Communicating Information
Radio Waves
Type
Frequency
Range
Surface
wave
<3MHz
<1000km
Sky wave
3–30MHz
Worldwide by
reflection at the
ionosphere & ground
Space wave
>30MHz
Line-of-sight including
between satellites
Half-Wave dipole
•
Half-wave dipole
radio aerial of length
/2
• Alternating signal
• Decelerating
electrons radiate
energy as e-m waves
Microwave transmission
• Microwaves (SHF) = 3 – 30GHz
centimetres
• Used for directional
point-to-Transmitter
Parabolic reflecting dish
P. Lovatt & Hari.R
Fibre Optics
www.goalfinder.com/product.asp?productid=33
Fibre-Optics
• Critical angle for glass
42
0
Fibre-Optics
r
=
i
• If
i
(incident angle)
>
c
(critical angle)
Fibre-Optics
Fibre-Optics Advantages
• Fibre optics can carry much more
information in a much smaller cable
• No interference from electromagnet fields
means "clearer" connections
• No electrical resistance
• No electrical hazard
Syllabus Objectives
30.4 Channels of Information
Candidates should be able to:
(j) describe the use of satellites in communication.
(k) recall the relative merits of both geostationary and
polar orbiting satellites for communicating information.
(l) recall the frequencies and wavelengths used in
different channels of communication.
(m)
understand and use signal attenuation
expressed in dB and dB per unit length.
Satellite Communications
Up
link
fre
que
ncy
, f
upDow
nlin
k fr
equ
enc
y, f
downf
down
<
f
up
Why not use SW?
Satellites at 6/4GHz are preferred to SW because:
• Long-distance communication using SW is
unreliable due to changing ionospheric conditions
• The SW band is already filled by existing
broadcasting stations
Global Positioning (GPS)
24 satellites in 6
orbits at 22,200m
4 satellites are
required to fix
your position in
3D (including
altitude)
Global Positioning (GPS)
science.nasa.gov/Realtime/jtrack/3d/JTrack3D.html
• Low altitude
• Short period
90 mins
• Appear to move rapidly across the sky
• Will pass every point of the Earth’s surface
• Each pass takes place 360
0
x 1.5/24 = 22.5
0
to
the west of the previous orbit
• Used to monitor weather, surveying surface
P. Lovatt & Hari.R
Satellites – Polar Orbit
Indian TES in 560km polar orbit at 0552GMT on
14/01/08. Provides remote sensing to 1-metre
• Orbit the equator
• Appear stationary in the sky
• Same direction as Earth’s rotation
• Period = 24hrs
Geostationary
Satellites
m
sat
v
2
/r = Gm
sat
m
e
/r
2
v = 2
r/T
4
2
r
3
/T
2
= Gm
e
r = (Gm
e
T
2
/4
2
)
1/3
r =
4.23 x 10
7
m
Geostationary Satellites
Signal Attenuation
All signals lose power over distance:
• Electrical signals
heat due to resistance
in wires
• Light pulses
scattering and impurities in
glass fibres
Signal Attenuation
• An adequate signal must have a high enough
signal-to-noise ratio (SNR)
Typically, 100 times
•
Repeater
amplifiers may
have a
gain
of 10
5
• Satellite received power may be
10
19
smaller than Earth
transmitted power
V
t
‘
Noise
’
level
Signal Level
The
bel
:
B = log
10
(P
1
/P
2
)
The
decibel
:
Example 1 - Gain
The gain of an amplifier = 45dB.
Calculate the output power P
out
if the
input power P
in
= 2.0
W.
45 = 10 log
10
(P
out
/2.0 x 10
-6
)
10
4.5
= P
out
/2.0 x 10
-6
P
out
= 10
4.5
x 2.0 x 10
-6
Example 2 - Attenuation
The input power to a cable of length 25km is 500mW. The
attenuation per unit length of the cable is 2dBkm
-1. Calculate the
output power of the signal from the cable.
Signal loss in cable = 2 x 25 = 50dB
50 = 10 log
10
(500 x 10
-3
/P
out
)
5 = log
10
(500 x 10
-3
/P
out
)
Example 3 – Signal-to-Noise ratio
The background noise in a signal is
5 x 10
-13W
and the minimum
permissible Signal-to-Noise Ratio is
20dB
. Calculate the minimum
signal power,
P
min.
20 = 10 log
10
(P
min
/5 x 10
-13
)
2 = log
10
(P
min
/5 x 10
-13
)
P
min
= 5 x 10
-13
x 10
2
P
min
= 50pW
Example 4 – Cable length
Calculate the maximum length of cable for a
500mW
input signal
if the minimum power out is
50pW
and the attenuation in the cable
is
2dBkm
-1.
Max loss in dB
= 10log
10
(500 x 10
-3
/5 x 10
-11
)
= 120dB
Example 5 - Hearing
SAQ5.2
(Page 37, telecommunications)
The threshold for human hearing is
10
-12
Wm
-2
(the least intense sound it is possible to hear).
If a person’s voice is detected by the ear at an
intensity of
5 x 10
-7
Wm
-2
, what is the intensity
level in decibels?
dB level = 10log
10
(5 x 10
-7
/10
-12
)
Example 6 - Concert
dB level = 10log
10
(0.2/10
-12
)
= 113dB
(Page 37, telecommunications)
A musical concert produces a sound intensity
at the back of the hall of
0.2Wm
-2
. Relative to
Human Hearing
www.mccofnsw.org.au/a/117.html
Musical Concert (example 6)
Talking (example 5)
Gunshot or firecracker 140 dB
Rock concert or jet engine 120 dB Car horn 110 dB Lawnmower 90 dB Normal conversation 60 dB Whisper 15 dB
Human Hearing
Decibel level
relative to the
threshold of
human hearing:
Question
SAQ 5.3 (
page 39, Telecommunications
)
A
200mW
signal enters a cable system of length
100km
.
The cable has an attenuation of
8dBkm
-1. Amplifiers of
gain
41dB
are located at
5km
intervals. Calculate:
(a) The total power loss of the signal travelling through the
cable
(b) The total signal gain as a result of passing through the
amplifiers in the system
(c) The signal power passing through the final amplifier at the
end of the system.
Signal Attenuation
The attenuation of the signal power with
distance is not linear. However, by taking
logs the attenuation in
decibels
is a linear
function of the distance travelled.
P = P
0
e
-
x
Questions
Questions 1-4
Syllabus Objectives
30.5 The mobile phone network
Candidates should be able to:
(o) understand that in a mobile-phone system, the public
switched telephone network (PSTN) is linked to base
stations via a cellular exchange.
(p) understand the need for an area to be divided into a
number of cells, each cell served by a base station.
(q) understand the role of the base station and the cellular
exchange during the making of a call from a mobile
phone handset.
PSTN Networks
Public Switched Telephone Network (PSTN)
Trunk
exchange
Trunk
exchange
7
4
3
5
2
6
7
4
5
1
2
6
2
6
7
4
4
3
5
1
5
1
2
6
Mobile Cellular Network
Each cell is
allocated a
range of
frequencies
that are not
shared by
Mobile Base Station Network
Mobile Base Station Network
Cellular
exchange
PSTN
Mobile
Communications
• Base stations use UHF
– Limited range
– Low power
– Requires a shorter receiver aerial
• Due to cell overlap, different carrier
Making a Mobile Call
Allocating a base station with the strongest signal
Cellular
exchange
PSTN
Making a Mobile Call
1. Handset
emits an identification signal
2. Signal received by a number of
base stations
3. Signal transferred to a
cellular exchange
4. Cellular exchange allocates
• a base station with the strongest signal
• a
carrier frequency
for communication
5. During a call the cellular exchange monitors
Tuner
Demodulator
RF Amplifier
DAC
Serial-to-parallel
speaker
AF
amplifier
Digital to
analogue
Mobile
Handset
microphone
AF
amplifier
ADC
Parallel-to-serial
Modulator
Switch
Oscillator
RF Amplifier
transmitter
Samples
the signal
Converts the sampled
voltage to a series of
bits
Adds the signal
bits to the carrier
wave
Mobile Handset
• Caller speaks into microphone
• Audio signal amplified
• Analogue signal converted to a series of bits
using an ADC and a parallel-to-serial
converter
• Signal added to carrier wave (modulation)
• Signal is amplified
Radiowave.swf
M
ult
ip
le
Final Message
000001000010
010100010010
A 1 000001 B 2 000010 C 3 000011 D 4 000100 E 5 000101 F 6 000110 G 7 000111 H 8 001000 I 9 001001 J 10 001010 K 11 001011 L 12 001100 M 13 001101 N 14 001110 O 15 001111 P 16 010000 R 17 010001 S 18 010010 T 19 010011 U 20 010100 V 21 010101
