Chapter 26 : Filters

Example No. 26_1 Page No. 824

In [2]:
from math import pi,sqrt,atan
# Calculate (a)the cutoff frequency fc# (b)Vout at fc# (c)Theta at fc (Assume Vin = 10 Vpp for all frequencies)

# Given data

R = 10.*10**3#        # Resistor=10 kOhms
C = 0.01*10**-6#     # Capacitor=0.01 uFarad
Vin = 10.#           # Input Voltage=10Vpp
# To calculate fc

fc = 1./(2*pi*R*C)#
print 'The Cutoff Frequency = %0.2f Hertz'%fc
print 'i.e 1.592 kHz'

# To calculate Vout at fc

Xc = 1./(2*pi*fc*C)#

Zt = sqrt((R*R)+(Xc*Xc))#

Vout = Vin*(Xc/Zt)#
print 'The Output Voltage = %0.2f Vpp'%Vout

# To calculate Theta

Theta = atan(-(R/Xc))*180/pi
print 'The Phase angle (Theta z) = %0.2f Degree'%Theta

The Cutoff Frequency = 1591.55 Hertz
i.e 1.592 kHz
The Output Voltage = 7.07 Vpp
The Phase angle (Theta z) = -45.00 Degree

Example No. 26_2 Page No. 825

In [3]:
from math import pi,sqrt,atan
# Calculate (a)the cutoff frequency fc# (b)Vout at 1 kHz# (c)Theta at 1 kHz (Assume Vin = 10 Vpp for all frequencies)

# Given data

R = 1.*10**3#         # Resistor=1 kOhms
L = 50.*10**-3        # Inductor=50 mHenry
Vin = 10.#           # Input Voltage=10Vpp
f = 1.*10**3#         # Frequency=1 kHz
# To calculate fc

fc = R/(2*pi*L)#
print 'The Cutoff Frequency = %0.2f Hertz'%fc,
print 'i.e 3.183 kHz'

# To calculate Vout at fc

Xl = 2*pi*f*L#

Zt = sqrt((R*R)+(Xl*Xl))#

Vout = Vin*(R/Zt)#
print 'The Output Voltage = %0.2f Vpp'%Vout
print 'approx 9.52 Volts(p-p)'

# To calculate Theta

Theta = atan(-(Xl/R))*180/pi
print 'The Phase angle (Theta z) = %0.2f Degree'%Theta

The Cutoff Frequency = 3183.10 Hertz i.e 3.183 kHz
The Output Voltage = 9.54 Vpp
approx 9.52 Volts(p-p)
The Phase angle (Theta z) = -17.44 Degree

Example No. 26_3 Page No. 826

In [4]:
from math import pi,sqrt
# Calculate the cutoff frequency for (a) the RC high-pass filter# (b) the RL high-pass filter

# Given data

R = 1.5*10**3#       # Resistor=1.5 kOhms
L = 100.*10**-3       # Inductor=100 mHenry
C = 0.01*10**-6#     # Capacitor=0.01 uFarad

# To calculate fc for RC high-pass filter

fc = 1./(2*pi*R*C)#
print 'The Cutoff Frequency for RC High-Pass Filter = %0.2f Hertz'%fc
print 'i.e 10.61 kHz'

# To calculate fc for RL high-pass filter

fc1 = R/(2*pi*L)#
print 'The Cutoff Frequency for RL High-Pass Filter = %0.2f Hertz'%fc1
print 'approx 2.39 kHz'

The Cutoff Frequency for RC High-Pass Filter = 10610.33 Hertz
i.e 10.61 kHz
The Cutoff Frequency for RL High-Pass Filter = 2387.32 Hertz
approx 2.39 kHz

Example No. 26_4 Page No. 827

In [5]:
from math import pi
# Calculate the cutoff frequencies fc1 and fc2.

#Given data

R1 = 1.*10**3#        # Resistor 1=1 kOhms
C1 = 1.*10**-6#       # Capacitor 1=1 uFarad
R2 = 100.*10**3#      # Resistor 2=100 kOhms
C2 = 0.001*10**-6#   # Capacitor 2=0.001 uFarad

# To calculate fc1 for RC high-pass filter

fc1 = 1/(2*pi*R1*C1)#
print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc1
print 'i.e 159 Hz'

# To calculate fc2 for RC high-pass filter

fc2 = 1/(2*pi*R2*C2)#
print 'The Cutoff Frequency for RC High-Pass filter = %0.2f Hertz'%fc2
print 'i.e 1.59 kHz'

The Cutoff Frequency for RC High-Pass filter = 159.15 Hertz
i.e 159 Hz
The Cutoff Frequency for RC High-Pass filter = 1591.55 Hertz
i.e 1.59 kHz

Example No. 26_5 Page No. 828

In [8]:
from math import pi
# Calculate the notch frequency fn if R1 is 1 kOhms and C1 is 0.01 uF. Also, calculate the required values for 2R1 and 2C1 in the low-pass filter.

# Given data

R1 = 1.*10**3#        # Resistor 1=1 kOhms
C1 = 0.01*10**-6#    # Capacitor 1=0.01 uFarad

# To calculate Notch frequency fn for RC low-pass filter

fn = 1/(4*pi*R1*C1)#
print 'The Notch Frequency for RC Low-Pass filter = %0.2f Hertz'%fn
print 'i.e 7.96 kHz'

A = 2*R1#
print 'The Required Value of 2R1 = %0.2f Ohms'%A
print 'i.e 2 kohms'

B = 2*C1#
print 'The Required Value of 2C1 = %0.2e Ohms'%B
print '0.02 uF'

The Notch Frequency for RC Low-Pass filter = 7957.75 Hertz
i.e 7.96 kHz
The Required Value of 2R1 = 2000.00 Ohms
i.e 2 kohms
The Required Value of 2C1 = 2.00e-08 Ohms
0.02 uF

Example No. 26_6 Page No. 829

In [9]:
from math import log10
# A certain amplifier has an input power of 1 W and an output power of 100 W.Calculate the dB power gain of the amplifier.

# Given data

Pi = 1.#     # Input power=1 Watts
Po = 100.#   # Output power=100 Watts

N = 10*log10(Po/Pi)#
print 'The Power Gain of Amplifier = %0.2f dB'%N

The Power Gain of Amplifier = 20.00 dB

Example No. 26_7 Page No. 830

In [10]:
from math import log10
# The input power to a filter is 100 mW, and the output power is 5 mW. Calculate the attenuation, in decibels, offered by the filter.

# Given data

Pi = 100.*10**-3#    # Input power=1 Watts
Po = 5.*10**-3#      # Output power=100 Watts

N = 10*log10(Po/Pi)#
print 'The Attenuation offered by the Filter = %0.2f dB'%N

The Attenuation offered by the Filter = -13.01 dB

Example No. 26_8 Page No. 832

In [13]:
from math import pi,log10,sqrt
# Calculate the attenuation, in decibels, at the following frequencies: (a) 0 Hz# (b) 1.592 kHz# (c) 15.92 kHz. (Assume that Vin is 10 V p-p at all frequencies.)

# Given data

f1 = 0#                 # Frequency 1=0 Hz
f2 = 1.592*10**3#        # Frequency 2=1.592 kHz (cutoff frequency)
f3 = 15.92*10**3#        # Frequency 3=15.92 kHz
Vi = 10.#                # Voltage input=10 Volts(p-p)
R = 10.*10**3#            # Resistor 1=10 kOhms
C = 0.01*10**-6#         # Capacitor 1=0.01 uFarad 

Vo1 = Vi#
Vo2 = 0.707*Vi#

# At 0 Hz

N1 = 20*log10(Vo1/Vi)#
print 'The Attenuation at 0 Hz = %0.f dB'%N1

#At 1.592 kHz (cutoff frequency)

N2 = 20*log10(Vo2/Vi)#
print 'The Attenuation at 1.592 kHz = %0.2f dB'%N2

# At 15.92 kHz

Xc = 1./(2*pi*f3*C)#

A = R*R#
B = Xc*Xc#

Zt = sqrt (A+B)#

N3 = 20*log10(Xc/Zt)#
print 'The Attenuation at 15.92 kHz = %0.2f dB'%N3

The Attenuation at 0 Hz = 0 dB
The Attenuation at 1.592 kHz = -3.01 dB
The Attenuation at 15.92 kHz = -20.05 dB