Chapter 13 : Multiplexing and Multiple Access Techniques

Example 1 : pg 437

In [1]:
 
# page no 437
# prob no 13_1
#calculate the no. of signals in all cases
from math  import log
#given
freq_band=1.*10**6;
# A)For SSBSC AM, the bandwidth is the same as the maximunm modulating freq.
fmax=4.*10**3;
#calculations and results
B=fmax;
no_of_signal=freq_band/B;
print 'a)The number of signals are ',no_of_signal
# B)For DSB AM, the bandwidth is twice the maximunm modulating freq.
B=2*fmax;
no_of_signal=freq_band/B;
print 'b)The number of signals are ',no_of_signal
# C)Using Carson's Rule to approximate the bandwidth
f_max=15.*10**3; deviation=75.*10**3;
B=2*(deviation + f_max);
no_of_signal=freq_band/B;
print 'c)The number of signals are ',round(no_of_signal)
# D)Use Shannon-Hartley theorem to find the bandwidth
C=56.*10**3;M=4.;#for QPSK
B=C/(2*log(M) /log(2));
no_of_signal=freq_band/B;
print 'd)The number of signals are ',round(no_of_signal)

a)The number of signals are  250.0
b)The number of signals are  125.0
c)The number of signals are  6.0
d)The number of signals are  71.0

Example 2 : pg 444

In [2]:
 
# page no 444
# prob no 13_2
#calculate the value of SNR and noise power in both cases
#Voice transmisssion occupies 30 kHz.Spread spectrum is used to increase BW to 10MHz
from math import log10
#given
B1=30.*10**3;#BW is 30 kHz
B2=10.*10**6;#BW is 10 MHz
T=300.;#noise temp at i/p
PN=-110.;#signal has total signal power of -110 dBm at receiver
k=1.38*10**-23;#Boltzmann's const in J/K
#calculations and results
#Determination of noise power at B1=30kHz
PN1=10*(log10(k*B1*T/10**-3));
print 'The noise power at BW=30 kHz is',round(PN1,3),'dBm'
#Determination of noise power at B2=10MHz
PN2=10*(log10(k*B2*T/10**-3));
print 'The noise power at BW=10 MHz is',round(PN2,3),'dBm'
#Determination of SNR for 30kHz BW
SNR1=PN-PN1;
print 'The value of SNR for BW=30 kHz is',round(SNR1,3),'dB'
#Determination of SNR for 10MHz BW
SNR2=PN-PN2;
print 'The value of SNR for BW=10 MHz is',round(SNR2,3),'dB'

The noise power at BW=30 kHz is -129.059 dBm
The noise power at BW=10 MHz is -103.83 dBm
The value of SNR for BW=30 kHz is 19.059 dB
The value of SNR for BW=10 MHz is -6.17 dB

Example 3 : pg 445

In [3]:
 
# page no 445
# prob no 13_3
#calculate the time required
#given
no_of_freq_hops =100.; total_time_req=10.;
#calculations
time_for_each_freq = total_time_req  / no_of_freq_hops;
#results
print 'Time required for each freq',time_for_each_freq,'sec/hop'

Time required for each freq 0.1 sec/hop

Example 4 : pg 446

In [4]:
 
# page no 446
# prob no 13_4
#calculate the no. of signal changes
from math import log
#given
bit_rate=16.*10**3;#in bps
#chip_rate =10:1;
no_of_chip=10.;
#calculations
total_bit_rate=no_of_chip*bit_rate;
m=4;n=log(m)/log(2);
symbol_rate = total_bit_rate/n;
#results
print 'The no of signal changes i.e. symbol rate is ',symbol_rate,'baud'

The no of signal changes i.e. symbol rate is  80000.0 baud

Example 5 : pg 447

In [5]:
 
# page no 447
# prob no 13_5
#calculate the value of BW, SNR
#signal with bandwidth Bbb=200 kHz & SNR=20 dB spred at chip rate 50:1
from math import log10
#given
Bbb=200.*10**3;#Bandwidth
Gp=50.;#chip rate
SNR_in=20.;#SNR is 20 dB without spreading
#calculations and results
#Determination of BW after spreading
Brf=Gp*Bbb;
print 'The value of BW after spreading',Brf/10**6,'MHz'
#Converting into dB 
Gp_dB=10*log10(Gp);
print 'The value of processing gain',round(Gp_dB,3),'dB'
#Determination of SNR after spreadng
SNR_out=SNR_in-Gp_dB;
print 'The value of SNR after spreading in dB',round(SNR_out,3),'dB'

The value of BW after spreading 10.0 MHz
The value of processing gain 16.99 dB
The value of SNR after spreading in dB 3.01 dB