Transcript The Beauty of Mathematics in Communications

The Beauty of Mathematics
in Communications
R. C. T. Lee
Dept. of Information Management &
Dept. of Computer Science
National Chi Nan University
1
Operating systems and compilers
Can be built without mathematics.
Most drugs were invented without
Mathematics.
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Can communications systems be
built without mathematics?
Ans: Absolutely no.
Modern communication systems
are totally based upon mathematics.
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For computer scientists, data are stored
in memory as bits, either 1 or 0.
How are the data transmitted?
They are transmitted as pulses: A pulse
represents a 1 and no pulse represents
a 0.
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Fig. 1
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Is this possible when the transmission is
done in a wireless environment?
Impossible.
Fact: Wireless communication is done
every day.
How is this possible?
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Can we mix together two bits and send out?
Impossible if the two bits are represented as
pulses.
Fact: We often mix 256 bits together and
send them at the same time.
How is this done?
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Is an antenna open-circuited?
Yes, it must be. You can easily prove
this by looking at your mobile phone
antenna.
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If an antenna is open-circuited, then
there must be no current on it.
How can it induce electromagnetic
fields without any current?
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Can we broadcast our voice signals
directly through some antenna?
Impossible. Some kind of modulation
must be done.
Why?
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All of these questions can be answered
by mathematics and only by
mathematics.
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f(t)
4
3
2
1
0
-1
-2
-3
-4
0
0.1
0.2
0.3
0.4
0.5
Time
Fig. 2
0.6
0.7
0.8
0.9
t
1
12
4
3.5
3
2.5
2
1.5
1
0.5
0
0
10
20
30
Frequency
40
50
60
f
Fig. 3.The Discrete Fourier Transform Spectrum of
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the Signal in Fig. 2 after Sampling.
f(t)
150
100
50
0
-50
-100
-150
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
Fig. 4 A Signal with Some Noise.
t
1
14
70
60
50
40
30
20
10
0
0
50
100
150
Frequency
200
f
250
Fig. 5 The Discrete Fourier Transform of
the Signal in Fig. 4 after Sampling.
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f(t)
150
100
50
0
-50
-100
-150
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
1
t
Fig. 6 The Signal Obtained by Filtering Out the Noise.
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0.3
0.2
0.1
0
-0.1
-0.2
-0.3
-0.4
0
0.1
0.2
0.3
0.4
0.5
Time
0.6
0.7
0.8
0.9
1
t
Fig. 7 A Music Signal Lasting 1 Second.
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150
100
50
0
0
2000
4000
6000
10000
8000
Frequency
12000
14000
16000
Fig. 8 A Discrete Fourier Transform Spectrum of
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the Signal in Fig. 7.
f
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By using Fourier transform, we can see
that the frequency components in our
human voice are roughly contained in
3k Hertz.
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For a signal with frequency f, its
wavelength  can be found as follows:
v

f

where v is the velocity of light.
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3 10
5
If f  310 ,  
 10 m  100km .
3
3 10
8

3
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It can also be proved that the

length of an antenna is around
.
2
For human voice, this means that
the wavelength is 50km.
No antenna can be that long.
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What can we do?
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Answer: By amplitude modulation.
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Let x(t ) be a signal.

The amplitude modulation is defined
as follows:x(t ) cos( 2f ct )

where fc is the carrier frequency?
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What is the Fourier transform of
x(t ) cos( 2f ct ) ?
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Fig. 9
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The effect of amplitude modulation is to lift
the baseband frequency to the carrier
frequency level, a much higher one.
Once the frequency becomes higher, its
corresponding wavelength becomes smaller.
An antenna is now possible.
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After we receive s(t )  x(t ) cos( 2f ct ),
how can we take x(t ) out of it?
Answer: Multiply s (t ) by cos( 2f ct ) .
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s (t ) cos( 2f c t )  x(t ) cos 2 (2f c t )
1
 ( x(t )  x(t ) cos( 2 (2 f c )t ))
2

Thus x(t ) is recovered.
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Fig. 10
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Our next question: How is a bit
transmitted?
Answer: A bit is usually represented by
a cosine function.
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Let us assume that bit 1 is represented
by cos( 2mf0t ) and bit 0 is represented
by cos(2nf 0t ) .
When the receiver receives a bit, how
can it detect whether 1, or 0, is sent?
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The basic scheme behind the detection
is the inner product property of cosine
functions:
T
cos(2mf0t ), cos(2nf 0t )   cos(2mf0t ) cos(2nf 0t )dt
0

where T  1 f .
0
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It can be proved that
cos(2mf0t ), cos(2nf 0t )  0
T
cos( 2mf0t ), cos( 2nf 0t ) 
2
if m  n
if m  n
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This inner product property gives us the
fundamental mechanism of detecting 1 or 0.
Let the sent signal be denoted as s (t ) .
We perform two inner products:
2
y1 (t )   s (t ), cos( 2mf0t )
T
2
and y2 (t )   s(t ), cos( 2nf 0t )
T

Decision rule: If y1 (t )  1 , say that 1 is sent.
If y2 (t )  1 , say that 0 is sent.
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Suppose that we have two bits to send.
Can we bundle them together and send
the bundled result at the same time?
Answer: Of course, we can.
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Let the two bits be demoted as m1 and m2 .
m1  1 or 0.
m2  1 or 0.
Let s1  1(0) if m1  1(0)
Let s2  1(0) if m2  1(0)
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The sent signal is
s(t )  s1 cos( 2mf0t )  s2 cos( 2nf 0t )
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Our job is to determine the values of
s1 and s2 .
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We perform inner product again.
and
2
y1 (t )   s (t ), cos( 2mf0t )  s1
T
2
y2 (t )   s (t ), cos( 2nf 0t )  s2
T
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Can we bundle 256 bits together and
send them at the same time?
Answer: Yes, as along as the signals
are orthogonal to one another.
This is the basic principle of ADSL:
OFDM (Orthogonal Frequency Division
Method).
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Can we extend the above idea to two
users case?
Answer: We can.
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Let User 1 use 1 (t ) to represent 1 and
 1 (t ) to represent 0.
Let User 2 use 2 (t ) to represent 1 and
 2 (t ) to represent 0.
i (t ),  j (t )  1 if i=j and
i (t ),  j (t )  0 if i≠j.
1 (t ) and 2 (t ) are orthogonal.
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The sent signal is denoted as
s(t )  s11 (t )  s22 (t ) where s1  1
and s2  1 .
To determine, we perform inner products:
 s(t ), 1 (t )  s1
 s(t ), 2 (t )  s2
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Fig. 11 Signature Signals Generated from Hadamand matrix H8.
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All of the signals are orthogonal to each other.
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We can also view the problem as a
vector analysis problem.
Assume that User 6 sends 1 and User 8
sends 0.
s6  (1,1,1,1,1,1,1,1)
 s8  (1,1,1,1,1,1,1,1)
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V6=(1,-1,1,-1,-1,,1,-1,1)
V8=(1,1,-1,-1,-1,-1,1,1)
The inner product of v6 and v8 is
1-1-1+1+1-1-1+1=0
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The sent signal is
s  s6  (s8 )  (0,2,2,0,0,2,2,0)

1
1
s  s 6  ( 2  2  2  2)  1
8
8
1 is sent.
1
1
s  ( s8 )  (2  2  2  2)  1 0 is sent.

8
8
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This is the principle of CDMA (code division
multiple access).
It can be extended to more than two users.
It was used by the military as an
encryption method before.
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Suppose we send a signal entirely in
digital form, can we say that this signal
is an analog signal?
Yes, we can because according to
Fourier series analysis, a pulse also
contains a set of cosine functions.
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X(f )
x(t )
t
1/100
f
1
50
X(f )
x(t )
t
1/200
f
1
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Obviously, the smaller the pulse-width is, the more
frequency components it contains. One may even
say that the smaller the pulse-width is, the more
information it may contain.
Note that if a pulse has a small pulse-width, it means
that within a second, a large number of pulses can
be sent. This corresponds to “high bit rate”.
Now, we know why a wire which has a high bit rate
may be called broadband.
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It is important to observe the following:
Bits are represented by analog signals.
There are no digital signals in the
world.
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I
t
i
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Maxwell’s Equations
Equations Concerning with
Electromagnetic Waves
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Electric Field Induced by Charges
Fig. 12 Coulomb’s Law .
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Magnetic Field Induced by a Current Segment
Fig. 13 Magnetic Flux Density Induced by a Current.
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Do the electric field and magnetic field
affect each other?
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No, not in the static field.
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Yes, if the fields are time-varying.
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The curl of a vector.
 Az Ay 
 Ay Ax 
 Ax Az 


yˆ  

▽  A 
 xˆ  
 zˆ

z 
x 
y 
 z
 y
 x
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Faraday’s law
B
E
t
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Voltmeter
N
S
Fig. 14 The Voltage Caused by the Movement of a
Magnet Inside a Coil
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The changing of magnetic field with time
causes an electric field.
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
Ampere’s law
▽
H  J
Fig. 15 Magnetic Flux Density Induced by a Current.
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The Ampere’s law modified by Maxwell
D
▽ H  J 
t
The changing of electric field with time will
induce a magnetic field.
Maxwell modified Ampere’s law without
performing any experiments.
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Maxwell’s Equations
Differential form
Integral form

▽
B
E  
t
▽
D
H  J 
t
B
cE  dl   s t  ds
D
cH  dl  sJ  ds  s t  ds
▽  D  v
 D  ds    dv
▽ B  0
 B  ds  0
s
s
v
v
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Plane Electromagnetic Waves
With specical boundary conditions, Maxwell’s
equations reduce to
 Ex
 Ex
 
2
2
z
t
2
2
E x  E0 cos k ( z  vt)
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
The speed of the wave: v  1 
  4 (10 )
7
  10 9 / 36
v

1

 9  (1016 )  3  10 8 m / sec
Implication: The electromagnetic
waves travel with the speed of light.
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
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Maxwell was not able to prove his theory.
Hertz proved the correctness of
Maxwell’s equations.
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E
t=0
t=t
z=vt
z
Fig. 16 The Traveling of a Wave.
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Fig. 17 The Electric and Magnetic Fields in a Plane
Electromagnetic Wave.
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E
z

Fig. 18 The Wavelength.

v
We can prove that  
.
f
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


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Transmission Line:
Any electric wire which carries currents
with high frequency can be considered
as a transmission line.
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a
b
Fig. 19 A Co-Axial Cable Transmission Line
y
z
8
F
i
i
d
w
i
x
Fig. 20 Twin-Strip Parallel Plate Transmission Line
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dv
v
dx
Fig. 21 An Equivalent Circuit of a Lossless Transmission Line.
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 2V
 2V
 LC 2
2
x
t
2I
2I
 LC 2
2
x
t
The above equations show that there are
waves on the transmission line.
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RS
Vf

LCX  t
If

LCX  t

LCX  t

LCX  t

Vb

Ib


RL
vva
X=0
X=A
Fig. 22 The Waves on a Transmission Line.
It can be proved that the velocity of
the waves is roughly the speed of light.
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Standing Waves
V ( y)
I ( y)
y

3
4

2

4
Fig. 23 The Case of Open-Circuited Load
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I

2
I
Fig. 24 A Half Wave Dipole Antenna
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In ancient times, human beings built spectacular
buildings.
But, modern communications systems were
possible only recently.
Why?
Answer: Modern communication systems can
not exist without sophisticated mathematics.
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Thank you.
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