RC Low and High Pass Filters
First Order Low Pass RC Filter
The diagram illustrates a First Order Low Pass RC Filter. As previously mentioned a low pass filter passes
low frequency AC signals (this refers to the pass band region of the filter) and attenuates high frequency
signals (in the stop band region). There is also a phase shift between the input and output signals in the
stop band area. The corner, cut-off or critical frequency of the filter fC provides a boundary between
the pass band and stop band.
The frequency response for the voltage gain is illustrated.
The filter is called a first order filter because of the presence of a single reactive component C. Higher
order filters have more than 1 reactive component (L or C). The order of a filter specifies the shape of
the pass band response, the roll-off rate and phase shift in the stop band.
Corner Frequency
The corner frequency of the filter is defined as the frequency where the voltage gain has dropped to
0.707 (linear value) or -3 dB (log value) of what the gain was at zero frequency.
1
The corner frequency is found as:
𝑓𝐶 =
First Order Low Pass RC Filter Analysis
1
𝐻𝑧 𝑜𝑟 𝑎𝑠
2𝜋𝑅𝐶
The capacitive reactance of C is expressed as XC and is given as
The voltage gain AV of the filter is derived using Voltage Divider.
𝐴𝑉
From 𝜔𝐶 =
1
𝑅𝐶
𝐴𝑉 =
𝜔𝐶 =
1
𝑅𝐶
𝑟𝑎𝑑/𝑠𝑒𝑐
1
𝑋𝐶 = −𝑗𝜔𝐶 = 𝑗𝜔𝐶
1
1
𝑉𝑂𝑈𝑇
𝑋𝐶
1
𝑗𝜔𝐶
𝑗𝜔𝐶
=
=
=
=
=
1
𝑗𝜔𝐶𝑅
+
1
1 + 𝑗𝜔𝐶𝑅
𝑉𝐼𝑁
𝑅 + 𝑋𝐶
𝑅 + 𝑗𝜔𝐶
𝑗𝜔𝐶
we can get 𝑅𝐶 =
1
1
𝜔𝐶
Substitute this into the expression for voltage gain
𝜔 𝑎𝑛𝑑 𝑡ℎ𝑖𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑙𝑠𝑜 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠
1 + 𝑗 �𝜔𝑐�
𝐴𝑉 =
1
1+𝑗�
𝑓
�
𝑓𝑐
This expression for voltage gain is a complex number and has a magnitude and phase angle that can be
found for its real and imaginary parts.
For simplicity use a substitution of A = f/fC and then rationalize the denominator of the resulting
expression to create real and imaginary parts of the complex number.
𝐴𝑉 =
1
1 − 𝑗𝐴
1 − 𝑗𝐴
1
𝐴
𝑥
=
=
−
𝑗
1 + 𝑗𝐴 1 − 𝑗𝐴
(1 − (𝑗 2 )𝐴2 )
1 + 𝐴2
1 + 𝐴2
This is now a complex number expressed in rectangular form as:
c = a + jb
The magnitude of the voltage gain is:
|AV|
= ��
1
2
𝐴
2
1
� + �1+𝐴2 � = �(1+𝐴2 )2 +
1+𝐴2
Substitute A = f/fC
Gain of the filter
(𝐴2 )
(1+𝐴2 )2
(1+𝐴2 )
1
= � (1+𝐴2)2 = � (1+𝐴2)
back into the equation. This gives an expression for the Linear Voltage
𝐴𝑉 = �
1
𝑓
(1 + ( )2 )
𝑓𝐶
2
=
1
2
�(1 + � 𝑓 � )
𝑓𝐶
This is the Linear Voltage Gain as a function of frequency
Logarithmic Voltage Gain
The voltage gain of the circuit expressed using dBs is found by taking the quantity 20 Log(Linear Voltage
gain).
𝐴𝑉 = 20 𝑙𝑜𝑔
1
�(1+� 𝑓 �
𝑓𝐶
2
)
The phase angle of the filter circuit is given as
Bode Plots
𝑓 2
= −10 log �1 + �𝑓𝑐� �
𝑓
𝜃 = − tan−1 ( ).
𝑓𝐶
Bode plots are specifically named graphs of Voltage Gain (Bode magnitude plot) vs frequency and Phase
Angle (Bode phase plot) vs frequency. These are shown below.
3
Calculations
The voltage gain (linear and log) values and phase angle are calculated at a variety of
frequencies that are multiples of f/fc using the equations for voltage gain and phase angle
derived above.
Frequency
Av (dB)
Phase Angle (°)
f/fc = 0.1
-0.04
-5.7
f/fc = 1.0
-3.03
-45
f/fc = 10
-20
-84
f/fc = 100
-40
-89
f/fc = 1000
-60
-89.9
The data from this table can be plotted in two graphs – one for voltage gain and one for phase angle.
The voltage gain of the filter at low frequency is essentially 0 dB (linear gain = 1). The phase angle for
the filter is 0 degrees at low frequency and asymptotically approaches -90 ° at high frequency. The
phase angle is -45 ° at the corner frequency. If plotted these graphs would be essentially the same as
those shown above.
Roll Off Rates in Stop Band
The slope of the frequency response graph in the Stop Band is called the Roll-Off Rate of the filter. For a
first order filter the Roll-Off rate is
-20dB/Decade
where a decade is a change in frequency of a factor of 10
or - 6dB/Octave
where an octave is a change in frequency of a factor of 2
Examples
1.
A first order RC filter has a voltage gain of -9.3 dB at 20 kHz. What is the voltage gain at 40 kHz
and 200 kHz?
A frequency of 40 kHz is one octave higher than 20 kHz. The voltage gain will then decrease by 6 dB to –
15.3 dB.
4
A frequency of 200 kHz is one decade higher than 20 kHz. The voltage gain will then decrease by 20 dB
to – 29.3 dB.
Higher Order Filters
The order of a filter is determined by the number of reactive components (L and C) in the filter. For
example a second order filter can be constructed using an Inductor and a Capacitor as shown. The
analysis and design of second and higher order filters is complex process that uses the concept of
families of filters ( Bessel, Butterworth and Chebyshev) with different response characteristics being
introduced.
Roll-Off Rates of Higher Order Filters
Each increase in order of a filter adds an additional 20 dB/decade (or 6 dB/octave) to the roll-off rate.
Filter Order
Roll-Off rate (dB/decade)
Roll-Off rate (dB/octave)
Second order
40 dB/decade
12 dB/octave
Third order
60 dB/decade
18 dB/octave
Forth order
80 dB/decade
24 dB/octave
These increasing roll-off rates are shown in the diagram.
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First Order High Pass RC Filter
The diagram illustrates a First Order High Pass RC Filter. As previously mentioned a high pass filter
passes high frequency AC signals (this refers to the pass band region of the filter) and attenuates low
frequency signals (in the stop band region). There is also a phase shift between the input and output
signals in the stop band area. The corner, cut-off or critical frequency of the filter fC again provides a
boundary between the pass band and stop band.
Corner Frequency
The corner frequency of the filter is defined as the frequency where the voltage gain has dropped to
0.707 (linear value) or -3 dB (log value) of what the gain was at high frequency.
The corner frequency is found as:
𝑓𝐶 =
First Order High Pass RC Filter Analysis
1
2𝜋𝑅𝐶
𝐻𝑧 𝑜𝑟 𝑎𝑠
The capacitive reactance of C is expressed as XC and is given as
The voltage gain AV of the filter is derived using Voltage Divider.
6
1
𝜔𝐶 = 𝑅𝐶 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑋𝐶 = −𝑗𝜔𝐶 =
1
𝑗𝜔𝐶
𝐴𝑉 =
From 𝜔𝐶 =
1
𝑅𝐶
arranging gives
𝑉𝑂𝑈𝑇
𝑅
𝑅
𝑅
𝑗𝜔𝐶𝑅
=
=
=
=
1
𝑗𝜔𝐶𝑅 + 1
𝑅 + 𝑋𝐶
1 + 𝑗𝜔𝐶𝑅
𝑉𝐼𝑁
𝑅 + 𝑗𝜔𝐶
𝑗𝜔𝐶
we can get 𝑅𝐶 =
1
𝐴𝑉 =
𝜔
1 + 𝑗 � 𝜔𝐶 �
1
𝜔𝐶
Substitute this into the expression for voltage gain and re-
𝑎𝑛𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑙𝑠𝑜 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠
𝐴𝑉 =
1
𝑓
1 + 𝑗 � 𝐶�
𝑓
This expression for voltage gain is a complex number and has a magnitude and phase angle that can be
found for its real and imaginary parts.
For simplicity use a substitution of A = fC/f and then rationalize the denominator of the resulting
expression to create real and imaginary parts of the complex number of the form c = a + jb
𝐴𝑉 =
1
1 − 𝑗𝐴
1 − 𝑗𝐴
1
𝐴
𝑥
=
=
−
𝑗
1 + 𝑗𝐴 1 − 𝑗𝐴
(1 − (𝑗 2 )𝐴2 )
1 + 𝐴2
1 + 𝐴2
This is now a complex number expressed in rectangular form
The magnitude of the voltage gain is:
|AV| = ��
1
2
𝐴
2
1
� + �1+𝐴2 � = �(1+𝐴2 )2 +
1+𝐴2
Substitute A = fC/f
Gain of the filter
(𝐴2 )
(1+𝐴2 )2
(1+𝐴2 )
1
= � (1+𝐴2)2 = � (1+𝐴2)2
back into the equation. This gives an expression for the Linear Voltage
𝐴𝑉 =
�
1
2
𝑓
�1 + � 𝑐 � �
𝑓
=
1
2
𝑓
�(1 + � 𝐶 � )
𝑓
This is the Linear Voltage Gain as a function of frequency. Notice the similarities and
differences this expression and the equivalent expression for the Low Pass filter.
Logarithmic Voltage Gain
The voltage gain of the circuit expressed using dBs is found by taking the quantity 20 Log(Linear Voltage
gain).
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𝐴𝑉 (𝑑𝐵) = 20 𝑙𝑜𝑔
1
� �1+�𝑓𝑐�2�
𝑓
The phase angle of the high pass filter circuit is given as
Bode Plots
𝑓
𝑓
𝜃 = tan−1( 𝑓𝐶 ).
The Bode Plots for the First Order High Pass are shown below.
8
2
= −10 log �1 + � 𝑓𝐶 � �
Calculations
The voltage gain (linear and log) values and phase angle are calculated at a variety of
frequencies that are multiples of fC/f using the equations for voltage gain and phase angle
derived above.
Frequency
Av (dB)
Phase Angle (°)
fC/f = 1000
-60
+89.9
fC/f = 100
-40
+89
fC/f = 10
-20
+84
fC/f = 1.0
-3.03
+45
fC/f = 0.1
-0.04
+5.7
Question: Does fC/f = 0.1 represent a low or high frequency?
The data from this table can be plotted in two graphs – one for voltage gain and one for phase angle.
The voltage gain of the filter at high frequency is essentially 0 dB (linear gain = 1). The phase angle for
the filter is+90 degrees at low frequency and asymptotically approaches 0 ° at high frequency. The
phase angle is +45 ° at the corner frequency. If plotted these graphs would be essentially the same as
those shown above.
Roll Off Rates in Stop Band
As with the low pass filter the slope of the frequency response graph in the Stop Band is called the RollOff Rate of the filter. For a first order filter the Roll-Off rate is
-20dB/Decade
where a decade is a change in frequency of a factor of 10
or - 6dB/Octave
where an octave is a change in frequency of a factor of 2
Application Note
Written by David Lloyd
Computer Engineering Program
Humber College
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