Chapter 5 : Transmitters

Example 2 : pg 179

In [1]:
 
#page no 179
#prob no. 5.2
#calculate the power in both cases
#A transmitter  with carrier power o/p 10W at efficiency 70% at 100% modulatn
#given
Po=10.;eta=0.7;
#calculations
#Determination of dc power o/p
Ps=Po/eta;
Pa=0.5*Ps;
#results
print 'The value of dc power input is',round(Ps,2),'W'
#Determination of audio power 
print 'The value of audio power is',round(Pa,2),'W'
The value of dc power input is 14.29 W
The value of audio power is 7.14 W

Example 3 : pg 181

In [2]:
 
#page no 181
#prob no. 5.3
#calculate the impedance of both transformers
#given
#A transmitter operates at 12V, with collector current 2A.Modulatn transformer has turn ratio 4:1
#Determination of impedance at transformer secondary  
Vcc=12.;Ic=2.;N1=4.;N2=1.;
#calculations
Za=Vcc/Ic;
Zp=Za*(N1/N2)**2;
#results
print 'The impedance of transformer secondary is',Za,'ohm'
#Determination of impedance of transformer primary 
print 'The impedance of transformer primary is',Zp,'ohm'
The impedance of transformer secondary is 6.0 ohm
The impedance of transformer primary is 96.0 ohm

Example 4 : pg 182

In [3]:
 
#page no 182
#prob no. 5.4
#calculate the o/p power, supply power and dissipated power
#Class C amplifier with carrier o/p power of 100W with efficiency of 70% & with 100% modulation
#given
Pc=100;eta=0.7;
#calculations and results
#Determination of o/p power
Po=1.5*Pc;
print 'The o/p power with 100% modulation is',Po,'W'
#Determination of supply power
Ps=Po/eta;
print 'The value of supply power is',round(Ps,3),'W'
#Determination of power dissipated Pd
Pd=Ps-Po;
print 'Power dissipated is',round(Pd,3),'W'
The o/p power with 100% modulation is 150.0 W
The value of supply power is 214.286 W
Power dissipated is 64.286 W

Example 5 : pg 184

In [4]:
 
#page no 184
#prob no. 5.5
#calculate the load impedance
#given
#An FM transmitter produce 10W of carrier power operating at 15V
Vcc=15.;Pc=10.;
#calculations
#Determination of load impedance  seen from collector
Rl=((Vcc)**2)/(2*Pc);
#results
print 'The load impedance is',Rl,'ohm'
The load impedance is 11.25 ohm

Example 6 : pg 193

In [5]:
 
#page no 193
#prob no. 5.6
#Refer fig. 5.13
#calculate the frequency required
#given
#Filter method SSB generator 
fc=5.*10**6;#filter centre freq.
BW=3.*10**3;#Filter bandwidth
#calculations
foc=4.9985*10**6;#carrier oscillator freq.
foc2 = foc + BW
#results
print 'a)The USB will be passed'#Since carrier freq is at low end of passband
print 'b)The carrier freq should be moved to the high end of filter at',foc2/10**6,'MHz'#To generate the LSB
a)The USB will be passed
b)The carrier freq should be moved to the high end of filter at 5.0015 MHz

Example 7 : pg 196

In [7]:
 
#page no 196
#prob no. 5.7
#calculate the freq of local oscillator
#SSB transmitter refering fig.5.17 to transmit USB signal at carrier freq 21.5MHz
#given
fo=21.5;#carrier freq in MHz
foc=8.9985;#carrier oscillator freq. in MHz
#calculations
#Determination of freq of local oscillator
flo=fo-foc;
#results
print 'The freq of local oscillator',flo,'MHz'
The freq of local oscillator 12.5015 MHz

Example 8 : pg 199

In [8]:
 
#page no 199
#prob no. 5.8
#given
#LSB transmitter refering fig.5.14 with new carrier freq 9.0015 MHz & local oscillator freq 12.5015MHz
fco=9.0015;#carrier oscillator freq
flo=12.5015;#local oscillator freq
#calculations
#Determination of new o/p freq
fo=fco+flo;
#results
print 'The o/p carrier freq',fo,'MHz'
The o/p carrier freq 21.503 MHz

Example 9 : pg 204

In [9]:
 
#page no 204
#prob no. 5.9
#A direct FM transmitter with kf=2kHz/V & max deviatn of 300Hz.
#calculate the freq, rms voltage
from math import sqrt
#given
kf=2.*10**3;tx_dev=300.;
print 'a)See fig.5.23 for this block diagram'
f_mul=3*2*3;#3 stage freq multiplier with tripler doubler and tripler
#b)Determination of max dev at oscillator 
dev_o=5*10**3;#Deviation at o/p
dev_osc=dev_o/f_mul;
if  dev_osc < tx_dev:
    print 'b)Transmitter is capable of 5kHz deviation'
else:
    print 'b)Transmitter is not capable of 5kHz deviation'
#c)Determination of oscillator freq
fo=150.;#carrier freq in MHz
fosc=fo/f_mul;
print 'c)The oscillator freq is',round(fosc,3),'MHz'
#d)Determination of audio voltage for full deviation
Vi_peak=dev_osc/kf;#dev at oscillator of 278Hz causes full 5kHz deviation
#converting peak voltage to rms voltage
Vi_RMS=Vi_peak/sqrt(2);
print 'The audio RMS voltage is',round(Vi_RMS*10**3,3),'mV'
a)See fig.5.23 for this block diagram
b)Transmitter is capable of 5kHz deviation
c)The oscillator freq is 8.333 MHz
The audio RMS voltage is 97.934 mV

Example 10 : pg 206

In [10]:
 
#page no 206
#prob no. 5.10
#Refer fig.5.24
#calculate the freq deviation
#given
#Till the antenna there are 2 doubler and 4 tripler
f_mul=18.*18;
dev_o=75*10**3;#o/p freq deviation is 75kHz
#calculations
#Determiantion of reqd freq deviation of oscillator
dev_osc=dev_o/f_mul;
#results
print 'Freq deviation of oscillator is',round(dev_osc,3),'Hz'
Freq deviation of oscillator is 231.481 Hz

Example 11 : pg 207

In [11]:
 
#page no 207
#prob no. 5.11
#calculate the carrier freq and rms voltage
#given
from math import sqrt
#A PLL FM generator refering fig.5.25 with
f_ref=100.*10**3;N=200.;kf=50.*10**3;#in Hz/V
#calculations and results
#a)Determination of carrier freq of o/p signal
fc=N*f_ref;
print 'The carrier freq of o/p signal',fc/10**6,'MHz'
#b)Determination of RMS modulating voltage for 10kHz deviation
dev=10.*10**3;
Vp=dev/kf;
#Converting peak voltage to RMS voltage
V_RMS=Vp/sqrt(2);
print 'The RMS voltage for needed deviation is ',round(V_RMS*1000,2),'mV'
The carrier freq of o/p signal 20.0 MHz
The RMS voltage for needed deviation is  141.42 mV