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Master of Computer Science 1 - MOB Mobile Internet
MOB TME Subject 1 – Signal Transmission
Data transmission on a physical support is the result of the propagation of a wave. The type of wave
depends on the physical support used:
 Electric wave: in the case of the voltage or electrical current ;
 Electromagnetic wave: in the case of a coupled electric and magnetic field (light, radio
wave, etc).
A wave is a function of both time variable and spaces. As an example, the electric field of a
electromagnetic wave that is propagating along an axis (x) can be represented by a sinusoidal
function of the form:
The parameter c represents the propagation speed of the wave, (c is equal to 300.000 km/s for an
electromagnetic wave which is propagated in the vacuum or the air). x is distance, x is measured by
metre (10-3km).
For a fixed value of x, this wave is thus represented by a sinusoidal signal Asin(20t +  ),
characterized by three parameters :
 Amplitude A : the peak value of the signal in the time, the average power is equal to
 Frequency f0 : the speed in which the signal repeats, expressed in cycles per second or in Hertz
(Hz) (the inverse of the frequency is called period T of signal and measured in seconds
 Phase φ : a measure of relative position in the time with a period of the signal, expressed in
radians (rad).
In practice, the propagating signal consists of several sinusoidal components of different
frequencies, amplitudes and phases.
As the waves propagate, some of them will be reflected or diffracted by obstacles in their path. The
received signal is thus composed of several copies of the emitted wave, with different delays and
attenuations compared to each other. This is called multi-path transmission. The combination of
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these waves at the receiver results in constructive and destructive interference and in fading of the
signal both in time and frequency.
Let us consider propagation with two paths.
a) Neglecting the attenuation, give the mathematic expression of the combination of received
E(x1,t)+E(x2,t) = A sin (2f0 (t-x1/c) + ) + A sin (2f0 (t-x2/c) + )
b) Explain why the multi-path attenuation is maximal when the waves arrive with a difference of
180° in their phase?
When the replica arrive with a phase of 180°, distances x1 and x2 done by tge 2 waves up to the reception
point are such that:
2f0 (t-x2/c) +  = 2f0 (t-x1/c) +  +   t
So :
E(x1,t)+E(x2,t) = A sin (2f0 (t-x1/c) + ) + A sin (2f0 (t-x1/c) +  + )
= A sin (2f0 (t-x1/c) + ) - A sin (2f0 (t-x1/c) +  ) car sin(+) = -sin
resulting signal is therefore null.
c) Which distances separating the two waves will maximize the multi-path attenuation on a 50km
microwave link at 1 GHz?
Interferences are destructives when the 2 replicas arrive with a phase of 180 °, so at the reception point
distances followed by each wave x2 and x1, are such that :
2f0 (t-x2/c) +  = 2f0 (t-x1/c) +  +   t
 2f0 (-x2/c) = 2f0 (-x1/c) + 
 2f0 ((x1-x2)/c) = 
 x1-x2 = c/(2f0)
=c/f0 with c = 3.108 m.s-2.
So : x1-x2 = /2.
As a wave propagates, it will be significantly attenuated. The quality of a connection depends
strongly on the power of the received electromagnetic wave. As the power at the transmitter and
received can vary by several orders of magnitude calculations of the power of the signal are
typically carried out on a logarithmic scale.
The power is Watts in a radio transmission is therefore typically expressed in the relative value of
decibels compared to a milliwatt (dBm). The relationship between the power in Watts and the power
in dBm is written as:
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The relationship between the power in Watts and the reference power of 1mW is a unit-less ratio,
expressed in dB. The basic unit of the dB is the “Bel”, but historically is unit was considered too
large and we typically use the unit decibel or dB with is one tenth of a “Bel”. For this reason the
calculation to obtain the power in decibels contains a factor of 10. The ratio of two powers is called
the gain :
It can be noted that this ratio in dB is positive when P2 / P1 > 1 and negative if P2 / P1 < 1.
The power is proportional to the square of the voltage, and so when the gain is calculated from
voltage, the calculation of gain in dB is expressed as the ratio of the square of the voltages, written
1. What does a negative value is dBm correspond to? (Answer: P(mW) < 1)
10 log10(P(mW)) < 0  P(mW) < 1. It corresponds to a power lesser than a mW
2. Let's consider an amplifier, which provides 8V in output for an input signal of 2V. Supposing
that the input and output stages of the amplifier both behave like a resistance of R = 50.
Recalling that the power passing through a resistance is written P = U2/ R, then
a) What is power at the input and output of the amplifier?
Poutput = 82 / 50 = 1.28 and Pinput = 22 / 50 = 0.08
Input power is 80mW and the output 1,28 W
b) What does the ratio Poutput/Pinput represent? Express this ratio in decibels.
Signal was amplified.
GdB = 10log10(P2/P1) = 10log10(1.28/0.08) = 12,04119983
c) Now express the formula for the gain in dB in terms of the voltages rather than the powers?
3. Fill in the table below:
Ratio (dB) Power ratio Voltage ratio
3 2 1.41
6 4 2
10 10 3,16
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20 100 10
30 1000 31,6
40 10 000 100
x 10 10(x/20)
4. What does a ratio of 16 between the powers and of 4 between the voltages correspond to:
5. If we attenuate a signal by 66dB below a voltage of 100V calculate the amplitude of the
attenuated signal :
log10(ab)= log10(a)+log10(b).
This is a good example of the simplifications brought about by the use of dBs, which are often used
in the calculation of cascaded gains (or losses).
An antenna can be defined as a conductor or a system of conductors used to radiate the
electromagnetic energy or collect it. To transmit a signal, the electric energy of the transmitter is
converted into electromagnetic energy by the antenna and radiated in the surrounding environment
(space, water, or the atmosphere). At the receiver, the electromagnetic energy reaching the antenna
is reconverted into electrical energy.
Antenna gain relates the intensity of an antenna in a given direction to the intensity that would be
produced by a hypothetical ideal antenna that radiates equally in all directions (isotropically) and
has no losses.
The gain constitutes the measurement of an antenna performance compared with the performance of
an isotropic antenna, or the measurement of its radiation capacity in a desired direction (directivity).
This gain is expressed in isotropic decibels (dBi). It is important to note that antennas do not
somehow magically create power but simply focus the radiated RF into narrower patterns such that
there appears to be more power coming from the antenna in the required direction.As can be seen,
"gain" is also "loss". The higher the gain of an antenna the smaller the effective angle of use.
Standard antennas provided with equipment have generally a very weak gain (2.14 dBi). It should
be noted that the gain of an antenna is the same of the gain at the reception and the emission.
The gain of an antenna depends on its effective surface which depends on its size and form. The
relation between the gain of an antenna and its effective surface is given by the formula:
2 2
G=4 πAe/λ or GdB=10 log10(4πAe/λ )
with G the gain of the antenna, Ae its effective surface and λ the wavelength.
For example, the effective surface of an isotropic radiator is λ2/4π (sphere). A parabolic antenna of
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surface A (=πr2) has an effective surface Ae of 0,56A.
The gain of an antenna is measured generally in dBi. However, the gain of certain antennas when
compared with the performance of a dipole antenna,is given in dBd. In this case, it will be necessary
to add 2.14 dB to have the gain in dBi.
Effective Isotropic Radiated Power (EIRP)
In radio communication systems, Equivalent isotropically radiated power (EIRP) or, alternatively,
Effective isotropically radiated power[1] is the amount of power that a theoretical isotropic antenna
(which evenly distributes power in all directions) would emit to produce the peak power density
observed in the direction of maximum antenna gain. EIRP can take into account the losses in
transmission line and connectors and includes the gain of the antenna. The EIRP is often stated in
terms of decibels over a reference power emitted by an isotropic radiator with an equivalent signal
strength. The EIRP allows comparisons between different emitters regardless of type, size or form.
From the EIRP, and with knowledge of a real antenna's gain, it is possible to calculate real power
and field strength values.
The EIRP is the value that regulations authorities such as the ETSI (European Telecommunications
Standards Institute) use to measure and fix the maximum emission power authorized for different
applications. This power is generally fixed in a standard. Therefore, there is a legal limitation on the
emission power for all wireless equipment IEEE 802.11, the telephones GSM, UMTS….
The EIRP is calculated by adding the transmission power (in dBm) to the antenna gain by removing
the losses in the cable (in dB):
EIRP (dBm) = power of transmitter (dBm) - loss in the cable (dB) + antenna gain (dBi)
When the calculation above is carried out by using the antenna gain in comparing with a dipole
antenna (gain in dBd), the result is called Effective Radiated Power (ERP).
1. Justify the above formula and explain how the addition of a transmitted power in dBm and a
gain expressed in dB can lead to an homogeneous value in dBm.
Pout (power of transmitter)  LdB (loss in cable)  Pinp (mW) (input power of the antenna) 
EIRP (mW) (power in the air)
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Pout(dBm) = 10 log10 (Pout (mW)) (Pout - power of transmitter)
LdB = 10 log10(Pout (mW) / Pinp (mW)) (LdB – loss in cable; Pinp (mW) - input power of the antenna)
GdB = 10 log10(EIRP (mW) / Pinp (mW)) (EIRP (mW) - power in the air)
EIRP = Pout(dBm) - LdB + GdB
= 10 log10 (Pout (mW)) - 10 log10(Pout (mW) / Pinp (mW)) + 10 log10(EIRP (mW) / Pinp (mW))
= 10 [log10 (Pout (mW)) - log10(Pout (mW) / Pinp (mW)) + log10(EIRP (mW) / Pinp (mW))]
= 10 { log10 (Pout (mW) / [Pout (mW) / Pinp (mW)] × [EIRP (mW) / Pinp (mW)]) }
= 10 { log10 (Pout (mW) × Pinp (mW) / Pout (mW) × EIRP (mW) / Pinp (mW)) }
= 10 log10 (EIRP (mW))
= EIRP (dBm)
dBm are homogeneous to a power mW whereas dB are homogeneous to ratio of power so
2. Let's consider an antenna whose gain is 6 dBd supplied by a signal of 1W. What will be the
gain of 6 dBd corresponds to 4 in power ratio (see table in page 3 in this subject). We therefore obtain an
output power of 4W.
Thus ERP = 4W
3. What is EIRP of a dipole fed with a power of 1 W?
The gain is 0 dBd. The dipole fed with 1 W power radiates 1 W ERP.
You need to add 2.14 dB to the gain in dBd to get the gain in dBi. (Reference: Chap 3: Large scale path loss,
page 72, Introduction to Wireless Communication … T.S. Rappaport)
2 ,14
2,14 = 10log10(EIRP / ERP)  EIRP / ERP = 10 10
= 1,639 and ERP = 1W.
The radiated power is therefore EIRP = 1,639 W
4. A microwaves transmitter has an output power of 0,1W at 2 GHz.
a) Suppose that this transmitter is used for a microwaves communication system. Transmitting
and receiving antennas parabolas of 1,2 m in diameter. What is the gain of each antenna in
isotropic decibels?
G = 4π × 0.56A / λ2 = 7A / λ2 = 7Af2 / c2= (7 × π × (0,6)2 × (2×109)2]/(3 × 108)2 = 351,85
(with 4×3,14×0,56  7, and A = π × (0,6)2, and λ = c/f)
So we get Gdb = 10 log10(351,85) = 25,46 dBi
b) If one takes the antenna gains into account, what is the EIRP of the transmitted signal?
EIRP = 0,1 W × 351,85 = 35,2 W
5. Certain antennas radio, known as "whip antenna", function as well as possible when their length
is equal to the wavelength of the radio wave. If the antennas length is between 1 cm and 5 m,
what is the frequency range concerned?
With f = c / λ
for λ=1 cm we get f = 30 GHz.
for λ=5 m we get f = 60 MHz.
Frequencies range from 60 MHz to 30 GHz.
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6. The frequencies spectrum of human voice is concentrated around 300 Hz. The antenna size for
required for this frequency range would of course be too large. To transfer human voice on a
radio link, the signal is converted into electromagnetic wave and modulated on a higher
frequency, for which the antenna is sized smaller.
a) What would be the required antenna size, if its size should be equal to a half of the
wavelength of 300 Hz?
λ = c/f = (3 × 108 m/s)/(300 Hz) = 106 m = 1 000 km. So λ/2 = 500 km.
b) Suppose that one wants an antenna of 1 m large, on which frequency would the signal be
modulated ?
The wavelength of the carrier frequency would correspond to λ = 1 m so the frequency would be
f = c/λ = (3 × 108 m/s)/(1 m) = 300 MHz.
7. Some stories relates to people who have received radio signals in their dental fillings. Suppose
that you have a filling of 2.5 mm width, which acts like as a radio antenna. If you assume the its
size would be equal to a half of the wavelength collected, which carrying frequency would you
then collect?
λ = 2 × 2,5 × 10–3 m = 5 × 10–3 m
f = c/λ = (3 × 108 m/s)/( 5 × 10-3 m) = 6 × 1010 Hz = 60 GHz
Antenna diversity
Several omnidirectional antennas can be associated with certain access points. The goal is to allow
the reception to receive signals coming from the same source but having followed different paths. It
also allows the access point to combine the different signals to retrieve the signal wave which comes
from a given direction only. This is about diversity mode.
1. Which relation exists between the distance between the antennas and the maximum delay
between the signals coming from only one source?
Max delay = distance / propagation speed
2. What is the maximum delay if the antennas are distant of 10 cm?
delay = 0,10/(3×108) = 0,33×10-9 s = 0,33 ns
Receive sensitivity (độ nhạy) indicates how faint an RF signal can be successfully received by the
receiver. The lower the power level that the receiver can successfully process, the better the receive
In Wi-Fi equipment, receive sensitivity is generally stated as a function of network speed. Vendors
will usually specify their receive sensitivity at one or more of the 11 Mbps, 5.5 Mbps, 2 Mbps or 1
Mbps data rates. For any given receiver, the higher the data rate, the less sensitive will be the
receiver because more power is required at the receiver to support the higher data rate.
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Example: The sensitivity of Wifi Orinoco PCMCIA Silver/Gold card and of CISCO Aironet 350
card (data manufacturer).
Flow Sensitivity of the Orinoco Sensitivity of the CISCO
PCMCIA Silver/Gold cards Aironet 350 cards
11 Mbps -82 dBm -85 dBm
5,5 Mbps -87 dBm -89 dBm
2 Mbps -91 dBm -91 dBm
1 Mbps -94 dBm -94 dBm
1. What does a sensitivity of -91 dBm corresponds to? Of -94 dBm?
Because receive sensitivity indicates how faint a signal can be successfully received by the receiver, the
lower power level, the better. This means that the larger the absolute value of the negative number, the better
the receive sensitivity. For example, a receive sensitivity of -94 dBm is better than a receive sensitivity of -91
dBm by 3 dB, or a factor of two. In other words, at a specified data rate, a receiver with a -94 dBm sensitivity
can hear signals that are half as strong as a receiver with a -91 dBm receive sensitivity.
2. Using the sensitivity specifications provided by the manufacturers, determine which Wifi card,
among these two models, enables to obtain better performances in terms of throughput? Justify.
Based on the given specifications the CISCO card performs better for higher data rates.
A link budget is the accounting of all the gains and losses from the transmitter, through the medium
(free space, cable, waveguide, fiber, etc.) to the receiver in a telecommunication system. It accounts
for the attenuation of the transmitted signal due to propagation, as well as the antenna gains, feed
line ans miscellaneous losses. Randomly varying channel gains such as fading are taken into
account by adding some margin depending on the anticipated severity of its effects. The amount of
margin required can be reduced by the use of mitigating () techniques such as antenna diversity or
frequency hopping
Received power (dBm) = EIRP (dBm) - Attenuation (dB) + Gain reception antenna (dBi) –
Cables receiver loss (dB)
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This received power (dBm) must be strictly higher than the sensitivity of the receiver card. This
sensitivity represents the minimum power (dBm) that the receiver must receive to be able to
interpret the received signal. One will take a safety margin to make reliable calculations, then obtain
the connection report according to:
EIRP - Attenuation + Gain reception antenna - Loss cables receiver > Sensitivity + Stroke
Caution: these calculations are theoretical and it is about maximum attainable (). In reality, the
atmospheric losses (moisture, dispersion, refraction), the reflection effects, the pointing defects of
antennas degrade again the level of received power. In addition, the interferences of other networks
(WLAN, Bluetooth…) and the industrial noises (microwaves for example) deteriorate the signal
report… It is always necessary to envisage a important margin to allow the connection functioning
Attenuation is modeled to allow overall, by the means of connection report, to estimate the received
power by an antenna and the cover that one can hope to obtain with an access point in fine. The
connection report thus finds its application primarily in deployment phase.
 In open space:
In telecommunication, free-space path loss (FSPL) is the loss in signal strength of an
electromagnetic wave that would result from a line-of-sight (LOS) path through free space,
with no obstacles nearby to cause reflection or diffraction. It does not include factors such as
the gain of the antennas used at the transmitter and receiver, nor any loss associated with
hardware imperfections.
A correspondence between attenuation in decibels (dB) and distances in kilometers (km) is
obtained by the formula of Friis:
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Alibre = 20 log10 (4πd/ λ)
with Alibre : attenuation in open space (in dB)
λ : wavelength (in m)
d : distance between transmitter – receiver (in m)
libre = free
 In a building :
The attenuation model is called indoor:
Ab = 20 log10(f) + γ log10(d) + Asol(n) – 28
with Ab : propagation loss in the building (in dB)
Asol (n) : loss related to the crossing of n stages in a building
γ : attenuation exponent
f : frequency in MHz
The exponent of attenuation allows accounting for additional attenuation induced by the
walls, the effects of mask (« shadowing ») …
Many other models exist to estimate the propagation attenuation. One can quote among the most
known empirical models Okumura - Hanta model and Walfish-Ikégami model. These models were
developed from the measurement fields.
It will be retained that the choice of a model depends enormously on the environment (rural, urban,
micro-cellular…) and on the frequency band.
The Friis formula for example applies only when the antennas are in clear line of sight. This
condition, known as propagation in free space, is only filled when at least 80% the definite volume
by the first ellipsoid of Fresnel (called Fresnel zone) is clear of obstacles .
The concept of Fresnel zone clearance may be used to analyze interference by obstacles () near the
path of a radio beam. The first zone must be kept largely free from obstructions to avoid interfering
with the radio reception. For establishing Fresnel zones, first determine the RF Line of Sight (RF
LoS), which in simple terms is a straight line between the transmitting and receiving antennas. Now
the zone surrounding the RF LoS is said to be the Fresnel zone. The equation for calculating r, the
first Fresnel zone radius at any point P in between the endpoints of the link is the following:
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