MOB Subject 3 Transmission Techniques

MOB Subject 3 Transmission Techniques

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:

– 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 nth 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 nappearing in this decomposition

are given by the integral following, define on an interval unspecified length T:

Exercise

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 x t dt

T

a

c

0 0



 T

T

a

0

0 ) 2 cos(

) (



 T

T

b

0

0 ) 2 sin(

) (

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 = 2n

Modulation rate is the number of symbols transmitted per second:

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 log2 (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 bby 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:

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/Tbbit/s

And Ts = log2M Tb (with M = 2n)

So: R =

s

T1 =

b T

M

2 log

M

2 log

b

T1 =

M

2 log

1  D.

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 = 106baud

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 = 106bit/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

on such a channel?

We get: 10 log10(S/N) = 35 dB S/N = 3162

And C = B log2 (1 +

N S ) = 4,5 × 106 × log2(1 + 3162) = 52.321.850 bit/s

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:

1000

10 1

mW

mW W

mW

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

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:

D

B N

S

N

0

The Shannon’s formula C = B log2 (1 +

N S ) can be writen as: 2B 1

C N

Substituing in above formula (with D = C), we have:















 2 1

0

B

C b

C

B

N

E

11 What is the minimum Eb/N0required to achieve a spectral effiency of 6 bps/Hz?

2 1 10,5 6

1 1

0





















 B

C b

C

B

N

E

12 Express the spectral effiency in Exercise 11 in dB unit?

We have: 10 log10(η) = 10 log106 = 7,78

13 Express the Eb/N0in 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?

We have:

D

B N

S N

0 Hence, the noise power in a signal with banwidth B isN = N0 B Substituing, we have:

D

N

S

N

E b

0

0

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

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 T s 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 f 0 according to the symbol value to transmit on each interval of

duration T s 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f 0 t + 0) for one "1" binary ;

 s(t) = - A cos(2f 0 t + 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f 0 t + 0) for one "1" binary;

 s(t) = A cos(2f 0 t + 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f 1 t + 0) for one "1" binary;

 s(t) = A cos(2f 2 t + 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f 0 t + 0) + A cos(2f 0 t + 0+/2) for a binary couple "11";

 s(t) = - A cos(2f 0 t + 0) + A cos(2f 0 t + 0 +/2) for a binary couple "01";

 s(t) = - A cos(2f 0 t + 0) - A cos(2f 0 t + 0 +/2) for a binary couple "00";

 s(t) = A cos(2f 0 t + 0) - A cos(2f 0 t + 0 +/2) for a binary couple "10"

That is also written:

 s(t) = A 2 cos(2f 0 t + 0 +/4) for a binary couple "11" ;

 s(t) = A 2 cos(2f 0 t + 0 + 3/4) for a binary couple "01" ;

 s(t) = A 2 cos(2f 0 t + 0 + 5/4) for a binary couple "00" ;

 s(t) = A 2 cos(2f 0 t + 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 itsbandwidth? 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 thebandwidth 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

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.

0