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 Δφ = 0 , a) Ex = Ey and b)

E x ≠ Ey

and

Δφ = 0 ,

c)

E x = Ey

and

Δφ = π / 2 ,

d)

E x = Ey

and

Δφ = −π / 2 ,

e)

E x = Ey

and

Δφ = π / 4 ,

f)

E x = Ey

and

Δφ = −π / 4 ,

g)

Ex = 0.5Ey

and

Δφ = π / 2 ,

h)

Ex = 0.5Ey

and

Δφ = −π / 2 .

In all cases justify the answer.

a)

Linear because +φ = 0 .

b)

Linear because +φ = 0 .

c)

Circular because Ex = Ey and +φ = π 2 , LHCP / CCW, AR = 1 .

d)

Circular because Ex = Ey and +φ = −π 2 , RHCP / CW, AR = 1 .

e)

Elliptical because Δφ is not multiples of π 2 , CCW, 1/ 2

OA = E 0 ⋅ [ 0.5 ( 1 + 1 + 2 ) ]

1/ 2

AR : Ex = Ey = E 0 : OB = E 0 ⋅ [ 0.5 ( 1 + 1 − 2 ) ] ⇒ AR =

= 1.30656 ⋅ E 0 ⎫⎪⎪ ⎪⎬ ⇒ = 0.541196 ⋅ E 0 ⎪⎪⎪ ⎭

1.30656 = 2.414 0.541196

1 1 ⎡ 2 ⋅ 1 ⋅ cos ( 45D ) ⎤ τ = 90D − ⋅ tan−1 ⎢ ⎥ = 90D − ⋅ ( 90D ) = 45D 2 1−1 2 ⎢⎣ ⎥⎦


f)


Elliptical because Δφ is not multiples of π 2 , CW,

AR = OA / OB :

OA = 1.30656 ⋅ E 0 ⎫ ⎪ 1.30656 ⎪ = 2.414 ⎬ ⇒ AR = 0.541196 OB = 0.541196 ⋅ E 0 ⎪ ⎪ ⎭

1 τ = 90D − ⋅ ( 90D ) = 45D 2

g)

Elliptical because: Ex ≠ Ey AND Δφ is not zero nor multiples of π . CCW,

AR = OA / OB . ⎫⎪ 1 ⎪⎪ =2 ⎬ ⇒ AR = 1/ 2 0.5 OB = Ey ⋅ [ 0.5 ( 0.25 + 1 − 0.75 ) ] = 0.5 ⋅ Ey ⎪⎪⎪ ⎭ 1 0 1 ⎡ ⎤ D D D τ = 90D − ⋅ tan−1 ⎢ ⎥ = 90 − ⋅ ( 0 ) = 90 2 0.75 2 − ⎣ ⎦

OA = Ey ⋅ [ 0.5 ( 0.25 + 1 + 0.75 ) ]1/ 2 = Ey

h)

Elliptical because: Ex ≠ Ey AND Δφ is not zero nor multiples of π . CW,

AR = OA / OB . OA = Ey

⎫ ⎪ 1 ⎪ =2 ⎬ ⇒ AR = 0.5 OB = 0.5 ⋅ Ey ⎪ ⎪ ⎭ 1 τ = 90D − ⋅ ( 0D ) = 90D . 2



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.

z z θ φ

x

a)

φ

y

y

x

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

G G G ρw = ax cos φw + ay sin φw G G G ρa = ax cos φa + ay sin φa G G PLF = ρwT ⋅ ρa

2

2 T G G G G = ( ax cos φw + ay sin φw ) ⋅ ( ax cos φa + ay sin φa )

PLF = cos φw ⋅ cos φa + sin φw ⋅ sin φa

b)

2

= cos ( φw − φa )

2

arbitrarily directed polarization vectors:

G G G G ρw = ax sin θw cos φw + ay sin θw sin φw + az cos θw G G G G ρa = ax sin θa cos φa + ay sin θa sin φa + az cos θa G G 2 PLF = ρwT ⋅ ρa T G G G = ( ax sin θw cos φw + ay sin θw sin φw + az cos θw ) ⋅ 2 G G G ⋅ ( ax sin θa cos φa + ay sin θa sin φa + az cos θa )

= sin θw cos φw sin θa cos φa + sin θw sin φw sin θa sin φa + cos θw cos θa = sin θw sin θa ⋅ ( cos φw cos φa + sin φw sin φa ) + cos θw cos θa = sin θw sin θa ⋅ cos ( φw − φa ) + cos θw cos θa

2

2

2



Problem 3.3 : A linearly polarized wave traveling in the negative z-direction has a tilt angle τ of 45°. It is G G 4ax + jay G ρa = 17 incident upon an antenna whose polarization characteristics are given by Find the polarization loss factor (PLF) dimensionless and in dB.

Polarization vector of the linearly polarized wave :

G G ax + ay G ρw = 2

Polarization vector of the elliptically polarized wave :

G G 4ax + jay G ρa = 17

2

G G 2 1 1 ⎛ 4⎞ 17 PLF = ρ wT ⋅ ρ a = (1 1) ⋅ ⎜ j ⎟ = 34 = 0.5 = −3dB 2 17 ⎝ ⎠

z τ

elliptical LH / CCW

y x



Problem 3.4 :

Problem taken from a previous exam

A right-hand circularly polarized antenna has normalized electric far-field pattern given as:

a) b) c)

a)

⎧ sin θ ⋅ cos φ 0D ≤ θ ≤ 180D, − 90D ≤ φ ≤ 90D ⎪ ⎪ E ( θ, φ ) = ⎨ ⎪ 0 elsewhere ⎪ ⎩ Calculate the direction (θ,φ) of maximum radiation and the exact maximum directivity in dB. Find the 3-dB beamwidths in azimuthal and elevation planes. 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.

2 1 E ( θ, φ ) 2η

U =

D0 =

Prad =

π 2

1

1 = 2η

π 2

θ = 0 φ =−

2 π 2

π

∫

sin2 θ ⋅ d θ ⋅

θ =0

Thus

4π ⋅

2

1 π

∫ π cos φ ⋅ d φ = 2η 2 ⋅ 2

φ =−

π = 2η

D0 =

π

∫ ∫ πU sin θ ⋅ d θ ⋅ d φ = 2η ∫ ∫ π sin2 θ ⋅ cos φ ⋅ d θ ⋅ d φ

θ = 0 φ =−

Prad

1 @ θ = 90D & φ = 0D 2η

4πU max Prad π

Prad

U max =

Æ

2

1 2η

=4 π 2η D0 = 10 ⋅ log 4 = 6.02 dB

b)

D Azimuthal plane ( θ = 90 ): U 1 = 1 @ φ1 = 0D

()

1 1 @ 0.5 = cos φ2 ⇒ φ2 = cos−1 = 60D 2 2 D D HPBW azimuthal = 2 ⋅ 60 = 120

U2 =

D Elevation plane ( φ = 0 ):


U1 = 1

@


θ1 = 90D

()

1 1 @ 0.5 = sin θ2 ⇒ θ2 = sin−1 = 30D 2 2 HPBW elevation = 2 ⋅ ( 90D − 30D ) = 120D

U2 =

c)

the polarization vector of the RHCP receiving antenna at origin, looking in +xdirection, is G G G a y − ja z ρa = 2 The polarization vector of the incident wave (-x-direction, CW elliptically polarized) is G G G 2a y + ja z ρw = 5 Therefore, the PLF reads G G G G G G 2 a y − ja z 2a y + ja z PLF = ρa ⋅ ρw = ⋅ 5 2

2

=

2+1 2 9 = 10 10

z circular RHCP

y elliptical

x

CW



Problem 3.5 : A λ / 2 dipole, with a total loss resistance of 1 Ω , is connected to a generator whose internal impedance is 50 + j 25 Ω . Assuming that the peak voltage of the generator is 2 V and the impedance of the dipole, excluding the loss resistance, is 73 + j 42.5 Ω , find the power. a) b) c)

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

a RL Vg

Ig

Rg Xg

Rrad

b XA

Rg = 50 Ω, X g = 25 Ω, RL = 1 Ω, Rrad = 73 Ω, X A = 42.5 Ω Ig =

2 2 A= A 124 + j 67.5 ( 50 + 1 + 73 ) + j ( 25 + 42.5 )

I g = ( 12.442 + j 6.7724 ) mA = 14.166 ⋅ e− j 28.56 mA D

a)

PS =

1 1 Re {Vg I g * } = Re { 2 ⋅ ( 12.442 + j 6.7724 ) ⋅ 10−3W } → PS = 12.442 ⋅ 10−3 W 2 2

b)

Prad =

c)

PL =

2 1 I g Rrad = 7.325 ⋅ 10−3 W 2

1 2 I g RL = 0.1003 ⋅ 10−3 W 2

The remaining supplied power is dissipated as heat in the internal resistor of the generator, or 1 2 Pg = I g Rg = 5.0169 ⋅ 10−3 W 2 Prad + PL + Pg = ( 7.325 + 0.1003 + 5.0169 ) ⋅ 10−3 W=12.442 ⋅ 10−3 W Thus, Prad + PL + Pg = PS



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) b)

Calculate the maximum effective aperture of the antenna with the assumption of a matched load and a matched polarization. A generator with an internal impedance of Z G = 50 + j 20 Ω and a peak generator voltage of VG = 4 V is connected to the antenna. Calculate the peak electric field strength at z = 1km .

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).

RL= 5Ω y

x

Rr= 45Ω

z

a)

⎛ Rr ⎞ ecd = ⎜ ⎟ = 0.9 ⎝ Rr + RL ⎠

b)

,

D0 = 10

⎛ 12dBi ⎞ ⎜ ⎟ ⎝ 10 ⎠

= 15.848

,

⎛ λ2 Aem = ecd ⎜ ⎝ 4π

⎞ 2 ⎟ D0 = 0.011m ⎠

The E-field strength at R = z = 1 km distance can be calculated from the local power E2 density Wt according to: Wt = (free-space wave impedance η). The radiated power Prad 2η 1 is related as Wt = Prad D0 , where D0 = 12 dBi = 15.85. Using ZG = Rg +jXg , the 4π R 2 2 ⎤ Vg ⎡ RR ⎢ ⎥ 34.62 mW , radiated power can be calculated: Prad = 2 ⎢ ( RR + RL + Rg )2 + ( X A + X g )2 ⎥ ⎣ ⎦ P D 2η V where XA = 0. Finally, E = rad 0 2 = 5.74 ⋅ 10−3 4π R m

c)

ρ t ρ r = cos2 450 = 0.5 2



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.

TA … background radiation weighted with antenna pattern Tcable … cable temperature ILcable … cable insertion loss

Tx … effective temperature at receiver input

Determine cable insertion loss from :

ILcable = exp(−2αL)

Calculate effective temperature according to :

Tx = TAe−2αL + Tcable ( 1 − e−2αL )

Transfrom Fahrenheit to Kelvin:

Tcable = 72D F =

5 (72 − 32) + 273 = 295.2 K 9

TA = 5 K Insertion loss over 100 ft :

−4[dB] = 10 log10 e −2α → α = 0.46[Nepers]

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



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-tonoise 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: f = 9 GHz → λ = 3 cm , R = 104 m , G0t = G0r , Pt = 10 W Pr = 10 μ W Friis’ transmission equation:

⎛ λ ⎞⎟2 2 Pr = ⎜⎜ ⎟⎟ G0 = 10−6 ⎜ Pt ⎝ 4πR ⎠ Results: G0 = 10−3 ⋅ 12π ⋅ 105 = 3769.9 G0dB = 10 log 3769.9 = 35.76 dBi

Solution 3 Antenna

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