Chapter 25 : Fiber Optic Systems

Example 1 : pg 919

In [1]:
 
# page no 919
# prob no 25_1
#calculate the system margin
#given
from math import log10
span_length = 40#in km
Pin_mW = 1.5
signal_strength_dBm = -25
fiber_length = 2.5#in km
loss_per_slice_dB = 0.25
f_loss_dB_per_km = 0.3
loss_connector_dB = 4
#calculations
Pin_dBm = 10 * log10(Pin_mW)
splices = span_length / fiber_length - 1
fiber_loss = span_length * f_loss_dB_per_km
splice_loss = splices * loss_per_slice_dB
T_loss = fiber_loss + splice_loss + loss_connector_dB
P_out = Pin_dBm - T_loss
sys_margin = P_out - signal_strength_dBm
#results
print 'The system margin is',round(sys_margin,3),'dB'
The system margin is 7.011 dB

Example 2 : pg 921

In [2]:
 
# page no 921
# prob no 25_2
#calculate the max permissible value
#given
L=45;#in km
dt=100.;#in ns
#calculations
#The maximum permissible value for the pulse-spreading constant is 
D=dt/L;
#results
print 'The maximum permissible value for the pulse-spreading constant is',round(D,2),'ns/km'
The maximum permissible value for the pulse-spreading constant is 2.22 ns/km

Example 3 : pg 922

In [3]:
 
# page no 922
# prob no 25_3
#calculate the max bitrate in both cases#given
#given
from math import sqrt
L=45.;
T_Rtx=50.; T_Rrx=75.; T_Rf=100.;
#calculations and results
T_RT=sqrt(T_Rtx**2 + T_Rrx**2 + T_Rf**2);
# a) for NRZ
fb=1/T_RT;
print 'a) The maximum bit rate for NRZ',round(fb,5),'GHz'
# b) for RZ
fb=1/(2*T_RT);
print 'b) The maximum bit rate for NRZ',round(fb,5),'GHz'
a) The maximum bit rate for NRZ 0.00743 GHz
b) The maximum bit rate for NRZ 0.00371 GHz

Example 4 : pg 924

In [4]:
 
# page no 924
# prob no 25_4
#calculate the total rise time and dispersion
#given
Bl=500.;#in MHz-km
L=5.;#in km
#calculations and results
# using the bandwidth-distance product formula dispersion is given as
D=500/Bl;
print 'Dispersion is',D,'ns/km'
# Total rise time is given as
T_rt= D*L;
print 'Total rise time is',T_rt,'ns'
Dispersion is 1.0 ns/km
Total rise time is 5.0 ns

Example 5 : pg 924

In [5]:
 
# page no 924
# prob no 25_5
#calculate the max acceptable dispersion and bandwidth-distance product
#given
from math import sqrt
T_Rrx=3.*10**-9;
T_Rtx=2.*10**-9;
fb=100.*10**6;#in bps
L=25;#in km
#calculations and results
T_RT = 1/(2*fb)
# we have to compute rise time therefore
T_rf= sqrt(T_RT**2 - T_Rtx**2 - T_Rrx**2)
# dispersion per km is
D= T_rf/L;
print 'The maximum acceptable dispersion is',round(D/10**-9,3),'ns/km'
# using the bandwidth-distance product
Bl=500/D;
print 'The bandwidth-distance product is',round(Bl*10**-9,3),'MHz-km'
The maximum acceptable dispersion is 0.139 ns/km
The bandwidth-distance product is 3608.439 MHz-km