Chapter 8 : The Telephone System

Example 3 : pg 323

In [1]:
 
# page no 323
# prob no 8.3
#calculate the net loss
#A telephone signal takes 3 ms to reach its destination
#given
t=2.;
#calculations
#Determination of net loss VNL reqd for acceptable amount of echo.
VNL=(0.2*t)+0.4;
#results
print 'The net loss is',VNL,'dB'
The net loss is 0.8 dB

Example 5 : pg 326

In [2]:
 
# page no 326
# prob no 8.5
#calculate the carrier freq value
#given
#Refering the fig.8.15 channel 12 has lowest carrierr freq 64 kHz
F=64.;
c_total=12.;
#Carrier freq goes up 4kHz per channel
f_up=4;
#Determination of carrier freq for channel 5
c=5;
#calculations
fc=F+(f_up*(c_total-c));
#results
print 'The value of carrier freq for channel 5 is',fc,'kHz'
The value of carrier freq for channel 5 is 92.0 kHz

Example 6 : pg 328

In [3]:
 
#page no 328
#prob no 8.6
#calculate the tone 
# 2kHz tone is present on channel 5 of group 3 of a supergroup
# refer to example 8.5, calculated fc=92kHz
#given
fc=92.;#in kHz
#calculations
# Here signal is lower sideband,the 2kHz baseband signal therefore will be
fg=fc-2;
#from fig 10.15,group 3 in the supergroup is moved to the range 408-456 kHz, with a suppressed carrier frequency of 516kHz.
fsc=516;# in kHz
#the modulation is lower sideband,so the supergroup o/p freq will be 90kHz lower than carrier freq
fsg=fsc-fg;
#results
print 'The tone appear in the supergroup output at frequency of',fsg,'kHz'
The tone appear in the supergroup output at frequency of 426.0 kHz