# Ch-8 Telecommunication Application¶

## Example 8.45 Page no 499¶

In [10]:
from __future__ import division
from math import sqrt, pi, log10

# Calculation of incident optical power.

#Given data
lamda=1.3*10**-6#                      # Wavelength
B=6*10**6#                             # Bandwidth
S=10**5#                               # Total system margin
n=1#                                 # Efficiency
v=3*10**14#
h=6.62*10**-34#                       # Planck constant

# Incident optical power
P=(2*S*v*h*B)/n#

P1=10*log10(P/10**-3)#

print"Incident optical power = %0.1f nW "%(P1)#

# The answers vary due to round off error

Incident optical power = -36.2 nW


## Example 8.46 Page no 500¶

In [11]:
# Calculation of maximum repeater spacing of a)ASK hetrodyne b)PSK homodyne

#Given data

S=0.2#                             # Split loss
c=3*10**8#                          # velocity of light
lamda=1.55*10**-6#                 # Wavelength
B1=50*10**6#                        # Speed of communication
h=6.63*10**-34                      # Planck constant
B2=1*10**9#                         # Speed of communication

# a)Maximum repeater spacing for ASK hetrodyne
P1=(36*h*c*B1)/lamda#
P1=10*log10(P1/10**-3)#
s1=4-P1#
R1=s1/S#
P2= (36*h*c*B2)/lamda#
P2=10*log10(P2/10**-3)#
s2=4-P2#
R2=s2/S#
#b)Maximum repeater spacing for PSK homodyne
K1= (9*h*c*B1)/lamda#
K1=10*log10(K1/10**-3)#
K1=4-K1#
R3=K1/S#
K2= (9*h*c*B2)/lamda#
K2=10*log10(K2/10**-3)#
K2=4-K2#
R4=K2/S#

print"Maximum repeater spacing = %0.3f km "%(R1)#
print"Maximum repeater spacing = %0.3f km "%(R2)#
print"Maximum repeater spacing = %0.3f km "%(R3)#
print"Maximum repeater spacing = %0.3f km "%(R4)#
# The answers vary due to round off error

Maximum repeater spacing = 351.821 km
Maximum repeater spacing = 286.770 km
Maximum repeater spacing = 381.924 km
Maximum repeater spacing = 316.873 km


## Example 8.47 Page no 499¶

In [12]:
# Calculation of incident optical power.

#Given data

h=6.62*10**-34#                       # Planck constant
c=3*10**8#                           # velocity of light
lamda=1.55*10**-6#                 # Wavelength
B=400*10**6#                        # Speed of communication

# Maximum repeater spacing
P=(36*h*c*B)/lamda#
P=10*log10(P/10**-3)#

print"Incident optical power = %0.3f nW "%(P)#

# The answers vary due to round off error

Incident optical power = -57.340 nW


## Example 8.48 Page no 502¶

In [13]:
# Calculation of optical power budget.

#Given data
M=-10#                      # Mean optical power
S=-25#                      # Split loss
TS=7#                       # Total system margin
SP=1.4#                     # Split loss
C=1.6#                     # Connector loss
SM=4#                      # Safety margin

# Net margin between LED and receiver
N=M-S#

# Total system loss
T=TS+SP+C+SM#
# Excess power margin
E=N-T#

print"Excess power margin = %0.0f dB "%(E)#

# The answers vary due to round off error

Excess power margin = 1 dB


## Example 8.49 Page no 503¶

In [14]:
# Calculation of viability of digital link.

#Given data
M=-10#                      # Mean optical power
TS=18.2#                    # Total system margin
SP=3#                       # Split loss
C=1.5#                     # Connector loss
SM=6#                       # Safety margin

# Net margin between LED and receiver
N=M-S#

# Total system loss
T=TS+SP+C+SM#
# Excess power margin
E=N-T#

print"Excess power margin = %0.1f dB "%(E)#

# The answers vary due to round off error

Excess power margin = 2.3 dB


## Example 8.50 Page no 499¶

In [15]:
# Calculation of signal to noise ratio.

#Given data

h=6.62*10**-34#                       # Planck constant
c=3*10**8#                           # velocity of light
lamda=1.55*10**-6#                 # Wavelength
B=400*10**6#                        # Speed of communication
s=2#
# S/N ratio

sn=(s*4.24)/(2**(1/2))#
i=(sn)**2#

print"Incident optical power = %0.2f nW "%(i)#

# The answer is wrong in the book

Incident optical power = 35.96 nW


## Example 8.52 Page no 504¶

In [16]:
# Calculation of a)Bit rate of communication system  b)Bit duration c)Time period

#Given data
f=8*10**3#                    # Power launched in port 1
P2=0.082*10**-6#              # Power launched in port 2
P3=47*10**-6#                # Power launched in port 3
P4=52*10**-6#                # Power launched in port 4

# a)Bit rate of communication system
c=32*8#
B=f*c#
B=B*10**-6#
# b)Bit duration
D=1/B#
D=D*10*10**2#
P=8*D#

# c)Time period
T=32*P#
T=T*10**-3#

print"Bit rate of communication system = %0.3f Mb/s "%(B)#
print"Bit duration = %0.0f ns "%(D)#
print"Time period = %0.0f micro sec "%(T)#

# The answers vary due to round off error

Bit rate of communication system = 2.048 Mb/s
Bit duration = 488 ns
Time period = 125 micro sec