# Chapter No.9 : Multiple access techniques for wireless communications¶

## Example no 9.1 Page no. 451¶

In [1]:
# To find the intermodulation frequencies generated

# Given data
f1=1930#                                         # First carrier frequency
f2=1932#                                         # second carrier frequency
F1=1920#                                         # Lower frequency of the band
F2=1940#                                         # Upper frequency of the band

for n in range(0,4):
x1=(2*n+1)*f1-2*n*f2
if x1 <= F2:
print '\n IF frequency %0.0f MHz lies inside the band'%(x1)#
else:
print '\n IF frequency %0.0f MHz lies outside the band'%(x1)#

for n in range(0,4):
x2=(2*n+2)*f1-(2*n+1)*f2
if x2 <= F2:
print '\n IF frequency %0.0f MHz lies inside the band'%(x2)#
else:
print '\n IF frequency %0.0f MHz lies outside the band'%(x2)#

for n in range(0,4):
x3=(2*n+1)*f2-2*n*f1
if x3 <= F2 :
print '\n IF frequency %0.0f MHz lies inside the band'%(x3)#
else:
print '\n IF frequency %0.0f MHz lies outside the band'%(x3)#

for n in range(0,4):
x4=(2*n+2)*f2-(2*n+1)*f1
if x4 <= F2:
print '\n IF frequency %0.0f MHz lies inside the band'%(x4)#
else:
print '\n IF frequency %0.0f MHz lies outside the band'%(x4)#


 IF frequency 1930 MHz lies inside the band

IF frequency 1926 MHz lies inside the band

IF frequency 1922 MHz lies inside the band

IF frequency 1918 MHz lies inside the band

IF frequency 1928 MHz lies inside the band

IF frequency 1924 MHz lies inside the band

IF frequency 1920 MHz lies inside the band

IF frequency 1916 MHz lies inside the band

IF frequency 1932 MHz lies inside the band

IF frequency 1936 MHz lies inside the band

IF frequency 1940 MHz lies inside the band

IF frequency 1944 MHz lies outside the band

IF frequency 1934 MHz lies inside the band

IF frequency 1938 MHz lies inside the band

IF frequency 1942 MHz lies outside the band

IF frequency 1946 MHz lies outside the band


## Example no 9.2 Page no. 452¶

In [2]:
# To find number of channels available

# Given data
Bt=12.5*10**6#                                            # Total spectrum allocation in Hz
Bguard=10*10**3#                                          # Guard band allocated in Hz
Bc=30*10**3#                                              # Channel bandwidth in Hz

# The number of channels available
N=(Bt-2*Bguard)/Bc#                                      # The number of channels available

# Displaying the result in command window
print '\n The number of channels available in FDMA system = %0.0f'%(N)#

 The number of channels available in FDMA system = 416


## Example no 9.3 Page no. 455¶

In [3]:
# To find number of simultaneous users accommodated in GSm

# Given data
m=8#                                                # Maximum speech channels supported by single radio channel
Bc=200*10**3#                                        # Radio channel bandwidth in Hz
Bt=25*10**6#                                         # Total spectrum allocated for forward link
Bguard=0#                                           # Guard band allocated in Hz

# The number of simultaneous users accommodated in GSm
N=(m*(Bt-2*Bguard))/Bc#                             # The number of simultaneous users

# Displaying the result in command window
print '\n The number of simultaneous users accommodated in GSM system = %0.0f'%(N)#

 The number of simultaneous users accommodated in GSM system = 1000


## Example no 9.4 Page no. 456¶

In [4]:
# To find a)the time duration of a bit b)the time duration of a slot c)the time duration of a frame d)how long must a user occupying single time slot wait between two successive transmission

# Given data
N=8#                                                        # Number of time slots in each frame
Nb=156.25#                                                  # Number of in each time slot
DR=270.833*10**3#                                            # Data rate of transmission in channel

# a)To find the time duration of a bit
Tb=1/DR#                                                   # The time duration of a bit in sec

# b)To find the time duration of a slot
Tslot=Nb*Tb#                                              # The time duration of a slot

# c)To find the time duration of a frame
Tf=N*Tslot#                                               # The time duration of a frame

#d) The waiting time between two successive transmission
Tw=Tf#                                                   # The arrival time of new frame for its next transmission

# Displaying the result in command window
print '\n The time duration of a bit = %0.3f microseconds'%(Tb*10**6)#
print '\n The time duration of a slot = %0.3f ms'%(Tslot*10**3)#
print '\n The time duration of a frame = %0.3f ms'%(Tf*10**3)#
print '\n The arrival time of new frame for its next transmission = %0.3f ms'%(Tw*10**3)#

 The time duration of a bit = 3.692 microseconds

The time duration of a slot = 0.577 ms

The time duration of a frame = 4.615 ms

The arrival time of new frame for its next transmission = 4.615 ms


## Example no 9.5 Page no. 456¶

In [5]:
# To find the frame efficiency

# Given data
Btrail=6#                                               # Number of trailing bits per slot
Bg=8.25#                                                # Number of guard bits per slot
Btrain=26#                                              # Number of training bits per slot
Nb=2#                                                   # Number of burst
Bburst=58#                                              # Number of bits in each burst
Nslot=8#                                                # Number of slots in each frame

N=Btrail+Bg+Btrain+2*Bburst#                            # Total number of bits in each slot
Nf=Nslot*N#                                             # Total number of bits in a frame
bOH=Nslot*Btrail+Nslot*Bg+Nslot*Btrain#                 # Number of overhead bits per frame

# To find the frame efficiency
nf=(1-(bOH/Nf))*100#                                   # Frame efficiency

# Displaying the result in command window
print '\n The frame efficiency = %0.2f percentage'%(nf)#

 The frame efficiency = 74.24 percentage


## Example no 9.6 Page no. 466¶

In [6]:
from __future__ import division
from math import exp
# To determine the maximum throughput using ALOHA and slotted ALOHA

#The maximum throughput using ALOHA
Rmax=1/2#                                            #Maximum rate of arrival calculated by equating ALOHA throughput formula derivative to zero
T=Rmax*exp(-1)#                                      #The maximum throughput using ALOHA

# Displaying the result in command window
print '\n The maximum throughput using ALOHA = %0.4f'%(T)#

#The maximum throughput using slotted ALOHA
Rmax=1#                                              #Maximum rate of arrival calculated by equating slotted ALOHA throughput formula derivative to zero
T=Rmax*exp(-1)#                                      #The maximum throughput using slotted ALOHA

# Displaying the result in command window
print '\n The maximum throughput using slotted ALOHA = %0.4f'%(T)#

 The maximum throughput using ALOHA = 0.1839

The maximum throughput using slotted ALOHA = 0.3679


## Example no 9.7 Page no. 472¶

In [7]:
from __future__ import division
from math import log10
# To evaluate 4 different radio standards and to choose the one with maximum capacity

# Given data
ABc=30*10**3#                                              # Channel bandwidth of system A
ACImin=18#                                                # The tolerable value of carrier to interference ratio for system A
BBc=25*10**3#                                              # Channel bandwidth of system B
BCImin=14#                                                # The tolerable value of carrier to interference ratio for system B
CBc=12.5*10**3#                                           # Channel bandwidth of system C
CCImin=12#                                                # The tolerable value of carrier to interference ratio for system C # Value of CCImin is given wrong in book
DBc=6.25*10**3#                                           # Channel bandwidth of system D
DCImin=9#                                                # The tolerable value of carrier to interference ratio for system D
Bc=6.25*10**3#                                            # Bandwidth of particular system

ACIeq=ACImin+20*log10(Bc/ABc)#                           # Minimum C/I for system A when compared to the (C/I)min for particular system
BCIeq=BCImin+20*log10(Bc/BBc)#                           # Minimum C/I for system B when compared to the (C/I)min for particular system
CCIeq=CCImin+20*log10(Bc/CBc)#                           # Minimum C/I for system C when compared to the (C/I)min for particular system
DCIeq=DCImin+20*log10(Bc/DBc)#                           # Minimum C/I for system D when compared to the (C/I)min for particular system

# Displaying the result in command window
print '\n Minimum C/I for system A when compared to the (C/I)min for particular system = %0.3f dB'%(ACIeq)#
print '\n Minimum C/I for system B when compared to the (C/I)min for particular system = %0.2f dB'%(BCIeq)#
print '\n Minimum C/I for system C when compared to the (C/I)min for particular system = %0.0f dB'%(CCIeq)#
print '\n Minimum C/I for system D when compared to the (C/I)min for particular system = %0.0f dB'%(DCIeq)#
print '\n \n Based on comparison, the smallest value of C/I should be selected for maximum capacity. So, System B offers the best capacity.'

 Minimum C/I for system A when compared to the (C/I)min for particular system = 4.375 dB

Minimum C/I for system B when compared to the (C/I)min for particular system = 1.96 dB

Minimum C/I for system C when compared to the (C/I)min for particular system = 6 dB

Minimum C/I for system D when compared to the (C/I)min for particular system = 9 dB

Based on comparison, the smallest value of C/I should be selected for maximum capacity. So, System B offers the best capacity.


## Example no 9.9 Page no. 472¶

In [8]:
# To determine the maximum number of users using a)omnidirectional base station antenna and no voice activity b)three-sectors at the base station and voice activity detection

# Given data
W=1.25*10**6#                                                                  # Total RF bandwidth in Hz
R=9600#                                                                       # Baseband information bit rate in bps
EbNo=10#                                                                      # Minimum acceptable SNR in dB

# a)Maximum number of users using omnidirectional base station antenna and no voice activity
N1=1+(W/R)/EbNo#                                                              # Maximum number of users using omnidirectional

# b)Maximum number of users using three-sectors at the base station antenna and voice activity with alpha=3/8
alpha=3/8#                                                                    # Voice activity factor
Ns=1+(1/alpha)*((W/R)/EbNo)#                                                  # Maximum number of users
N2=3*Ns#                                                                      # Maximum number of users using three-sectors

# Displaying the result in command window
print '\n Maximum number of users using omnidirectional base station antenna and no voice activity = %0.0f'%(N1)#
print '\n Maximum number of users using three-sectors at the base station antenna and voice activity (with alpha=3/8) = %0.0f'%(N2)#

 Maximum number of users using omnidirectional base station antenna and no voice activity = 14

Maximum number of users using three-sectors at the base station antenna and voice activity (with alpha=3/8) = 107