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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013 - Electromagnetics and Applications
Fall 2005
Lecture 21 - Receiving Antennas
Prof. Markus Zahn
December 6, 2005
I. Review of Transmitting Antennas (Short Dipoles)
A. Far fields (r � λ)
z
v
I (0)
dl eff
=
v
1
+ dl 2
I (0)
∫
v
I (z) dz
dl 2
Image by MIT OpenCourseWare.
ˆ eff k 2 η
Ê0
Idl
Êθ = ηĤφ =
sin(θ)e−jkr , Ê0 = −
,η =
jkr
4π
�
µ
�
B. Intensity �Sr �
�Sr � =
�
�
1
ˆ ∗ = 1 |E
¯ˆ × H̄
¯ˆ |2
Re E
2
2η
=
1 |Ê0 |2
sin2 (θ)
2η k 2 r 2
ˆ eff |2 k�42 η�2
1 |Idl
sin2 (θ)
2 r2
2η� 16π 2�
k�
ˆ eff |2 k 2 η
|Idl
=
sin2 (θ)
32π 2 r2
=
C. Total time average power �P �
� π �
�P � =
dθ
0
2π
dφ �Sr � r2 sin(θ)
0
ˆ eff |2 ηk 2
|Idl
=
12π
�
�
1 ˆ2
2πη dleff 2
= |I| R ⇒ R =
2
3
λ
radiation resistance
D. Gain
G(θ, φ) =
=
=
�Sr �
�P � /(4πr2 )
��
2 η sin2 (θ) · 12π(4πr�
2
ˆ�eff
|Idl
|2�
k�
�
� ��� )
�
��
2 |Idl
2
�
�
r�
|2 �
k�
32�
π 2�
�ˆ�eff
8�
3
sin2 (θ)
2
1
II. Receiving Antennas
Z
In absence of receiving antenna: Ēinc = Ē0 , H̄inc =
H̄0 . With d1 � λ, over size scale of antenna, Ē0 and
H̄0 are approximately spatially uniform. In presence
of receiving antenna, electric and magnetic fields are
¯ and normal H
¯ are zero
perturbed so that tangential E
along the perfectly conducting length of the antenna.
S1
c
d1
θ
Eo
+ a _I1
_1
V
_ b
y
e
ϕ
x
For d1 << λ
Image by MIT OpenCourseWare.
¯ = Ē0 + Ē1
E
¯ = H̄0 + H̄1
H
(1)
(2)
Surface S1 above intimately hugs the antenna so that
�
�
�
�
¯ × dā =
E
da Ē0 + Ē1 × n̄ = 0
S1
¯ = 0)
(tangential E
�
�
¯ =
da K
da n̄ × (H̄0 + H̄1 ) =
S1
�
S1
+d1 /2
dz I¯1 (z) = I1 deff īz
(4)
−d1 /2
Another useful relationship:
�
�
da (Ē0 × H̄0 ) · n̄ = (Ē0 × H̄0 ) ·
S1
da n̄ = 0
(5)
S1
Integral of normal over closed surface is zero:
�
�
�
dV �f =
da f n,
¯
Take f = 1, �f = 0 =
da n
¯=0
V
(3)
S1
S
(6)
S
Scalar Triple Product Identity:
(ā × ¯b) · c̄ = ā · (¯b × c̄)
(7)
(Interchange of cross and dot)
Complex power supplied by receiving antenna
�
¯ · n̄
P =
da Sˆ
(8)
S1
�
�
1 ˆ
ˆ
ˆ
ˆ
ˆ¯ ∗ + H
ˆ¯ ∗ )
¯×H
¯ ∗ ) = 1 (E
¯0 + E
¯1 ) × (H
Sˆ¯ = (E
0
1
2
2
�
1�ˆ
∗
ˆ
ˆ¯ ∗ ) + E
ˆ¯ × (H
ˆ¯ ∗ + H
ˆ¯ ∗ )
¯
¯
=
E0 × (H0 + H
(9)
1
1
0
1
��
�
�2
�
�
�
�
�
ˆ × (H̄
ˆ ∗ + H̄
ˆ ∗ ) · n̄ +
ˆ × (H̄
ˆ ∗ + H̄
ˆ ∗ ) · n̄
¯ · n̄ = 1
P =
da Sˆ
da Ē
da
Ē
(10)
0
1
0
1
0
1
2
S1
S1
S1
2
�
�
�
�
ˆ
ˆ∗ + H
ˆ ∗ ) · n̄ =
¯0 × (H̄
¯
da E
0
1
�
�
ˆ · (H̄
ˆ ∗ + H̄
ˆ ∗ ) × n̄
da Ē
0
0
1
S1
�
�
�
ˆ ·
ˆ¯ ∗ + H
ˆ¯ ∗ ) × n̄
= Ē
da (H
0
0
1
S1
S1
ˆ · Iˆ d ī
= −Ē
(from (4))
0
1 eff z
�
�
�
�
�
�
�
ˆ
ˆ
ˆ
ˆ × H̄
ˆ ∗ · n̄ +
ˆ × H̄
ˆ ∗ · n̄
¯1 × (H
¯ 0∗ + H
¯ 1∗ ) · n̄ =
da E
da Ē
da
Ē
1
1
0
1
S1
S1
� S1
��
�
�
�
(11)
(12)
|Iˆ1 |2 (R+jX)
where R is the radiation resistance
and X is the antenna reactance
�
�
�
�
�
∗
ˆ
ˆ
ˆ∗ × E
ˆ¯ · n̄
¯
da E1 × H̄0 · n̄ = −
da H̄
1
0
S1
� S1
�
�
ˆ¯ ∗ × (E
ˆ¯ + E
ˆ ) · n̄
¯
=−
da H
0
1
0
�S1
�
�
�
∗
∗
ˆ
ˆ
ˆ
¯
¯
=−
da H0 · (E0 + Ē1 ) × n̄ = −H̄0 ·
�
S1
�
�
ˆ¯ + Ē
ˆ × n̄
da E
0
1
S1
= 0 (from (3))
� 1
1 � ˆ¯ ˆ
P =
−E0 · I1 deff īz + |Iˆ¯1 |2 (R + jX) = V̂1 Iˆ1∗
2
2
(13)
Thevenin Equivalent Circuit
Image by MIT OpenCourseWare.
ˆ · d¯
V̂T H = V̂oc = −Ē
0
eff
(d¯eff = deff īz )
3
III. Transmitting and Receiving Antennas
A. Circuit Description
^
I1
^
I2
^
^
V1
V2
Two port network
Z12 =
Vˆ1 ��
�
Iˆ2 Iˆ1 =0
Image by MIT OpenCourseWare.
V̂1 = Iˆ1 Z11 + Iˆ2 Z12
V̂2 = Iˆ1 Z21 + Iˆ2 Z22
V̂2 ��
Z21 =
�
Iˆ1 Iˆ2 =0
Reciprocity Theorem: Z12 = Z21
Z11 = R1 + jX1
B. Antenna Thevenin Equivalent
I1 Circuits
Vth1
I2
Z*22
V1
Z22 = R2 + jX2
Vth2
V2
Receiver (Balanced load Z*22 to cancel
reactance X2)
Transmitter
Image by MIT OpenCourseWare.
V̂th1
= Iˆ2 Z12 = −
�
ˆ¯ · dl = −E
ˆ¯ · dl
E
1
1
eff
ˆ¯ · dl
V̂th2 = Iˆ1 Z21 = −E
2
eff
4
�P2 � =
1 |V̂th2 /2|2
1 |V̂th2 |2
1 |Ê2 dleff sin(θ)|2
=
=
�
�
2 R2
8 R2
8 2πη dleff 2
3
λ
�
�
2
2
2
2
ˆ�
1 |�
1 |E
2|
2 | sin (θ) · 3λ
�P2 � = Arec (θ, φ) �Sr � = Arec (θ, φ) · �
= �
8
2πη�
2 η�
2
2
λ
3
λ
Arec (θ, φ) = sin2 (θ)
= Grec (θ, φ)
2
4π
4π
C. Representative Parameters
1. Minimum received power ≈ 10−20 watts
For total transmitted power of 1 watt, how far away can the receiver be at f = 1
GHz?
�Prec � =
�Ptrans �
λ2
G
G
trans
rec
2
� 4πr ��
� � ��4π�
�Sr �
fλ = c ⇒ λ =
Arec (θ,φ)
108
3×
c
=
f
109
= .3 m
3
sin2 (θ) (for short dipoles) (identical transmitting and receiving antennas)
2
π
3
Take θ = ⇒ Gtrans = Grec =
2
2
� �2
Ptrans
λ
2
r =
Gtrans Grec
Prec
4π
� � � �2
1
9
.3
= −20
10
4
4π
Gtrans = Grec =
= 1.28 × 1017 m2
r = 3.58 × 108 m = 3.58 × 105 km ≈ 200, 000 miles
2. For data transmission, receivers need Eb > 4 × 10−20 Joules/bit
Power received = M Eb where M is the data rate, bits/s
−9
−9
10−9 watts received power allows M = 10Eb = 4×1010−20 = .25 × 1011 bits/s
1 CD = 700 × 106 bytes = 5600 × 106 bits (1 byte = 8 bits)
M = .25 × 1011 bits/sec ≈ 4.5 CD/sec
3. Distance is not a barrier to wireless communications
r = 1 lightyear = 3 × 108 m/s ·3 × 107 s/yr = 9 × 1015 m/yr
Ptrans =?
c
= .1 m
f
M = 1 bit/s, Eb = 4 × 10−20 Joules/bit
f = 3GHz ⇒ λ =
Prec = M Eb = 4 × 10−20 Watts
Gtrans = Grec = 107
5
Ptrans
� �2
Prec 4πr
λ
=
Gtrans Grec
�
�2
4π(9×1015 )
−20
4 × 10
.1
=
14
10
= 512 Watts
For M = 2.4 kb/s ⇒ Ptrans ≈ 1.2 MW (with a 1 year delay each way)
4. Optical Communications: E = hf, h = 6.625 × 10−34 Joule-sec (Planck’s Constant)
a. Radio Photons
f = 1 GHz ⇒ E = 6.625 × 10−25 Joules/Photon
Eb
EN = Eb ⇒ N =
photons/bit
E
4 × 10−20
=
≈ 6000 photons/bit
6.625 × 10−25
b. Optical Photons
c
3 × 108
=
≈ 6 × 1014 Hz
.5 × 10−6
.5 × 10−6
Eb
4 × 10−20
N=
=
≈ .1 photon/bit
hf
6.625 × 10−34 · 6 × 1014
λ = 0.5 µm ⇒ f =
6