Chapter 3 - Radio Transmitters

Example 1 - pg 132

In [2]:
#calculate the total radiated power
#given
Pm = 85.#percentage modulation
m = .85#modulation index
P_c = 50*10**3#carrier power in watts

#calculation
P_t = P_c*(1+(m**2/2));#total radiated power 
 
#result
print "Total radiated power =",round(P_t/1000.,2),"kW"

Total radiated power = 68.06 kW

Example 2 - pg 138

In [4]:
#calculate the carrier frequencies
#given from the figure
f = 20#frequency in hertz
#At point 1 from fig
f_c1 = 2*13.5#carrier frequency
deltaf1 = 2*8.5*10**-3#change in frequency

#calculations
f_max1 = f_c1 + deltaf1#maximum frequency at point 1 in fig
f_min1 = f_c1 - deltaf1#maximum frequency at point 1 in fig
f_d1 = f_max1 - f_c1;#frequency deviation at point 1 in fig
f_d2 = f_c1 - f_min1;#frequency deviation at point 1 in fig
#At point 2 from fig
f_c2 = 3*f_c1#carrier frequency
deltaf2 = 3*deltaf1#change in frequency
f_max2 = f_c2 + deltaf2#maximum frequency at point 2 in fig
f_min2 = f_c2 - deltaf2#minimum frequency at point 2 in fig
f_d3 = f_max2 - f_c2;#frequency deviation at point 2 in fig
f_d4 = f_c2 - f_min2;#frequency deviation at point 2 in fig
#At point 3 in fig
f_c3 = f_c2 + f;#carrier frequency at point 3 in fig
f_max3 = f_max2 + f#maximum frequency at point 3 in fig
f_min3 = f_min2+ f#minimum frequency at point 3 in fig
f_d5 = f_max3 - f_c3;#frequency deviation at the last point
f_d6 = f_c3 - f_min3 ;#frequency deviation at the last point

#results
print "i.a Carrier frequency at point 1 in fig = ",f_c1,"MHz"
print "  b Frequency deviation (kHz) =  ",f_d1*1000.
print "  c Frequency deviation  (kHz) = ",f_d2*1000.
print "ii.a Carrier frequency at point 2 in fig (MHz) = ",f_c2
print "    b Frequency deviation (kHz) = ",f_d3*1000.
print "    c Frequency deviation (kHz) =  ",f_d4*1000.
print "ii.a Carrier frequency at point 3 in fig (MHz) = ",f_c3
print "    b Frequency deviation (kHz) = ",f_d5*1000.
print "    c Frequency deviation  (kHz) = ",f_d6*1000.
print "Thus, in mixer, frequency deviation is not altered but only carrier frequency is increased"

i.a Carrier frequency at point 1 in fig =  27.0 MHz
  b Frequency deviation (kHz) =   17.0
  c Frequency deviation  (kHz) =  17.0
ii.a Carrier frequency at point 2 in fig (MHz) =  81.0
    b Frequency deviation (kHz) =  51.0
    c Frequency deviation (kHz) =   51.0
ii.a Carrier frequency at point 3 in fig (MHz) =  101.0
    b Frequency deviation (kHz) =  51.0
    c Frequency deviation  (kHz) =  51.0
Thus, in mixer, frequency deviation is not altered but only carrier frequency is increased

Example 3 - pg 138

In [6]:
#calculate the appropriate multiplier values
print "\tInput frequency deviation is 10Khz, while the output frequency deviation \nrequired is 60KHz. Thus, a frequency multiplication of 6*3*2 is required."
print "\tThe frequency multiplication of 6 will give the carrier frequency of\n9*6 = 54MHz only. Hence we have to use heterodyning. The two inputs to the \nmixer are the carrier frequency pf 54MHz and oscillator frequency. Assuming \nthat at the output of the mixer addition of imput frequencies is selected the \nrequired oscillator frequency, to have the final carrier output frequency of \n106MHz, comes out to be 52MHz.[52+54 =106]. "

	Input frequency deviation is 10Khz, while the output frequency deviation 
required is 60KHz. Thus, a frequency multiplication of 6*3*2 is required.
	The frequency multiplication of 6 will give the carrier frequency of
9*6 = 54MHz only. Hence we have to use heterodyning. The two inputs to the 
mixer are the carrier frequency pf 54MHz and oscillator frequency. Assuming 
that at the output of the mixer addition of imput frequencies is selected the 
required oscillator frequency, to have the final carrier output frequency of 
106MHz, comes out to be 52MHz.[52+54 =106].