# Chapter No.4 : Mobile radio propagation large scale path loss¶

## Example 4.1 Page No.109¶

In [3]:
from __future__ import division
# To find far field distance for antenna with maximum dimensions and operating frequency

# Given data
D=1#                   # Maximum dimension in m
f=900*10**6#            # Operating frequency in Hz
C=3*10**8#              # Speed of light in m/sec

lamda=C/f#            # Carrier wavelength in m

# To find far field distance
df=(2*D**2)/lamda#     #Far field distance

#Displaying the result in command window
print '\n Far field distance = %0.0f meter'%(df)#

 Far field distance = 6 meter


## Example 4.2 Page No.109¶

In [3]:
from __future__ import division
from math import log10,pi
# To find a)transmitter power in dBm b)Transmitter power in dBW and the received power of antenna in dBm at free space distance of 100m from antenna and 10km

# Given data
Pt=50#                                             # Transmitter power in W
fc=900*10**6#                                       # Carrier frequency in Hz
C=3*10**8#                                         # Speed of light in m/s

#a)Transmitter power in dBm
PtdBm=round(10*log10(Pt/(1*10**(-3))))#            #Transmitter power in dBm

# Displaying the result in command window
print '\n Transmitter power = %0.1f dBm'%(PtdBm)#

#b)Transmitter power in dBW
PtdBW=round(10*log10(Pt/1))#                     #Transmitter power in dBW

# Displaying the result in command window
print '\n Transmitter power = %0.1f dBW'%(PtdBW)#

# To find receiver power at 100m
Gt=1#                                            #Transmitter gain
d=100#                                           #Free space distance from antenna in m
L=1#                                             #System loss factor since no loss in system
lamda=C/fc#                                     #Carrier wavelength in m

#Displaying the result in command window
print '\n Receiver power = %0.1f dBm'%(PrdBm)#

#For Pr(10km)
d0=100#                                         #Reference distance
d=10000#                                        #Free space distance from antenna
Pr10km=PrdBm+20*log10(d0/d)#                    #Received power at 10km from antenna in dBm

#Displaying the result in command window
print '\n Receiver power at 10km from antenna = %0.1f dBm'%(Pr10km)#

 Transmitter power = 47.0 dBm

Transmitter power = 17.0 dBW

Receiver power at 10km from antenna = -64.5 dBm


## Example 4.3 Page no. 112¶

In [2]:
from __future__ import division
from math import sqrt,pi,log10
# To find a)power at receiver b)magnitude of E-field at receiver c)rms voltage applied to receiver input

# Given data
Pt=50#                                                                      # Transmitter power in Watt
fc=900*10**6#                                                                # Carrier frequency in Hz
Gt=1#                                                                       # Transmitter antenna gain
Rant=50#                                                                    # Receiver antenna resistance in ohm

d=10*10**3#                                                                  # Distance from antenna in meter
lamda=(3*10**8)/fc#                                                         # Carrier wavelength in meter
Prd1=10*log10((Pt*Gt*Gr*lamda**2)/((4*pi)**2*d**2))#                         # Power at transmitter in dBW
Prd=10*log10(((Pt*Gt*Gr*lamda**2)/((4*pi)**2*d**2))/(10**-3))#                # Power at transmitter in dBm

# Displaying the result in command window
print '\n Power at receiver = %0.1f dBW'%(Prd1)#
print '\n Power at receiver = %0.1f dBm'%(Prd)#

# b)Magnitude of E-field at receiver
Ae=(Gr*lamda**2)/(4*pi)#                                                   # Aperture gain
Pr=10**(Prd1/10)#                                                            # Receiver power in W
E=sqrt((Pr*120*pi)/Ae)#                                                    # Magnitude of E-field at receiver

# Displaying the result in command window
print '\n \n Magnitude of E-field at receiver = %0.4f V/m'%(E)#

# c)rms voltage applied to receiver input
Vant=sqrt(Pr*4*Rant)*10**3#                                                 # rms voltage applied to receiver input
#Answer is varrying due to round-off error

#Displaying the result in command window
print '\n \n RMS voltage applied to receiver input = %0.3f mV'%(Vant)#

 Power at receiver = -91.5 dBW

Power at receiver = -61.5 dBm

Magnitude of E-field at receiver = 0.0039 V/m

RMS voltage applied to receiver input = 0.375 mV


## Example no. 4.5 Page no. 119¶

In [2]:
from __future__ import division
from math import sqrt,asin,degrees
# To calculate the Brewster angle

# Given data
Er=4#                                 # Permittivity
x=sqrt((Er-1)/(Er**2-1))             # Sine of brewster angle
theta=degrees(asin(x))                       # Brewster angle
#Answer is varrying due to round off error

# Displaying the result in command window
print '\n Brewster angle = %0.2f degree'%(theta)#

 Brewster angle = 26.57 degree


## Example no 4.6 Page no. 125¶

In [2]:
from __future__ import division
from math import log10,sqrt,pi
# To find a)the length and effective aperture of receiving antenna b)the received power at mobile

# Given data
d=5*10**3#                                               # distance of mobile from base station in m
E0=1*10**-3#                                             # E-field at 1Km from transmitter in V/m
d0=1*10**3#                                              # Distance from transmitter in m
f=900*10**6#                                             # Carrier frequency used for the system in Hz
c=3*10**8#                                               # Speed of ligth in m/s
gain=2.55#                                              # Gain of receiving antenna in dB
G=10**(gain/10)#                                         # Gain of receiving antenna

# a)To find the length and effective aperture of receiving antenna
lamda=c/f#                                             # Wavelength
L=lamda/4#                                             # Length of antenna
Ae=(G*lamda**2)/(4*pi)#                                # Effective aperture of receiving antenna

# Displaying the result in command window
print '\n Length of antenna = %0.4f m'%(L)#
print ' = %0.2f cm'%(L*10**2)#
print '\n Effective aperture of receiving antenna = %0.3f m**2'%(Ae)#

# b)To find the received power at mobile
# Given data
ht=50#                                                  # Heigth of transmitting antenna
hr=1.5#                                                 # Heigth of receiving antenna
ERd=(2*E0*d0*2*pi*ht*hr)/(d**2*lamda)#                 # Electic field at distance d in V/m
Prd=((ERd**2/377)*Ae)#                                   # The received power at mobile in W
PrddB=10*log10(Prd)#                                    # The received power at mobile in dBW
PrddBm=10*log10(Prd/10**-3)#                             # The received power at mobile in dBm
Prd=((ERd**2/377)*Ae)*10**13#                             # The received power at mobile in 10**-13W

# Displaying the result in command window
print '\n \n The received power at mobile = %0.1f X 10**-13 W'%(Prd)#
print ' = %0.2f dBW'%(PrddB)#
print ' = %0.2f dBm'%(PrddBm)#

 Length of antenna = 0.0833 m
= 8.33 cm

Effective aperture of receiving antenna = 0.016 m**2

The received power at mobile = 5.4 X 10**-13 W
= -122.68 dBW
= -92.68 dBm


## Example no 4.7 Page no. 132¶

In [8]:
from __future__ import division
from math import log10,sqrt,pi
# To compute diffraction loss and identify Fresnel zone within which tip of obstruction lies for a)h=25m b)h=0 c)h=-25m

# Given data
lamda=1/3#                                         # Wavelength in meter
d1=1*10**3#                                          # Distance between transmitter and obstructing screen in m
d2=1*10**3#                                          # Distance between receiver and obstructing screen in m

# a) For h=25m
h=25#                                               # Effective heigth of obstruction screen in m
v=h*sqrt((2*(d1+d2))/(lamda*d1*d2))#               # Fresnel diffraction parameter
print '\n a) For h=25m Fresnel diffraction parameter v = %0.2f'%(v)#
print '\n From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is 22dB.'
Gd=-20*log10(0.225/v)#                               # Diffraction loss for v>2.4 in dB
print '\n Using numerical approximation, diffraction loss for v > 2.4  = %0.1f dB'%(Gd)#
delta=(h**2/2)*((d1+d2)/(d1*d2))#                    # Path length difference between direct and diffracted rays
n=(2*delta)/lamda#                                 # Number of Fresnel zones in which the obstruction lies
print '\n Fresnel zone within which tip of obstruction lies = %0.2f'%(n)#
print '\n Therefore, the tip of obstruction completely blocks the first three Fresnel zones.'

# b) For h=0
h=0#                                                # Effective heigth of obstruction screen in m
v=h*sqrt((2*(d1+d2))/(lamda*d1*d2))#               # Fresnel diffraction parameter
print '\n \n b) For h=0 Fresnel diffraction parameter v = %0.0f'%(v)#
print '\n From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is 6dB.'
Gd=-20*log10(0.5-0.62*v)#                           # Diffraction loss for v=0 in dB
print '\n Using numerical approximation, diffraction loss for v=0 = %0.0f dB'%(Gd)#
delta=(h**2/2)*((d1+d2)/(d1*d2))#                    # Path length difference between direct and diffracted rays
n=(2*delta)/lamda#                                 # Number of Fresnel zones in which the obstruction lies
print '\n Fresnel zone within which tip of obstruction lies = %0.0f'%(n)#
print '\n Therefore, the tip of obstruction lies in middle of first Fresnel zone.'

# c) For h=-25m
h=-25#                                                # Effective heigth of obstruction screen in m
v=h*sqrt((2*(d1+d2))/(lamda*d1*d2))#                 # Fresnel diffraction parameter
print '\n \n c) For h=-25m Fresnel diffraction parameter v = %0.2f'%(v)#
print '\n From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is approximately 1dB.'
Gd=0#                                                # Diffraction loss for v<-1 in dB
print '\n Using numerical approximation, diffraction loss for v < -1 = %0.0f in dB'%(Gd)#
delta=(h**2/2)*((d1+d2)/(d1*d2))#                    # Path length difference between direct and diffracted rays
n=(2*delta)/lamda#                                 # Number of Fresnel zones in which the obstruction lies
print '\n Fresnel zone within which tip of obstruction lies = %0.2f'%(n)#
print '\n Therefore, the tip of obstruction completely blocks the first three Fresnel zones but diffraction loss is negligible.'

 a) For h=25m Fresnel diffraction parameter v = 2.74

From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is 22dB.

Using numerical approximation, diffraction loss for v > 2.4  = 21.7 dB

Fresnel zone within which tip of obstruction lies = 3.75

Therefore, the tip of obstruction completely blocks the first three Fresnel zones.

b) For h=0 Fresnel diffraction parameter v = 0

From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is 6dB.

Using numerical approximation, diffraction loss for v=0 = 6 dB

Fresnel zone within which tip of obstruction lies = 0

Therefore, the tip of obstruction lies in middle of first Fresnel zone.

c) For h=-25m Fresnel diffraction parameter v = -2.74

From the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter, diffraction loss is approximately 1dB.

Using numerical approximation, diffraction loss for v < -1 = 0 in dB

Fresnel zone within which tip of obstruction lies = 3.75

Therefore, the tip of obstruction completely blocks the first three Fresnel zones but diffraction loss is negligible.


## Example no 4.8 Page no. 133¶

In [1]:
from __future__ import division
from math import log10,sqrt,pi,atan
# To determine a)the loss due to knife-edge diffraction b)the heigth of obstacle required to induce 6dB diffraction loss

# Given data
f=900*10**6#                                                  # Operating frequency in Hz
c=3*10**8#                                                    # Speed of ligth in m/s
hr=25#                                                       # Heigth of receiver in m
ht=50#                                                       # Heigth of transmitter in m
h=100#                                                      # Heigth of obstruction in m
d1=10*10**3#                                                 # Distance between transmitter and obstruction in m
d2=2*10**3#                                                  # Distance between receiver and obstruction in m

# a)To determine the loss due to knife-edge diffraction
lamda=c/f#                                                # Operating wavelength in m
ht=ht-hr#                                                  # Hegth of transmitter after subtracting smallest heigth (hr)
h=h-hr#                                                    # Heigth of obstruction after subtracting smallest heigth (hr)
bet=atan((h-ht)/d1)#                                       # From geometry of environment in rad
gamm=atan(h/d2)#                                           # From geometry of environment in rad
alpha=bet+gamm#                                            # From geometry of environment in rad
v=alpha*sqrt((2*d1*d2)/(lamda*(d1+d2)))#                  # Fresnel diffraction parameter

# the loss due to knife-edge diffraction
Gd=-20*log10(0.225/v)#                                     # Diffraction loss for v>2.4 in dB

# Displaying the result in command window
print '\n The loss due to knife-edge diffraction = %0.1f dB'%(Gd)#

# b)To determine the heigth of obstacle required to induce 6dB diffraction loss
Gd=6#                                                      # Diffraction loss in dB
v=0#                                                       # Fresnel diffraction parameter from the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter
# v=0 is possible only if alpha=0. Therefore bet=-gamm
# By considering this situation, the geometry of environment provides (h/d2)=(ht/(d1+d2))
h=(ht*d2)/(d1+d2)#                                         # the heigth of obstacle required to induce 6dB diffraction loss

# Displaying the result in command window
print '\n The heigth of obstacle required to induce 6dB diffraction loss = %0.2f m'%(h)#

 The loss due to knife-edge diffraction = 25.5 dB

The heigth of obstacle required to induce 6dB diffraction loss = 4.17 m


## Example no 4.9 Page no. 143¶

In [2]:
from __future__ import division
from math import log10,sqrt,erf
# To find a)the minimum mean square error b)the standard deviation about mean value c)received power at d=2 km d)the likelihood that the received signal level at 2 km e) the percentage of  area within 2 km

# Given data
d0=100#                                        # First receiver distance in meter
d1=200#                                        # Second receiver distance in meter
d2=1000#                                       # Third receiver distance in meter
d3=3000#                                       # Fourth receiver distance in meter
p0=0#                                          # Receved power of first receiver in dBm
p1=-20#                                        # Receved power of second receiver in dBm
p2=-35#                                        # Receved power of third receiver in dBm
p3=-70#                                        # Receved power of forth receiver in dBm

# a)To find the minimum mean square error
n=2887.8/654.306#                              # Loss exponent after differentiating and equating the squared error function with zero

# Displaying the result in command window
print '\n Loss exponent = %0.0f'%(n)#

# b)To find the standard deviation about mean value
P0=-10*n*log10(d0/100)#                        # The estimate of p0 with path loss model
P1=-10*n*log10(d1/100)#                        # The estimate of p1 with path loss model
P2=-10*n*log10(d2/100)#                        # The estimate of p2 with path loss model
P3=-10*n*log10(d3/100)#                        # The estimate of p3 with path loss model
J=(p0-P0)**2+(p1-P1)**2+(p2-P2)**2+(p3-P3)**2#     # Sum of squared error
SD=sqrt(J/4)#                                  # The standard deviation about mean value

# Displaying the result in command window
print '\n The standard deviation about mean value = %0.2f dB'%(SD)#
# The decimal point is not given in the answer given in book.

# c)To find received power at d=2 km
d=2000#                                         # The distance of receiver
P=-10*n*log10(d/100)#                           # The estimate of p2 with path loss model

# Displaying the result in command window
print '\n The received power (at d=2 km) = %0.2f dBm'%(P)#
# Answer is varying due to round off error

# d)To find the likelihood that the received signal level at 2 km
gam=-60#                                         # The received power at 2km will be greater than this power
z=(gam-P)/SD#
Pr=(1/2)*(1-erf(z/sqrt(2)))#                     # The probability that received signal will be greater than -60dBm

# Displaying the result in command window
print '\n The probability that received signal will be greater than -60dBm = %0.1f percent'%(Pr*100)#
# Answer is varying due to round off error

# e)To find the percentage of  area within 2 km
A=92#                                            # From figure 4.18, area receives coverage above -60dBm

# Displaying the result in command window
print '\n The percentage of  area within 2 km = %0.0f percent'%(A)#

 Loss exponent = 4

The standard deviation about mean value = 6.16 dB

The received power (at d=2 km) = -57.42 dBm

The probability that received signal will be greater than -60dBm = 66.2 percent

The percentage of  area within 2 km = 92 percent


## Example no 4.10 Page no. 152¶

In [3]:
from __future__ import division
from math import log10,pi
# To find the power at receiver

# Given data
d=50*10**3#                                                            # Distance between transmitter and receiver in m
hte=100#                                                              # Effective heigth of transmitter in m
hre=10#                                                               # Effective heigth of receiver in m
EIRP=1*10**3#                                                          # Radiated power in Watt
f=900*10**6#                                                           # Operating frequency in Hz
c=3*10**8#                                                             # Speed of ligth in m/s
lamda=c/f#                                                           # operating wavelength in m
EIRP=20*log10(EIRP)#                                                  # Radiated power in dB
Gr=0#                                                                 # Receiving gain in dB

Lf=-10*log10(lamda**2/(4*pi*d)**2)#                                   # Free space path loss in dB
Amu=43#                                                               # Attenuation relative to free space in dB from Okumuras curve
Garea=9#                                                              # Gain due to type of environment in dB from Okumuras curve
Ghte=20*log10(hte/200)#                                               # Base station antenna heigth gain factor for 1000m > hte > 30m
Ghre=20*log10(hre/3)#                                                 # Mobile antenna heigth gain factor for 10m > hre > 3m
L50=Lf+Amu-Ghte-Ghre-Garea#                                           # Total mean path loss

# The median reeived power
Pr=EIRP-L50+Gr#

#Displaying the result in command window
print '\n The power at receiver = %0.2f dBm'%(Pr)#

#Answer is varrying due to round-off error

 The power at receiver = -95.07 dBm


## Example no 4.11 Page no. 166¶

In [1]:
from math import log10
# To find the mean path loss

# Given data
d0=1#                                               # Reference distance in m
d=30#                                               # Distance from transmitter in m
nSF=3.27#                                           # Exponent value for same floor
nMF=5.22#                                           # Path loss exponent value for multiple floors
FAF=24.4#                                           # Floor attenuation factor for specified floor in dB
n=2#                                                # Number of blocks
PAF=13#                                             # Particular attenuation factor for paricular obstruction in dB
PLSFd0=31.5#                                        # Attenuation at reference distance for same floor in dB
PLMFd0=5.5#                                         # Attenuation at reference distance for multiple floor in dB

#Mean path loss at same floor
PL1=PLSFd0+10*nSF*log10(d/d0)+FAF+n*PAF#

#Mean path loss at multiple floor
PL2=PLMFd0+10*nMF*log10(d/d0)+n*PAF#

#Displaying the result in command window
print '\n The mean path loss at same floor = %0.1f dB'%(PL1)#
print '\n The mean path loss at multiple floor = %0.1f dB'%(PL2)#

 The mean path loss at same floor = 130.2 dB

The mean path loss at multiple floor = 108.6 dB


## Example no 4.12 Page no. 167¶

In [4]:
from __future__ import division
# To find far field distance for antenna with maximum dimensions and operating frequency

# Given data
D=1#                   # Maximum dimension in m
f=600*10**6#            # Operating frequency in Hz
C=3*10**8#              # Speed of light in m/sec

lamda=C/f#            # Carrier wavelength in m

# To find far field distance
df=(2*D**2)/lamda#     #Far field distance

#Displaying the result in command window
print '\n Far field distance = %0.0f meter'%(df)#

 Far field distance = 4 meter