ECE 480
Wireless Systems
Lecture 4
Propagation and Modulation
of RF Waves
1
Antenna Radiation Characteristics
Antenna pattern:
• Describes the far – field directional properties of
an antenna when measured at a fixed distance
from the antenna
• 3 – d plot that displays the strength of the radiated
field (or power density) as a function of direction
(spherical coordinates) specified by the zenith
angle  and the azimuth angle 
• From reciprocity, a receiving antenna has the
same directional antenna pattern as the pattern
that it exhibits when operated in the transmission
mode
2
The differential power
through an elemental
area dA is
d Prad  S av
S av
d A  S av
ˆ dA  S dA
R
always in the radial direction in the far – field region
3
d Prad  S dA
d A  R sin  d  d 
2
Define: Solid angle, 
dA
d   2  sin  d  d 
R
  4
steradians
for a spherical surface
4
d Prad  S dA
dA
d  2
R
d A R2 d
d Prad  R 2 S R , ,  d
The total power radiated by an antenna is given by
2
Prad  R 2

S  R , ,  sin  d  d 




0
0
5
2
Prad  R 2 S max

F  ,  sin  d  d 




0
0
Prad  R 2 S max
F  , 
  F  ,  d 
4
is the normalized
radiation intensity
F  ,   1
6
3 – D Pattern of a Narrow – Beam Antenna
7
Antenna Pattern
It is convenient to characterize the variation of
F ( , ) in two dimensions
Two principle planes of the
spherical coordinate system
Elevation Plane ( - plane)
Corresponds to a single value
of  ( = 0
x –z plane)
( = 90
y –z plane)
Azimuth Plane ( - plane)
Corresponds to  = 90 o
(x – y plane)
8
Clearer to express F in db for highly directive patterns
F dB   10 log F
 = 0 plane
9
Side lobes are undesirable
• Wasted energy
• Possible interference
10
Beam Dimensions
Define: Pattern solid angle  p
 p = Equivalent width of the main lobe
p 
  F  ,  d 
4
sr
For an isotropic antenna with F ( , ) = 1 in all directions:
 p  4
sr
11
Defines an equivalent cone over which all the
radiation of the actual antenna is concentrated with
equal intensity signal equal to the maximum of the
actual pattern
12
The half – power (3
dB) beamwidth, , is
defined as the angular
width of the main lobe
between the two
angles at which the
magnitude of F ( , )
is equal to half its
peak value
   2  1
13
F () is max at  = 90 o ,  2 = 135 0 ,  1 = 45 o ,
 = 135 o – 45 o = 90 o
14
Null Beamwidth,  null
Beamwidth between
the first nulls on
either side of the peak
15
Antenna Directivity
D

F max
Fav
1


1
Fav
4
1
  F  ,  d 
4
4
p
 p = Pattern solid angle
For an isotropic antenna,  p = 4 
D=1
16
D can also be expressed as
4  R 2 S max
D
Prad


S max
S av
Prad
4 R 2
S av  S iso
S iso = power density radiated by an isotropic antenna
D = ratio of the maximum power density radiated by
the antenna to the power density radiated by
an isotropic antenna
17
For an antenna with a single main lobe pointing in the
z direction:
p
D
 xz  yz
4
4
p
 xz  yz
18
Example – Antenna Radiation Properties
Determine:
a. The direction of maximum radiation
b. Pattern solid angle
c. directivity
d. half – power beamwidth
in the y-z plane for an antenna that radiates into only
the upper hemisphere and its normalized radiation
intensity is given by
F  ,   cos 
2
19
Solution
The statement
F  ,   cos 
2
in the upper hemisphere
can be written mathematically as
F  ,   F    cos 
2
0
0  

2
0    2
elsewhere
20
a. The function
F    cos 
2
is maximum when  = 0
b. The pattern solid angle is
given by
p 
Polar plot of
F    cos 
2
  F  ,  d 
4



2
2


    cos 2  sin  d   d 
  0   0


2

 cos 3   2
  
 d 
3 0
 0
2

0
1
2
d 
3
3
sr
21
c.
D 
4
p
 3 
 4 
6
 2 
D  d B   10 log  6   7.78 d B
d. The half – power by setting
F    cos   0.5
2
Polar plot of
F    cos 
2
 1   45
o
 2  45
o
   2   1  90 o
22
Example – Directivity of a Hertzian Dipole
For a Hertzian dipole:
F  ,   sin 
2
D

4

4
F  ,  sin  d  d 
4
2





0
0
4

 1.5
8
3
sin d  d 
3
23
Antenna Gain
P t = Transmitter power sent to the antenna
P rad = Power radiated into space
P loss = Power loss due to heat in the antenna
= P t – P rad
Define: Radiation Efficiency, 
 
Prad
Pt
 = 1 for a lossless antenna
24
Define: Antenna Gain, G
G
D
4  R 2 S max
Pt
4  R 2 S max
Prad
G D
Accounts for the losses in the antenna
25
Radiation Resistance
P loss = Power loss due to heat in the antenna
= P t – P rad
1 2
Pt  I 0 R ant  Ploss  Prad
2
1 2
Ploss  I 0 R loss
2
1 2
Prad  I 0 R rad
2
 
Prad
Pt

Prad
Prad  Ploss

R rad
R rad  R loss
26
To find the radiation resistance:
• Find the far – field power by integrating the far
– field power density over a sphere
• Equate to
Prad
1 2
 I 0 R rad
2
27
Example – Radiation Resistance and Efficiency of
a Hertzian Dipole
A 4 – cm long center – fed dipole is used as an
antenna at 75 MHz. The antenna wire is made of
copper and has a radius a = 0.4 mm. The loss
resistance of a circular wire is given by
R loss 
l
2 a
 f c
c
Calculate the radiation resistance and the
radiation efficiency of the dipole antenna
28
Solution
The parameters of copper are
r  1
 c  5.8  10
R loss 
l
2 a
7
S
m
 f c
c
0.04

2  0.004
  75  10 6  4   10  7


0.036
5.8  10
7
29
At 75 MHz:
c
3  10 8
 
4 m
6
f
75  10
l
4 cm

 0.01

4m
 This is a short dipole
From before,
Prad
4 R 2

S max
D
S max 
15  I
R
2
2
0
l
 

2
30
2
2
15

I
4 R
0 l 


 
2
D
R

2
Prad
l 
 40  I  

2
Prad
2
0
1 2
2 2 l 
 I 0 R rad  40  I 0  
2

2
R rad
 
2
2
l 
3  0.04 
 80     80  


 4 
 0.08 
R rad
R rad  R loss
2
2
0.08

 0.69  69 %
0.08  0.036
31
Half – Wave Dipole Antenna
i  t   I 0 cos  t cos k z  Re  I 0 cos k z e j  t 
In phasor form:
I  z   I 0 cos k z  

4
z

4
k 
2

32
For a short dipole
j I 0 l k 0 e  j k R
E 
sin 
4
R
H 
E
0
Expand these expressions to obtain similar
expressions for the half – wave dipole
33
Consider an
infinitesimal dipole
segment of length dz
excited by a current I  z 
and located a distance
from the observation
point
d E z 
d H  z 
j k0
4
e  jks
I  z  dz
sin  s
s
d E z
0
34
The far field due to radiation by the entire

antenna is given by
E 
4
  d E
z
4
Two assumptions:
1
s
1
(length factor)
R
s 
35
j k 0
e jks
d E z 
I z d z
sin  s
4
s
Note that "s" appears in the equation
twice – once for the distance away
and once for the phase factor
s
R
is not valid for the length factor
If Q is located at the top of the dipole, the phase

factor is
which is not acceptable
2
s
R  z cos 
36
d E z 
j k 0
4
I z d z
e
 jkR
R
sin  e j k z cos 



 cos  2 cos     e  j k R 


E   j 60 I 0 


sin 

 R 


H 
E
0
S  R ,  
E
2
2 0


2 
 cos  2 cos   





2
sin 




37
S  R ,  
E
2
2 0
S  R ,  
E

2
2 0
15 I 02
R2


2 
 cos  2 cos   



2
sin







2 
 cos  2 cos   


S0 
2
sin





38
S  R , 
is max when
S max  S 0 
 

2
15 I 02
 R2



cos  cos   

S  R , 
2




F   

S0
sin 




2
39
Directivity of Half – Wave Dipole
Need P rad and S (R , )
Prad  R 2
  S R ,  d 
4
2

15 I 02




cos   
2  2   cos 
2
  sin  d  d 

0 0  sin 



 36.6 I 02
S max 
D
4  R 2 S max
Prad

15 I 02
 R2
4 R
36.6 I 02
2
 15 I 02 
 1.64  2.15 d B 

2 


R


40
Radiation Resistance of Half – Wave Dipole
2 Prad
R rad 
I
2
0
2  36.6 I 02

I
2
0
73 
Recall: for the short dipole (l = 4 cm) at 75 MHz
R rad = 0.08 
R loss = 0.036 
 
R rad
R rad  R loss
0.08

 69 %
0.08  0.036
For the half – wave dipole (l = 4 m) at 75 MHz
R loss = 1.8 
 
R rad
R rad  R loss
73

 98 %
73  1.8
41
Effective Area of a Receiving Antenna
Assume an incident wave with a power density of S i
The effective area of the antenna, A e , is
Ae 
P int
Si
P int = Power intercepted by the antenna
It can be shown:
Pint  PL 
V oc
V oc
2
8 R rad
= Magnitude of the open – circuit voltage
developed across the antenna 42
The power density carried by the wave is
Si 
Ae 
Ei
2

2 0
Pint
Si

Ei
2
240 
30  V oc
R rad E i
2
2
For the short dipole
R rad
l
 80   

2
2
V oc  E i l
3 2
Ae 
8
43
In terms of D:
2D
Ae 
4
m2
Valid for any antenna
Example: Antenna Area
The effective area of an antenna is 9 m 2. What is
its directivity in db at 3 GHz?
c 3  10
 
 0.1
9
f
3  10
8
D
4 Ae
2
4  9


2
 0.1
 1.13  10 4  40.53 d B
44
Friis Transmission Formula
Assumptions:
• Each antenna is in the far – field region of the other
• Peak of the radiation pattern of each antenna is
aligned with the other
• Transmission is lossless
45
For an isotropic antenna:
S iso 
Pt
4 R
2
In the practical case, S r  G t S iso   t D t S iso
(ideal)
 t D t Pt

4 R 2
In terms of the effective area A t of the transmitting antenna
 t At Pt
Sr  2 2
 R
46
On the receiving side, Pint
 t At A r Pt
 S r Ar 
2R2
Prec   r Pint
 t  r At Ar
  

 Gt Gr 

2
2
Pt
 R
 4 R 
Prec
Friis transmission formula
2
47
When the antennas are not aligned
(More general expression)
 t  r At Ar
  

 Gt Gr 
 Ft  t ,  t  F r  r ,  r 
2
2
Pt
 R
 4 R 
Prec
2
48
Homework
1. Determine the following:
a. The direction of maximum radiation
b. Directivity
c. Beam solid angle
d. Half – power beamwidth in the x – z plane
for an antenna whose normalized radiation intensity is
given by:
F  ,   1 for 0    60 o , 0    2 
 0 elsewhere
Hint: Sketch the pattern first
49
2. An antenna with a pattern solid angle of 1.5 (sr)
radiates 30 W of power. At a range of 1 km, what is
the maximum power density radiated by the antenna?
3. The radiation pattern of a circular parabolic –
reflector antenna consists of a circular major lobe
with a half – power beamwidth of 2 o and a few minor
lobes. Ignoring the minor lobes, obtain an estimate
for the antenna directivity in dB.
50
Analog Modulation
• High frequencies require smaller antennas
• Modulation impresses a lower frequency onto
a higher frequency for easier transmission
• The signal is modulated at the transmission
end and demodulated at the receiving end
Several basic types
• Amplitude modulation (AM)
• Frequency modulation (FM)
• Pulse code modulation (PCM)
• Pulse width modulation (PWM)
51
Amplitude Modulation
Carrier wave – High frequency signal that transports
the intelligence
Signal wave – Low frequency signal that contains the
intelligence
52
AM transmitter
• DC shifts the modulating signal
• Multiplies it with the carrier wave using a
frequency mixer
• Mixer must be nonlinear
• Output is a signal with the same frequency as the
carrier with peaks and valleys that vary in
proportion to the strength of the modulating signal
• Signal is amplified and sent to the antenna
53
The mixer is usually a "square law" device,
such as a diode or B – E junction of a transistor
Output  input 2
Suppose that we apply the following signals to a
square law device
f 1  t   A1 cos  1 t
f 2  t   A 2 cos  2 t
The output will be

f o  t   A o A1 cos  1 t  A 2 cos  2 t

2
54
Homework

f o  t   A o A1 cos  1 t  A 2 cos  2 t

2
Determine all possible output frequencies
55
Advantages
• Simplicity
• Cost
Disadvantages
• Susceptible to atmospheric interference (static)
• Narrow bandwidth (550 – 1500 KHz)
56
AM Receiver
• Tunable filter
• Envelope detector (diode)
• Capacitor is used to eliminate the carrier and to
undo the DC shift
• Will generally include
some form of automatic
gain control (AGC)
57
Forms of Amplitude Modulation
In the most basic form, an AM signal in the
frequency domain consists of
• The carrier signal
• Information at f c + f m (upper sideband)
• Information at f c - f m (lower sideband)
(US and LS are mirror images)
This wastes transmission power
• Carrier contains no information
• Information is all contained in only one of the
sidebands
58
Frequently, in communications systems, the carrier
and/or one of the sidebands is suppressed or
reduced
• If only the carrier is reduced or suppressed, the
process is called "Double – Sideband Suppressed
(Reduced) Carrier" (DSSC or DSRC)
• If the carrier and one of the sidebands is suppressed
or reduced, the process is called "Single – Sideband
Suppressed (Reduced) Carrier" (SSSC or SSRC)
• Often, the carrier and one of the sidebands is totally
suppressed. This process is simply called "Single
Sideband"
• The carrier must be regenerated at the receiver end
59
Example
Consider a carrier with a frequency  c

c  t   C sin  c t

Suppose we want to modulate the carrier with a signal

m  t   M sin  m t  

The signal is amplitude – modulated by adding m(t) to C
The expression for this signal is


 
y  t   C  M sin  m t sin  c t

Expanding this expression


y  t   C sin  c t  M
 
  cos   
M
cos    m   c t
2
2
m

 c t
60
Convert to frequency domain by taking the
Fourier Transform

y  t   C  m  t   cos  c t
y  t   C  m  t 
e
j c t

e
2
 j c t
Take Fourier Transform

Y     C     c
 x






1
1
 M   c   C   c  M   c
2
2
= Unit impulse function
61

Eff = 100 %
Eff = 33 %
Eff = 100 %
62
Modulation Index
Measure of the modulating signal wrt the
carrier signal
h
peak value of m  t 
C
63
64
65