x-

direction, is

2y z

aa ja

ρ−

=

The polarization vector of the incident wave (-x-direction, CW elliptically polarized)

is 2

5y z

wa ja

ρ+

=

Therefore, the PLF reads

2 22 2 2 1 9PLF

5 102 10y z y z

a wa ja a ja

ρ ρ− + +

= ⋅ = ⋅ = =

x

y

z

circular

RHCP

elliptical

CW


Problem 3.5 : A /2λ dipole, with a total loss resistance of 1 Ω , is connected to a generator whose internal impedance is 50 25j+ Ω . Assuming that the peak voltage of the generator is 2 V and the

impedance of the dipole, excluding the loss resistance, is 73 42.5j+ Ω , find the power.

a) supplied by the source (real), b) radiated by the antenna, c) dissipated by the antenna.

50 , 25 , 1 , 73 , 42.5g g L rad AR X R R X= Ω = Ω = Ω = Ω = Ω

( ) ( )( ) 28.56

2 2A A124 67.550 1 73 25 42.5

12.442 6.7724 mA 14.166 mA

g

jg

Ijj

I j e−

= =++ + + +

= + = ⋅

a)

( ) * 3 31 1Re Re 2 12.442 6.7724 10 12.442 10 W2 2S g g SP V I j W P− −= = ⋅ + ⋅ → = ⋅

b) 2 31 7.325 10 W

2rad g radP I R −= = ⋅

c) 2 31 0.1003 10 W2L g LP I R −= = ⋅

The remaining supplied power is dissipated as heat in the internal resistor of the generator, or

2 31 5.0169 10 W2g g gP I R −= = ⋅

Thus, ( ) 3 37.325 0.1003 5.0169 10 W=12.442 10 Wrad L g

rad L g S

P P P

P P P P

− −+ + = + + ⋅ ⋅

+ + =

RL

XA

Rrad

Xg

Rg

Vg Ig

a

b


Problem 3.6 : Problem taken from a previous exam Assume a horn antenna located at the origin of the coordinate system. Its E-plane lies in the yz-plane and its H-plane lies in the xz-plane. The antenna is operated at 3 GHz. The figure below shows the orientation of the antenna and its equivalent circuit. The radiation resistance is RR = 45 Ω and the loss resistance is RL = 5 Ω. The maximum directivity of the antenna appears in the +z-direction and is D0 = 12 dBi. a) Calculate the maximum effective aperture of the antenna with the assumption of a

matched load and a matched polarization. b) A generator with an internal impedance of G 50 20Z j= + Ω and a peak generator

voltage of G 4 VV = is connected to the antenna. Calculate the peak electric field

strength at 1kmz = .

c) A lossless dipole antenna is located at z = 1 km and lies in the plane parallel to the xy-plane. The E-field of the antenna is parallel to the xy-plane and assumes an angle of 45° with respect to the x-axis. Calculate the polarization loss factor (PLF).

y

x

z

RL= 5Ω

Rr= 45Ω

a) cd 0.9r

r L

ReR R

⎛ ⎞= =⎜ ⎟+⎝ ⎠

, 12dBi

100 10 15.848D

⎛ ⎞⎜ ⎟⎝ ⎠= = ,

22

em cd 0 0.011m4

A e Dλπ

⎛ ⎞= =⎜ ⎟

⎝ ⎠

b) The E-field strength at R = z = 1 km distance can be calculated from the local power

density Wt according to: 2

2tEWη

= (free-space wave impedance η). The radiated power Prad

is related as rad 0 2

14tW P D

Rπ= , where D0 = 12 dBi = 15.85. Using ZG = Rg +jXg , the

radiated power can be calculated: ( ) ( )

2

rad 2 2 34.62 mW2g R

R L g A g

V RPR R R X X

⎡ ⎤⎢ ⎥=⎢ ⎥+ + + +⎣ ⎦

,

where XA = 0. Finally, 3rad 02

2 V5.74 104 m

P DER

ηπ

−= = ⋅

c) 2 2 0cos 45 0.5t rρ ρ = =


Problem 3.7 : The effective antenna temperature of an antenna looking towards zenith is approximately 5 K. Assuming that the temperature of the connected transmission line (waveguide) is 72 °F, find the effective temperature at the receiver terminals when the attenuation of the transmission line is given as 4 dB / 100 ft, and its length is : a) 2 ft, b) 100 ft.

Determine cable insertion loss from : exp( 2 )cableIL Lα= − Calculate effective temperature according to : ( )2 21L L

x A cableT T e T eα α− −= + −

Transfrom Fahrenheit to Kelvin: 572 (72 32) 273 295.2 K9cableT F= = − + =

5 KAT =

Insertion loss over 100 ft : 2104[dB] 10 log 0.46[Nepers]e α α−− = → =

Insertion loss for 1 ft : α = 0.0046 Np / ft a) L = 2 ft : Tx = 4.91 K + 5.38 K = 10.3 K b) L = 100 ft : Tx = 1.99 K + 177.5 K = 179.5 K

TA … background radiation weighted with antenna pattern Tcable … cable temperature

ILcable … cable insertion loss

Tx … effective temperature at receiver input


Problem 3.8 : In a long-range microwave communication system operating at 9 GHz, the transmitting and receiving antennas are identical, and they are separated by 10 000 m. To meet the signal-to-noise ratio of the receiver, the received power must be at least Pr = 10 μW. Assuming the two antennas are aligned for a maximum reception to each other, including being polarized matched, what should the gains (in dB) of the transmitting and receiving antennas be when the input power to the transmitting antenna is 10 W ? Data given: 9 GHz 3 cmf λ= → = , 410 mR = , 0t 0G rG= , 10 WtP = 10 WrP μ= Friis’ transmission equation:

22 60 10

4r

t

PG

P Rλπ

−⎛ ⎞⎟⎜= =⎟⎜ ⎟⎟⎜⎝ ⎠

Results:

3 50

0dB

10 12 10 3769.910 log 3769.9 35.76dBi

GG

π−= ⋅ ⋅ == =

solution 3 antenna

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