Chapter 6 : Receivers

Example 1 : pg 227

In [1]:
 
#page no 227
#prob no. 6.1
#calculate the bandwidth
#given
from math import sqrt
#A tuned ckt with broadast band (540 to 1700 kHz).Bw=10kHz at 540 kHz
BW1=10.*10**3;f1=540.*10**3;f2=1700.*10**3;#all in Hz
#calculations
#Determination of BW at 1700kHz
BW2=BW1*sqrt(f2/f1);
#results
print 'The Bandwidth at 1700kHz is ',round(BW2/1000,3),'kHz'
The Bandwidth at 1700kHz is  17.743 kHz

Example 4 : pg 236

In [2]:
 
#page no 236
#prob no. 6.4
#calculate the voltage value of the signal
#given
#A receiver with sensitivity 0.5uV & blocking dynamic range 70dB.
#Determination of vpltage signal V1 
P1_P2=70.;V2=0.5*10**-6;#let
#calculations
V1=V2*10**(P1_P2/20);
#results
print 'The voltage value of signal is',round(V1*1000,3),'mV'
The voltage value of signal is 1.581 mV

Example 5 : pg 238

In [3]:
 
#page no 238
#prob no. 6.5
#calculate the image freq and rejection
#Refer the fig 6.5
from math import sqrt, log10
#A receiver tuned to station at 590kHz 
#given
f_if=455.*10**3;#Intermediate freq
f_sig=590.*10**3;
#calculations and results
#a)Determintion of image freq
f_image=f_sig+2*f_if;
print 'a)The image freq is',f_image/1000,'kHz'
Q=40.;#Q_factor
#b)Determination of image rejection 
x=(f_image/f_sig)-(f_sig/f_image);
Asig_Aimage=sqrt(1+(Q*x)**2);#image rejection
#converting in dB
IR_dB=20*log10(Asig_Aimage);
print 'b)The image rejection is',round(IR_dB,3),'dB'
a)The image freq is 1500.0 kHz
b)The image rejection is 38.687 dB

Example 6 : pg 239

In [4]:
 
#page no 239
#prob no. 6.6
#calculate the frequencies in all cases
#given
import numpy
#An AM high-freq receiver with IF=1.8MHz tuned at freq 10MHz
f_sig = 10
f_if = 1.8#All freq in MHz
#calculations
#Determination of local oscillator freq f_lo
f_lo = f_sig + f_if
#determination of freq.  that cause IF response
m = ([1, 1, 2, 2])#values of m that are integer
n = ([1, 2, 1, 2])#values of n that are integer
fs1 = fs2 = numpy.zeros(4);
for i in range(0,3):
    fs1[i] = ((m[i] / n[i]) * (f_lo)) + ((f_if) / n[i])

for i in range(0,3):
    fs2[i] = ((m[i] / n[i]) * (f_lo)) - ((f_if) / n[i])

print 'All freqs are in MHz.','The different freqs are',fs1,fs2
All freqs are in MHz. The different freqs are [ 10.   -0.9  21.8   0. ] [ 10.   -0.9  21.8   0. ]

Example 7 : pg 245

In [5]:
 
#page no 245
#prob no. 6.7
#calculate the sensitivity of detector
#An FM detector produce Vpp=1.2V with dev=10kHz
#given
Vpp=1.2;dev=10.*10**3;
#calculations
#Determination of detector sensitivity
Vp=Vpp/2;#Peak voltage
kd=Vp/dev;
#results
print 'the sensitivity of detector is',kd*10**6,'uV/Hz'
the sensitivity of detector is 60.0 uV/Hz

Example 8 : pg 249

In [6]:
 
#page no 249
#prob no. 6.8
#calculate the rms voltage
from math import sqrt
#given
#A PLL FM detector with kf=100kHz/V & dev=75kHz
kf=100.*10**3;dev=75.*10**3;
#calculations
#Determination of RMS voltage
Vp_op=dev/kf;
#Converting peak voltage in RMS voltage
V_RMS=Vp_op/sqrt(2);
#results
print 'The RMS voltage is',round(V_RMS,3),'V'
The RMS voltage is 0.53 V

Example 9 : pg 258

In [9]:
 
#page no 258
#prob no. 6.9
#calculate the critical and optimum coupling factor, BW
from math import sqrt
#given
#An IF transformer at 455kHz & primary ckt has Qp=40 & secondary Q=30
fo=455.*10**3;Qp=40.;Qs=30.;
#calculations and results
#a)Determination of critical coupling factor
kc=1/sqrt(Qp*Qs);
print 'a)The critical coupling factor is',round(kc,3)
#b)Determination of optimum coupling factor
Kopt=1.5*kc;
print 'b)The optimum coupling factor is',round(Kopt,3)
#c)Determination of optimum coupling factor
B=Kopt*fo;
print 'c)The BW using optimum coupling factor is',round(B/1000,3),'kHz'
a)The critical coupling factor is 0.029
b)The optimum coupling factor is 0.043
c)The BW using optimum coupling factor is 19.702 kHz

Example 10 : pg 261

In [10]:
 
#page no 261
#prob no. 6.10
#calculate the first and second IF
#given
#Receiver refering in fig.6.28
f_sig=25*10**6;#signal i/p freq
f_lo1=29.5*10**6;#Ist local oscillator freq
#calculations and results
#determination of Ist IF which uses high side injection
f_IF1=f_lo1-f_sig;#high side injection
print 'The first IF is',f_IF1/10**6,'MHz'
#Determination of IInd IF which uses low side injection
f_lo2=4*10**6;#IInd local oscillator freq
f_IF2=f_IF1-f_lo2;
print 'The second IF is',f_IF2/10**3,'kHz'
The first IF is 4.5 MHz
The second IF is 500.0 kHz

Example 11 : pg 265

In [11]:
 
#page no 265
#prob no. 6.11
#calculate the signal strength at receiver i/p
#An S-meter is given
#given
V1=50*10**-6;#signal strength at transmitter in V
P=18.;#18 dB power
#calculations
V2=V1/(10**(P/20.));
#results
print 'Signal strength at receiver i/p is' ,round(V2*10**6,3),'uV'
Signal strength at receiver i/p is 6.295 uV