# Chapter 4: An Overview of Digital Communication and Transmission¶

## Example 4.1, Page 93¶

In [2]:
#Variable declaration
Fm=20;  # in KHz

#Calculations
print "An Engineering version of the Nyquist sampling rate : fs>=2.2*fm."
print 'Therefore sampling rate of >= %d ksps should be used '%(2.2*Fm);
print "The sampling rate for a compact disc digital audio player = 44.1 ksps and for a studio quality audio player = 48 ksps are used."

An Engineering version of the Nyquist sampling rate : fs>=2.2*fm.
Therefore sampling rate of >= 44 ksps should be used
The sampling rate for a compact disc digital audio player = 44.1 ksps and for a studio quality audio player = 48 ksps are used.


## Example 4.2, Page 96¶

In [5]:
import math

#Variable declaration
Rt=1; #Resistance(ohm)
#L= Number of quantization values
L1=32;
L2=64;
L3=128;
L4=256;

#Calculations
# L=2**R i.e R=math.log2(L);
R1=math.log(L1,2);
R2=math.log(L2,2);
R3=math.log(L3,2);
R4=math.log(L4,2);

#P=A**2/2; #average power of signal
#sig**2=0.333*A**2*2**(-2*Rt); #Avg quantization noise power
#SNR=P/sig**2;
# SNR(dB)=1.8+ 6R;

SNR1=1.8+6*R1;
SNR2=1.8+6*R2;
SNR3=1.8+6*R3;
SNR4=1.8+6*R4;

#Result
print 'For L=32, SNR is %.1f dB\n '%SNR1
print 'For L=64, SNR is %.1f dB\n '%SNR2
print 'For L=128, SNR is %.1f dB\n '%SNR3
print 'For L=256, SNR is %.1f dB\n '%SNR4

For L=32, SNR is 31.8 dB

For L=64, SNR is 37.8 dB

For L=128, SNR is 43.8 dB

For L=256, SNR is 49.8 dB



## Example 4.3, Page 99¶

In [6]:
#Variable declaration
Fs=8*10**3;  #in Hz
Fm=3.4*10**3; # in Hz
VCH=24; #voice channels
SCH=1; #sunchronization channel
PDur=1;  #extra pulse duration in microsec

#Calculations&Results
Ts=1/(Fs);
TimeCH=Ts/(VCH+SCH)*10**6;   # in microsec
print 'Time between the pulses is %d microsec\n'%(TimeCH-PDur);
#Now by using the engineering version of Nyquist rate sampling
NyquistRate=2.2*Fm;
Ts1_microsec=1/NyquistRate*10**6;
Tc=round(Ts1_microsec)/(VCH+SCH);
print "Time between the pulses by using engineering version of Nyquist rate sampling  is %.2f microsec\n"%(Tc-PDur);

Time between the pulses is -1 microsec

Time between the pulses by using engineering version of Nyquist rate sampling  is 4.36 microsec



## Example 4.4, Page 101¶

In [7]:
import math

#Variable declaration
Fm=3000; #highest modulating frequency in signal(Hz)
M=32; # number of pulse levels
b=5; #bits per symbol
p=0.01; #Quantization distortion

#Calculations
#2**R = L >= 1/2P
# where R is the number of bits required to represent quantization levels L
R=math.log10(1./(2*p))/math.log10(2);
Fs=2*Fm;  # Nyquist sampling criteria (samples per second)
fs=round(R)*Fs;
Rs=fs/b;

#Result
print 'The minimum number of bits/sample or bits/PCM word that should be used are %d'%(round(R));
print 'The minimum sampling rate is %d samples per second\n '%Fs;
print 'The resulting transmission rate is %d bps\n '%fs;
print 'The PCM pulse or symbol transmission rate is %d symbols/sec\n'%Rs

The minimum number of bits/sample or bits/PCM word that should be used are 6
The minimum sampling rate is 6000 samples per second

The resulting transmission rate is 36000 bps

The PCM pulse or symbol transmission rate is 7200 symbols/sec



## Example 4.5, Page 110¶

In [13]:
import math

#Variable declaration
S_No=53.; #dB-Hz
R=9.6*10**3; #bps
BW=4.8*10**3; #Khz
Pb=10**-5; #BER<=10**-5;

#Calculations
print "Since the required data rate of 9.6 kbps is more than the available bandwidth of 4.8 kHz, the channel is bandwidth-limited."
Eb_No=S_No-10*math.log10(R); #dB
# Try for 8-PSK modulation scheme
M=8;
Ps=math.log(M,2)*Pb; #Max ps
Es_No=math.log(M,2)*10**(0.1*Eb_No);
#Ps(8)=2*Q(math.sqrt(2*Es_No)*sin(math.pi/8));
#2*Q(math.sqrt(2*Eb_No))=erfc(math.sqrt(Eb_No));  #Refer EQn C(7) from appendix C

Ps8=math.erfc(math.sqrt(Es_No)*math.sin(math.pi/8));

#Result
print 'Symbol error rate is given as %.5f \n '%Ps
print 'The ratio of signal energy to noise is %.2f \n '%Es_No;
print 'Symbol error rate for 8-PSK  is %.5f \n '%Ps8;
print "As symbol error rate for 8-PSK modulation is lower than threshold value. so, We can use 8-PSK modulation."

Since the required data rate of 9.6 kbps is more than the available bandwidth of 4.8 kHz, the channel is bandwidth-limited.
Symbol error rate is given as 0.00003

The ratio of signal energy to noise is 62.35

Symbol error rate for 8-PSK  is 0.00002

As symbol error rate for 8-PSK modulation is lower than threshold value. so, We can use 8-PSK modulation.


## Example 4.6, Page 111¶

In [15]:
import math

#Variable declaration
SNR=48.; #dB-Hz
BW=45.*10**3; #in Hz
R=9.6*10**3; #bps
Pb=10**-5; #Bit error rate
e=2.71828; #Natural exponent e

#Calculations&Results
print "since the available bandwidth of 45 kHz is more than adequate to support the required data rate of 9.6 kbps.";
print "So, the channel is not bandwidth limited ";
Eb_No=SNR-10*math.log10(R);
#We try the 16-FSK modulation scheme
M=16;

Es_No=math.log(M,2)*Eb_No;
Ps=(M-1)/2*e**(-Es_No/2);
#For orthogonal signalling
Ps16=(2**M-1)/(2**(M-1))*Pb;
print ""
print 'The maximum symbol error probability is %0.5f \n '%Ps16
print 'The symbol error probability achieved by 16-PSK is %.9f \n '%Ps;
print "As achieved symbol error probability is far less than maximum tolerable value";
print "So, we can meet the given speciﬁcations for this power-limited channel with a 16-FSK modulation scheme without any error-correction coding"

since the available bandwidth of 45 kHz is more than adequate to support the required data rate of 9.6 kbps.
So, the channel is not bandwidth limited

The maximum symbol error probability is 0.00001

The symbol error probability achieved by 16-PSK is 0.000000553

As achieved symbol error probability is far less than maximum tolerable value
So, we can meet the given speciﬁcations for this power-limited channel with a 16-FSK modulation scheme without any error-correction coding