1. The Beauty of Mathematics in
Communications
R. C. T. Lee
Dept. of Information Management
and
Dept. of Computer Science
National Chi Nan University

2. We often hear the term “modulation”.
For instance, we are all familiar with the term
“amplitude modulation”.
Can we transmit our voice or music without
going through the process of modulation?

3. A Music Signal Lasting 1 Second

4. What is inside the signal?
Fourier transform.
for t = 0, 1, 2, …, n-1.

5. A Discrete Fourier Transform Spectrum of the Signal

6. It can be seen that the frequencies of the
signal are in the range of 100 to 3000 Hertz.

7. A Signal

8. The Discrete Fourier Transform Spectrum of the Signal

9. The above analysis shows that the signal
is as follows:

10. Through Fourier transform, we can find out
that the frequencies of our human voice
are in the range of 300 to 3000 Hertz.
Through Maxwell’s equations, we know
that the length of an antenna to transmit a signal
with frequency f has to be
where v is the speed of light and f is the frequency
of the signal.
If f =3000 Hertz, the wavelength is 100km.
No such an antenna is possible.

11. How can we lift the base band frequency to
a higher one such that it can be transmitted
through an antenna?
Through the use of “modulation” which
effectively lifts the frequencies of the base
band.

12. Modulation: Multiplying the base band signal
by a cosine signal.
m(t): the base band signal.
Carrier wave signal:

13. From Fourier transform theory, every signal
component in m(t) with frequency f will have
frequencies fc+f and fc-f after m(t) is multiplied by .
This can be easily understood as follows:
Consider any cosine function with frequency
f :

14. Now, after multiplication with , we
have
Thus, every signal with frequency f, which is
relatively low, becomes two signals with
frequencies fc+f and fc-f which are relatively
high because fc is a high frequency.

15. We shall show later that high frequency means
small wavelength.
Because the frequencies of the signals
are high now, their wavelengths are now
relatively small.
They can now be transmitted by an antenna
with reasonable length.

16. Illustrating the Amplitude Modulation Process.
(a) Baseband Signal m(t). (b) AM Wave where
for All t.

17. Spectrum of the Message Signal.
(b) Spectrum of the DSB-SC Modulated Signal.

18. But the signal we receive with frequencies
fc+f and fc-f are not comprehensible because
they are of high frequencies.
We must have a scheme to recover the base
band signal thoroughly.
How?
Through demodulation.

19. The modulation is done by multiplication of.
The demodulation is also done by multiplication.
Consider any base band signal component with
frequency f.

20. After modulation, the sent signal is

21. After demodulation, we have

22. Thus, after demodulation, we have two signals:
and

23. Illustrating the Spectrum of the Demodulation Process.

24. The receiver receives two functions,
and
Do we really need both of them to recover the
sent signal ?
Ans: No.

25. Let us assume that the receiver uses a filter
so that only is passed.
Then let us demodulate this signal by
multiplying it by
The result is

26. But , after demodulation,
Thus the sent signal is recovered.
This is the principle of Single Side Band.

27. Suppose we do not want to use a filter, what
can we do?
We can do the following trick:
Let m(t) be
Let be

28. Then we send the following signal:

29. This time, only one signal, namely
is sent.
But, in a signal such as human voice, there
are a lot of cosine functions, how can we
transform each cosine function into a
sine function?

30. This is done by Fourier transform.
Note that
and

31. For the Fourier transform coefficient of f,
we multiply it by 2j and for that of -f, we
multiply it by –2j.
We then perform an inverse Fourier transform,
the result is:
Each signal in the original signal undergoes a
phase shift.

32. Digital Modulation
Binary Phase Shift Keying

33. Signals of BPSK.

34. Binary Frequency Shift Keying
for some fixed integer nc and i=1,2
If we want to send 1(0), we send s1(t)(s2(t)).

35. We can prove that
and
s1(t) and s2(t) are orthogonal.

36. To detect s1(t), we multiply the sent signal by s1(t).
If 1 is transmitted,
If 0 is transmitted,

37. Coherent Detector for FSK System.

38. Signals for an FSK System.

39. Can we mix two bits together and send them
out at the same time?
Yes, QPSK.

41. To detect si1, we multiply si(t) by .

42. In QPSK, we transmit 2 bits together.
Can we further extend the idea into, say, sending
n bits together?
Yes, OFDM.

43. To compute s(t), we can use the fast Fourier
transform.
OFDM is used in our ADSL system.

44. Multiple Access Communications

45. FDMA

46. Spectrum of an FDMA System

47. TDMA

48. CDMA

49. m1=+1 or -1
m2=+1 or -1.

50. We can prove that s1(t) and s2(t) are orthogonal.
Let us represent s1(t) by (1,1) and s2(t) by (1,-1).
We can see that
( 1, 1)( 1, -1)=0

51. To detect m1, we multiply r(t) by s1(t) and
integrate.
To detect m2, we multiply r(t) by s2(t) and

53. Signature Signals Generated from H8.

54. Electromagnetic Theories
Some basic notations:
The cross product of two vectors. Let A and B
be two vectors.
Then

55. Then

56. E: electric field
D: electric flux density
B: magnetic flux density
H: magnetic field
Maxwell’s Equations

57. The plane EM wave:

58. Let
In free space, and
Thus,
This means that the EM wave travels in
free space with the velocity of light.

59. The Electric and Magnetic Fields in a Plane Electromagnetic Wave

60. It can be proved that the wavelength of the
wave is as follows:
where f is the frequency of the signal.
Let f=60 Hertz.

61. If the frequency is very high, the wavelength
becomes smaller.
The length of an antenna is usually
This is why we always have to elevate the
signal frequency into a higher one so that
some antenna can be used.

62. Transmission Lines
If the wavelength of the signal on a line
is small as compared with the length of the
line, this line is considered a transmission
line.
Therefore, the power line in our houses is
not a transmission line because the frequency
of the signal is only 60 Hertz ( =5000 km).

63. A Co-Axial Cable Transmission Line
Twin-Strip Parallel Plate Transmission Line

65. There are two voltage waves and two current
waves on the transmission line.

66. The speed of the waves v is .
Therefore the waves travel with the speed of
light in a transmission line.

67. Standing Waves

68. Voltage and Current with Respect to Time along the Transmission Line
A Lumped Circuit whose length Is Way Short as
Compared with the Wavelength of the Signal

69. But, is it true that we are now sending digital
signals (pulses) in the transmission lines?

70. If we want to send a large number of bits per
second, the width of the pulses will become
smaller.
This in turn means that the corresponding
frequency spectrum will become wider.
This is why we call the advanced transmission
lines which transmits pulses with high speed
wide band transmission lines.

71. Antennas: Consider any transmission line
From very far away, there is no current.
No current, no radiation.
This is not an antenna.

72. Simple Dipole Antenna

73. What happens if the length of the antenna
is larger than ?
No good at all.

74. Conclusion:
We can not have modern communications
without mathematics.