# Chapter No.7 : Equalization diversity and channel coding¶

## Example no 7.3 Page no. 373¶

In [2]:
from __future__ import division
from math import pi,sqrt
# To determine a)the maximum Doppler shift b)the coherence time of the channel c)the maximum number of symbolsthat could be transmitted

#Given data
f=900*10**6#                                                  # Carrier frequency in Hz
c=3*10**8#                                                    # Speed of ligth in air (m/s)
v=80#                                                        # Velocity of mobile in km/hr
v=v*(5/18)#                                                  # Velocity of mobile in m/s
lamda=c/f#                                                  # Carrier wavelength in meter

# a)To determine the maximum Doppler shift
fd=v/lamda#                                                 # The maximum Doppler shift in Hz

# b)To determine the coherence time of the channel
Tc=sqrt(9/(16*pi*fd**2))#                                   # The coherence time of the channel
# Answer is varrying due to round-off error

# c)To determine the maximum number of symbols that could be transmitted with symbol rate 24.3 ksymbols/sec
Rs=24.3*10**3#                                              # Symbol rate in symbols/sec
Nb=Tc*Rs#                                                  # The maximum number of transmitted symbols

# Displaying the result in command window
print '\n The maximum Doppler shift = %0.2f Hz'%(fd)#
print '\n The coherence time of the channel = %0.2f ms'%(Tc*10**3)#
print '\n The maximum number of symbols that could be transmitted with symbol rate 24.3 ksymbols/sec = %0.0f symbols'%(Nb)#

 The maximum Doppler shift = 66.67 Hz

The coherence time of the channel = 6.35 ms

The maximum number of symbols that could be transmitted with symbol rate 24.3 ksymbols/sec = 154 symbols


## Example no 7.4 Page no. 383¶

In [4]:
from __future__ import division
from math import exp
# To determine probability that the SNR will drop below threshold SNR

# Given data
M1=4#                                     # Number of branch diversity
M2=1#                                     # Number of branch diversity
gamm=10#                                  # Specified SNR threshold in dB
gamm=10**(gamm/10)#                        # Specified SNR threshold
Gamma=20#                                 # Average SNR in dB
Gamma=10**(Gamma/10)#                      # Average SNR

# Probability that the SNR will drop below 10dB when 4 branch diversity is used
P4=(1-exp(-gamm/Gamma))**M1#              # Probability that the SNR will drop below 10dB

# Probability that the SNR will drop below 10dB when single branch diversity is used
P1=(1-exp(-gamm/Gamma))**M2#              # Probability that the SNR will drop below 10dB

# Displaying the result in command window
print '\n Probability that the SNR will drop below 10dB when 4 branch diversity is used = %0.6f'%(P4)#
print '\n Probability that the SNR will drop below 10dB when single branch diversity is used = %0.3f'%(P1)#
print '\n \n From above results, it is observed that without diversity the SNR drops below the specified threshold with a probability that is three orders of magnitude greater \n than if four branch diversity is used.'

 Probability that the SNR will drop below 10dB when 4 branch diversity is used = 0.000082

Probability that the SNR will drop below 10dB when single branch diversity is used = 0.095

From above results, it is observed that without diversity the SNR drops below the specified threshold with a probability that is three orders of magnitude greater
than if four branch diversity is used.