# Chapter 3: Radiowave Propagation in Troposphere¶

## Example 3.1, page 61¶

In [1]:
#Initialisation
h=2                             #height in Km
h1=5                            #height in Km

#Calculation
t2=290-(6.5*h)                         #Proposed formula for height h=2Km
p2=950-117*h
e2=8-3*h
t21=294.98-5.22*h-0.007*h**2
p21=1012.82-111.56*h+3.86*h**2
p=14.35*2.72**(-0.42*h-0.02*h*h+0.001*h**3)

t5=290-6.5*h1                        #Proposed formula for height h=5Km
p5=950-117*h1
e5=8-3*h1
t51=294.98-5.22*h1-0.007*h1**2
p51=1012.82-111.56*h1+3.86*h1**2
p1=14.35*2.72**(-0.42*h1-0.02*h1**2+0.001*h1**3)

#Results
print'T(2) = %.1f K'%t21
print'P(2) = %.2f hpa'%p21
print'p(2) = %.2f hpa\n'%p
print'T(5) = %.1f K'%t51
print'P(5) = %.2f hpa'%p51
print'p(5) = %.2f hpa'%p1

T(2) = 284.5 K
P(2) = 805.14 hpa
p(2) = 5.76 hpa

T(5) = 268.7 K
P(5) = 551.52 hpa
p(5) = 1.21 hpa


## Example 3.2, page 63¶

In [23]:
#Initialisation
h=2                                 #Height in Km
T=277                               #Tempreture in Kelvin
p=716
e=2

#Calculation
er=1+(151.1/T)*(p+(4810*h/T))*10**-6
n=er**(0.515)                         #refractive index of the air
N=(n-1)*10**6                       #refractivity number

#Results
print'n = %.5f'%n
print'N = %d'%N

n = 1.00021
N = 210


## Example 3.3, page 67¶

In [30]:
#Initialisation
er=1.001                                   #relative permittivity of a medium
dn=35*10**-6                               #vertical gradient of refractive index
Re=6370                                    #actual redius of earth
d=20                                       #transmitter and receiver distance in Km
d1=5
d2=15
K1=1.3333                                  #standard atmosphere condition

#Calculation
R=(er**0.5)/dn
K=R/(R-Re)
hm=(125*d**2)/(K*Re)                        #Earth bulge value in the middle of the path
h1=(500*d1*d2)/(K*Re)                       #h1
h2=(500*d1*d2)/(K1*Re)                      #h2

#Results
print'Bulge value  = %.1f m'%hm
print'Bulge value, h1  = %.2f m'%h1
print'Bulge value, h2  = %.2f m'%h2

Bulge value  = 6.1 m
Bulge value, h1  = 4.58 m
Bulge value, h2  = 4.42 m


## Example 3.4, page 68¶

In [35]:
#Initialisation
K=1.33
d1=24                                     #heigth in Km
d2=15                                      #heigth in Km
K1=1
Re=6370                                    #actual redius of earth

#Calculation
R=4.12*(d1**0.5+d2**0.5)
R1=K1*Re
Rrh=(2*R1*d1)**0.5+(2*R1*d2)**0.5

#Results
print'K=1.33'
print'Rrh = %.1f km\n'%R
print'K=1'
print'Rrh = %.1f km'%Rrh                     #wrong answer in textbook

K=1.33
Rrh = 36.1 km

K=1
Rrh = 990.1 km


## Example 3.5, page 74¶

In [16]:
import math
#Initialisation
No=1                                          #index of refraction
N1=1.3*10**-7
h=20                                          #height

#Calculation
wc=math.asin(((4*No)/((4*No)+((h**2)*N1))))     #critical angle

#Results
print'Critical angle = %f'%wc                    #ans is not written in textbook

Critical angle = 1.565697


## Example 3.6, page 76¶

In [18]:
import math

#Initialisation
d=20                                  #height
b=0.074                               #elevation angle from graph 3.10
f=7                                   #frequency in Ghz from graph 3.11
c=2*10**-6

#Calculation
t=0.156                                #1000/6370
dm=dn+t
a=(-c*dm*d)**0.5                      #elevation angle of waves

#Results
print'Elevation angle of waves = %.2f mrad'%(a*10**3)
print'Elevation angle = %.3f'%b                                    #from graph 3.10
print'Minimum frequency of coupling waves into the duct = %d Ghz'%f     #from graph 3.11

Elevation angle of waves = 1.33 mrad
Elevation angle = 0.074
Minimum frequency of coupling waves into the duct = 7 Ghz


## Example 3.7, page 80¶

In [7]:
import math

#Initialisation
f=18                                #frequency in GHz
d=30                                #in km
R=25                                #rainfall intensity in mm

#Using Table 3.3
av15=1.128
av20=1.065
av18=1.09
kv15=0.0335
kv20=0.0691
kv18=0.0587

#Calculation
yr=kv18*R**av18                        #rain specific attenuation
de=(90*(90+d)**-1)*d
A=de*yr                                #Maximum rain attenuation

#Results
print'(1) Rain specific attenuation = %.2f dB/km'%yr
print'(2) Maximum rain attenuation = %.1f dB'%A

(1) Rain specific attenuation = 1.96 dB/km
(2) Maximum rain attenuation = 44.1 dB


## Example 3.9, page 89¶

In [11]:
import math

#Initialisation
rh=-1
s=4                                    #sigma in S/m
f=5*10**9                              #frequency in Hz
eo=8.85*10**-12                        #permitivity of free space
er=75                                  #permitivity of medium

#Calculation
w=2*math.pi*f
x=s*(w*eo)**-1
a=math.sin(w1)-math.sqrt((er-x)-math.cos(w1)**2)
a1=math.sin(w1)+math.sqrt((er-x)-math.cos(w1)**2)
rh1=a/a1
b1=(er-x)*math.sin(w1)-math.sqrt((er-x)-math.cos(w1)**2)
b2=(er-x)*math.sin(w1)+math.sqrt((er-x)-math.cos(w1)**2)
rv=-b1/b2

#Results
print'(2) X = %.1f'%x
print'(3) Rh = %.3f'%rh1
print'    Rv = %.1f'%rv

(2) X = 14.4
(3) Rh = -0.879
Rv = -0.6


## Example 3.10, page 92¶

In [26]:
import math

#Initialisation
f=5*10**9                                                         #frequency in Hz
c=3*10**8                                                         #speed of light
h1=6                                                              #in metre
h2=2                                                              #in metre

#Calculation
h=c*f**-1                                                       #wavelength
w1=w*180*math.pi**-1                                        #grazing angle in degree
a=((2*math.pi*h1*h2)*(h*300)**-1)*3.14*180**-1
e=math.sin(a)
F=e*2*180*math.pi**-1                                              #PGF value (wrong calculated in textbook)
LR=20*math.log10(F)                                              #Decrease in received signal level

#Results
print'(1) Grazing angle = %.2f degree'%w1
print'(2) PGF value = %f'%F                                              #wrong value calculated in textbook
print'(3) Decrease in received signal level = %.2f dB'%LR                #wrong value calculated in textbook

(1) Grazing angle = 0.15 degree
(2) PGF value = 8.365884
(3) Decrease in received signal level = 18.45 dB


## Example 3.11, page 98¶

In [14]:
#Initialisation
h=12.5*10**-2           #in meter
d1=10*10**3             #in meter
d2=15*10**3              #in meter
d3=12.5*10**3            #in meter
d4=12.5*10**3               #in meter
h=1.25                    #in Kilometer

#Calculation
r1=(((d1*d2)/(d1+d2))*h)**0.5        #radius of first and fourth Fresnel zones
r4=r1*(4)**0.5
R1=(((d3*d4)/(d3+d4))*h)**0.5        #radius of first and fourth ellipse zones
R4=R1*(4)**0.5

#Results
print'Radius of first fresnel zones, r1 = %.2f m'%r1
print'Radius of Second fresnel zones, r4= %.2f m'%r4
print'h = %.2f x 10^-4 Km'%h
print'Radius of first ellipse, R1 = %.2f m'%R1
print'Radius of second ellipse, R4 = %.1f m'%R4

Radius of first fresnel zones, r1 = 86.60 m
Radius of Second fresnel zones, r4= 173.21 m
h = 1.25 x 10^-4 Km
Radius of first ellipse, R1 = 88.39 m
Radius of second ellipse, R4 = 176.8 m


## Example 3.12, page 105¶

In [93]:
import math
%matplotlib inline
import matplotlib
import numpy as np
import matplotlib.pyplot as plt

#Initialisation
L=13200                                       #L parameter in m
H=10240                                       #H parameter
Re=6370000                                    #actual redius of earth
ht=30                                         #height in m
hr=20                                         #height in m
re1=8453000                                    # in metre
h1=30000                                     # in metre
h2=20000                                    # in metre
dt1=22.5
f=10*10**9                                  #frequency in Hz
c=3*10**8                                   #speed of light
d=30000                                     #distance in m
pt=30                                       #transmitter antenna power
gt=40                                       #transmitter antenna gain
F3=-3
H=-34
D=0.75

#Calculation
dt=math.sqrt(2*re1*ht)
X=3*dt*L**-1
Z1=h1*H**-1
Z2=h2*H**-1
vx=10**-3.5                              #from fig 3.26
z1=10**0.95                              #from fig 3.27
z2=10**0.65                              #from fig 3.27

#for d=3dt
lv=20*math.log10(vx)
lz1=20*math.log10(z1)
lz2=20*math.log10(z2)
F=(lv+lz1+lz2)*20**-1
F1=10**(F)
F11=20*math.log10(F1)
X1=2*dt*L**-1
d3=3
f3=-F11

vx1=10**-2.35                             #from fig 3.26
lv1=20*math.log10(vx1)

#for d=2dt
F4=1+D
F5=20*math.log10(F4)
d2=2
f2=-F5

#for d=1.1dt
F6=math.sqrt(1+D**2)
F7=20*math.log10(F6)
d11=1.1
f11=-F7

#for d=dt
d1=1
f1=0.2

#for plotting graph in terms of points

for N in range(0,4,1):
a=plot(1,0.2,'-o')
a1=plot(1.1,-1.9,'-o')
a2=plot(2,-4.8,'-o')
a3=plot(3,-38,'-o')

plt.annotate('d/dt',xy=(1,2),xytext=(1, 2))
plt.annotate('1.1d/dt',xy=(1.2,0.7),xytext=(1.2, 0.7))
plt.annotate('2d/dt',xy=(2,-0.7),xytext=(2, -0.7))
plt.annotate('3d/dt',xy=(3,-35),xytext=(3, -35))
xlabel("$d/dt$")
ylabel("$20 log(F)$ (dB)")
title("Path gain F")
grid()
show(a)
show(a1)
show(a2)
show(a3)

h=c*f**-1                                                                       #wavelength

#Results
print'(1) Effective receiver path gain F = %.4f'%F11
print'(2) Path gain F plot is shown'
print'(3) Received signal power Pr = %.1f dBm'%Pr

(1) Effective receiver path gain F = -38.0000
(2) Path gain F plot is shown
(3) Received signal power Pr = -27.5 dBm


## Example 3.13, page 109¶

In [41]:
import math

#Initialisation
eirp=800                                    #in KW
d=24                                        #in Km
d1=22                                       #in Km
d2=2                                        #in Km
h=0.4*10**-3                                #wavelength in m
Er=45                                       #in microvolt

#Calculation
E=104.8+10*math.log10(eirp)-20*math.log10(d)                #field intensity
V=a*math.sqrt((2*d2*d1)/((d1+d2)*h))                      #knife edge obstacle attenuation
Lke=23                                                    #from table 3.4
er=10**(Er*20**-1)

#Results
print'(1) Electric field intensity = %.3f microV/m'%er

(1) Electric field intensity = 177.828 microV/m


## Example 3.14, page 115¶

In [6]:
#Initialisation
f1=430                                    #upper frequency band
f2=410                                    #lower frequency band
d=80                                      #distance in meter

#Calculation
Yv=0.1                                     #Specific attenuation obtained from graph fig 3.34
Lv=Yv*d                                   #loss of forest trees
Am=((f1+f2)/2)**0.5                        #maximum value for trees excess loss.

#Results
print'Specific attenuation index, Yv = %.1f dB/m'%Yv
print'Loss of forest trees, Lv = %.1f dB'%Lv
print'Maximum value for trees excess loss = %.1f dB'%Am

Specific attenuation index, Yv = 0.1 dB/m
Loss of forest trees, Lv = 8.0 dB
Maximum value for trees excess loss = 20.5 dB


## Example 3.15, page 118¶

In [3]:
#Initialisation
d=40                                           #length in meter
Am=2                                           #area in square meter
f=10*10**9                                    #frequency in hertz

#Calculation
As=40                                      #using graph fig 3.36, As can be obtained
As1=30                                     #using graph fig 3.37, As can be obtained

#Results
print'Loss in the summer for trees with leaves, As = %d dB'%As
print'Loss in winter for trees without leaves, As = %d dB'%As1

Loss in the summer for trees with leaves, As = 40 dB
Loss in winter for trees without leaves, As = 30 dB