Dr. Dung Trinh
HCMUT / 2014
Antennas and Propagation Chapter 2: Fundamental Parameters of Antenna Dr. Dung Trinh
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Dr. Dung Trinh
HCMUT / 2014
Far field Radiation The electric field and magnetic field can be written in terms of magnetic vector potential A 1 and 𝐻 =𝛻×𝐴 𝐸 = −𝑗𝜔𝜇𝐴 − 𝛻 𝛻⋅𝐴 𝑗𝜔𝜀 ′
where: 𝐴(𝑟) = 𝑉
−𝑗𝑘|𝑟−𝑟 | 𝑒 𝐽(𝑟 ′ ) 𝑑𝑟 ′ ′ 4𝜋|𝑟 − 𝑟 |
For antenna and scattering problems, we usually interested in the spherical components of the E and H fields. Therefore the vector potential of A in spherical coordinates can be written as: 𝐴 = 𝐴𝑟 𝑟, 𝜃, 𝜙 𝑎𝑟 + 𝐴𝜃 𝑟, 𝜃, 𝜙 𝑎𝜃 + 𝐴𝜙 𝑟, 𝜃, 𝜙 𝑎𝜙
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Dr. Dung Trinh
HCMUT / 2014
Far field Radiation Near field
𝑅 = 𝑟 − 𝑟′ → |𝑅| = → |𝑅| =
𝑟 2 + 𝑟′2 − 2𝑟𝑟′ 𝑟 2 + 𝑟′2 − 2𝑟𝑟′cos𝜓
Far field
For phase variation: 𝑅 ≃ 𝑟 − 𝑟′cos𝜓 For amplitude variation:
where: 𝜓 = 𝑓(𝜃, 𝜃′, 𝜙, 𝜙′)
𝑅≃𝑟 3
Dr. Dung Trinh
HCMUT / 2014
Far field Radiation In the spherical coordinate:
Far field
Field point: 𝑥, 𝑦, 𝑧) = 𝑟sin𝜃cos𝜙, 𝑟sin𝜃sin𝜙, 𝑟cos𝜃
Source point: 𝑥′, 𝑦′, 𝑧′) = 𝑟′sin𝜃′cos𝜙′, 𝑟′sin𝜃′sin𝜙′, 𝑟′cos𝜃′ → cos𝜓 =
𝑟𝑟′ = cos𝜃cos𝜃′ + sin𝜃sin𝜃′cos 𝜙 − 𝜙′ 𝑟 ⋅ 𝑟′
→ 𝜓 = 𝑓(𝜃, 𝜃′, 𝜙, 𝜙′) ′
→ 𝐴(𝑟) = 𝑉
𝑒 −𝑗𝑘|𝑟−𝑟 | ′ 𝑒 −𝑗𝑘𝑟 ′ 𝐽(𝑟 ) 𝑑𝑟 ≃ 4𝜋|𝑟 − 𝑟 ′ | 4𝜋𝑟
𝐽(𝑟 ′ )𝑒 𝑗𝑘𝑟′cos𝜓 𝑑𝑟 ′ 𝑉
𝑒 −𝑗𝑘𝑟 ≃ 𝑁𝜃 𝜃, 𝜙 𝜃 + 𝑁𝜙 𝜃, 𝜙 𝑟
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Dr. Dung Trinh
HCMUT / 2014
Far field Radiation Then the electric field E in the far-field region can be written as: 𝐸 = −𝑗𝜔𝜇𝐴 −
1 𝛻 𝛻 ⋅ 𝐴 ≈ −𝑗𝜔𝜇𝐴 ≈ −𝑗𝜔𝜇 𝐴𝜃 𝑟 𝜃 + 𝐴𝜙 𝑟 𝜙 𝑗𝜔𝜀
𝑒 −𝑗𝑘𝑟 →𝐸≈ 𝐹𝜃 𝜃, 𝜙 𝜃 + 𝐹𝜙 𝜃, 𝜙 𝜙 𝑟 where:
𝐹𝜃 𝜃, 𝜙 = − 𝑗𝜔𝜇
𝐹𝜙 𝜃, 𝜙 = − 𝑗𝜔𝜇
1 4𝜋
1 4𝜋
𝐽(𝑟 ′ )𝑒 𝑗𝑘𝑟′cos𝜓 𝑑𝑟 ′ 𝜃 𝑉
𝐽(𝑟 ′ )𝑒 𝑗𝑘𝑟′cos𝜓 𝑑𝑟 ′ 𝜙 𝑉
And the magnetic field H in the far-field region can be written as: 1 𝐻 ≃ 𝑟×𝐸 𝑟 𝜂
1 𝑒 −𝑗𝑘𝑟 = 𝐹𝜃 𝜃, 𝜙 𝜙 − 𝐹𝜙 𝜃, 𝜙 𝜃 𝜂 𝑟 5
Dr. Dung Trinh
HCMUT / 2014
Radiation power density Electromagnetic waves are used to transport information through a wireless medium or a guiding structure. It is nature to assume that power and energy are associated with EM fields. The quantity used to describe the power associated with an EM wave is the instantaneous Poyting vector: 𝑆 =ℇ×ℋ 𝑆: instantaneous Poyting vector (W/m2) ℇ: instantaneous electric field intensity (V/m) ℋ: instantaneous magnetic field intensity (A/m)
It is often more desirable to find the average power density. For timeharmonic variations of the form ejwt: ℇ 𝑥, 𝑦, 𝑧, 𝑡 = 𝑅𝑒 𝐸 𝑥, 𝑦, 𝑧 𝑒 𝑗𝜔𝑡 ℋ 𝑥, 𝑦, 𝑧, 𝑡 = 𝑅𝑒 𝐻 𝑥, 𝑦, 𝑧 𝑒 𝑗𝜔𝑡 1 𝑗𝜔𝑡 = 𝐸𝑒 𝑗𝜔𝑡 + 𝐸 ∗ 𝑒 −𝑗𝜔𝑡 , we can write: Using the identity 𝑅𝑒 𝐸𝑒 2 𝟏 𝟏 𝑺 = ℇ × 𝓗 = 𝑹𝒆 𝑬 × 𝑯∗ + 𝑹𝒆 𝑬 × 𝑯𝒆𝒋𝟐𝝎𝒕 𝟐 𝟐 6
Dr. Dung Trinh
HCMUT / 2014
Radiation power density Then the average Poyting vector can be written as: 𝟏 𝑾 = 𝑹𝒆 𝑬 × 𝑯∗ 𝟐
Using the far-field form of E and H, we get: 𝑒 −𝑗.𝑘.𝑟 1 𝑒 𝑗𝑘𝑟 𝑊 (𝑟) = Re 𝐹𝜃 (𝜃, 𝜙). 𝜃 + 𝐹𝜙 (𝜃, 𝜙). 𝜙 × . 𝐹𝜃 (𝜃, 𝜙). 𝜙 − 𝐹𝜙 (𝜃, 𝜙). 𝜃 𝑟 𝜂 𝑟 𝑊 (𝑟) =
∗
1 |𝐹𝜃 (𝜃, 𝜙)|2 + |𝐹𝜙 (𝜃, 𝜙 |2 𝑟 2 2. 𝜂. 𝑟
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Dr. Dung Trinh
HCMUT / 2014
Radiation Intensity Radiation intensity U in a given direction is defined as “the power radiated from an antenna per unit solid angle”.
𝑑𝑆 = (𝑟. 𝑑𝜃)(𝑟. sin𝜃. 𝑑𝜙). 𝑟
Power radiated over an area dS is: 𝑊(𝑟). 𝑑𝑆 = 𝑊(𝑟). 𝑑𝛺. 𝑟 2 ⇒ 𝑈(𝑟) = 𝑟 2 . 𝑊(𝑟)
Radiation intensity does not depend on r, only depends on 𝜃 𝑎𝑛𝑑 𝜙.
The total power radiated is given by: 𝑃𝑅 =
𝑊 (𝑟). 𝑑𝑆 𝑆
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Dr. Dung Trinh
HCMUT / 2014
Total radiation power 𝑃𝑅 =
𝑊 (𝑟). 𝑑𝑆 𝑆
dS (r.d )(r.sin .d ).rˆ
𝑃𝑅 =
𝑊(𝑟). 𝑟. 𝑑𝑆 𝑆
2𝜋
𝜋
rˆ
𝑊(𝑟). 𝑟 . (𝑟. 𝑑𝜃)(𝑟. sin𝜃. 𝑑𝜙). 𝑟
𝑃𝑅 = 𝜙=0 2𝜋
𝜃=0
r
𝑊(𝑟). 𝑟 2 . sin𝜃. 𝑑𝜃. 𝑑𝜙 2𝜋
𝜋
𝑈(𝜃, 𝜙). sin𝜃. 𝑑𝜃. 𝑑𝜙
𝑃𝑅 =
ˆ
𝜃=0
𝑃𝑅 = 𝜙=0
M
𝜋
𝑃𝑅 = 𝜙=0
ˆ
𝜃=0
𝑈(𝜃, 𝜙). 𝑑𝛺 𝑆
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Dr. Dung Trinh
HCMUT / 2014
Total radiation power Example: An antenna radiated a field given by: 𝐹𝜃 𝜃, 𝜙 = sin𝜃
Calculate: a. Power density (Poyting) vector W(r)? b. Power intensity? c. Total radiation power?
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Dr. Dung Trinh
HCMUT / 2014
Polarization Polarization of an antenna in a given direction is defined as “the polarization of the wave transmitted (radiated) by the antenna. The polarization of a plane wave refers to the direction of the electric z field vector in the time domain.
y
E ( z, t ) x
We assume that the wave is traveling in the positive z direction. 11
Dr. Dung Trinh
HCMUT / 2014
Polarization Consider a plane wave with both x and y components y
Ey x
Ex Phasor domain: Assume:
E( z ) ( xˆ E x yˆ E y ) e jkz E x a real number E y be j
(In general, = phase of E y phase of E x ) 12
Dr. Dung Trinh
HCMUT / 2014
Polarization y
Time Domain:
E (t)
At z = 0: x
Ex Re ae jt a cos t E y Re be j e jt b cos t Depending on b/aonand three cases arise: Depending b/ab,and ,different three different cases arise: Linear polarization Circular polarization Elliptical polarization 13
Dr. Dung Trinh
HCMUT / 2014
Linear Polarization 0
or
E x a cos t
E y b cos t
E x a cos t
At z = 0:
E y b cos t
+ sign: = 0 - sign: =
E xˆ a yˆ b cos t y
x
y b
E
This is simply a “tilted” plane wave.
a
(shown for = 0 x 14
Dr. Dung Trinh
HCMUT / 2014
Circular Polarization b = a AND /2
At z = 0:
E x a cos t
Ex a cos t
E y b cos t
E y a cos(t / 2) a sin t
y
EE convention E (t ) a
/ 2
LHCP x
/ 2 RHCP
E Ex2 E y2 a 2 cos2 t a 2 sin2 t a 2 2
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Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Rotation in space vs. rotation in time Examine how the field varies in both space and time:
E( z ) xˆ a yˆ be j e jkz
Phasor domain
E ( z, t ) xˆ a cos t kz yˆ b cos t kz
Time domain
There is opposite rotation in space and time.
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Dr. Dung Trinh
HCMUT / 2014
Circular Polarization A snapshot of the electric field vector, showing the vector at different points. Notice that the rotation in space matches the left hand!
RHCP
z
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Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Animation of LHCP wave
http://en.wikipedia.org/wiki/Circular_polarization Circular polarization is most often use on satellite communications. This is particularly desired since the polarization of a linear polarized radio wave may be rotated as the signal passes through any anomalies (such as Faraday rotation) in the ionosphere.
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Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Circular polarization is often used in wireless communications to avoid problems with signal loss due to polarization mismatch. Misalignment of transmit and receive antennas Reflections off of building Propagation through the ionosphere Receive antenna
z
The receive antenna will always receive a signal, no matter how it is rotated about the z axis. However, for the same incident power density, an optimum linearly-polarized wave will give the maximum output signal from this linearly-polarized antenna (3 dB higher than from an incident CP wave). 19
Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Two ways in which circular polarization can be obtained: Method 1: Use two identical antenna rotated by 90o, and fed 90o out of phase. y
Antenna 2
/ 2
Antenna 1
Vy j 1 90
o
x
+
+
Vx 1
This antenna will radiate a RHCP signal in the positive z direction, and LHCP in the negative z direction. 20
Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Method 2: Use an antenna that inherently radiates circular polarization.
Helical antenna for WLAN communication at 2.4 GHz
Helical antennas on a GPS satellite
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Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Summary of possible scenarios 1) Transmit antenna is LP, receive antenna is LP Simple, works good if both antennas are aligned. The received signal is less if there is a misalignment.
2) Transmit antenna is CP, receive antenna is LP
Signal can be received no matter what the alignment is. The received signal is 3 dB less then for two aligned LP antennas. 3) Transmit antenna is CP, receive antenna is CP Signal can be received no matter what the alignment is. There is never a loss of signal, no matter what the alignment is. The system is now more complicated. 22
Dr. Dung Trinh
HCMUT / 2014
Circular Polarization Example: The electric field generated by an antenna in the far field has the form of: e j .k .r ˆ j.sin .ˆ E (r ) sin .cos . r Identify the polarization of the antenna in the direction: a. x+ b. xc. y+ d. y a. In the x+ direction: , 0, r x; ˆ zˆ, ˆ yˆ 2 j .k . x e E (r , t ) Re E (r ).e j t ˆ yˆ E (r ) zˆ j. yˆ z x jkx j. e jkx j t e j t 2 ˆ E (r , t ) Re zˆ. .e y.e . .e x x cos( t kx ) y cos( t kx) 2 E (r , t ) zˆ. yˆ. x x
r
x
E (r , t )
LHCP ˆ zˆ 23
Dr. Dung Trinh
HCMUT / 2014
Radiation Pattern
𝑒 −𝑗𝑘0 𝑟 𝐸 𝐹 𝜃, 𝜙 The far field always has the following form: 𝐸 𝑟, 𝜃, 𝜙 = 𝑟 𝐸 𝐹 𝜃, 𝜙 ≡ 𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑧𝑒𝑑 𝑓𝑎𝑟 − 𝑓𝑖𝑒𝑙𝑑 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑓𝑖𝑒𝑙𝑑
In dB:
|𝐸 𝐹 𝜃, 𝜙 | dB 𝜃, 𝜙 = 20log10 |𝐸𝐹 𝜃𝑚 , 𝜙𝑚 |
𝜃𝑚 , 𝜙𝑚 = 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛
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Dr. Dung Trinh
HCMUT / 2014
Radiation Pattern
The far-field pattern is usually shown vs. the angle (for a fixed angle ) in polar coordinates.
E F , dB , 20 log10 F E m , m z 30° 30°
0 60°
60°
-10 dB
A “pattern cut”
-20 dB
0 dB
m
The subscript “m” denotes the beam maximum.
-30 dB 120°
120°
150°
150° 25
Dr. Dung Trinh
HCMUT / 2014
Beamwidths The beamwidth of a pattern is defined as the angular separation between two identical points on opposite side of the pattern maximum: • Half Power beamwidth (HPBW): In a plane containing the direction of the maximum of a beam, the angle between the two directions in which the radiation intensity is one-half value of the beam. (IEEE) • First Null Bandwidth.
Three- and two-dimensional power patterns (in linear scale) of U(θ) = cos2(θ) cos2(3θ).
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Dr. Dung Trinh
HCMUT / 2014
Beamwidths Example: The normalized radiation intensity of an antenna is represented by 𝑈(𝜃) = cos2(𝜃) cos2(3𝜃), (0 ≤ 𝜃 ≤ 90◦ ≤ 𝜑 ≤ 360◦ ) Find the: a. Half-power beamwidth HPBW (in radians and degrees). b. First-null beamwidth FNBW (in radians and degrees)
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Dr. Dung Trinh
HCMUT / 2014
Beamwidths Solution: Since the 𝑈(𝜃) represents the power pattern, to find the halfpower beamwidth we set the function equal to half of its maximum, or a. HPBW 𝑈(𝜃)|𝜃=𝜃ℎ = cos 2 (𝜃)cos 2 (3𝜃)|𝜃=𝜃ℎ = 0.5 ⇒ cos 𝜃ℎ cos3𝜃ℎ = 0.707 → 𝜽𝒉 ≈ 0.25 radians = 14.325𝟎 → 𝑯𝑷𝑩𝑾 = 𝟐𝜽𝒉 = 𝟐𝟖. 𝟔𝟓𝟎 b. FNBW 𝑈(𝜃)|𝜃=𝜃𝑛 = cos 2 (𝜃)cos 2 (3𝜃)|𝜃=𝜃𝑛 = 0 This leads to two solutions for 𝜃𝑛 : cos 𝜃𝑛 = 0 ⇒ 𝜽𝒏 = 𝑐𝑜𝑠 −1 (0) = 𝜋/2 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 = 𝟗𝟎𝟎 cos 3𝜃𝑛 = 0 ⇒ 𝜽𝒏 = 1/3𝑐𝑜𝑠 −1 (0) = 𝜋/6 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 = 𝟑𝟎𝟎 28
Dr. Dung Trinh
HCMUT / 2014
Directivity Directivity of an antenna defined as “the ratio of the radiation intensity in a given direction from the antenna to the radiation intensity averaged over all directions. 𝑼 𝜽, 𝝓 𝟒𝝅𝑼 𝜽, 𝝓 𝑫 𝜽, 𝝓 ≡ = 𝑼𝟎 𝑷𝒓𝒂𝒅
If the direction is not specified, it implies the direction of maximum radiation intensity (maximum directivity) expressed as 𝑫𝒎𝒂𝒙 = 𝑫𝟎 =
𝑼𝒎𝒂𝒙 𝟒𝝅𝑼𝒎𝒂𝒙 = 𝑼𝟎 𝑷𝒓𝒂𝒅
D = directivity (dimensionless) D0 = maximum directivity (dimensionless) U = radiation intensity (W/unit solid angle) Umax = maximum radiation intensity (W/unit solid angle) U0 = radiation intensity of isotropic source (W/unit solid angle) Prad = total radiated power (W) 29
Dr. Dung Trinh
HCMUT / 2014
Directivity Example: Find the maximum directivity of the antenna whose radiation intensity is: a. 𝑾𝒓𝒂𝒅 = 𝒂𝑟 𝑊𝑟 = 𝒂𝑟 𝐴0 b. 𝑾𝒓𝒂𝒅 = 𝒂𝑟 𝑊𝑟 =
𝑠𝑖𝑛𝜃 𝑟2
sin2 𝜃 𝒂𝑟 𝐴0 2 𝑟
(Infinitesimal antenna).
a. The radiation intensity is given by: 𝑈 = 𝑊𝑟𝑎𝑑 𝑟 2 = 𝐴0 𝑠𝑖𝑛𝜃 The radiation power is: 2𝜋
𝜋
𝑃𝑅 = 𝜙=0
𝜃=0
𝜋 𝑈(𝜃, 𝜙). sin𝜃. 𝑑𝜃. 𝑑𝜙 = 2𝜋 𝐴0 = 𝐴0 𝜋 2 2
The maximum directivity is equal to 𝑫𝒎𝒂𝒙 =
𝑼𝒎𝒂𝒙 𝟒𝝅𝑼𝒎𝒂𝒙 𝟒𝝅𝐴0 𝟒 = = = = 𝟏. 𝟐𝟕 𝑼𝟎 𝑷𝒓𝒂𝒅 𝐴0 𝜋 2 𝝅 30
Dr. Dung Trinh
HCMUT / 2014
Directivity Example: Find the maximum directivity of the antenna whose radiation intensity is: b. 𝑾𝒓𝒂𝒅 = 𝒂𝑟 𝑊𝑟 =
sin2 𝜃 𝒂𝑟 𝐴0 2 𝑟
(Infinitesimal antenna).
The radiation intensity is given by: 𝑈 = 𝑊𝑟𝑎𝑑 𝑟 2 = 𝐴0 sin2 𝜃 The radiation power is: 2𝜋
𝜋
𝑃𝑅 = 𝜙=0
𝜃=0
8𝜋 𝑈(𝜃, 𝜙). sin𝜃. 𝑑𝜃. 𝑑𝜙 = 𝐴 3 0
The maximum directivity is equal to 𝑫𝒎𝒂𝒙 =
𝑼𝒎𝒂𝒙 𝟒𝝅𝑼𝒎𝒂𝒙 𝟒𝝅𝐴0 = = = 𝟏. 𝟓 8𝜋 𝑼𝟎 𝑷𝒓𝒂𝒅 3 𝐴0 31
Dr. Dung Trinh
HCMUT / 2014
Input Impedance Input impedance is defined as “the impedance presented by an antenna at its terminals or the ratio of the voltage to current at a pair of terminals or the ratio of the appropriate components of the electric to magnetic fields at a point.” RS
VS
~
jX S
IA
𝑍𝐴 = 𝑅𝐴 + 𝑗𝑋𝐴 RA
VA jX A
RA - Antenna resistance [(dissipation ) ohmic losses + radiation] XA - Antenna reactance [(energy storage) antenna near field]
PA is the power delivered to the antenna: 1 𝑃𝐴 = 𝑅𝑒 𝑉𝐴 𝐼𝐴∗ 2
𝑍𝐴 𝑍𝐴 + 𝑍𝑠 𝑉𝑠 𝐼𝐴 = 𝑍𝐴 + 𝑍𝑠
𝑉𝐴 = 𝑉𝑠 where:
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Dr. Dung Trinh
HCMUT / 2014
Input Impedance Then we have:
1 1 𝑉𝑠 𝑍𝐴 𝑉𝑠∗ ∗ 𝑃𝐴 = 𝑅𝑒 𝑉𝐴 𝐼𝐴 = 𝑅𝑒 2 2 𝑍𝐴 + 𝑍𝑠 𝑍𝐴 + 𝑍𝑠
∗
1 𝑉𝑠 2 𝑅𝐴 = 2 𝑍𝐴 + 𝑍𝑠 2
From the equivalent circuit for the generator/antenna system, we see that maximum power transfer occurs when: 𝑍𝐴 = 𝑍𝑠∗ (𝑅𝐴 = 𝑅𝑠 𝑎𝑛𝑑 𝑋𝐴 = −𝑋𝑠 ). Then the maximum power radiated by the antenna is: 𝑃𝐴,𝑚
1 𝑉𝑠 2 = 8 𝑅𝑠
Note that the power available from the generator source is 2 1 1 𝑉 𝑠 𝑃 = 𝑅𝑒 𝑉𝑠 𝐼𝐴∗ = 2 4 𝑅𝑠
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Dr. Dung Trinh
HCMUT / 2014
Antenna Radiation Efficiency RS VS , Z S
~
jX S
IA
Anten
RR VS
Rr - Antenna radiation resistance (radiation)
~
VA
RD
Z A RA jX A
Z A RR RD jX
jX A
RL - Antenna loss resistance (ohmic loss)
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Dr. Dung Trinh
HCMUT / 2014
Antenna Efficiency The power radiated by the antenna (Pr) can be written as 𝑃𝑅 = 𝑒𝑃𝐴
e: antenna efficiency.
The power dissipated by the antenna is: 𝑃𝐷 = 1 − 𝑒 𝑃𝐴 We have: 𝑃𝐴 =
1 𝑅 𝐼 2 𝐴 𝐴
2
1 𝑃𝐷 = 𝑅𝐷 𝐼𝐴 2
2
1 𝑅 𝐼 2 𝑅 𝐴
2
𝑃𝑅 =
𝒆=
𝑹𝑹 𝑹𝑹 = 𝑹𝑨 𝑹𝑫 + 𝑹 𝑹
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Dr. Dung Trinh
HCMUT / 2014
Antenna Gain Directivity of an antenna:
𝑼 𝜽, 𝝓 𝟒𝝅𝑼 𝜽, 𝝓 𝑫 𝜽, 𝝓 ≡ = 𝑼𝟎 𝑷𝒓𝒂𝒅
The gain of an antenna in the directions (, ) is defined as: 𝟒𝝅𝑼 𝜽, 𝝓 𝑮 𝜽, 𝝓 = = 𝒆𝑫 𝜽, 𝝓 𝑷𝑨
Example: A lossless half wavelength dipole antenna with input impedance 73 is connected to a transmission line whose characteristic impedace is 50. Assuming that the pattern of the antenna is given by: : 𝑼 = 𝐵0 sin3 𝜃 Find the maximum absolute gain of this antenna. 36
Dr. Dung Trinh
HCMUT / 2014
Antenna Gain Solution: At first we compute the maximum directivity of the antenna 𝑈𝑚𝑎𝑥 = 𝐵0 The radiation power is: 2𝜋
𝜋
𝑃𝑅 = 𝜙=0
𝜃=0
3𝜋 2 𝑈(𝜃, 𝜙). sin𝜃. 𝑑𝜃. 𝑑𝜙 = 𝐵0 4
The maximum directivity is equal to 𝑫𝒎𝒂𝒙 =
𝑼𝒎𝒂𝒙 𝟒𝝅𝑼𝒎𝒂𝒙 𝟒𝝅𝐵0 = = 2 = 𝟏. 𝟔𝟗𝟕 3𝜋 𝑼𝟎 𝑷𝒓𝒂𝒅 4 𝐵0
𝑫𝒎𝒂𝒙 𝒅𝑩 = 𝟐. 𝟐𝟗𝟕
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Dr. Dung Trinh
HCMUT / 2014
Antenna Gain Solution (cont): There is a loss due to reflection or mismatch losses between antenna and the transmission line. The loss is equal to: 𝒆𝒓 = 𝟏 − 𝚪
𝟐
𝟕𝟑 − 𝟓𝟎 = 𝟏− 𝟕𝟑 + 𝟓𝟎
𝟐
= 𝟎. 𝟗𝟔𝟓
𝒆𝒓 𝒅𝑩 = − 𝟎. 𝟏𝟓𝟓
The absolute gain is equal to: 𝑮𝒎𝒂𝒙 = 𝒆𝒓 𝑫𝒎𝒂𝒙 = 𝟎. 𝟗𝟔𝟓 × 𝟏. 𝟔𝟗𝟕 = 𝟏. 𝟔𝟑𝟕𝟔
𝑮𝒎𝒂𝒙 𝒅𝑩 = 𝟐. 𝟏𝟒𝟐
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Dr. Dung Trinh
HCMUT / 2014
Receiving Antenna E in c ZL
Anten
Taûi
( , )
When conjugate matching happens:
𝑍𝐴 = 𝑍∗𝐿
IL RA RL
Power delivered to the load is:
|𝑉𝐶 |2 𝑃𝐿 = 𝑃𝐶 = 8𝑅𝐴
VL
jX A VC
~
jX L
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Dr. Dung Trinh
HCMUT / 2014
Antenna Effective Area One way to characterize an antenna is with the effective area. Power delivered to the receiver (Pc) may be defined in terms of the antenna effective aperture (Ae) as: 𝑷𝑪 = 𝑨𝒆 . 𝑺𝒊𝒏𝒄
E in c Anten
( , )
ZL
Taûi
It can be proved* that: 𝐀𝒆 𝜽, 𝝓, 𝒑𝒊𝒏𝒄
𝝀𝟐 𝟐 = . 𝑮 𝜽, 𝝓 . |𝒑 𝜽, 𝝓 . 𝒑𝒊𝒏𝒄 | 𝟒𝝅
𝝀𝟐 𝐀𝒆𝒎 = 𝑮 𝟒𝝅 Where 𝑺𝒊𝒏𝒄 is power density of the incident wave. 𝑨𝒆 is the antenna effective area. 𝒑𝒊𝒏𝒄 is the * : C. A. Balanis, Antenna Engineering, Chapter 2, Section 2.16. 40
Dr. Dung Trinh
HCMUT / 2014
Antenna Link – Friis Equation
The Friis transmission equation defines the relationship between transmitted power and received power in an arbitrary transmit/receive antenna system. Given arbitrarily oriented transmitting and receiving antennas, the power density at the receiving antenna (Wr) is: 𝟏 𝑾 𝒓 = 𝑭𝜽 (𝜽, 𝝓) 𝟐 𝟐. 𝜼. 𝒓
𝟐
+ 𝑭𝝓 (𝜽, 𝝓
𝟐
𝟏 = 𝟐 . 𝑼𝒕 𝜽𝒕 , 𝝓𝒕 𝒓 41
Dr. Dung Trinh
HCMUT / 2014
Antenna Link – Friis Equation We note that:
𝑮 𝜽, 𝝓 =
𝟒𝝅𝑼 𝜽, 𝝓 = 𝒆𝑫 𝜽, 𝝓 𝑷𝑨
→ 𝑺𝒊𝒏𝒄 =𝑾 𝒓 = 𝒕
𝟏 𝒆𝒕 𝑫 𝜽𝒕 , 𝝓𝒕 𝑷𝑻 . 𝑼 𝜽 , 𝝓 = 𝒓𝟐 𝒕 𝒕 𝒕 𝟒𝝅𝒓𝟐
The total received power is then 𝑷𝑹 = 𝑨𝒆 𝜽𝒓 , 𝝓𝒓 , 𝒑𝒕 . 𝑺𝒊𝒏𝒄 𝒕
Where: 𝑨𝒆 𝜽𝒓 , 𝝓𝒓 , 𝒑𝒕
𝝀𝟐 = . 𝑮 𝜽𝒓 , 𝝓𝒓 . |𝒑 𝜽, 𝝓 . 𝒑𝒊𝒏𝒄 |𝟐 𝟒𝝅 𝟐
𝝀𝟐 . 𝑮𝒓 𝜽𝒓 , 𝝓𝒓 . |𝒑𝒓 𝜽𝒓 , 𝝓𝒓 . 𝒑𝒕 𝜽𝒕 , 𝝓𝒕 | . 𝑮𝒕 𝜽𝒕 , 𝝓𝒕 . 𝑷𝑻 → 𝑷𝑹 = 𝟒𝝅. 𝒓 𝟐
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Dr. Dung Trinh
HCMUT / 2014
Antenna Link – Friis Equation Example: Two dipoles whose directivity is given by: 𝐷 𝜃, 𝜙 = 1.5𝑠𝑖𝑛2 𝜃 are located at a distance of 100m. Parameters of the system: f = 600 MHz. Antenna efficiency: 2%. 𝑍𝐴 = 10Ω, 𝑍𝑆 = 𝑍𝐿 = 50Ω. 𝑉𝑆 = 20𝑉. Assume that there is no polarization mismatch. Find total received power PR?
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Dr. Dung Trinh
HCMUT / 2014
Q&A Reading: Balanis’s book - Chapter 2
Thank you for your attention
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