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Master of Computer Science 1 - MOB Mobile Internet and Surrounding
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 f0 and the second with the frequency 3f0:
Show graphically that s(t) is periodic of period T = 1/f0.
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 f0 is the fundamental frequency (opposite of the period T), an and bn are the cosine and
sine amplitudes of nth harmonic and c0 is the continues component of signal. Such of
decomposition is called Fourier series. The ratio c0, an and bn 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 an, bn and c0 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
1
T 0
c0  a0  x(t )dt
T
2
T 0
an  x(t ) cos(2f 0 t )dt
T
2
bn   x(t ) sin( 2f 0 t )dt
T 0
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/Tb (in bit/s)
with Tb 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 = 2n.
Modulation rate is the number of symbols transmitted per second:
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1
R= (in symb/s or bauds)
TS
with Ts 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:
S
C = B log2 (1 + ) (in bit/s)
N
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 Eb/N0 (without unit) is equal with the report of consumed energy Eb by transmitted bit
(in J) and of the spectral density of noise power N0 (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?
1
We get: R = baud
TS
And D = 1/Tb bit/s
And Ts = log2M  Tb (with M = 2n)
1 1 1 1 1
So: R = = =  =  D.
Ts log 2 M  Tb log 2 M Tb log 2 M
Thus, the relation between bit rate (D) and modulation rate (R) is:
D = R  log2M = R  log22n = 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 = 106 baud
Because the signal is digital signal (with n = 1, then Ts = log2M  Tb = log22n  Tb = n  Tb =
Tb), so Bit rate is equal to Modulation rate
Thus, D = 106 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  2n = 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 log10(S/N) = 35 dB  S/N = 3162
S
And C = B log2 (1 + ) = 4,5 × 106 × log2(1 + 3162) = 52.321.850 bit/s
N
Thus, η = D/B = 52.321.850 / (4,5 × 106) = 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 SNRdB ?
The values of SNR and SNRdB can be calculated as follows:
10mW 10mW
SNR =   0,01
1W 1000mW
SNRdB = 10log100,01 = 10 log1010-2 = 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 = log22 = 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 = log24 = 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 × log2M  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 log2(1 + SNR) = 3000 × log2(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 SNRdB = 24 dB. How many signal levels
are required?
We get: B = 4 MHz  3 MHz = 1 MHz
And SNRdB = 24 dB = 10 log10(SNR)  SNR = 251
Using Shannon’s formula, so C = 106  log2(1 + 251)  106  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  log2M  106  8 = 2  106  log2M  log2M = 4  M = 16.
10. What is the relates between the achievable spectral effiency η = D/B to Eb/N0?
We have:
Eb S B
 
N0 N D
C
S S
The Shannon’s formula C = B log2 (1 + ) can be writen as:  2 B  1.
N N
Substituing in above formula (with D = C), we have:
Eb B  CB 
  2  1
N0 C  
11. What is the minimum Eb/N0 required to achieve a spectral effiency of 6 bps/Hz?
E b B  CB  1
 
  2  1  2 6  1  10,5
N0 C   6
12. Express the spectral effiency in Exercise 11 in dB unit?
We have: 10 log10(η) = 10 log106 = 7,78
13. Express the Eb/N0 in Exercise 11 in dB unit?
We get: 10 log10(Eb/N0) = 10 log10(10,5)  10,21 dB.
14. What is the relates between the useful signal power S, and the data rate D to Eb/N0?
E S B
We have: b  
N0 N D
Hence, the noise power in a signal with banwidth B is N = N0  B. Substituing, we have:
Eb S S 1
  D 
N0 N0 D N0 E b
N0
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.
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The modern transmission systems call upon the digital techniques which allow between
others offering a better transmission quality. The modulating signal comes then from the form
setting of a digital message with assistance of a online code. The online codes used for a
base-band transmission are varied (NRZ, Manchester code; bipolar code…). Their choice
depends on the spectral characteristics of the signal produced with the exit of online coding
(spectral occupation, presence of one continuous component, facility of rhythm recovery on the
reception…). The online code used when the signal is destined to be transmitted on carrier
frequency is practically always the NRZ code M-surface. The modulating signal is thus a
multi-levels signal constituted by the succession of duration level Ts and of proportional
amplitude to the symbol value to emit. The signal g(t) met higher can be considered like
resulting of a binary NRZ coding of the binary digital message "10101". The modulation
operation, when it is realized with a modulating digital signal, is then called digital modulation.
1.2.1 Digital modulations definition
The modulations consist in modifying the amplitude, the phase or the frequency of a
sinusoid carrier of frequency f0 according to the symbol value to transmit on each interval of
duration Ts. There are thus three great modulation types: Amplitude Shift Keying (ASK), Phase
Shift Keying (PSK), and Frequency Shift Keying (FSK).
The figure associated with the initials of one modulation corresponds to the total number of
possible states of the modulated signal. Thus, a PSK-4 corresponds to a phase modulation where
the modulated signal could be in one of the four states of possible phase of the modulation. Each
state is associated with a symbol. The number of states is thus equal to the value of M. The
increase of the number of states of modulation allows decreasing the band-width of the resulting
signal, but at the price of a greater sensitivity with the noise.
Amplitude Shift Keying (ASK) associates each symbol to transmit different amplitude.
This is about the simplest technique and most natural to modulate a sinusoid carrier. In the case
of the ASK-2, the signal can be written over the first symbol duration:
 s(t) = A cos(2f0t + 0) for one "1" binary ;
 s(t) = - A cos(2f0t + 0) for one "0" binary.
Phase Shift Keying (PSK) associates each symbol to transmit a different phase. A
modulation with two levels PSK-2, Binary Phase Shift Keying (BPSK), can be defined on the
first symbol duration by the following resulting signal:
 s(t) = A cos(2f0t + 0) for one "1" binary;
 s(t) = A cos(2f0t + 0 +  ) for one "0" binary.
With this chose of phase states (0 and ), PSK-2 and ASK-2 are rigorously () identical.
Frequency Shift Keying (FSK) associates each symbol to transmit a different frequency. A
modulation with two states FSK-2, Binary Frequency Shift Keying (BFSK), can be
represented over the first symbol duration by the following resulting signal:
 s(t) = A cos(2f1t + 0) for one "1" binary;
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 s(t) = A cos(2f2t + 0) for one "0" binary.
The modulated signals in phase or frequency are less sensitive with the noise than the
amplitude modulated signals. The counterpart is that the transmitter and the receiver are more
complex and thus more expensive.
Quadrature Amplitude Modulation (QAM) is a hybrid technique of modulation which
employs a combination of amplitude modulation and phase modulation. It realized while
separating symbols to transmit into two parallel flows. Each flow is then transmitted in
amplitude modulation on each way thus made up. On one of these two channels, called “in
phase” channel, a sinusoid carrier is modulated in MDA- M . On the second channel, called “in
quadrature” channel , a MDA- M is realized in using this time a version shift phased from π/2
of same sinusoid carrier. The orthogonality of two channels is exploited to the reception to
allow the symbols decoding. The figure 1 represents the principle diagram of a modulator
QAM-4. The resulting modulated signal can also express itself from a same sinusoid function
which the amplitude and the phase dependent on bits to emit. It is the reason why a QAM can be
seen like the combination of an amplitude modulation and a phase modulation.
Fig. 1 – QAM Coder
A modulation of four QAM-4 states can be represented over the duration of first symbol by
the resulting signal following:
 s(t) = A cos(2f0t + 0) + A cos(2f0t + 0 + /2) for a binary couple "11";
 s(t) = - A cos(2f0t + 0) + A cos(2f0t + 0 + /2) for a binary couple "01";
 s(t) = - A cos(2f0t + 0) - A cos(2f0t + 0 + /2) for a binary couple "00";
 s(t) = A cos(2f0t + 0) - A cos(2f0t + 0 + /2) for a binary couple "10".
That is also written:
 s(t) = A 2 cos(2f0t + 0 + /4) for a binary couple "11" ;
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 s(t) = A 2 cos(2f0t + 0 + 3/4) for a binary couple "01" ;
 s(t) = A 2 cos(2f0t + 0 + 5/4) for a binary couple "00" ;
 s(t) = A 2 cos(2f0t + 0 + 7/4) for a binary couple "10".
A QAM-4 is thus nothing of other but a PSK-4 (QPSK) where the phase states are selected equal
to π/4, 3π/4, 5π/4, 7π/4.
The QAM is very largely employed by the modems because it allows doubling the flow
obtained with a ASK using the same band, all in preserving the same performances, with
condition of having a signal-with-noise report sufficient at the reception.
Exercises
1. A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and
radians?
We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is: 1/6  360° = 60° = /3 rad
2. If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500,
700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have
a maximum amplitude of 10 V.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then:
B = fh – fl = 900 – 100 = 800 Hz
3. One wishes to transmit the binary digital message "0010011100…"
a) One chooses to transmit this sequence with help of a modem which uses a FSK-2.
Draw the signal transmitted by the modem.
0 T 2T 3T 4T 5T 6T 7T 8T 9T 10T
b) One wishes to transmit our binary digital message with help of an ASK-4. Propose a
binary coding rule for M-surface. Draw the signal transmitted by the modem.
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c) One wishes finally to transmit our digital message with help of a QAM-4. Describe
the constitution mode the “in phase” and “in quadrature” channels. Give the duration
of a symbol. Propose a binary coding rule for M-surface and represent the transmitted
signal by the modem.
01 A 11
3/4 /4
5/4 7/4
00 10
4. Generic expression of a modulated signal
a) Which generic formula allows expressing the modulated signal m(t) whatever the
type of modulation?
Let the carrier be s(t) = Acsin (Ωct), then modulating signal can be m(t) = Amsin (ωmt)
b) Which way the information is carried by in ASK, PSK and FSK?
Mapping of the quantized amplitudes, frequencies or phases to codewords (bit groups).
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5. One wishes to transmit a digital message of bit rate 4 Mbit/s. The necessary band-width to
transmit a modulated signal being approximately equals to the modulation rate, let calculate
the necessary frequency band.
a) in ASK-16?
M = 16  n = 4
D=Rn
 R = D / n = 4  106 / 4 = 106 baud
The necessary band-width to transmit a modulated signal being approximately equals to
the modulation rate, so we get: f = 106 Hz = 1 MHz.
b) in QAM-16?
M = 16  n = 4.
So the result will look like the case of using ASK-16.
We get: f = 1 MHz.
c) Which is one of these two modulations preferable to use?
QAM is less susceptible to error than ASK
d) One wishes to double the bit rate transmitted in the frequency band that one employs
above. How to proceed?
D=Rn
and D’ = 2D  R  n’ = 2 (R  n) = 2  R  n  n’ = 2  n.
So we need to double the number of bits used to represent the signal.
1.2.2 Constellation
The combinations of possible value on the “in phase” channel and the “in quadrature”
channel of a modulation are commonly represented by a constellation of points, where each
point represents a symbol associated with a bit group to transmit. The distance of a constellation
point in comparing with the origin indicates the amplitude of the modulated signal; its angle
indicates the shift of phase. The figure 2 represents the constellation QAM-16 used on the
WLAN 802.11a.
One will note the binary coding for M-surface used at the same time on the “in phase”
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channel (symbols ak) and on the “in quadrature” channel (symbols bk). An error of detection on
an ak at the reception is do not induce the error on two bits of weak weight, coded independently
by the values of the symbols bk. A code of Gray is in addition used on each channel.
Fig. 2 – QAM-16 Constellation
Exercise
1. Which modulation does a constellation diagram correspond to two co-ordinates points
(1, 0) and (+1, 0)?
The basic BPSK scheme only uses one possible phase shift of 180°.
(-1; 0) (+1; 0)
2. The same question if all points of the diagram are on same ring centered on the origin.
The quadrature PSK (QPSK), one of the most common PSK schemes (sometimes also called
quaternary PSK). Here, higher bit rates can be achieved for the same bandwidth by coding
two bits into one phase shift.
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3. Can one meet a constellation diagram of 11 points?
No, it can not, because the number of points must be a even number.
4. Give a generic formula giving the number of points of a modulation QAM-M.
The number of points of a modulation QAM-M is M
1.2.3. Spectral characteristic of a modulated signal - Root Nyquist Filtering
The signal modulating NRZ is filtered before modulation. This operation limits the occupied
band by the base band signal. The digital communications theory show that this filtering must be
realized by a root Nyquist filter at the emission and at the reception to minimize the noise effect
at the reception and to avoid the interference between symbols (IES). The chain of processing
used at the emission is represented in Fig. 3.
Fig.3 – Coder QAM with filtering.
The figure 4 gives the frequential representation of a modulated signal QAM by the chain of
processing represented in Fig. 1. The figure 5 gives the frequential representation of the same
pre-filtered signal by a root Nyquist filter in according to the chain of processing represented in
Fig. 3.
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