MOB Subject 3 Transmission Techniques

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MOB Subject 3
Transmission Techniques
1. Signal transmission techniques - Practical
The local radio-electric networks use radio waves or infrared in order to transmit data. The
technique used with the origin for the radio transmissions is called narrow band transmission. It
consists in sharing the frequency band in relatively low sub-bandwidth in order to constitute
distinct channels. The different communications then passed on these channels called to narrow
band.
The radio transmissions however are subjected to many constraints which return the narrow
band transmission sometimes insufficient. It is the case in particular when the radio wave
propagation is effected by multi traffics. The transmission techniques like the spreading out of
spectrum dispersion and OFDM (Orthogonal Frequency Division Multiplexing) have been
developed to fight against the elimination problems which result of these multi traffics. These
techniques then use all the bandwidth available. One has then a transmission known as broad
band.
This practical aims to study these different aspects.
1.1. Definitions and basic concepts
1.1.1 Message, signal and of transmission modes
Message is constituted by the information elements or data that the user wishes to transmit.
One distinguishes two types of message:
- Analog message: data are analog. They are presented in the form of a function f(t)
continuous and with continuous time (ex: voice, video);
- Digital message: data are digital. They are presented in the form of a series {ik} of data
elements being able to take one among a discrete values unit called alphabet (ex:
characters of a text, entireties).
An analog message can be digitized by a sampling operation (discretization of the axis of times),
followed by a quantification operation (discretization of the values taken by the sampled analog
data).
Signals are the physic representation of the message to transmit. They are presented

generally in the form of a electric size (tension, current) which is then converted into an electric
or electromagnetic wave to be transmitted. Again, one makes the distinction between:
- Analog signal: signal associated an analog message;
- Digital signal: signal resulting of the setting in form of a digital message. A digital signal
is presented in the form of a succession of wave form being able to get one among
possibilities finished unit used to code information.
Transmission is the operation which consists in transmitting the signal of a machine towards
another, on a support given. It can be carried out in base band or on frequency carrier:
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– Base band transmission: it corresponds to transmission on channel of the low-pass type
of a centered spectrum signal around the null frequency. It is the case of the majority of
the signals associates the analog messages (voice, music, video…). It is also the case of
multilevel digital signals of "square" form.
– Carrier frequency transmission: when the channel is of band-pass type as in the case of
the hertzian transmission, it is essential to transpose the signal spectrum around one
carrier frequency located in the center of the frequency band envisaged to transmit the
signal. This operation is called modulation. It is carried out in modifying one of the
characteristics (amplitude, phase, instantaneous frequency) of one sinusoid carrier while
using of the information carrying signal to transmit. When this signal is analog, one talks
about analog modulation. If the signal is digital, the modulation is known as digital.
1.1.2. Frequential components of a signal - Periodic signals example
A signal can be seen like a sum of sinusoids, amplitude and various phase frequency. A
periodic signal has frequential components with multi-frequency of a same value, called
fundamental frequency, equal with the reverse period.
Exercise
Consider the example following according to the signal s(t) which has two frequential
components, one with the frequency f
0
and the second with the frequency 3f

0
:
Show graphically that s(t) is periodic of period T = 1/f
0
.
Decomposition in Fourier series
In the 19th century, Jean-Baptiste Fourier shows how all periodic function g(t) of period T
can decompose in a sum (possibly infinite) of sine and cosine functions:
where f
0
is the fundamental frequency (opposite of the period T), a
n
and b
n
are the cosine and
sine amplitudes of n
th
harmonic and c
0
is the continues component of signal. Such of
decomposition is called Fourier series. The ratio c
0
, a
n
and b
n
appearing in this decomposition
are given by the integral following, define on an interval unspecified length T:
Exercise
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One considers the signal periodic of period T of period motif:
This signal is a coded digital signal on two levels: +1 V and -1 V. It is supposed that an
impulse of +1 V codes one "1" and that an impulse of -1 V codes one "0". This coding type is
called NRZ (Non Return to Zero).
1. Trace the corresponding function g(t).
2. Find the value of the coefficients a
n
, b
n
and c
0
of g(t). Deduce the decomposition
expression in Fourier series of g(t) and draw the frequencies spectrum signal for the first
five harmonics.


T
dttx
T
ac
0
00
)(
1


T
n
dttftx

T
a
0
0
)2cos()(
2



T
n
dttftx
T
b
0
0
)2sin()(
2

3. Only the first five harmonics of the signal are transmitted by the channel, the following
being attenuated completely (low-pass channel). What is the form of the received signal?
A decoding of the transmitted data seems it possible?
1 1.3 Band-width, capacity, and signal-with-noise ratio
One considers subsequently the transmission of digital signals. One considers without
restriction of general information that the numerical message to transmit was coded in a suitable
way using binary characters.
Bit rate D is the number of binary bits transmitted per second:
D = 1/T
b
(in bit/s)

with T
b
duration of a bit.
The binary bits to transmit are often gathered in n-uplets, themselves coded by symbols. It is
the binary coding operation to M-surface. M is the number of symbols necessary to code all
value possible of one n-uplets of binary bits: M = 2
n
.
Modulation rate is the number of symbols transmitted per second:
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R =
T
S
1
(in symb/s or bauds)
3
with T
s
duration of a symbol.
Nyquist’s law imposes a limit on the modulation speed of a low-pass channel of band-width
B:
R ≤ 2B
Shannon’s theorem provides the fundamental limit on the bit rate maximum, still called
capacity, of a transmission channel (low-pass or band-pass) of bandwidth B, when this one is
subject to additive Gaussian white noise. This capacity express in the way following:
C = B log
2
(1 +
N

S
) (in bit/s)
where S/N represents the signal-with-noise report of the channel (report of the useful signal
power on the noise power, expressed in mW).
This limit, resulting from the information theory, provides a theoretical boundary with the
designers of communication system. The most advanced systems try to approach of this
boundary without still reaching it at the present time.
Nyquist’s law and the Shannon’s theorem have as a main interested thing to give a good idea
of the size order of the bit rate which one can hope to reach on a channel given.
Ratio E
b
/N
0
(without unit) is equal with the report of consumed energy E
b
by transmitted bit
(in J) and of the spectral density of noise power N
0
(in W/Hz). This ratio can be expresses
according to the band-width B of the signal, of the bit rate D and of the signal-with-noise report:
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The performances of the communication systems depend naturally very strongly on this ratio.
Spectral efficacy η express in bit/s/Hz. It characterizes the capacity of a communication
system to transfer a flow D in a bandwidth B given.
Shannon’s theorem provides in an implicit way the theoretical maximum spectral efficacy:
Exercise
1. Establish the relation between bit rate and modulation rate. Does the Nyquist’s law
impose a limitation of bit rate which one can transmit in a band B?
We get: R =

T
S
1
baud
And D = 1/T
b
bit/s
And T
s
= log
2
M  T
b
(with M = 2
n
)
So: R =
s
T
1
=
b
TM 
2
log
1
=
M
2
log

1

b
T
1
=
M
2
log
1
 D.
Thus, the relation between bit rate (D) and modulation rate (R) is:
D = R  log
2
M = R  log
2
2
n
= R  n
Does the Nyquist’s law impose a limitation of bit rate which one can transmit in a band B?
Yes, because we get R ≤ 2B  D/n ≤ 2B  D ≤ 2nB
2. It is considered that the signal frequency is 1 MHz. What is the flow of the digital signal
corresponding (in bits per second)?
We get: R = 1/T = f = 10
6
baud
Because the signal is digital signal (with n = 1, then T
s
= log
2

M  T
b
= log
2
2
n
 T
b
= n  T
b
=
T
b
), so Bit rate is equal to Modulation rate
Thus, D = 10
6
bit/s = 1 Mbps
3. One wishes to transmit the digital video having the following characteristics: matrix of
480×640 pixels where each pixel corresponds to an intensity which can take 32 different
values. The required speed is of 25 images per second.
a) What is the bit rate of the video source?
We get: M = 32  2
n
= 32  n = 5. On other hand: One pixel needs 5 bit.
Bits of an image = 480 × 640 × 5 = 1536000 bit/image
The speed is 25 images per second, so we get 25 × 1536000 = 38.400.000 bit/s = 38,4 Mbps
b) The channel of low-pass type, has a band-width of 4,5 MHz and a
signal-with-noise report from 35 dB. Can one transfer the video signal on this
channel? What is the maximum spectral efficacy that one can theoretically reach
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on such a channel?
We get: 10 log
10
(S/N) = 35 dB  S/N = 3162
And C = B log
2
(1 +
N
S
) = 4,5 × 10
6
× log
2
(1 + 3162) = 52.321.850 bit/s
Thus, η = D/B = 52.321.850 / (4,5 × 10
6
) = 11,63 bit/s/Hz
4. The power of a signal is 10 mW and the power of the noise is 1 ́W; what are the values of
SNR and SNR
dB
?
The values of SNR and SNR
dB
can be calculated as follows:
SNR =
01,0
1000
10
1

10

mW
mW
W
mW
SNR
dB
= 10log
10
0,01 = 10 log
10
10
-2
= 20 dB
5. Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two
signal levels. The maximum bit rate can be?
R = 2B = 6000 baud
Because a signal transmitting with two signal levels, so n = log
2
2 = 1, then:
D = R = 6000 bps
6. Consider the noiseless channel with a bandwidth of 3000 Hz transmitting a signal with
four signal levels (for each level, we send 2 bits). The maximum bit rate can be?
R = 2B = 6000 baud
Because a signal transmitting with four signal levels, so n = log
2
4 = 2, then:
D = R × 2 = 12000 bps
7. We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How

many signal levels do we need?
We can use the Nyquist formula as shown:
265000 = 2 × 20000 × log
2
M  M = 99
The Nyquist formula tells us how many signal levels we need.
8. We can calculate the theoretical highest bit rate of a regular telephone line. A telephone
line normally has a bandwidth of 3000Hz. The signal-to-noise ratio is usually 3162. For
this channel the capacity is?
We can use the Shannon formula as shown:
C = B log
2
(1 + SNR) = 3000 × log
2
(1 + 3162) = 3000 × 11,62 = 34.860 bps
The Shannon gives us the upper limit capacity.
9. Consider the relates the Nyquist and Shannon formulations. Suppose that the spectrum
of a channel is between 3 MHz and 4 MHz and SNR
dB
= 24 dB. How many signal levels
are required?
We get: B = 4 MHz  3 MHz = 1 MHz
And SNR
dB
= 24 dB = 10 log
10
(SNR)  SNR = 251
Using Shannon’s formula, so C = 10
6
 log

2
(1 + 251)  10
6
 8 = 8 Mbps.
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This is a theoretical, and we can not reached, but assume we can. Based on Nyquist’s
formula, we have: C = 2  B  log
2
M  10
6
 8 = 2  10
6
 log
2
M  log
2
M = 4  M = 16.
10. What is the relates between the achievable spectral effiency η = D/B to E
b
/N
0
?
We have:
D
B
N
S
N
E

b

0
The Shannon’s formula C = B log
2
(1 +
N
S
) can be writen as:
12 
B
C
N
S
.
Substituing in above formula (with D = C), we have:








 12
0
B
C
b
C

B
N
E
11. What is the minimum E
b
/N
0
required to achieve a spectral effiency of 6 bps/Hz?
 
5,1012
6
1
12
6
0










B
C
b
C
B

N
E
12. Express the spectral effiency in Exercise 11 in dB unit?
We have: 10 log
10
(η) = 10 log
10
6 = 7,78
13. Express the E
b
/N
0
in Exercise 11 in dB unit?
We get: 10 log
10
(E
b
/N
0
) = 10 log
10
(10,5)  10,21 dB.
14. What is the relates between the useful signal power S, and the data rate D to E
b
/N
0
?
We have:
D
B

N
S
N
E
b

0
Hence, the noise power in a signal with banwidth B is N = N
0
 B. Substituing, we have:
DN
S
N
E
b
00


0
0
1
N
E
N
S
D
b

1.2 The digital modulations
When the channel is of band-pass type like in the case of the hertzian transmission, it is

essential to transpose the signal spectrum around a carrier frequency located in the center of the
available frequency band on the channel. One realizes thus a transmission on carrier frequency.
The modulation technique allows shifting the signal spectrum, naturally located around the
null frequency (base-band signal), around one carrier frequency. It is carried out while
modifying one of the characteristic (amplitude, phase, instantaneous frequency) of one
sinusoid carrier while using the information carrying signal to transmit called modulating signal.
When the modulating signal is analog, one talks about analog modulation.