68 Wireless Communications

The simplest method of relating the received signal

ire power to the distance is to state that the received signal
power Pr is proportional to the distance between the
transmitter and receiver ‘r’, raised to an exponent γ,
referred to as path-loss exponent or distance-power
ell si e M ile gradient, that is,
r nsmi er
Pr = P0 r−γ (3.1)
Fig. 3.1 Radio propagation in a mobile where Pr is the received signal strength at ‘r’ metres
communication system
from the cell site transmitter, and P0 is the received
signal strength at a reference distance (usually one metre in indoor environments and 100 m or 1 km in out-
door environments depending on the cell size in practical systems using low-gain antennas). The path-loss
exponent, γ is valuable since it shows the rate of increase of path loss with respect to distance.
Converting it into decibel form, the expression can be written as
10 log (Pr) = 10 log (P0) − 10 γ log (r) (3.2)
The term 10 γ log (r) represents the power loss in dB with respect to the received power at reference distance,
say one metre.
The path loss at a distance of one metre is defined as
L0 (dB) = 10 log (Pt) −10 log (P0) (3.3)
Then the total path loss Lp at a given distance ‘r’ is given by
Lp (dB) = L0 (dB) + 10 γ log (r) (3.4)
This presents the total path-loss in the first metre plus the path-loss relative to the power received at one metre.
In the simplest form, the received power in dB is the transmitted power in dB minus the total path loss.
Thus, path loss can be defined as the ratio of the transmitted to received power usually expressed
in decibels.
A detailed understanding of radio propagation models is essential for appropriate design, deployment, and
optimised performance in any wireless communication system. Signal propagation through a wireless media
can vary significantly depending on the operating terrain, frequency of transmission, sources of interference,
speed of the mobile unit, and many other dynamic factors, and so it is heavily location-specific. As the path
loss encountered along any radio link serves as the dominant factor for characterisation of signal propagation,
radio-propagation models typically focus on realisation of the path loss with the objective of predicting the radio
coverage for a base station transmitter or modeling the distribution of signals over different types of regions.
Each individual wireless communication link generally
Facts to Know !
encounters different paths, terrain, obstructions, atmospheric
The propagation models rely conditions and other phenomena. So it is intractable to formu-
on computing the median path late the exact propagation path loss for all types of wireless
loss for a wireless link under a communication systems in a single mathematical equation. As
certain probability that the con- a result, different models exist for different types of radio links
sidered conditions will occur.
under different conditions.

3.2 FREE-SPACE PROPAGATION MODEL

Free-space propagation model is the fundamental for all propagation path-loss models for any wireless com-
munication application. A free-space propagation model is used to predict the received signal strength when
the transmitter and receiver has a clear unobstructed line-of-sight signal path between them. The free space
 The Propagation Models 69

environment could be categorised as one having less clutter and foliage. The propagation path-loss models
are extensively used for deployment of cellular communication networks. The coverage of a cell-site depends
on the power of the transmitted signal and the path loss. Each mobile unit has a specified receiver sensitiv-
ity, for example, it can only detect and decode signals with signal strength larger than its sensitivity level.
Because the signal strength decays with distance, the radio coverage can be estimated using the transmitter
power, the path-loss model and the receiver sensitivity.
In most operating environments, it is observed that the radio signal strength decays as some power γ of the
distance, called the power-distance gradient or path-loss exponent. That is, if the transmitted power is Pt, the
signal strength at a distance ‘r’ in metres will be proportional to Pt r−γ. In free space propagation, there is no
loss of energy as a radio wave propagates in free space, but there is attenuation due to the spreading of the
electromagnetic waves.
Since an isotropic radiator radiates equally in all Facts to Know !
directions, the power density is simply the transmit- Sensitivity is a receiver parameter that indi-
ted power divided by the surface area of the sphere. cates the minimum signal level required at
Expressing it mathematically, the input of receiver unit (at the output of
receiver antenna terminal) in order to pro-
PD = Pt / (4 π r2) (3.5) vide reliable communications. It is often expressed in
dBm, that is, the power in mW expressed in decibels.
where PD is the power density in watts/m2, Pt is the
transmitter power in watts, r is the distance from
the transmitting antenna in metres, and 4 π r2 is Facts to Know !
surface area of the sphere. If a sphere were drawn at any distance
It is emphasised here that the attenuation in the from the transmitting source and con-
signal is not due to any loss of energy in the free centric with it, all the signal energy from
space, but only due to the spreading out of the energy the source would pass through the sur-
as it moves away from the radiating source. face of the sphere.
Practical antennas do not radiate equally in all
directions as that of in case of isotropic antennas and, therefore, the transmitting antenna has gain, Gt. In
fact, antennas are passive devices and do not have actual power gain. They achieve a greater power density in
certain directions at the expense of reduced power density in other directions.
So Eq. (3.5) can be modified to include the transmitting antenna gain, Gt
That is, PD = (Pt Gt) / (4 π r2) (3.6)
Usually, antenna gain is specified in dBi, where the ‘i’ indicates gain with respect to an isotropic radiator.
The transmitting antenna gain Gt must be converted to a power ratio to be used in Eq. (3.6).
The effective isotropic radiated power (EIRP) of a transmitting system in a given direction is defined as
the transmitter power that would be needed with an isotropic radiator, to produce the same power density in
the given direction. That means,
EIRP = Pt Gt (3.7)
Equation (3.6) can be modified for use with EIRP as
PD = (EIRP) / (4 π r2) (3.8)

EXAMPLE 3.1 EIRP and power-density calculations

A wireless communication transmitter having RF power of 113 W is used with an antenna of 5 dBi gain. Calculate the EIRP
and the power density at a distance of 11 km.
70 Wireless Communications

Solution
Step 1. To convert antenna gain of 5 dBi to a power ratio
Gt = 10 (5/10) = 3.16
Step 2. To calculate the EIRP
EIRP = Pt Gt
EIRP = 113 × 3.16 = 357.1 W
Step 3. To calculate the power density
PD = EIRP / (4 π r2)
PD = 357.1W / 4π (11 × 103m)2 = 235 nW/m2

A receiving antenna absorbs some of the signal energy from radio waves that pass through it. Since
the signal energy in the radio wave is directly proportional to the area through which it passes, a receiving
antenna having large area will intercept more energy than a smaller one. Receiving antennas are also more
efficient at absorbing signal power from some directions than from other directions. That simply means
receiver antennas too have gain. In fact, the antenna gain is same whether the antenna is used for transmitting
or receiving RF signals.
The power extracted from the radio wave by a receiving antenna depends on its physical size as well as its
gain. The effective area of a receiving antenna can be defined as
Aeff = Pr / PD (3.9)
2
where Aeff = effective area of the receiving antenna in m
Pr = power delivered by a receiving antenna to the receiver in watts
PD = power density of the radio wave in watts/m2
It implies that the effective area of a receiving antenna is the area from which all the power in the incident
radio wave is extracted and delivered to the receiver unit.
Rewriting Eq. (3.9),
Pr = PD Aeff
Substituting PD from Eq. (3.6), we get
Pr = (Pt Gt) / (4 π r2) Aeff (3.10)
The effective area of a receiving antenna depends on its gain, Gr as well as the wavelength of the incident
radio wave λc, and is given as
Aeff = Gr λc2 / (4 π) (3.11)
Substituting Aeff into Eq. (3.10), the received power Pr in free space is given by
Pr = (Pt Gt) / (4 π r2) Gr λc 2 / (4 π)
2
Or, r= t t r λc / (4 π r ) (3.12)
where ‘r’ is the distance between the transmitter and receiver and is of same units as λc.
This equation is known as riis’ free-space equation and is used to estimate the signal power received by
a receiver antenna separated from a transmitting antenna by a distance ‘r’ in the free space, neglecting the
system losses.
Free-space propagation path loss, Lpf, is defined as
Lpf = Pt / Pr = 1 / (Gt Gr) (4 π r / λc) 2
 The Propagation Models 71

When Gt = Gr = 1 (unity gain), then free-space path loss is given by

f = (4 π r / λc ) 2 (3.13)

Or, Lpf = (4 π r fc / 3 × 108 ) 2 (where r is in metres)

Or, Lpf (dB) = 20 log (4 π / 3 × 108 ) + 20 log r (m) + 20 log fc (Hz)

Or, Lpf (dB) = −147.56 + 20 log r (m) + 20 log fc (Hz)

Or, f (dB) = 32.44 + 20 log r (km) + 20 log fc (MHz) (3.14)

This is called free-space path loss equation.
Figure 3.2 illustrates the plot between f (dB) and distance r’ at specified frequencies fc, based on free
space path loss equation.
As evident from Eq. (3.14) also, the plot of free space propagation path loss versus distance indicates that
as the frequency increases, the free space path loss also increases at the same distance from the transmit-
ter. Free-space path loss increases with frequency of transmission because the effective area of a receiving
antenna with a given gain decreases with frequency. However, for an ideal isotropic radiator and a receiving
antenna of constant effective area, there is no dependence of free-space attenuation on frequency. In the
free-space path-loss expression, the square-law attenuation in the signal due to distance is represented by its
logarithmic equivalent, that is, 20 log r (km). There is no term for antenna heights of cell-site or mobile unit,
since this is irrelevant in free space.

1 0

170
= 300
1 0

150
= 30
140
l ss d

130
=3
120
i n

110
= 300 M
100
Pr

0
= 30 M
0

70

0
1 5 10 50 100
is n e m

Fig. 3.2 Plot of free-space propagation path-loss vs distance

72 Wireless Communications

EXAMPLE 3.2 Free-space path loss

Calculate the free-space path loss for a signal transmitted at a frequency of 900 MHz, the distance between the transmitter
and receiver being 1 km.
Solution
The frequency of transmission, fc = 900 MHz or 900 × 106 Hz (given)
The distance between Tx and Rx, r = 1 km or 1000 m (given)
Step 1. To find the wavelength, λc
We know that λc = c / fc
Therefore, λc = (3 × 108 m/s) / (900 × 106 Hz) = 0.33 m
Step 2. To calculate the free-space path loss, Lpf
We know that free-space path loss, Lpf = (4 π r / λc ) 2
Therefore, Lpf = (4 π × 1000 / 0.33 ) 2
Or, Lpf (dB) = 10 log (4 π × 1000 / 0.33 ) 2 = 20 log (4 π × 1000 / 0.33 )
Hence, the free-space path loss f (dB) = 91.6 dB

Using Eq. (3.13) in Eq. (3.12), we get

r= t t r / f (3.15)
Or, r (dBm) = t (dBm) + t (dBi) + r (dBi) − f (dB) (3.16)
Attenuation due to transmission-line losses or mismatch at the transmitter and receiver, if any, should be
included in this equation to obtain the actual received power.
At the first metre (r = 1 m), let Pr = P0; then from Eq. (3.12)
P0 = Pt Gt Gr (λc / 4 π ) 2 (3.17)
Then from Eq. (3.12) and Eq. (3.17),
Pr = P0 / r2 (3.18)
In dB, this equation takes the form
10 log (Pr) = 10 log (P0) − 20 log (r) (3.19)
This means that there is a 20-dB per decade or 6-dB per octave loss in signal strength as a function of
distance in free space.

EXAMPLE 3.3 Range of a base station transmitter

Determine the radio coverage range of a base station that transmits a RF signal at 100 W, given that the receiver threshold
level is −100 dBm. Assume that the path loss at the first metre is 30 dB in a mobile radio propagation condition (γ = 4).
Solution
Step 1. To convert transmitter power in dBm units.
Transmitter power, Pt = 100 W or 100,000 mW (given)
We know that Pt (dBm) = 10 log Pt (mW)
Therefore, Pt (dBm) = 10 log Pt (100,000) = +50 dBm
Step 2. To find path loss, Lp (dB)
Receiver threshold level, Pr (dBm) = −100 dBm (given)
We know that path loss, Lp (dB) = Pt (dBm) − Pr (dBm)
Or, path loss, Lp (dB) = +50 dBm −(−100 dBm) = 150 dB
 The Propagation Models 73

Step 3. To determine the radio coverage range, r Facts to Know !

Path loss at first metre, L0 = 30 dB (given)
Propagation path constant, γ = 4 (given) Converting dBm to mW does
not change the relative mea-
We know that path loss, Lp (dB) = L0 (dB) + 10 log (r)γ
surement characteristics of
where r is the radio coverage range of the base station
decibel. It simply means that
transmitter in metres. there is a reference point for the values used
Therefore, Lp (dB) = L0 (dB) + 10 log (r)4 in the calculation; any dB values can be added
Or, Lp (dB) = L0 (dB) + 40 log (r) or subtracted as if they are represented in the
same units.
Or, 40 log (r) = Lp (dB) − L0 (dB)
Or, 40 log (r) = 150 − 30 = 120 dB
Hence, r = 10 (120 / 40) = 10 3 = 1000 metres or 1 km
Hence, the radio coverage range r = 1 km

EXAMPLE 3.4 Received power in free-space propagation

A wireless communication transmitter has an output power of 165 watts at a carrier frequency of 325 MHz. It is connected
to an antenna with a gain of 12 dBi. The receiving antenna is 15 km away and has a gain of 6 dBi. Calculate the power
delivered to the receiver, considering free-space propagation. Assume that there are no other losses or mismatches in the
system.
Solution: Refer Fig. 3.3.
Transmitter output power, = 12 d i = d i
Pt = 165W or 165,000 mW (given)
Transmitter antenna gain, Gt = 12 dBi (given) = 325 M
Receiver antenna gain, Gr = 6 dBi (given) = 15 m
Carrier frequency, fc = 325 MHz (given) r nsmi er e eiver
Distance between transmitter and receiver, =1 5 =
r = 15 km (given)
Fig. 3.3 Illustration of communication link for
Step 1. To convert Pt (W) in Pt (dBm) Example 3.4
We know that Pt (dBm) = 10 log Pt (mW)
Therefore, Pt (dBm) = 10 log 165,000 = + 52.17 dBm
Step 2. To find free-space path loss, Lpf (dB)
Lpf (dB) = 32.44 + 20 log r (km) + 20 log fc (MHz)
Lpf (dB) = 32.44 + 20 log 15 (km) + 20 log 325 (MHz)
Lpf (dB) = 32.44 + 23.52 + 50.24 = 106.20 dB
Step 3. To calculate power delivered to the receiver, Pr (dBm)
Pr(dBm) = Pt (dBm) + Gt (dBi) + Gr (dBi) − Lpf (dB)
Pr(dBm) = 52.17 + 12 + 6 − 106.20 = 36.03 dBm
Step 4. To express Pr(dBm) in Pr(watts)
We know Pr (dBm) = 10 log Pr (mW)
Therefore, −36.03 = 10 log Pr (mW)
Or, Pr (mW) = 10 (−36.03/10) = 249.46 × 10 −6 mW
Hence, Pr (W) = 249.46 × 10 −9 watts
74 Wireless Communications

The transmission delay or time taken by a wave from transmitter to a point ‘r’ metre away from it, can
be given by
=r/c where c is the speed of electromagnetic waves in space
Or, = r(m) / (3 × 108 m/s)
≈ 3.33 r ns or 3.33 ns per metre of distance (3.20)

EXAMPLE 3.5 Path loss, received power and transmission delay

A wireless communication base station transmits 10 watts of power at a carrier frequency of 1 GHz. If the receiver station is
at a distance of 1.6 km from the base station, then determine
(a) the propagation path loss (in dB) in a free space environment
(b) the received signal power (in dBm)
(c) the transmission delay in ns
Assume that the transmitter and receiver antenna gains are 1.6 each.
Solution: Refer Fig. 3.4.
Transmitter output power, Pt = 10W or 10,000 mW (given)
Transmitter antenna gain, Gt = 1.6 (given)
Receiver antenna gain, Gr = 1.6 (given)
Carrier frequency, fc = 1 GHz or 1000 MHz (given)
Distance between transmitter and receiver, r = 1.6 km (given)
Step 1. To convert Pt (W) in Pt (dBm)
= 1. = 1. We know that Pt (dBm) = 10 log Pt (mW)
Therefore, Pt (dBm) = 10 log 10,000
d = +40 dBm
=1
r nsmi er = 1. m e eiver Step 2. To convert Gt (ratio) in Gt (dB)
= 10 d m
We know that Gt (dB) = 10 log Gt (ratio)
Therefore, Gt (dB) = 10 log 1.6 = 2 dB
Fig. 3.4 Illustration of communic ation link
Step 3. To convert Gr (ratio) in Gr (dB)
for Example 3.5
We know that Gr (dB) = 10 log Gr (ratio)
Therefore, Gr (dB) = 10 log 1.6 = 2 dB
(a) To determine free-space path loss, Lpf (dB)
Lpf (dB) = 32.44 + 20 log r (km) + 20 log fc (MHz)
Lpf (dB) = 32.44 + 20 log 1.6 (km) + 20 log 1000 (MHz)
Lpf (dB) = 32.44 + 4.08 + 60 = 96.5 dB
(b) To determine the received signal power, Pr (dBm)
Pr (dBm) = Pt (dBm) + Gt (dB) + Gr (dB) − Lpf (dB)
Pr (dBm) = 40 + 2 + 2 − 96.5 = − 52.5 dBm
(c) To determine tansmission delay, (ns)
Transmission delay, = 3.33 ns per metre of distance
For 1600 m distance, = 3.33 × 1600 = 5.33 s

EXAMPLE 3.6 Power delivered to the receiver

A base-station transmitter has a power output of 10 watts operating at a frequency of 250 MHz. The transmitter is
connected by 20 m of an RF coaxial cable, which has a loss of 3-dB/100 m specification, to an antenna that has a gain
 The Propagation Models 75

of 9 dBi. The receiving antenna is 25 km away and has a gain of 4 dBi. There is negligible loss in the receiver feeder
line, but the receiver is mismatched; the receiving antenna and feeder cable are designed for a 50-Ω impedance, but
the receiver input has 75-Ω impedance, resulting into a mismatch loss of about 0.2 dB˜. Calculate the power delivered to
the receiver, assuming free-space propagation.
Solution: Refer Fig. 3.5.
Transmitter output power, Pt = 10 W or 10,000 mW (given)
Transmitter antenna gain, Gt = 9 dBi (given)
Receiver antenna gain, Gr = 4 dBi (given)
Carrier frequency, fc = 250 MHz (given)
Distance between transmitter and receiver, r = 25 km (given)
Step 1. To convert Pt (W) in Pt (dBm)
We know that Pt (dBm) = 10 log Pt (mW) = d i =4d i
Therefore, Pt (dBm) = 10 log 10,000 = +40 dBm
= 250 M
Step 2. To determine free-space path loss, Lpf (dB) = 25 m d m
Lpf (dB) = 32.44 + 20 log r (km) + 20 log fc (MHz) r nsmi er e eiver
= 0.2 d
= 32.44 + 20 log 25 (km) + 20 log = 10 Tx eedline
250 (MHz) en = 20 m
ss = 3 d /100 m
= 108.3 dB
Fig. 3.5 Illustration of wireless link for Example 3.6
Step 3. To find the Tx antenna RF cable loss, Lt (dB)
Cable length = 20 m (given)
Cable attenuation = 3 dB/100m (given)
Therefore, Tx antenna RF cable loss, Lt = 20 m × 3 dB/100m
Tx antenna R cable loss, Lt = 0.6 dB
Step 4. To calculate the power delivered to the receiver, Pr (dBm)
Rx antenna R cable loss, Lr = 0.2 dB ( given)
Pr (dBm) = Pt (dBm) Lt (dB) + Gt (dB) Lpf (dB) + Gr (dB) – Lr (dB)
Pr (dBm) = 40 0.6 + 9 – 108.3 + 4 – 0.2 = 56.1 dBm
Hence power delivered to the receiver is 56.1 dBm.

3.3 MOBILE POINT-TO-POINT PROPAGATION MODEL

The free-space propagation model does not apply in a mobile radio environment. Propagation path loss depends
on distance of the mobile from its serving cell-site, carrier frequency of transmission fc (or wavelength λc ), the
antenna heights of cell-site and mobile unit, and the local terrain characteristics such as buildings and hills.
Mobile radio propagation path loss is uniquely different from the kind of path losses experienced in other com-
munication media. The signal received by a mobile unit remains constant only over a small operating area and
varies as the mobile unit moves. This is mainly due to the variations in the terrain conditions and presence of
man-made structures surrounding the mobile unit.
Facts to Know !
Area-to-area signal prediction models are not useful for cel-
lular communication systems because of the large uncertainty Radio propagation in mobile
of the prediction. The area-to-area prediction model usually environment is mainly charac-
provides an accuracy of prediction with a measured standard terised by three partially related
deviation of about 8 dB, which means that 68 per cent of the actual effects known as multipath
fading, shadowing, and path loss.
path-loss data are within 8 dB of the predicted path-loss value.