Chapter 2 : Radio Frequency Circuits

Example 1 : pg 50

In [1]:
#page no 50
#calculate the resonant frequency and bandwidth
#prob no 2.1
from math import pi,sqrt
#Given:
#Refer the fig 2.6 of page 50. L1=25uH;C1=50pF
L1=25.*10**-6;C1=50.*10**-12;Q=15;
#calculations
#A) The resonent freqency is given as
fo=(1/(2*pi*sqrt(L1*C1)));
#B) The bandwidth is given as 
B=fo/Q;
#results
print 'a)The resonant frequency is ',round(fo/10**6,2),'MHz'
print 'The bandwidth is ',round(B/1000.),'kHz'
a)The resonant frequency is  4.5 MHz
The bandwidth is  300.0 kHz

Example 2 : pg 62

In [2]:
#page no 62
# prob no. 2.2
#calculate the operating frequency, gain
from math import pi,sqrt
# Given : Hartley oscillators L=10uH; C=100pF
L=10.*10**-6; C=100.*10**-12;N1=10;N2=100.
#calculations and results
# A)The operating frequency is 
fo=1/(2.*pi*sqrt(L*C));
print '1)The operating frequency is',round(fo/10**6,2),'MHz'
# The feedback fraction is given by
B=-N1/N2;
#Operating gain is given as 
A=1/B;
print '2)Operating gain',A
print 'The -ve sign denotes a phase inversion'
#B) The operating frequency is same as in part A)
N1=20;N2=80.;
# The feedback fraction is given by
B=(N1+N2)/N1;
#Operating gain is given as 
A=1./B;
print '3)Operating gain',A
1)The operating frequency is 5.03 MHz
2)Operating gain -10.0
The -ve sign denotes a phase inversion
3)Operating gain 0.2

Example 3 : pg 66

In [3]:
# page no 66
#prob no 2.3
#calculate the operating frequency and feedback fraction
from math import pi,sqrt
#Given:
C1=10.*10**-12; C2=100.*10**-12; L=1*10**-6;
#calculations and results
# The effective capacitance is 
CT=(C1*C2)/(C1+C2);
# The operating frequency is 
f0=1/(2*pi*sqrt(L*CT));
print 'The effective capacitance (pF)=',round(CT*10**12,2);
print '1)The operating frequency is',round(f0/10**6,1),'Hz'
# The feedback fraction is given approximately by
B=-C1/C2; 
print 'The feedback fraction is',B
# For the common-base ckt, the op-freq is same but the feedback fraction willbe different.
C1=100*10**-12; C2=10*10**-12;
# It is given by 
B=C2/(C1+C2);
print 'The feedback fraction is',round(B,4)
The effective capacitance (pF)= 9.09
1)The operating frequency is 52.8 Hz
The feedback fraction is -0.1
The feedback fraction is 0.0909

Example 4 : pg 68

In [4]:
# page no 68
#prob no 2.4
#caclculate the effective total capacitance and operating frequency
#Refer fig 2.22
from math import pi,sqrt
#Given:
c1=1000.;c2=100.;c3=10.;#all values are in pf
#calculations and results
#The effective total capacitance
Ct=1/((1/c1)+(1/c2)+(1/c3));
print 'The effective total capacitance is',round(Ct,2),'pF'
CT=Ct*10**-12;L=10**-6;
#The operating freq is
f0=1/(2*pi*sqrt(L*CT));
print 'The operating freq is',round(f0/10**6,2),'MHz'
The effective total capacitance is 9.01 pF
The operating freq is 53.03 MHz

Example 5 : pg 70

In [5]:
# page no 70
#prob no 2.5
#calculate the resonant freq and tuning voltage
from math import sqrt,pi
#Given:
C=80*10**-12; L= 100*10**-6;
#calculations
#Part a) The resonent frequency is 
f0=1/(2*pi*sqrt(L*C));
# Part b) In this part the circuit is resonate on doubling the frequency,therefore
f1=2*f0;
# from the equation of resonent frequency 
C1=1/(4*(pi*f1)**2*L);
# Now for tuning voltage  we have to use equation C1=Co/sqrt(1+2V)
Co=C;
# after solving the expression
v=((Co/C1)**2 -1)/2;
#results
print 'The resonent freq is',round(f0/10**6,2),'MHz'
print 'The tuning voltage is ',v,'V'
The resonent freq is 1.78 MHz
The tuning voltage is  7.5 V

Example 7 : pg 76

In [7]:
#page no 76
#calculate the output frequencies
#problem 2.7
#Given:
# all frequencies are in MHz
f1=11.;f2=10;
#calculations
# output frequencies at the output of square-law mixer
a=f1+f2;
b=f1-f2;
#results
print 'The output frequencies at the output of square-law mixer are : ',a,'MHz ',b,'MHz'
The output frequencies at the output of square-law mixer are :  21.0 MHz  1.0 MHz

Example 8 : pg 85

In [9]:
#page no 85
#calculate the capture range, lock range, frequency
#problem no. 2.8
# all the frequencies are in MHz
#Given:
freq_free_run =12.; 
freq_lock1 =10.;
freq_lock2 =16.;
#calculations and results
# capture range is approximately twice the difference between the free-running freq and the freq at which lock is first achieved
capture_range =2*(freq_free_run - freq_lock1 );
print 'The capture range is ',capture_range,'MHz'
# lock range is approximately twice the the difference between the freq where lock is lost and free-running freq
lock_range = 2*(freq_lock2 - freq_free_run);
print 'The lock range is ',lock_range,'MHz'
# The PLL freq response id approximate symmetrical.
#This means the free-running freq is in the center of the lock range and capture range. Therefore
freq_lock_acquired = freq_free_run + (capture_range/2);
freq_lock_lost = freq_free_run - capture_range
print 'The freq at which the lock is acquired, moving downward in freq is ',freq_lock_acquired,'MHz'
print 'Lock will be lost on the way down at',freq_lock_lost,'MHz'
The capture range is  4.0 MHz
The lock range is  8.0 MHz
The freq at which the lock is acquired, moving downward in freq is  14.0 MHz
Lock will be lost on the way down at 8.0 MHz

Example 9 : pg 86

In [10]:
#page no 86
#calculate the values of N at high and low ends
# prob no 2.9
# refer fig 2.38
#Given:
#Here we are using a 10MHz crystal, it will be necessar to devide it by a factor to get 10kHz
f_osc = 10.*10**6; f_ref=10.*10**3;f0_1=540.*10**3;f0_2=1700.*10**3;
#calculations
Q=f_osc/f_ref;
# we have to specify the range of values of N. Find N at each and of the tuning range
N1=f0_1/f_ref;
N2=f0_2/f_ref;
#results
print 'The values of N at high end is',N2
print 'The values of N at low end is',N1
The values of N at high end is 170.0
The values of N at low end is 54.0

Example 10 : pg 89

In [12]:
#page no 89
# prob no 2.10
# refer fig 2.40
#Given:
P=10.;f_ref=10.*10**3;M=10.;
#consider 
N=1.;
#calculations
# With a fixed-modulus prescalar, the min freq step is 
step_size=M*f_ref;
# With the two-modulus system, let the main divider modulus N remain constant & 
#increase the modulus m to (m+1) to find how much the freq changes.
# for 1st case, o/p freq 
fo=(M+N*P)*f_ref;
# for 2nd case where leave N alone but changes M to M+1, new o/p freq 
fo_=(M+1+N*P)*f_ref;
# The difference is 
f= fo_-fo;
#results
print 'The step size that would have been obtained without prescaling',f,'Hz'
The step size that would have been obtained without prescaling 10000.0 Hz

Example 11 : pg 91

In [13]:
#page no 91
#prob no 2.11
#refer fig 2.42
#Given:
f_ref= 20.*10**3;
f_osc= 10.*10**6;
N1=10;N2=100.;
#calculations
f0=(N1*f_ref) + f_osc;
f1=(N2*f_ref) + f_osc;
step_size=(f1-f0)/(N2-N1);
#results
print 'The output frequencies are',f0/10**6,'MHz',f1/10**6,'MHz'
print 'The step size is',step_size/1000.,'kHz'
The output frequencies are 10.2 MHz 12.0 MHz
The step size is 20.0 kHz