7. Lecture 07 - Active Filter Circuits.pdf
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About This Presentation
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Slide Content
Lecture # 7Active Filter Circuits
BASS TREBLEHigh-pass
Low-pass
Power
amplifiers
Treble
Bass vin
+
-
R1A
R2A
Rf1
Ra
+
-
R2BR1B
Rf2
Rb
C1B
C2B
C2A
C1A
Treble
Bass
Two-pole low-
pass circuit
Two-pole high-
pass circuit
A
B
Mai Linh, PhD
Email: [email protected] ; [email protected]
BASS TREBLEHigh-pass
Low-pass
Power
amplifiers
Treble
Bass vin
+
-
R1A
R2A
Rf1
Ra
+
-
R2BR1B
Rf2
Rb
C1B
C2B
C2A
C1A
Treble
Bass
Two-pole low-
pass circuit
Two-pole high-
pass circuit
A
B
Mai Linh, PhD
Email: [email protected] ; [email protected]
2
Introduction: Categories
Filters
Digital Filters Analog Filters
Analog Continuous
Filters
Analog sampled
data Filters
Passive Filters
(RC & RLC)
Active Filters
(R + C + Op-Amp)
Switched
Capacitor Filters
Switched
Current Filters
Mai Linh, PhD
Introduction: Categories
Filters
Digital Filters Analog Filters
Analog Continuous
Filters
Analog sampled
data Filters
Passive Filters
(RC & RLC)
Active Filters
(R + C + Op-Amp)
Switched
Capacitor Filters
Switched
Current Filters
Mai Linh, PhD
3
Introduction: Definitions & Applications
A filter is a device that passes electric signals at certain frequencies or frequency ranges while
preventing the passage of others.
Filter Circuits Applications:
Filters are basic building blocks in all Communication Systems such as RF, IF and LF filters
❖Low Frequency band-pass filters are used in the audio frequency range (0 kHz to 20 kHz) for
modems and speech processing.
❖High-frequency band-pass filters (several hundred MHz) are used for channel selection in
telephone central offices.
❖Data acquisition systems usually require anti-aliasing low-pass filters as well as low-pass noise
filters in their preceding signal conditioning stages.
❖System power supplies often use band-rejection filters to suppress the 60 (or 50)-Hz line
frequency and high frequency transients.
❖In addition, there are filters that do not filter any frequencies of a complex input signal, but just add
a linear phase shift to each frequency component, thus contributing to a constant time delay. These
are called all-pass filters.
Mai Linh, PhD
Introduction: Definitions & Applications
A filter is a device that passes electric signals at certain frequencies or frequency ranges while
preventing the passage of others.
Filter Circuits Applications:
Filters are basic building blocks in all Communication Systems such as RF, IF and LF filters
❖Low Frequency band-pass filters are used in the audio frequency range (0 kHz to 20 kHz) for
modems and speech processing.
❖High-frequency band-pass filters (several hundred MHz) are used for channel selection in
telephone central offices.
❖Data acquisition systems usually require anti-aliasing low-pass filters as well as low-pass noise
filters in their preceding signal conditioning stages.
❖System power supplies often use band-rejection filters to suppress the 60 (or 50)-Hz line
frequency and high frequency transients.
❖In addition, there are filters that do not filter any frequencies of a complex input signal, but just add
a linear phase shift to each frequency component, thus contributing to a constant time delay. These
are called all-pass filters.
Mai Linh, PhD
Introduction
❖Filter circuits consist of R, L, C
❖Inductors are usually large, heavy and costly,
and may introduce electromagnetic field effects
→ disadvantage
❖Filter circuits consist of op-amps
❖Filter circuits without using inductor
❖Maximum magnitude does not exceed 1
❖Cut-off frequency and passband magnitude were
altered with the addition of a resistive load at the
output of the filter.
❖Filter circuits provide a control over amplification.
❖Does not affected → use active filter circuits to
implement filter designs when gain, load variation
and physical size are important parameters in the
design specifications.
Passive filter Active filter
4Mai Linh, PhD
❖Filter circuits consist of R, L, C
❖Inductors are usually large, heavy and costly,
and may introduce electromagnetic field effects
→ disadvantage
❖Filter circuits consist of op-amps
❖Filter circuits without using inductor
❖Maximum magnitude does not exceed 1
❖Cut-off frequency and passband magnitude were
altered with the addition of a resistive load at the
output of the filter.
❖Filter circuits provide a control over amplification.
❖Does not affected → use active filter circuits to
implement filter designs when gain, load variation
and physical size are important parameters in the
design specifications.
Passive filter Active filter
4Mai Linh, PhD
5
Introduction:
Figure: Ideal transmission characteristics of the four major filter types: (a) low-pass (LP), (b)
high-pass (HP), (c) bandpass (BP), and (d) bandstop (BS).
Ideal Filter Response Characteristics |H| |H|
|H| |H|
Mai Linh, PhD
Introduction:
Figure: Ideal transmission characteristics of the four major filter types: (a) low-pass (LP), (b)
high-pass (HP), (c) bandpass (BP), and (d) bandstop (BS).
Ideal Filter Response Characteristics |H| |H|
|H| |H|
Mai Linh, PhD
6
Introduction: Filter Specification
•Passband edge:
p Stopband edge :
s
•Maximum allowed variation in passband transmission : A
max
•Minimum required stopband attenuation : A
min
• The first step of filter design is to determine the filter specifications.
• Then find a transfer function H(s) whose magnitude meets the specifications.
• The process of obtaining a transfer function that meets given specifications is called filter approximation.
Fig.: Spec. of the transmission characteristics of a LPF Fig.: Transmission specifications for a BPF |H|, |H|,
Mai Linh, PhD
Introduction: Filter Specification
•Passband edge:
p Stopband edge :
s
•Maximum allowed variation in passband transmission : A
max
•Minimum required stopband attenuation : A
min
• The first step of filter design is to determine the filter specifications.
• Then find a transfer function H(s) whose magnitude meets the specifications.
• The process of obtaining a transfer function that meets given specifications is called filter approximation.
Fig.: Spec. of the transmission characteristics of a LPF Fig.: Transmission specifications for a BPF |H|, |H|,
Mai Linh, PhD
7
Introduction
Filter Transfer Function
✓ A filter is a linear 2-port network represented by the ratio of the output to input voltage.
✓ Transfer function H(s) = V
o(s) / V
i(s).
✓ Transmission: evaluating H(s) for physical frequency s = j → H(j) = |H(j)|.e
j()
.
Gain function: G() 20log|H(j)| (dB)
Attenuation function: A() - 20log|H(j)| (dB)
✓ Output frequency spectrum : |V
o(s)| = |H(s)||V
i(s)|.
Figure: The filters studied in this chapter are linear circuits represented by the general
two-port network shown. The filter transfer function H(s) ≡ V
o(s) ⁄ V
i(s).
Mai Linh, PhD
Introduction
Filter Transfer Function
✓ A filter is a linear 2-port network represented by the ratio of the output to input voltage.
✓ Transfer function H(s) = V
o(s) / V
i(s).
✓ Transmission: evaluating H(s) for physical frequency s = j → H(j) = |H(j)|.e
j()
.
Gain function: G() 20log|H(j)| (dB)
Attenuation function: A() - 20log|H(j)| (dB)
✓ Output frequency spectrum : |V
o(s)| = |H(s)||V
i(s)|.
Figure: The filters studied in this chapter are linear circuits represented by the general
two-port network shown. The filter transfer function H(s) ≡ V
o(s) ⁄ V
i(s).
Mai Linh, PhD
8
Introduction: General Frequency Response of a Circuit
Generally, a circuit’s transfer function (frequency dependent gain expression) can be
written as the ratio of polynomials:()( )( )
( )( )( )
1 2 3
1 2 3
11...
111...
z z zout
in p p p
sssv
A
v sss
++
=
+++ ()() ()
() () ()
22
1 2 3
2 2 2
1 2 3
1 1...
1 1 1...
z z z
out
in
p p p
v
A
v
−−
=
− − −
s = j
Complex Roots of the numerator polynomial are called “zeros” while complex roots
of the denominator polynomial are called “poles”
➢Each zero causes the transfer function to “break to higher gain” (slope increases by
20 dB/decade)
➢Each pole causes the transfer function to “break to lower gain” (slope decreases by
20 dB/decade)
Mai Linh, PhD
Introduction: General Frequency Response of a Circuit
Generally, a circuit’s transfer function (frequency dependent gain expression) can be
written as the ratio of polynomials:()( )( )
( )( )( )
1 2 3
1 2 3
11...
111...
z z zout
in p p p
sssv
A
v sss
++
=
+++ ()() ()
() () ()
22
1 2 3
2 2 2
1 2 3
1 1...
1 1 1...
z z z
out
in
p p p
v
A
v
−−
=
− − −
s = j
Complex Roots of the numerator polynomial are called “zeros” while complex roots
of the denominator polynomial are called “poles”
➢Each zero causes the transfer function to “break to higher gain” (slope increases by
20 dB/decade)
➢Each pole causes the transfer function to “break to lower gain” (slope decreases by
20 dB/decade)
Mai Linh, PhD
9
Introduction: General Frequency Response of a Circuit20log
out
in
v
v
0 dB/Decade
0 dB/Decade2
1
z
= 1
1
p
= 2
1
p
= 3
1
p
= 3
1
z
=
Typically, = RC
Mai Linh, PhD
Introduction: General Frequency Response of a Circuit20log
out
in
v
v
0 dB/Decade
0 dB/Decade2
1
z
= 1
1
p
= 2
1
p
= 3
1
p
= 3
1
z
=
Typically, = RC
Mai Linh, PhD
10
Assume the open loop gain, A
Open Loop, of the op
amp is constant at all frequencies.
Real Op amps have a frequency dependant
open loop gain()
OB T
OpenLoop
BB
A
As
ss
==
++
s = j
A
O = Open loop gain at DC
B = Open loop bandwidth
c Unity - gain frequency (frequency
where |A(s) = 1|
Mai Linh, PhD
Assume the open loop gain, A
Open Loop, of the op
amp is constant at all frequencies.
Real Op amps have a frequency dependant
open loop gain()
OB T
OpenLoop
BB
A
As
ss
==
++
s = j
A
O = Open loop gain at DC
B = Open loop bandwidth
c Unity - gain frequency (frequency
where |A(s) = 1|
Mai Linh, PhD
11
Introduction:FILTER TRANSFER FUNCTION
❖The filter TF is expressed as the ratio of 2 polynomials:1
1
1
1
...
()
...
mm
m m o
nn
n n o
a s a s a
Hs
a s a s b
−
−
−
−
+ + +
=
+ + + ( )( )( )
( )( )( )
12
12
...
(s)
...
mm
n
a s z s z s z
H
s p s p s p
− − −
=
− − −
✓The degree of the denominator → filter order.
✓To ensure the stability of the filter → n ≥ m
✓The coefficients a
i and b
j are real numbers.
❖The TF can be factored and expressed as:
✓Zeros: z
1, z
2, …, z
m and (n - m) zeros at infinity.
✓Poles: p
1, p
2,…, p
n
✓Zeros and poles can be either a real or a complex number.
✓Complex zeros and poles must occur in conjugate pairs.
Poles values of s that cause the denominator to become zero (roots of the denominator
polynomial).
Zeros values of s that cause the numerator to become zero.
Mai Linh, PhD
Introduction:FILTER TRANSFER FUNCTION
❖The filter TF is expressed as the ratio of 2 polynomials:1
1
1
1
...
()
...
mm
m m o
nn
n n o
a s a s a
Hs
a s a s b
−
−
−
−
+ + +
=
+ + + ( )( )( )
( )( )( )
12
12
...
(s)
...
mm
n
a s z s z s z
H
s p s p s p
− − −
=
− − −
✓The degree of the denominator → filter order.
✓To ensure the stability of the filter → n ≥ m
✓The coefficients a
i and b
j are real numbers.
❖The TF can be factored and expressed as:
✓Zeros: z
1, z
2, …, z
m and (n - m) zeros at infinity.
✓Poles: p
1, p
2,…, p
n
✓Zeros and poles can be either a real or a complex number.
✓Complex zeros and poles must occur in conjugate pairs.
Poles values of s that cause the denominator to become zero (roots of the denominator
polynomial).
Zeros values of s that cause the numerator to become zero.
Mai Linh, PhD
12
Introduction:TF Examples
➢Low-Pass Filter (with Stopband Ripple):
➢Low-Pass Filter (without Stopband Ripple):
➢Band-Pass Filter:( )( )
2 2 2 2
4 1 2
5 4 3 2
4 3 2 1 0
(s)
ll
a s s
H
s b s b s b s b s b
++
=
+ + + + + 5 4 3 2
4 3 2 1 0
(s)
m
a
H
s b s b s b s b s b
=
+ + + + + ( )( )
2 2 2 2
5 1 2
6 5 4 3 2
5 4 3 2 1 0
(s)
ll
a s s s
H
s b s b s b s b s b s b
++
=
+ + + + + + |H|
|H| |H|
Mai Linh, PhD
Introduction:TF Examples
➢Low-Pass Filter (with Stopband Ripple):
➢Low-Pass Filter (without Stopband Ripple):
➢Band-Pass Filter:( )( )
2 2 2 2
4 1 2
5 4 3 2
4 3 2 1 0
(s)
ll
a s s
H
s b s b s b s b s b
++
=
+ + + + + 5 4 3 2
4 3 2 1 0
(s)
m
a
H
s b s b s b s b s b
=
+ + + + + ( )( )
2 2 2 2
5 1 2
6 5 4 3 2
5 4 3 2 1 0
(s)
ll
a s s s
H
s b s b s b s b s b s b
++
=
+ + + + + + |H|
|H| |H|
Mai Linh, PhD
13
Fundamentals of Low-Pass Filters
The simplest low-pass filter is the passive RC low-pass network. Its
transfer function is: ()
1
1
11
sC
Hs
sRC
R
sC
==
+
+
For a normalized presentation of the transfer function, s is referred to the filter’s corner (cut-off)
frequency, or –3 dB frequency, ω
C, and has these relationshipsC C C
s j f
s j j
f
= = = =
With the corner frequency of the low-pass in Fig. being f
C = 1/2πRC, s becomes s = sRC and the
transfer function H(s) results in:
Fig.: 1st-Order RC Low-Pass
And the magnitude of the gain response is:()
2
1
1
H=
+
For frequencies >> 1, the rolloff is 20 dB/decade.1
()
1
Hs
s
=
+
Mai Linh, PhD
Fundamentals of Low-Pass Filters
The simplest low-pass filter is the passive RC low-pass network. Its
transfer function is: ()
1
1
11
sC
Hs
sRC
R
sC
==
+
+
For a normalized presentation of the transfer function, s is referred to the filter’s corner (cut-off)
frequency, or –3 dB frequency, ω
C, and has these relationshipsC C C
s j f
s j j
f
= = = =
With the corner frequency of the low-pass in Fig. being f
C = 1/2πRC, s becomes s = sRC and the
transfer function H(s) results in:
Fig.: 1st-Order RC Low-Pass
And the magnitude of the gain response is:()
2
1
1
H=
+
For frequencies >> 1, the rolloff is 20 dB/decade.1
()
1
Hs
s
=
+
Mai Linh, PhD
14
First-order low-pass filter
(or Single-pole active low-pass filter)
1st-Order Noninverting Low-Pass Filter 1st-Order Noninverting Low-Pass Filter with Amplification
The amplifier is configured as a voltage-
follower (Buffer) giving it a DC gain: A
v = +1
as opposed to the previous passive RC filter
which has a DC gain of less than unity.
The DC voltage gain for the filter:2
1
out F
v
in
C
V A
A
V
f
f
==
+
The voltage gain:
A
F = the pass band gain of the filter,
(1 + R
2/R
1)
ƒ = the freq. of the input signal in Hz
ƒ
C = the cut-off freq. in Hz
Mai Linh, PhD
First-order low-pass filter
(or Single-pole active low-pass filter)
1st-Order Noninverting Low-Pass Filter 1st-Order Noninverting Low-Pass Filter with Amplification
The amplifier is configured as a voltage-
follower (Buffer) giving it a DC gain: A
v = +1
as opposed to the previous passive RC filter
which has a DC gain of less than unity.
The DC voltage gain for the filter:2
1
out F
v
in
C
V A
A
V
f
f
==
+
The voltage gain:
A
F = the pass band gain of the filter,
(1 + R
2/R
1)
ƒ = the freq. of the input signal in Hz
ƒ
C = the cut-off freq. in Hz
Mai Linh, PhD
First-order low-pass filter
(or Single-pole active low-pass filter)
Transfer function of the low-pass filter:()
()
1
2/1||
R
sCR
Z
Z
sH
i
f
−=−= ()
c
c
H s A
s
=−
+ 2
1
R
A
R
= CR
c
2
1
=
The transfer function has the same form as for passive
low-pass filter except the gain A in the pass-pand.
15
First order Inverting Low-passfilter
A general op amp circuit.
Feedback path
The gain in the passband, A, is set by the ratio R
2/R
1
Mai Linh, PhD
(or Single-pole active low-pass filter)
Transfer function of the low-pass filter:()
()
1
2/1||
R
sCR
Z
Z
sH
i
f
−=−= ()
c
c
H s A
s
=−
+ 2
1
R
A
R
= CR
c
2
1
=
The transfer function has the same form as for passive
low-pass filter except the gain A in the pass-pand.
15
First order Inverting Low-passfilter
A general op amp circuit.
Feedback path
The gain in the passband, A, is set by the ratio R
2/R
1
Mai Linh, PhD
Review The Decibelin
out
in
out
in
out
i
i
v
v
p
p
1010
10
20log20log decibels ofNumber
10log decibels ofNumber
==
=
Bode plots differ from the frequency response plots in two important ways:
✓First, instead of using a linear axis for the frequency values, a Bode plot uses a logarithmic axis ➔
plot a wider range of frequencies of interest.
✓Second, instead of plotting the absolute magnitude of the transfer function versus frequency, the
Bode magnitude is plotted in decibels (dB) versus the log of the frequency.
A
dB = 20log
10|H(j)|
16
When A
dB = 0, the transfer function magnitude is 1, since 20log
10(1) = 0.
When A
dB< 0, the transfer function magnitude is between 0 and 1, and
when, A
dB > 0, the transfer function magnitude is greater than 1.
Mai Linh, PhD
out
in
out
in
out
i
i
v
v
p
p
1010
10
20log20log decibels ofNumber
10log decibels ofNumber
==
=
Bode plots differ from the frequency response plots in two important ways:
✓First, instead of using a linear axis for the frequency values, a Bode plot uses a logarithmic axis ➔
plot a wider range of frequencies of interest.
✓Second, instead of plotting the absolute magnitude of the transfer function versus frequency, the
Bode magnitude is plotted in decibels (dB) versus the log of the frequency.
A
dB = 20log
10|H(j)|
16
When A
dB = 0, the transfer function magnitude is 1, since 20log
10(1) = 0.
When A
dB< 0, the transfer function magnitude is between 0 and 1, and
when, A
dB > 0, the transfer function magnitude is greater than 1.
Mai Linh, PhD
17
Logarithmic axisReview
Fig. (c) Semilog plane showing
log scale on the x-axis and linear
scale on the y-axis.
Fig. (a) showing log scale on x axis.
A logarithmic scale (or log scale) is a way of displaying
numerical data over a very wide range of values in a compact
way. As opposed to a linear number line in which every unit of
distance corresponds to adding by the same amount, on a
logarithmic scale, every unit of length corresponds to
multiplying the previous value by the same amount. Hence, such
a scale is nonlinear.
Fig. (b) frequency is showing log scale on x axis.
Mai Linh, PhD
Logarithmic axisReview
Fig. (c) Semilog plane showing
log scale on the x-axis and linear
scale on the y-axis.
Fig. (a) showing log scale on x axis.
A logarithmic scale (or log scale) is a way of displaying
numerical data over a very wide range of values in a compact
way. As opposed to a linear number line in which every unit of
distance corresponds to adding by the same amount, on a
logarithmic scale, every unit of length corresponds to
multiplying the previous value by the same amount. Hence, such
a scale is nonlinear.
Fig. (b) frequency is showing log scale on x axis.
Mai Linh, PhD
18
Review Bode Plots1
1
out
in
C
V
jV
=
+ 10
20log
1
1
1
3
2
out
in
out C
in C
out
in
out
in
V
dB
V
V
V
V
Vj
V
dB
V
==
= = −
=
+
= = −
➢If <<
C
➢If >>
C
➢If =
C
C
()
Mai Linh, PhD
Review Bode Plots1
1
out
in
C
V
jV
=
+ 10
20log
1
1
1
3
2
out
in
out C
in C
out
in
out
in
V
dB
V
V
V
V
Vj
V
dB
V
==
= = −
=
+
= = −
➢If <<
C
➢If >>
C
➢If =
C
C
()
Mai Linh, PhD
19
Review Bode Plots (/C)
(/C)
(/C) 10
20log
1
1
1
3
2
out
in
out C
in C
out
in
out
in
V
dB
V
V
V
V
Vj
V
dB
V
==
= = −
=
+
= = −
➢If <<
C
➢If >>
C
➢If =
C1
1
out
in
C
V
jV
=
+
Mai Linh, PhD
Review Bode Plots (/C)
(/C)
(/C) 10
20log
1
1
1
3
2
out
in
out C
in C
out
in
out
in
V
dB
V
V
V
V
Vj
V
dB
V
==
= = −
=
+
= = −
➢If <<
C
➢If >>
C
➢If =
C1
1
out
in
C
V
jV
=
+
Mai Linh, PhD
20
Review Bode Plots
→ log(|H()|) = - log(|1 + j/
1|) - log(|1 + j/
2|)
The first term on the right is zero for
1
The second term on the right is zero for
2
The magnitude of H() is zero for
1, then decreases by 20 dB per
decade after
1. After
2 the second term on the right kicks in and the
magnitude decreases by another 20 dB per decade to 40 dB per decade.
Mai Linh, PhD
Review Bode Plots
→ log(|H()|) = - log(|1 + j/
1|) - log(|1 + j/
2|)
The first term on the right is zero for
1
The second term on the right is zero for
2
The magnitude of H() is zero for
1, then decreases by 20 dB per
decade after
1. After
2 the second term on the right kicks in and the
magnitude decreases by another 20 dB per decade to 40 dB per decade.
Mai Linh, PhD
21 1
101
2
102
20log(/1)
−20log(/2)
−20log(/1) − 20log(/2)
1 2
dB
The bode plots for
each factor add to
give the bode plot
for the function.
Review Bode Plots
Mai Linh, PhD
101
2
102
20log(/1)
−20log(/2)
−20log(/1) − 20log(/2)
1 2
dB
The bode plots for
each factor add to
give the bode plot
for the function.
Review Bode Plots
Mai Linh, PhD
22
Example
Designing a low-pass Op Amp Filter
Using the circuit shown in Fig., calculate values for C and R
2 that, together with
R
1 = 1 Ω, produce a low-pass filter having a gain of 1 (i.e. A = 1) in the passband
and a cutoff frequency of 1 rad/s. Construct the transfer function for this filter
and use it to sketch a Bode magnitude plot of the filter’s frequency response.
Solution
Calculate the required value of R
2:
R
2 = AR
1 = 1 × 1 = 1 (Ω)
Calculate C to meet the specified cutoff frequency
The transfer function for the low-pass filter is2
11
1 ( )
11
C
CF
R
= = =
1
()
1
C
C
H s A
ss
−
= − =
++
Mai Linh, PhD
Example
Designing a low-pass Op Amp Filter
Using the circuit shown in Fig., calculate values for C and R
2 that, together with
R
1 = 1 Ω, produce a low-pass filter having a gain of 1 (i.e. A = 1) in the passband
and a cutoff frequency of 1 rad/s. Construct the transfer function for this filter
and use it to sketch a Bode magnitude plot of the filter’s frequency response.
Solution
Calculate the required value of R
2:
R
2 = AR
1 = 1 × 1 = 1 (Ω)
Calculate C to meet the specified cutoff frequency
The transfer function for the low-pass filter is2
11
1 ( )
11
C
CF
R
= = =
1
()
1
C
C
H s A
ss
−
= − =
++
Mai Linh, PhD
Example
Designing a low-pass Op Amp Filter
23
The Bode plot of |H(j)| is shown in
Fig. This is the so-called prototype
low-pass op amp filter, because it uses
a resistorvalue of 1 Ωand a capacitor
value of 1 F, and it provides a cutoff
frequency of 1 rad/s.
>> w=0.1:.1:10;
>> h=20*log10(abs(1./(1+j*w)));
>> semilogx(w,h);
>> grid on;
>> xlabel('\omega(rad/s)');
>> ylabel('|H(j\omega)|dB');
A
dB = 20log
10|H(j)|
Mai Linh, PhD
Designing a low-pass Op Amp Filter
23
The Bode plot of |H(j)| is shown in
Fig. This is the so-called prototype
low-pass op amp filter, because it uses
a resistorvalue of 1 Ωand a capacitor
value of 1 F, and it provides a cutoff
frequency of 1 rad/s.
>> w=0.1:.1:10;
>> h=20*log10(abs(1./(1+j*w)));
>> semilogx(w,h);
>> grid on;
>> xlabel('\omega(rad/s)');
>> ylabel('|H(j\omega)|dB');
A
dB = 20log
10|H(j)|
Mai Linh, PhD
First-order high-pass filter()
( )
2
1
1/
f
i
Z R
Hs
Z R sC
= − = −
+
Transfer function of the low-pass
filter:()
c
s
H s A
s
=−
+ 2
1
R
A
R
= CR
c
1
1
=
It also has the same form as
passive high-pass filter, except
for the gain.
24
Mai Linh, PhD
( )
2
1
1/
f
i
Z R
Hs
Z R sC
= − = −
+
Transfer function of the low-pass
filter:()
c
s
H s A
s
=−
+ 2
1
R
A
R
= CR
c
1
1
=
It also has the same form as
passive high-pass filter, except
for the gain.
24
Mai Linh, PhD
Example
Designing a high-pass Op Amp Filter
We have the Bode magnitude plot of a
high-pass filter. Calculate values of R
1
and R
2 that produce the desired
magnitude response.
Given C = 0.1 μF.
25
First-order high-pass filter
Mai Linh, PhD
Designing a high-pass Op Amp Filter
We have the Bode magnitude plot of a
high-pass filter. Calculate values of R
1
and R
2 that produce the desired
magnitude response.
Given C = 0.1 μF.
25
First-order high-pass filter
Mai Linh, PhD
Example
Designing a high-pass Op Amp Filter
26
Solution:
Note that the gain in the passband is 20 dB; therefore,
A = 10. Also note that the 3 dB point is 500 rad/s.
The transfer function that has the magnitude response
shown in the Bode plot is given by
so
Equating the numerators and denominators and then
simplifying we get two equations:10
()
500
s
Hs
s
−
=
+ ( )
( )
21
1
/10
()
500 1/
R R ss
Hs
s s RC
−−
==
++ 2
11
12
1
10 , 500
20 , 200 .
R
R R C
R k R k
==
= =
Mai Linh, PhD
Designing a high-pass Op Amp Filter
26
Solution:
Note that the gain in the passband is 20 dB; therefore,
A = 10. Also note that the 3 dB point is 500 rad/s.
The transfer function that has the magnitude response
shown in the Bode plot is given by
so
Equating the numerators and denominators and then
simplifying we get two equations:10
()
500
s
Hs
s
−
=
+ ( )
( )
21
1
/10
()
500 1/
R R ss
Hs
s s RC
−−
==
++ 2
11
12
1
10 , 500
20 , 200 .
R
R R C
R k R k
==
= =
Mai Linh, PhD
27
Example
Designing a high-pass Op Amp Filter Solution:
The circuit is shown in Fig.
Because we have assumed that the opamp in this high-pass filter circuit is ideal,
the addition of any load resistor, regardless of its resistance, has no effect on the
behavior of the opamp. Thus, the magnitude response of a high-pass filter with a
load resistor is the same as that of a high-pass filter with no load resistor.
>> w=1:10000;
>> h=20*log10(10*(abs((j*w/500)./(1+j*w/500))));
>> semilogx(w,h);
>> grid on;
>> xlabel('\omega(rad/s)');
>> ylabel('|H(j\omega)|dB');
Mai Linh, PhD
Example
Designing a high-pass Op Amp Filter Solution:
The circuit is shown in Fig.
Because we have assumed that the opamp in this high-pass filter circuit is ideal,
the addition of any load resistor, regardless of its resistance, has no effect on the
behavior of the opamp. Thus, the magnitude response of a high-pass filter with a
load resistor is the same as that of a high-pass filter with no load resistor.
>> w=1:10000;
>> h=20*log10(10*(abs((j*w/500)./(1+j*w/500))));
>> semilogx(w,h);
>> grid on;
>> xlabel('\omega(rad/s)');
>> ylabel('|H(j\omega)|dB');
Mai Linh, PhD
Problem 1
Designing a high-pass Op Amp Filter
Compute the values for R
2 and C that yield a high-pass filter with a passband
gain of 1 and a cutoff frequency of 1 rad/s if R
1 is 1. (Note: This is the
prototype high-pass filter.)
28
Mai Linh, PhD
Designing a high-pass Op Amp Filter
Compute the values for R
2 and C that yield a high-pass filter with a passband
gain of 1 and a cutoff frequency of 1 rad/s if R
1 is 1. (Note: This is the
prototype high-pass filter.)
28
Mai Linh, PhD
Solution:
29
Problem 1
Designing a high-pass Op Amp Filter( )
( )
21
1
1
1
2
21
1
prototype
/
()
1/
1
1 / ; 1 , 1
1; 1
()
1
R R s
Hs
s R C
rad s R C F
RC
R
RR
R
s
Hs
s
−
=
+
= = =
= = =
−
=
+
Mai Linh, PhD
29
Problem 1
Designing a high-pass Op Amp Filter( )
( )
21
1
1
1
2
21
1
prototype
/
()
1/
1
1 / ; 1 , 1
1; 1
()
1
R R s
Hs
s R C
rad s R C F
RC
R
RR
R
s
Hs
s
−
=
+
= = =
= = =
−
=
+
Mai Linh, PhD
Problem 2
Designing a low-pass Op Amp Filter
Compute the resistor values needed for the low-pass filter circuit in Fig. to
produce transfer function5000
20000
)(
+
−
=
s
sH
Assume that C = 5 F
30
Mai Linh, PhD
Designing a low-pass Op Amp Filter
Compute the resistor values needed for the low-pass filter circuit in Fig. to
produce transfer function5000
20000
)(
+
−
=
s
sH
Assume that C = 5 F
30
Mai Linh, PhD
Solution:
31
Problem 2
Designing a low-pass Op Amp Filter( )
( )
1
2
1
1 6
2 6
2
1/ 20000
()
1/ 5000
1
20000; 5
1
10 ( )
20000 (5 10 )
11
5000 40 ( )
5000 (5 10 )
RC
Hs
s R C s
CF
RC
R
R
RC
−
−
− −
==
++
==
= =
= = =
Mai Linh, PhD
31
Problem 2
Designing a low-pass Op Amp Filter( )
( )
1
2
1
1 6
2 6
2
1/ 20000
()
1/ 5000
1
20000; 5
1
10 ( )
20000 (5 10 )
11
5000 40 ( )
5000 (5 10 )
RC
Hs
s R C s
CF
RC
R
R
RC
−
−
− −
==
++
==
= =
= = =
Mai Linh, PhD
Scaling
32
The designer can transform the convenient values into realistic values using the
process known as scaling.
There are two types of scaling magnitude and frequency.
Magnitude scaling: Scale a circuit in magnitude by multiplying the impedance at a
given frequency by the scale factor k
m. Thus we multiply all resistors and inductors
by k
m and all capacitors by 1/k
m.
R’ = k
mR; L’ = k
mL; C’ = C/k
m
Frequency scaling: we change the circuit parameters so that at the new frequency,
the impedance of each element is the same as it was at the original frequency. If we
let k
f denote the frequency scale factor. Thus for frequency scaling:
R’ = R; L’ = L/k
f; C’ = C/k
f
Mai Linh, PhD
32
The designer can transform the convenient values into realistic values using the
process known as scaling.
There are two types of scaling magnitude and frequency.
Magnitude scaling: Scale a circuit in magnitude by multiplying the impedance at a
given frequency by the scale factor k
m. Thus we multiply all resistors and inductors
by k
m and all capacitors by 1/k
m.
R’ = k
mR; L’ = k
mL; C’ = C/k
m
Frequency scaling: we change the circuit parameters so that at the new frequency,
the impedance of each element is the same as it was at the original frequency. If we
let k
f denote the frequency scale factor. Thus for frequency scaling:
R’ = R; L’ = L/k
f; C’ = C/k
f
Mai Linh, PhD
33
Example:
A circuit can be scaled simultaneously in both magnitude and frequency. The
scaled values (primed) in terms of (unprimed) are
Scaling a Series RLC passive Circuit
Scaling 1
;;
m
m
f m f
k
R k R L L C C
k k k
= = = 1
+
1 H
1 F
vi
_
vs
Mai Linh, PhD
The series RLC circuit shown in Fig. has a center freq. of
(LC)
-0.5
= 1 rad/s, a bandwidth of R/L = 1 rad/s, and thus
a quality factor of 1. Use scaling to compute new values
of R and L that yield a circuit with the same quality
factor but with a center frequency of 500 Hz. Use a 2 F
capacitor.
Example:
A circuit can be scaled simultaneously in both magnitude and frequency. The
scaled values (primed) in terms of (unprimed) are
Scaling a Series RLC passive Circuit
Scaling 1
;;
m
m
f m f
k
R k R L L C C
k k k
= = = 1
+
1 H
1 F
vi
_
vs
Mai Linh, PhD
The series RLC circuit shown in Fig. has a center freq. of
(LC)
-0.5
= 1 rad/s, a bandwidth of R/L = 1 rad/s, and thus
a quality factor of 1. Use scaling to compute new values
of R and L that yield a circuit with the same quality
factor but with a center frequency of 500 Hz. Use a 2 F
capacitor.
34
Solution:
Let computing the frequency scale factor that will shift the center frequency from 1 rad/s
to 500 Hz. The unprimed variables represent values before scaling, whereas the primed
variables represent values after scaling
The magnitude scale factor:
Scaling 2 (500)
3141.59
1
o
f
o
k
= = = 6
11
159.155
(3141.59)(2 10 )
m
f
C
k
kC
−
= = =
159.155 ( ),
50.66 ( )
m
m
f
R k R
k
L L mH
k
= =
==
Mai Linh, PhD
Solution:
Let computing the frequency scale factor that will shift the center frequency from 1 rad/s
to 500 Hz. The unprimed variables represent values before scaling, whereas the primed
variables represent values after scaling
The magnitude scale factor:
Scaling 2 (500)
3141.59
1
o
f
o
k
= = = 6
11
159.155
(3141.59)(2 10 )
m
f
C
k
kC
−
= = =
159.155 ( ),
50.66 ( )
m
m
f
R k R
k
L L mH
k
= =
==
Mai Linh, PhD
Problem 3
What magnitude and frequency scale factor will transform the prototype
high-pass filter into a high pass filter with a 0.5 pF capacitor and a cutoff
frequency of 10 kHz?
35
Scaling
Mai Linh, PhD
What magnitude and frequency scale factor will transform the prototype
high-pass filter into a high pass filter with a 0.5 pF capacitor and a cutoff
frequency of 10 kHz?
35
Scaling
Mai Linh, PhD
Solution:
36
Scaling ( )( )
4
6
6
2 2 10 20000 ( / )
20000 62831.85
1
0.5 10
1
31.83
0.5 10 62831.85
cc
f
f m m f
m
f rad s
k
C
C
k k k k
k
−
−
= = =
= =
= =
= =
Mai Linh, PhD
36
Scaling ( )( )
4
6
6
2 2 10 20000 ( / )
20000 62831.85
1
0.5 10
1
31.83
0.5 10 62831.85
cc
f
f m m f
m
f rad s
k
C
C
k k k k
k
−
−
= = =
= =
= =
= =
Mai Linh, PhD
First-order band-pass filter
Bode magnitude plot of a bandpass filter
The band-pass filter consists of 3 separate components:
1.ω
c2 – cut-off frequency of unity-gain low-pass filter
2.ω
c1 – cut-off frequency of unity-gain high-pass filter
3.Gain component to provide the desired level of gain in the
passband.
Requirement for two cut-off frequencies in
broadband bandpass filter (BPF):2
1
2
c
c
37
These three components are cascaded in series. So:
Pass-band/Bandwidth
Mai Linh, PhD
Bode magnitude plot of a bandpass filter
The band-pass filter consists of 3 separate components:
1.ω
c2 – cut-off frequency of unity-gain low-pass filter
2.ω
c1 – cut-off frequency of unity-gain high-pass filter
3.Gain component to provide the desired level of gain in the
passband.
Requirement for two cut-off frequencies in
broadband bandpass filter (BPF):2
1
2
c
c
37
These three components are cascaded in series. So:
Pass-band/Bandwidth
Mai Linh, PhD
Construct the BPF
38
We can construct a circuit that provides each of the three components by cascading a low-pass
op amp filter, a high-pass op amp filter, and an inverting amplifier, as shown in Fig. below.
Transfer function of the band-pass filter is the product of the transfer functions of the three
cascaded components:()
( )( )
22
2 1 1 2
(1)
fo c c
i c c i c c
RV A s s
Hs
V s s R s s
− −− −
= = =
+ + + +
LL
c
CR
1
2
= 1
1
,
c
HH
RC
= f
i
R
A
R
=
First-order band-pass filter
❖The standard form for the transfer function of a BPF must be:22
( ) (2)
o
s
Hs
ss
=
++
➢Need to convert (1) into the form of (2) for a BPF → we
require that:
c2 >>
c1
where
Note: is bandwidth, =
c2 –
c1
→ (
c2 +
c1)
c2 vi Low-pass filterHigh-pass filter Inverting amplifiervo
Mai Linh, PhD
38
We can construct a circuit that provides each of the three components by cascading a low-pass
op amp filter, a high-pass op amp filter, and an inverting amplifier, as shown in Fig. below.
Transfer function of the band-pass filter is the product of the transfer functions of the three
cascaded components:()
( )( )
22
2 1 1 2
(1)
fo c c
i c c i c c
RV A s s
Hs
V s s R s s
− −− −
= = =
+ + + +
LL
c
CR
1
2
= 1
1
,
c
HH
RC
= f
i
R
A
R
=
First-order band-pass filter
❖The standard form for the transfer function of a BPF must be:22
( ) (2)
o
s
Hs
ss
=
++
➢Need to convert (1) into the form of (2) for a BPF → we
require that:
c2 >>
c1
where
Note: is bandwidth, =
c2 –
c1
→ (
c2 +
c1)
c2 vi Low-pass filterHigh-pass filter Inverting amplifiervo
Mai Linh, PhD
39
The bandpass filter circuit
➔ the transfer function for the
cascaded BPF:
➢Compute the values of R
L and C
L in the low-pass filter to give us the desired upper cutoff
frequency,
c2:
First-order band-pass filter2
2
2 1 2
()
c
c c c
As
Hs
ss
−
=
++ 2
1
c
LL
RC
=
➢Compute the values of R
H and C
H in the high-pass filter to give us the desired lower cutoff
frequency,
c1:1
1
c
HH
RC
=
Mai Linh, PhD
The bandpass filter circuit
➔ the transfer function for the
cascaded BPF:
➢Compute the values of R
L and C
L in the low-pass filter to give us the desired upper cutoff
frequency,
c2:
First-order band-pass filter2
2
2 1 2
()
c
c c c
As
Hs
ss
−
=
++ 2
1
c
LL
RC
=
➢Compute the values of R
H and C
H in the high-pass filter to give us the desired lower cutoff
frequency,
c1:1
1
c
HH
RC
=
Mai Linh, PhD
40
Compute the values of R
i and R
f in the inverting amplifier to provide the desired passband gain. To
do this, we consider the magnitude of the BPF's transfer function, evaluated at the center frequency
o
Besides, the gain A of the inverting amplifier
Any choice of resistors that
satisfies Eq. (*) will produce
the desired passband gain.
First-order band-pass filter22
2
2 2 1 2
()
()
( ) ( )
c o c
o
o c o c c c
A j A
H j A
jj
−
= = =
++ ( ) (*)
f
o
i
R
Hj
R
= 2
2
1
2
2
2
,
22
22
co
co
= − + +
= + + +
2
1
2
2
11
1,
22
11
1
22
co
co
QQ
QQ
= − + +
= + +
22
() (2)
o
As
Hs
ss
=
++
Mai Linh, PhD
Compute the values of R
i and R
f in the inverting amplifier to provide the desired passband gain. To
do this, we consider the magnitude of the BPF's transfer function, evaluated at the center frequency
o
Besides, the gain A of the inverting amplifier
Any choice of resistors that
satisfies Eq. (*) will produce
the desired passband gain.
First-order band-pass filter22
2
2 2 1 2
()
()
( ) ( )
c o c
o
o c o c c c
A j A
H j A
jj
−
= = =
++ ( ) (*)
f
o
i
R
Hj
R
= 2
2
1
2
2
2
,
22
22
co
co
= − + +
= + + +
2
1
2
2
11
1,
22
11
1
22
co
co
= − + +
= + +
22
() (2)
o
As
Hs
ss
=
++
Mai Linh, PhD
Example
Design a BPF for a graphic equalizer to
provide an amplification of 2 within the
band of frequencies between 100 and
10000 Hz. Use 0.2 µF capacitors.
41
Solution
We design each subcircuit in the cascade and meet the specified cutoff frequency values. In
this case
c2 = 100
c1 so we can say that
c2 >>
c1
Begin with the Low-pass stage.
First-order band-pass filter2
6
1
2 (10000),
1
80 ( )
2 10000 0.2 10
c
LL
L
RC
R
−
==
=
Mai Linh, PhD
Design a BPF for a graphic equalizer to
provide an amplification of 2 within the
band of frequencies between 100 and
10000 Hz. Use 0.2 µF capacitors.
41
Solution
We design each subcircuit in the cascade and meet the specified cutoff frequency values. In
this case
c2 = 100
c1 so we can say that
c2 >>
c1
Begin with the Low-pass stage.
First-order band-pass filter2
6
1
2 (10000),
1
80 ( )
2 10000 0.2 10
c
LL
L
RC
R
−
==
=
Mai Linh, PhD
42
Next, calculate for the high-pass stage
For the gain stage: one of the resistors can be selected arbitrarily. Let's select a 1 kΩ resistor
for R
i . So with A = R
f/R
i ➔ R
f = AR
i = 2(1000) = 2 (kΩ)
The resulting circuit is shown in Fig.
First-order band-pass filter1 6
11
2 (100), 7958 ( )
2 100 0.2 10
cH
HH
R
RC
−
= = =
Solution
Mai Linh, PhD
Next, calculate for the high-pass stage
For the gain stage: one of the resistors can be selected arbitrarily. Let's select a 1 kΩ resistor
for R
i . So with A = R
f/R
i ➔ R
f = AR
i = 2(1000) = 2 (kΩ)
The resulting circuit is shown in Fig.
First-order band-pass filter1 6
11
2 (100), 7958 ( )
2 100 0.2 10
cH
HH
R
RC
−
= = =
Solution
Mai Linh, PhD
First-order band-reject filter
Bode magnitude plot of a band-reject filter
The band-reject filter consists of 3 separate components:
1.ω
c1 – cut-off frequency of unity-gain low-pass filter
2.ω
c2 – cut-off frequency of unity-gain high-pass filter
3.Gain component to provide the desired level of gain in
the passband.
These 3 components cannot be cascaded
in series, but we have to use a parallel
connection and a summing amplifier to
construct band-reject filter.
43
We can use a component approach to the design of op amp bandreject filters too.
Mai Linh, PhD
Bode magnitude plot of a band-reject filter
The band-reject filter consists of 3 separate components:
1.ω
c1 – cut-off frequency of unity-gain low-pass filter
2.ω
c2 – cut-off frequency of unity-gain high-pass filter
3.Gain component to provide the desired level of gain in
the passband.
These 3 components cannot be cascaded
in series, but we have to use a parallel
connection and a summing amplifier to
construct band-reject filter.
43
We can use a component approach to the design of op amp bandreject filters too.
Mai Linh, PhD
Construct the band-reject filter
44
A parallel op amp
bandreject filter
Block diagram
circuit
The most important difference is
that these three components
cannot be cascaded in series
because they do not combine
additively on the Bode plot.
Instead, we use a parallel
connection and a summing
amplifier
First-order band-reject filtervi
Low-pass filter
High-pass filter
Inverting Amp.vo
Mai Linh, PhD
44
A parallel op amp
bandreject filter
Block diagram
circuit
The most important difference is
that these three components
cannot be cascaded in series
because they do not combine
additively on the Bode plot.
Instead, we use a parallel
connection and a summing
amplifier
First-order band-reject filtervi
Low-pass filter
High-pass filter
Inverting Amp.vo
Mai Linh, PhD
()
+
−
+
+
−
−=
21
1
cc
c
i
f
s
s
sR
R
sH
()
( )( )
++
++
=
21
211
2
2
cc
ccc
i
f
ss
ss
R
R
sH
LL
c
CR
1
1
= HH
c
CR
1
2
= 45
Assume that the two cutoff frequencies are widely separated → the resulting design is a broadband
bandreject filter, and
c2 >>
c1. Then each component of the parallel design car be created
independently, and the cutoff frequency specifications will be satisfied. The transfer function of the
resulting circuit is the sum of the low-pass and high-pass filter transfer functions
The cutoff frequencies are
In the two passbands (as s → 0 and s → ∞), the gain of
the transfer function is R
f/R
i.
Therefore K = R
f/R
i
First-order band-reject filter
Mai Linh, PhD
+
−
+
+
−
−=
21
1
cc
c
i
f
s
s
sR
R
sH
()
( )( )
++
++
=
21
211
2
2
cc
ccc
i
f
ss
ss
R
R
sH
LL
c
CR
1
1
= HH
c
CR
1
2
= 45
Assume that the two cutoff frequencies are widely separated → the resulting design is a broadband
bandreject filter, and
c2 >>
c1. Then each component of the parallel design car be created
independently, and the cutoff frequency specifications will be satisfied. The transfer function of the
resulting circuit is the sum of the low-pass and high-pass filter transfer functions
The cutoff frequencies are
In the two passbands (as s → 0 and s → ∞), the gain of
the transfer function is R
f/R
i.
Therefore K = R
f/R
i
First-order band-reject filter
Mai Linh, PhD
Example
Design a circuit based on the parallel band-reject op-amp
filter as in the figure. Use 0.5 µF capacitors for the
design.
46
Solution
From the Bode figure, we see that the bandreject filter has
cutoff frequencies of 100 rad/s and 2000 rad/s and a gain of 3
in the pass bands →
c2 = 20
c1
So, assume that:
c2 >>
c1
From low-pass filter model and use scaling to meet the
specifications for cutoff frequency and capacitor value. The
frequency scale factor k
f = 100 → shifts the cutoff frequency
from 1 rad/s to 100 rad/s. The magnitude scale factor f
m is
20000, which permits the use of a 0.5 F capacitor
First-order band-reject filter
Mai Linh, PhD
Design a circuit based on the parallel band-reject op-amp
filter as in the figure. Use 0.5 µF capacitors for the
design.
46
Solution
From the Bode figure, we see that the bandreject filter has
cutoff frequencies of 100 rad/s and 2000 rad/s and a gain of 3
in the pass bands →
c2 = 20
c1
So, assume that:
c2 >>
c1
From low-pass filter model and use scaling to meet the
specifications for cutoff frequency and capacitor value. The
frequency scale factor k
f = 100 → shifts the cutoff frequency
from 1 rad/s to 100 rad/s. The magnitude scale factor f
m is
20000, which permits the use of a 0.5 F capacitor
First-order band-reject filter
Mai Linh, PhD
47
Cutoff frequency of the low pass filter
Similarly, to design high pass filter, from the high-pass op amp filter model. We
have k
f = 2000, and k
m = 1000, resulting in the following scaled component values
Finally, because the cutoff frequencies are widely separated, we can use the ratio R
f/R
i to establish
the desired passband gain of 3.
Choose R
i = 1 kΩ, (as for R
H). ➔ R
f = 3kΩ, and K = R
f/R
i = 3000/1000 = 3
The resulting parallel op amp bandreject filter circuit is shown in Fig.20 , 0.5 .
LL
R k C F = = 1 36
11
100 ( / )
20 10 0.5 10
c
LL
rad s
RC
−
= = =
1 , 0.5 .
HH
R k C F = =
First-order band-reject filter
Solution
Mai Linh, PhD
Cutoff frequency of the low pass filter
Similarly, to design high pass filter, from the high-pass op amp filter model. We
have k
f = 2000, and k
m = 1000, resulting in the following scaled component values
Finally, because the cutoff frequencies are widely separated, we can use the ratio R
f/R
i to establish
the desired passband gain of 3.
Choose R
i = 1 kΩ, (as for R
H). ➔ R
f = 3kΩ, and K = R
f/R
i = 3000/1000 = 3
The resulting parallel op amp bandreject filter circuit is shown in Fig.20 , 0.5 .
LL
R k C F = = 1 36
11
100 ( / )
20 10 0.5 10
c
LL
rad s
RC
−
= = =
1 , 0.5 .
HH
R k C F = =
First-order band-reject filter
Solution
Mai Linh, PhD
>> f=10:80000;
>> w=2*pi*f;
>> H=(((-2*pi*100)./(j*w+2*pi*100))+((-j*w)./(j*w+2*pi*2000)))*(-5);
>> A=20*log10(abs(H));
>> semilogx(f,A);
>> grid on;
>> xlabel ('F (Hz)');
>> ylabel ('A_{dB}');
First-order band-reject filter
Solution
481
12
()
fc
c c i
Rj
Hj
j j R
− −
= + −
++
()
10
20log
dB
A H j =
Mai Linh, PhD
>> w=2*pi*f;
>> H=(((-2*pi*100)./(j*w+2*pi*100))+((-j*w)./(j*w+2*pi*2000)))*(-5);
>> A=20*log10(abs(H));
>> semilogx(f,A);
>> grid on;
>> xlabel ('F (Hz)');
>> ylabel ('A_{dB}');
First-order band-reject filter
Solution
481
12
()
fc
c c i
Rj
Hj
j j R
− −
= + −
++
()
10
20log
dB
A H j =
Mai Linh, PhD
49
The resulting bandreject filter circuit design example
First-order band-reject filter
Solution
Mai Linh, PhD
The resulting bandreject filter circuit design example
First-order band-reject filter
Solution
Mai Linh, PhD
High Order Op Amp Filters
50
This figure shows the Bode magnitude plots of a
cascade of identical prototype low-pass filters and
includes plots of just one filter, two in cascade,
three in cascade, and four in cascade.
The more poles filter has, the faster its roll-off rate.
As the order of the low-pass filter is increased by
adding prototype low-pass filters to the cascade, the
cutoff frequency also change
The Order (number of poles) of the filter
A pole (single pole) is simply one resistor & one capacitor.
Case of Low-Pass Filter
Mai Linh, PhD
50
This figure shows the Bode magnitude plots of a
cascade of identical prototype low-pass filters and
includes plots of just one filter, two in cascade,
three in cascade, and four in cascade.
The more poles filter has, the faster its roll-off rate.
As the order of the low-pass filter is increased by
adding prototype low-pass filters to the cascade, the
cutoff frequency also change
The Order (number of poles) of the filter
A pole (single pole) is simply one resistor & one capacitor.
Case of Low-Pass Filter
Mai Linh, PhD
51
➢The number of poles determines the roll-off rate of the filter. For example, a
Butterworth response produces -20dB/decade/pole.
This means that:
✓ One-pole (first-order) filter has a roll-off of -20 dB/decade
✓ Two-pole (second-order) filter has a roll-off of -40 dB/decade
✓ Three-pole (third-order) filter has a roll-off of -60 dB/decade
High Order Op Amp Filters Case of Low-Pass Filter
Mai Linh, PhD
➢The number of poles determines the roll-off rate of the filter. For example, a
Butterworth response produces -20dB/decade/pole.
This means that:
✓ One-pole (first-order) filter has a roll-off of -20 dB/decade
✓ Two-pole (second-order) filter has a roll-off of -40 dB/decade
✓ Three-pole (third-order) filter has a roll-off of -60 dB/decade
High Order Op Amp Filters Case of Low-Pass Filter
Mai Linh, PhD
52
➢ The number of filter poles can be increased by cascading. To obtain a filter
with three poles, cascade a two-pole with one-pole filters.
Fig.: example of Three-pole (third-order) low-pass filter.
High Order Op Amp Filters Case of Low-Pass Filter
Mai Linh, PhD
➢ The number of filter poles can be increased by cascading. To obtain a filter
with three poles, cascade a two-pole with one-pole filters.
Fig.: example of Three-pole (third-order) low-pass filter.
High Order Op Amp Filters Case of Low-Pass Filter
Mai Linh, PhD
53
In general, an n-element cascade of identical low-pass filter will transition from the
passband to the stopband with a slope of 20n dB/decade. Both the block diagram and
the circuit diagram for such a cascade are shown in Fig. below:
High Order Op Amp Filters Case of Low-Pass Filtervi Low-pass filter voLow-pass filter Low-pass filter
Mai Linh, PhD
In general, an n-element cascade of identical low-pass filter will transition from the
passband to the stopband with a slope of 20n dB/decade. Both the block diagram and
the circuit diagram for such a cascade are shown in Fig. below:
High Order Op Amp Filters Case of Low-Pass Filtervi Low-pass filter voLow-pass filter Low-pass filter
Mai Linh, PhD
54
The transfer function for a cascade of n prototype low-pass filters
❖The order of a filter is determined by the number of poles in its transfer function
High Order Op Amp Filters Case of Low-Pass Filter
➢To calculate the cutoff frequency (
c) of the higher order filter formed in the
cascade of 1
st
-order filter, we can use frequency scaling to calculate component
values that move the
c to it specified location.
➢By solving for the value of
c that results in |H(j)| = Τ12()
1 1 1 1
...
1 1 1 1
n
Hs
s s s s
− − − −
==
+ + + + () ()
( )
1 11
,
1 21
n
cn n
cn
Hs Hj
s j
−
= = =
+ + 2/
2
2
11
2 1 2 1
1 2
n
nn
cn cn
cn
= = + = −
+
The transfer function for a cascade of n prototype low-pass filters
❖The order of a filter is determined by the number of poles in its transfer function
High Order Op Amp Filters Case of Low-Pass Filter
➢To calculate the cutoff frequency (
c) of the higher order filter formed in the
cascade of 1
st
-order filter, we can use frequency scaling to calculate component
values that move the
c to it specified location.
➢By solving for the value of
c that results in |H(j)| = Τ12()
1 1 1 1
...
1 1 1 1
n
Hs
s s s s
− − − −
==
+ + + + () ()
( )
1 11
,
1 21
n
cn n
cn
Hs Hj
s j
−
= = =
+ + 2/
2
2
11
2 1 2 1
1 2
n
nn
cn cn
cn
= = + = −
+
55
For example, n = 4:
Thus we can design a fourth order low-pass filter with any arbitrary cutoff
frequency by starting with a 4th-order cascade consisting of prototype low-pass
filters and then scaling the components by k
f =
c/0.435 to place the cutoff
frequency at any value of
c desired.
Example: Designing a 4th-order low-pass op amp fitter with a cutoff frequency
of 500 Hz and a passband gain of 10. Use 1 F capacitors. Sketch the Bode
magnitude plot for this filter.
High Order Op Amp Filters4
4
2 1 0.435 ( / )
c
rad s= − Case of Low-Pass Filter
Mai Linh, PhD
For example, n = 4:
Thus we can design a fourth order low-pass filter with any arbitrary cutoff
frequency by starting with a 4th-order cascade consisting of prototype low-pass
filters and then scaling the components by k
f =
c/0.435 to place the cutoff
frequency at any value of
c desired.
Example: Designing a 4th-order low-pass op amp fitter with a cutoff frequency
of 500 Hz and a passband gain of 10. Use 1 F capacitors. Sketch the Bode
magnitude plot for this filter.
High Order Op Amp Filters4
4
2 1 0.435 ( / )
c
rad s= − Case of Low-Pass Filter
Mai Linh, PhD
56
ExampleSolution
We have already used Eq. to calculate the cutoff frequency for the resulting
fourth-order low-pass filter as 0.435 rad/s.
A frequency scale factor of k
f = 7222.39 will scale the component values to give a 500 Hz cutoff
frequency. A magnitude scale factor of k
m = 138.46 permits the use of 1 F capacitors. The scaled
component values are thus: R = 138.46 ; C = 1 F
Finally, add an inverting amplifier stage with a gain of R
f/R
i = 10. As usual, we can arbitrarily select one
of the two resistor values. Because we are already using 138.46 resistors, let R
i = 138.46 ; then,
R
f = 10R
i = 1384.6
The circuit for this cascaded the 4
th
-order low-pass filter is shown in Fig. (next slide).
It has the transfer function:
High Order Op Amp Filters4
4
2 1 0.435
c
rad
s
= −
Case of Low-Pass Filter()
4
7222.39
10
7222.39
Hs
s
=−
+
Mai Linh, PhD
ExampleSolution
We have already used Eq. to calculate the cutoff frequency for the resulting
fourth-order low-pass filter as 0.435 rad/s.
A frequency scale factor of k
f = 7222.39 will scale the component values to give a 500 Hz cutoff
frequency. A magnitude scale factor of k
m = 138.46 permits the use of 1 F capacitors. The scaled
component values are thus: R = 138.46 ; C = 1 F
Finally, add an inverting amplifier stage with a gain of R
f/R
i = 10. As usual, we can arbitrarily select one
of the two resistor values. Because we are already using 138.46 resistors, let R
i = 138.46 ; then,
R
f = 10R
i = 1384.6
The circuit for this cascaded the 4
th
-order low-pass filter is shown in Fig. (next slide).
It has the transfer function:
High Order Op Amp Filters4
4
2 1 0.435
c
rad
s
= −
Case of Low-Pass Filter()
4
7222.39
10
7222.39
Hs
s
=−
+
Mai Linh, PhD
57 1384.6
Fig.: The cascade circuit for the fourth
order low-pass filter design in example
Mai Linh, PhD
Fig.: The cascade circuit for the fourth
order low-pass filter design in example
Mai Linh, PhD
The Bode magnitude plot for this transfer function
58
>> f=10:1:1000;
>> w=2*pi*f;
>> wc=7222.39;
>> h=20*log10(10)-80*log10(abs(1+j*w/wc));
>> semilogx(f,h);
>> grid on;
>> xlabel(‘F[Hz]');
>> ylabel('|H(jf)| dB');4
4
1
(j ) ( 10) ( 10)
1
c
c
c
H
jj
− −
= − = −
+ +
10 10
20log (10) 20log 1
dB
c
j
A
= + +
Mai Linh, PhD
58
>> f=10:1:1000;
>> w=2*pi*f;
>> wc=7222.39;
>> h=20*log10(10)-80*log10(abs(1+j*w/wc));
>> semilogx(f,h);
>> grid on;
>> xlabel(‘F[Hz]');
>> ylabel('|H(jf)| dB');4
4
1
(j ) ( 10) ( 10)
1
c
c
c
H
jj
− −
= − = −
+ +
10 10
20log (10) 20log 1
dB
c
j
A
= + +
Mai Linh, PhD
59
➢There are 3 characteristics of filter
response :
i) Butterworth characteristic
ii) Chebyshev characteristic
iii) Bessel characteristic.
➢Each of the characteristics is
identified by the shape of the
response curve Comparative plots of three types of filter
response characteristics.
Filter response characteristics
Mai Linh, PhD
➢There are 3 characteristics of filter
response :
i) Butterworth characteristic
ii) Chebyshev characteristic
iii) Bessel characteristic.
➢Each of the characteristics is
identified by the shape of the
response curve Comparative plots of three types of filter
response characteristics.
Filter response characteristics
Mai Linh, PhD
60
➢Filter response is characterized by flat
amplitude response in the passband.
➢Provides a roll-off rate of -20
dB/decade/pole.
➢Filters with the Butterworth response
are normally used when all frequencies
in the passband must have the same
gain.
Filter response characteristics
Butterworth characteristic
Mai Linh, PhD
➢Filter response is characterized by flat
amplitude response in the passband.
➢Provides a roll-off rate of -20
dB/decade/pole.
➢Filters with the Butterworth response
are normally used when all frequencies
in the passband must have the same
gain.
Filter response characteristics
Butterworth characteristic
Mai Linh, PhD
61
Filter response characteristics
Butterworth characteristic
Fig.: Butterworth Response
Unity gain
Passband
➢In some cases, we need to design
Butterworth filters just have the unity gain
in the passband (H(s) = 1 or A = 0 dB).
Mai Linh, PhD
Filter response characteristics
Butterworth characteristic
Fig.: Butterworth Response
Unity gain
Passband
➢In some cases, we need to design
Butterworth filters just have the unity gain
in the passband (H(s) = 1 or A = 0 dB).
Mai Linh, PhD
62
Filter response characteristics
Chebyshev characteristic
➢Filter response is characterized by overshoot or ripples in
the passband.
➢ Provides a roll-off rate greater than -20 dB/decade/pole.
➢Filters with the Chebyshev response can be implemented
with fewer poles and less complex circuitry for a given roll-
off rate.
Mai Linh, PhD
Filter response characteristics
Chebyshev characteristic
➢Filter response is characterized by overshoot or ripples in
the passband.
➢ Provides a roll-off rate greater than -20 dB/decade/pole.
➢Filters with the Chebyshev response can be implemented
with fewer poles and less complex circuitry for a given roll-
off rate.
Mai Linh, PhD
63
Filter response characteristics
Bessel characteristic
➢Filter response is characterized by a linear
characteristic, meaning that the phase shift
increases linearly with frequency.
➢Filters with the Bessel response are used
for filtering pulse waveforms without
distorting the shape of waveform.
Mai Linh, PhD
Filter response characteristics
Bessel characteristic
➢Filter response is characterized by a linear
characteristic, meaning that the phase shift
increases linearly with frequency.
➢Filters with the Bessel response are used
for filtering pulse waveforms without
distorting the shape of waveform.
Mai Linh, PhD
64
➢This active filter consists of an amplifier, a negative feedback
circuit & RC circuit.
➢The amplifier and feedback & connected in a non-inverting
configuration.
➢DF (≡ ) is determined by the negative feedback and defined as: 1
2
2
R
DF
R
=−
Fig.: General diagram of active filter
Damping factor (DF)
The damping factor (DF) of an active filter determines which
response characteristic the filter exhibits. If the value for the
damping factor is precisely set, then this tends to make the response
curve flat in the passband of the filter.
By advanced mathematics, values of DF (≡ ) have been derived for various orders of filters to
achieve the maximally flat response of the Butterworth characteristic.
Example: to achieve a 2nd-order Butterworth response, the damping factor must be 2 (1.414).
Mai Linh, PhD
➢This active filter consists of an amplifier, a negative feedback
circuit & RC circuit.
➢The amplifier and feedback & connected in a non-inverting
configuration.
➢DF (≡ ) is determined by the negative feedback and defined as: 1
2
2
R
DF
R
=−
Fig.: General diagram of active filter
Damping factor (DF)
The damping factor (DF) of an active filter determines which
response characteristic the filter exhibits. If the value for the
damping factor is precisely set, then this tends to make the response
curve flat in the passband of the filter.
By advanced mathematics, values of DF (≡ ) have been derived for various orders of filters to
achieve the maximally flat response of the Butterworth characteristic.
Example: to achieve a 2nd-order Butterworth response, the damping factor must be 2 (1.414).
Mai Linh, PhD
65
Sallen-Key filters
The Sallen-Key topology, also known as a voltage control voltage source (VCVS), was first
introduced in 1955 by R. P. Sallen & E. L. Key of MIT’s Lincoln Labs, and hence they are generally
known as Sallen-Key filters. The topology is used to implement second-order active filters that
easily enhance the Q of the filter using controlled positive feedback. The topology is more simple in
comparison to other active filter topologies.
TheSallen and Key Filtertopology which we can use as the basic building blocks for implementing
higher order filter circuits, such as low-pass (LPF), high-pass (HPF) and band-pass (BPF) filter
circuits.
Main advantages:
✓Simplicity and Understanding of their Basic Design
✓The use of a Non-inverting Amplifier to Increase Voltage Gain
✓Can be Easily Cascaded Together
✓Low-pass and High-pass stages can be Cascaded Together
✓Replication of RC Components and Amplifiers
✓Second-order Sallen-key Stages have Steep 40 dB/decade roll-off
than cascaded RC
Mai Linh, PhD
Sallen-Key filters
The Sallen-Key topology, also known as a voltage control voltage source (VCVS), was first
introduced in 1955 by R. P. Sallen & E. L. Key of MIT’s Lincoln Labs, and hence they are generally
known as Sallen-Key filters. The topology is used to implement second-order active filters that
easily enhance the Q of the filter using controlled positive feedback. The topology is more simple in
comparison to other active filter topologies.
TheSallen and Key Filtertopology which we can use as the basic building blocks for implementing
higher order filter circuits, such as low-pass (LPF), high-pass (HPF) and band-pass (BPF) filter
circuits.
Main advantages:
✓Simplicity and Understanding of their Basic Design
✓The use of a Non-inverting Amplifier to Increase Voltage Gain
✓Can be Easily Cascaded Together
✓Low-pass and High-pass stages can be Cascaded Together
✓Replication of RC Components and Amplifiers
✓Second-order Sallen-key Stages have Steep 40 dB/decade roll-off
than cascaded RC
Mai Linh, PhD
66
Fig. (a): Low-Pass Sallen-Key Ckt. Fig. (b): High-Pass Sallen-Key Ckt. Fig. (c): Band-Pass Sallen-Key Ckt.
Fig.: Generalized Sallen-Key Circuit
Sallen-Key filters+
-
R2R1
Rf2
Rb
C1
C2
Two-pole low-
pass circuit
vin
vo vin
+
-
R1
R2
Rf1
Ra
C2
C1
Two-pole high-
pass circuit
vo R3
-
Z4
+
Vf
R4
R3
vin
vO
Z1 Z2 Vp
Vn R5
-
R2
+
C1
R4
R3
C2
R1
vO
vin
Mai Linh, PhD
Fig. (a): Low-Pass Sallen-Key Ckt. Fig. (b): High-Pass Sallen-Key Ckt. Fig. (c): Band-Pass Sallen-Key Ckt.
Fig.: Generalized Sallen-Key Circuit
Sallen-Key filters+
-
R2R1
Rf2
Rb
C1
C2
Two-pole low-
pass circuit
vin
vo vin
+
-
R1
R2
Rf1
Ra
C2
C1
Two-pole high-
pass circuit
vo R3
-
Z4
+
Vf
R4
R3
vin
vO
Z1 Z2 Vp
Vn R5
-
R2
+
C1
R4
R3
C2
R1
vO
vin
Mai Linh, PhD
67
Butterworth Filters
A unity-gain Butterworlh low-pass filter has a transfer function whose magnitude is given by
where n is an integer that denotes the order of the filter
✓The cutoff frequency is
C rad/s for all values of n.
✓If n is large enough, the denominator is always close to unity when <
C
✓In the expression for |H(j)|, the exponent of /
C is always even.
Given an equation for the magnitude of the transfer function, how do we find H(s)?
1)Set
C equal to 1 rad/s in Eq. (*)
2)We use scaling to transform the prototype filter to a filter that meets the given filtering
specifications. Note that s = j
|H(j)|
2
= H(j)H(-j) = H(s)H(-s)() ()
2
*
1
1
C
n
Hj
=
+
❑Butterworth Low-Pass Filter
Mai Linh, PhD
Butterworth Filters
A unity-gain Butterworlh low-pass filter has a transfer function whose magnitude is given by
where n is an integer that denotes the order of the filter
✓The cutoff frequency is
C rad/s for all values of n.
✓If n is large enough, the denominator is always close to unity when <
C
✓In the expression for |H(j)|, the exponent of /
C is always even.
Given an equation for the magnitude of the transfer function, how do we find H(s)?
1)Set
C equal to 1 rad/s in Eq. (*)
2)We use scaling to transform the prototype filter to a filter that meets the given filtering
specifications. Note that s = j
|H(j)|
2
= H(j)H(-j) = H(s)H(-s)() ()
2
*
1
1
C
n
Hj
=
+
❑Butterworth Low-Pass Filter
Mai Linh, PhD
68
With s
2
=
2
➔
Butterworth Filters2
2 2 2 2
1 1 1 1
(j )
1 1 ( ) 1 ( ) 1 ( 1)
n n n n n
H
ss
= = = =
+ + + − + − 2
1
( ) ( )
1 ( 1)
nn
H s H s
s
−=
+−
The procedure for finding H(s) for a given value of n is as follows:
1. Find the roots of the polynomial
1 + (-1)
n
s
2n
= 0
2.Assign the left-half plane roots to H(s) and the right-half plane roots to H(-s).
3. Combine terms in the denominator of H(s) to form 1st- and 2nd-order factors.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
With s
2
=
2
➔
Butterworth Filters2
2 2 2 2
1 1 1 1
(j )
1 1 ( ) 1 ( ) 1 ( 1)
n n n n n
H
ss
= = = =
+ + + − + − 2
1
( ) ( )
1 ( 1)
nn
H s H s
s
−=
+−
The procedure for finding H(s) for a given value of n is as follows:
1. Find the roots of the polynomial
1 + (-1)
n
s
2n
= 0
2.Assign the left-half plane roots to H(s) and the right-half plane roots to H(-s).
3. Combine terms in the denominator of H(s) to form 1st- and 2nd-order factors.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
69
Example: Find the Butterworth transfer functions for n = 2 and n = 3.
Solution
For n = 2, we find the roots of the polynomial: 1 + (-1)
2
s
4
= 0
Rearranging terms we find
➢Therefore, the four roots are:
Roots s
2 and s
3 are in the left-half plane. So:
Butterworth Filters4
1 1 180s= − = 1
2
3
4
1 45 1/ 2 / 2
1 135 1/ 2 / 2
1 225 1/ 2 / 2
1 315 1/ 2 / 2
sj
sj
sj
sj
= = +
= = − +
= = − + −
= = + − 2
1
(s)
( 1/ 2 / 2)( 1/ 2 / 2)
1
21
H
s j s j
ss
=
+ − + +
=
++
❑Butterworth Low-Pass Filter
Mai Linh, PhD
Example: Find the Butterworth transfer functions for n = 2 and n = 3.
Solution
For n = 2, we find the roots of the polynomial: 1 + (-1)
2
s
4
= 0
Rearranging terms we find
➢Therefore, the four roots are:
Roots s
2 and s
3 are in the left-half plane. So:
Butterworth Filters4
1 1 180s= − = 1
2
3
4
1 45 1/ 2 / 2
1 135 1/ 2 / 2
1 225 1/ 2 / 2
1 315 1/ 2 / 2
sj
sj
sj
sj
= = +
= = − +
= = − + −
= = + − 2
1
(s)
( 1/ 2 / 2)( 1/ 2 / 2)
1
21
H
s j s j
ss
=
+ − + +
=
++
❑Butterworth Low-Pass Filter
Mai Linh, PhD
SolutionFor n = 3, we find the roots of the polynomial 1 + (-1)
3
s
6
= 0
➢Therefore, the six roots are:
Roots s
3, s
4, and s
5 are in the left half plane. Thus
Note that in passing that the roots of the Butterworth polynomial are always equally spaced
around the unit circle in the s plane.
70
Butterworth Filters6
1 0 1 360s= = 1
2
3
1 0 1
1 60 1/ 2 3 / 2
1 120 1/ 2 3 / 2
s
sj
sj
= =
= = +
= = − + 4
5
6
1 180 1 0
1 240 1/ 2 3 / 2
1 300 1/ 2 3 / 2
sj
sj
sj
= = − +
= = − + −
= = + − 2
11
()
( 1)( 1)( 1)( 1/ 2 3 / 2)( 1/ 2 3 / 2)
Hs
s s ss s j s j
==
+ + ++ + − + +
❑Butterworth Low-Pass Filter
3
s
6
= 0
➢Therefore, the six roots are:
Roots s
3, s
4, and s
5 are in the left half plane. Thus
Note that in passing that the roots of the Butterworth polynomial are always equally spaced
around the unit circle in the s plane.
70
Butterworth Filters6
1 0 1 360s= = 1
2
3
1 0 1
1 60 1/ 2 3 / 2
1 120 1/ 2 3 / 2
s
sj
sj
= =
= = +
= = − + 4
5
6
1 180 1 0
1 240 1/ 2 3 / 2
1 300 1/ 2 3 / 2
sj
sj
sj
= = − +
= = − + −
= = + − 2
11
()
( 1)( 1)( 1)( 1/ 2 3 / 2)( 1/ 2 3 / 2)
Hs
s s ss s j s j
==
+ + ++ + − + +
❑Butterworth Low-Pass Filter
To assist in the design of Butterworth filters, Table 1 lists the Butterworth polynomials up to
n = 8.
71
Butterworth Filters
N (order) Normalized Denominator Polynomials in Factored Form
1 (s + 1)
2
3
4
5
6
7
8
Table 1 Normalized (so that
c = 1 rad/s) Butterworth polynomials up to 8
th
order.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
n = 8.
71
Butterworth Filters
N (order) Normalized Denominator Polynomials in Factored Form
1 (s + 1)
2
3
4
5
6
7
8
Table 1 Normalized (so that
c = 1 rad/s) Butterworth polynomials up to 8
th
order.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
72
Butterworth Filters
Higher order Butterworth filters are obtained by cascading first and second order Butterworth
filters. This can be shown as follows:
❑Butterworth Low-Pass Filter1
st
order
a = 1
First order 2
nd
order
a1, b1
Second order 1
st
order
a1
Third order
2
nd
order
a2, b2 2
nd
order
a1, b1
Fourth order
2
nd
order
a2, b2 1
st
order
a1
Fifth order
2
nd
order
a2, b2
2
nd
order
a3, b3 2
nd
order
a1, b1
Sixth order
2
nd
order
a2, b2
2
nd
order
a3, b3
Where a
n& b
nare pre-
determined filter coefficients
and these are used to
generate the required transfer
functions.
Mai Linh, PhD
Butterworth Filters
Higher order Butterworth filters are obtained by cascading first and second order Butterworth
filters. This can be shown as follows:
❑Butterworth Low-Pass Filter1
st
order
a = 1
First order 2
nd
order
a1, b1
Second order 1
st
order
a1
Third order
2
nd
order
a2, b2 2
nd
order
a1, b1
Fourth order
2
nd
order
a2, b2 1
st
order
a1
Fifth order
2
nd
order
a2, b2
2
nd
order
a3, b3 2
nd
order
a1, b1
Sixth order
2
nd
order
a2, b2
2
nd
order
a3, b3
Where a
n& b
nare pre-
determined filter coefficients
and these are used to
generate the required transfer
functions.
Mai Linh, PhD
The Butterworth polynomials in Table 1 are the product of 1st- and 2nd-order factors ➔ we
can construct a circuit whose transfer function has a Butterworth polynomial in its denominator
by cascading op amp circuits, each of which provides one of the needed factors.
Example: A block diagram of a cascade is shown in Fig. below, using a 5th-order Butterworth
polynomial (a cascade of 1st-and 2nd-order circuits yielding a fifth-order low-pass
Butterworth filter).
All odd-order Butterworth polynomials include
the factor (s + 1) → all odd-order Butterworth
filter circuits must have a subcircuit that provides
the transfer function H(s) = 1/(s + 1). This is the
transfer function of the 1
st
-order prototype low-
pass op amp filter.
73
Butterworth Filters ❑Butterworth Low-Pass Filter
Mai Linh, PhD
can construct a circuit whose transfer function has a Butterworth polynomial in its denominator
by cascading op amp circuits, each of which provides one of the needed factors.
Example: A block diagram of a cascade is shown in Fig. below, using a 5th-order Butterworth
polynomial (a cascade of 1st-and 2nd-order circuits yielding a fifth-order low-pass
Butterworth filter).
All odd-order Butterworth polynomials include
the factor (s + 1) → all odd-order Butterworth
filter circuits must have a subcircuit that provides
the transfer function H(s) = 1/(s + 1). This is the
transfer function of the 1
st
-order prototype low-
pass op amp filter.
73
Butterworth Filters ❑Butterworth Low-Pass Filter
Mai Linh, PhD
Now, we need to find a circuit that provides a transfer
function of the form: H(s) = 1/(s
2
+ b
1s + 1) (1)
❖The s-domain nodal equations at the noninverting
terminal of the op amp and at the node labeled V
a:
✓Using Cramer's rule
Fig.: This circuit provides the 2nd-order transfer
function (1) for the Butterworth low-pass filter cascade
74
Butterworth Filters11
2
(2 ) (1 )
(1 ) 0
a o i
ao
RC s V RC s V V
V RC s V
+ − + =
− + + = 1
22
11 1 2 2
2
2
10
2 (1 ) 21
11
i
i
o
RC s V
V
V
RC s RC s R C C s RC s
RC s
+
−
==
+ − + ++
−+ 1
2
( ) 0
0
a i a o
ao
oa
o
V V V V
V V sC
RR
VV
V sC
R
−−
+ − + =
−
+=
The Sallen-Key low-pass is a 2nd-order (two-pole) filter. Notice that there are two lowpass RC circuits that
provide a roll-off of -40dB/decade above the cutoff frequency.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
function of the form: H(s) = 1/(s
2
+ b
1s + 1) (1)
❖The s-domain nodal equations at the noninverting
terminal of the op amp and at the node labeled V
a:
✓Using Cramer's rule
Fig.: This circuit provides the 2nd-order transfer
function (1) for the Butterworth low-pass filter cascade
74
Butterworth Filters11
2
(2 ) (1 )
(1 ) 0
a o i
ao
RC s V RC s V V
V RC s V
+ − + =
− + + = 1
22
11 1 2 2
2
2
10
2 (1 ) 21
11
i
i
o
RC s V
V
V
RC s RC s R C C s RC s
RC s
+
−
==
+ − + ++
−+ 1
2
( ) 0
0
a i a o
ao
oa
o
V V V V
V V sC
RR
VV
V sC
R
−−
+ − + =
−
+=
The Sallen-Key low-pass is a 2nd-order (two-pole) filter. Notice that there are two lowpass RC circuits that
provide a roll-off of -40dB/decade above the cutoff frequency.
❑Butterworth Low-Pass Filter
Mai Linh, PhD
❖The transfer function for the circuit:
Finally, set R = 1
❑The Eq. (2) required for the 2nd-order circuit in the Butterworth cascade or the form
Choose capacitor values so that:
The procedure for designing an nth-order Butterworth low-pass filter circuit with a cutoff
frequency of
c = 1 rad/s and a gain of 1 in the passband: Use frequency scaling to calculate
revised capacitor values that yield any other cutoff frequency; and use magnitude scaling to
provide more realistic or practical component values in our design.
➔ We can cascade an inverting amplifier circuit to provide a gain other than 1 in the passband.
75
Butterworth Filters2
12
2
2
1 1 2
1
()
21
o
i
V R C C
Hs
V
ss
RC R C C
==
++ 12
2
1 1 2
1
()
21
CC
Hs
ss
C C C
=
++ 2
1
1
(s)
1
H
s b s
=
++ 1
1 1 2
21
&1b
C C C
==
(2)
❑Butterworth Low-Pass Filter
Mai Linh, PhD
Finally, set R = 1
❑The Eq. (2) required for the 2nd-order circuit in the Butterworth cascade or the form
Choose capacitor values so that:
The procedure for designing an nth-order Butterworth low-pass filter circuit with a cutoff
frequency of
c = 1 rad/s and a gain of 1 in the passband: Use frequency scaling to calculate
revised capacitor values that yield any other cutoff frequency; and use magnitude scaling to
provide more realistic or practical component values in our design.
➔ We can cascade an inverting amplifier circuit to provide a gain other than 1 in the passband.
75
Butterworth Filters2
12
2
2
1 1 2
1
()
21
o
i
V R C C
Hs
V
ss
RC R C C
==
++ 12
2
1 1 2
1
()
21
CC
Hs
ss
C C C
=
++ 2
1
1
(s)
1
H
s b s
=
++ 1
1 1 2
21
&1b
C C C
==
(2)
❑Butterworth Low-Pass Filter
Mai Linh, PhD
76
Example: Design a 4th-order Butterworth low-pass filter with a cutoff frequency of 500 Hz
and a passband gain of 10. Use as many 1 k resistors as possible. Compare the Bode magnitude
plot for this Butterworth filter with that of the identical cascade filter in previous example.
From the Table 1 (slide #65), we find that the 4
th
-order Butterworth polynomial is
(s
2
+ 0.765s + 1)(s
2
+ 1.848s + 1)
Solution:
Need a cascade of two 2nd-order filters to yield the 4th-order transfer function plus an inverting
amplifier circuit for the passband gain of 10. The circuit is shown in Fig. below
Butterworth Filters
❑Butterworth Low-Pass Filter
Mai Linh, PhD
Example: Design a 4th-order Butterworth low-pass filter with a cutoff frequency of 500 Hz
and a passband gain of 10. Use as many 1 k resistors as possible. Compare the Bode magnitude
plot for this Butterworth filter with that of the identical cascade filter in previous example.
From the Table 1 (slide #65), we find that the 4
th
-order Butterworth polynomial is
(s
2
+ 0.765s + 1)(s
2
+ 1.848s + 1)
Solution:
Need a cascade of two 2nd-order filters to yield the 4th-order transfer function plus an inverting
amplifier circuit for the passband gain of 10. The circuit is shown in Fig. below
Butterworth Filters
❑Butterworth Low-Pass Filter
Mai Linh, PhD
Solution (cont.)
➢Let the second stage of the cascade
implement the transfer function for
the polynomial (s
2
+ 1.848s + 1).
➢Let the first stage of the cascade implement the transfer function for the polynomial
(s
2
+ 0.765s + 1).
Use Eq.
The preceding values for C
1, C
2, C
3, & C
4
yield a 4th-order Butterworth filter with a
c
of 1 rad/s. k
f = 3141.6 will move f
c to 500 Hz.
A magnitude scale factor of k
m = 1000 will
permit the use of 1 k resistor in place of 1
resistors.
➢The resulting scaled component values:
R = 1 k; C
1 = 831 nF; C
2 = 121 nF; C
3 = 344 nF; C
4 = 294 nF
77
➢Finally, we need to specify the resistor values in the inverting amplifier stage to yield a
passband gain of 10. Let R
1 = 1 k; then R
f = 10R
1 = 10 k
Butterworth Filters1
1 1 2
21
&1b
C C C
==
C
1 = 2.61 F
C
2 = 0.38 F
C
3 = 1.08 F
C
4 = 0.924 F
❑Butterworth Low-Pass Filter
Mai Linh, PhD
➢Let the second stage of the cascade
implement the transfer function for
the polynomial (s
2
+ 1.848s + 1).
➢Let the first stage of the cascade implement the transfer function for the polynomial
(s
2
+ 0.765s + 1).
Use Eq.
The preceding values for C
1, C
2, C
3, & C
4
yield a 4th-order Butterworth filter with a
c
of 1 rad/s. k
f = 3141.6 will move f
c to 500 Hz.
A magnitude scale factor of k
m = 1000 will
permit the use of 1 k resistor in place of 1
resistors.
➢The resulting scaled component values:
R = 1 k; C
1 = 831 nF; C
2 = 121 nF; C
3 = 344 nF; C
4 = 294 nF
77
➢Finally, we need to specify the resistor values in the inverting amplifier stage to yield a
passband gain of 10. Let R
1 = 1 k; then R
f = 10R
1 = 10 k
Butterworth Filters1
1 1 2
21
&1b
C C C
==
C
1 = 2.61 F
C
2 = 0.38 F
C
3 = 1.08 F
C
4 = 0.924 F
❑Butterworth Low-Pass Filter
Mai Linh, PhD
78
The figure compares the magnitude
responses of the 4th-order identical
cascade low-pass filter from previous
Example and the Butterworth low-pass
filter we just designed. Note that both
filters provide a passband gain of 10 (20
dB) and a cutoff frequency of 500 Hz, but
the Butterworth filter is closer to an ideal
low-pass filter due to its flatter passband
and steeper roll off at the cutoff frequency.
Thus, the Butterworth design is preferred
over the identical cascade design.
Butterworth Filters
❑Butterworth Low-Pass Filter
Mai Linh, PhD
The figure compares the magnitude
responses of the 4th-order identical
cascade low-pass filter from previous
Example and the Butterworth low-pass
filter we just designed. Note that both
filters provide a passband gain of 10 (20
dB) and a cutoff frequency of 500 Hz, but
the Butterworth filter is closer to an ideal
low-pass filter due to its flatter passband
and steeper roll off at the cutoff frequency.
Thus, the Butterworth design is preferred
over the identical cascade design.
Butterworth Filters
❑Butterworth Low-Pass Filter
Mai Linh, PhD
79
Butterworth Filters
❑Butterworth High-Pass Filter
An nth-order Butterworth high-pass filter has a transfer function with the nth-
order Butterworth polynomial in the denominator, just like the nth-order
Butterworth low-pass filter, but in the high-pass filter, the numerator is s
n
Again, we use a cascade approach in designing the Butterworth high-pass
filter. The 1st-order factor is achieved by including a prototype high-pass
filter (Fig. (a)).
To produce the 2nd-order factors in the Butterworth polynomial, we have a
circuit (Fig. (b)) with a transfer function of the form:2
2
1
(s) (3)
1
s
H
s b s
=
++
Fig. (b): A second-order
Butterworth high-pass filter circuit.()
2
2
2
2 1 2
(4)
21
o
i
V s
Hs
V
ss
R C R R C
==
++
Fig. (a): 1st-order prototype filter
circuit.
❖Analyze the circuit in Fig. (b) we obtain:
Mai Linh, PhD
Butterworth Filters
❑Butterworth High-Pass Filter
An nth-order Butterworth high-pass filter has a transfer function with the nth-
order Butterworth polynomial in the denominator, just like the nth-order
Butterworth low-pass filter, but in the high-pass filter, the numerator is s
n
Again, we use a cascade approach in designing the Butterworth high-pass
filter. The 1st-order factor is achieved by including a prototype high-pass
filter (Fig. (a)).
To produce the 2nd-order factors in the Butterworth polynomial, we have a
circuit (Fig. (b)) with a transfer function of the form:2
2
1
(s) (3)
1
s
H
s b s
=
++
Fig. (b): A second-order
Butterworth high-pass filter circuit.()
2
2
2
2 1 2
(4)
21
o
i
V s
Hs
V
ss
R C R R C
==
++
Fig. (a): 1st-order prototype filter
circuit.
❖Analyze the circuit in Fig. (b) we obtain:
Mai Linh, PhD
80
Butterworth Filters
❑Butterworth High-Pass
Fig. (b): A second-order
Butterworth high-pass filter circuit.
➢Let C = 1 F yields:()
2
2
2 1 2
(5)
21
o
i
V s
Hs
V
ss
R R R
==
++
❖Thus, we can realize any second-order factor in a Butterworth
polynomial of the form (s
2
+ b
1s + 1) by including in the
cascade the second-order circuit in Fig. (b) with resistor values
that satisfy:1
2 1 2
21
, and 1 (6)b
R R R
==
➢We can use frequency and magnitude scaling to design a Butterworth high-
pass filter with practical component values and a cutoff frequency other than
1 rad/s. Adding an inverting amplifier to the cascade will accommodate
designs with nonunity passband gains.
Mai Linh, PhD
Butterworth Filters
❑Butterworth High-Pass
Fig. (b): A second-order
Butterworth high-pass filter circuit.
➢Let C = 1 F yields:()
2
2
2 1 2
(5)
21
o
i
V s
Hs
V
ss
R R R
==
++
❖Thus, we can realize any second-order factor in a Butterworth
polynomial of the form (s
2
+ b
1s + 1) by including in the
cascade the second-order circuit in Fig. (b) with resistor values
that satisfy:1
2 1 2
21
, and 1 (6)b
R R R
==
➢We can use frequency and magnitude scaling to design a Butterworth high-
pass filter with practical component values and a cutoff frequency other than
1 rad/s. Adding an inverting amplifier to the cascade will accommodate
designs with nonunity passband gains.
Mai Linh, PhD
81
Narrow Bandpass & Bandreject Filters
❖Case of broadband filters or low Q filters that constructed from simple high-pass & low-
pass filters.
This broadband filter with the transfer functions for cascaded bandpass & parallel bandreject
fillers that have discrete real poles. The synthesis techniques work best for cutoff frequencies that
are widely separated and therefore yield the lowest quality factors (Q).
✓Consider the transfer function that results:
This is standard form of the transfer function of a bandpass filter
The bandwidth and center frequency
o for the broadband BPF:
Thus with discrete real poles, the highest quality bandpass filter (or bandreject filter) we can
achieve has Q = 1/2.2 2 2 2
0.5
()
2
cc
c c c c c
sss
Hs
s s s s s s
− −
= = =
+ + + + + +
22
2
c
oc
=
= 1
22
oc
c
Q
= = =
Mai Linh, PhD
Narrow Bandpass & Bandreject Filters
❖Case of broadband filters or low Q filters that constructed from simple high-pass & low-
pass filters.
This broadband filter with the transfer functions for cascaded bandpass & parallel bandreject
fillers that have discrete real poles. The synthesis techniques work best for cutoff frequencies that
are widely separated and therefore yield the lowest quality factors (Q).
✓Consider the transfer function that results:
This is standard form of the transfer function of a bandpass filter
The bandwidth and center frequency
o for the broadband BPF:
Thus with discrete real poles, the highest quality bandpass filter (or bandreject filter) we can
achieve has Q = 1/2.2 2 2 2
0.5
()
2
cc
c c c c c
sss
Hs
s s s s s s
− −
= = =
+ + + + + +
22
2
c
oc
=
= 1
22
oc
c
Q
= = =
Mai Linh, PhD
82
❖Narrow Bandpass: (high Q values) we need an Op-Amp circuit that can produce a transfer
function with complex conjugate poles (in the figure).
@ inverting input
@ node a:
(transfer function V
0 /V
i)
The standard form of the transfer function for a bandpass filter:
Narrow Bandpass & Bandreject Filters2
2
3 1 3
2 1 1
;;
o
eq
A
RCRCRRC
= = = ()
22
o
As
Hs
ss
−
=
++ 12
12
12
||
eq
RR
R R R
RR
==
+ ()
1
2
2
33
21
eq
s
RC
Hs
ss
R C R R C
−
=
++ ( )
1 1 2 1
1 2 /
i a o
V sRC R R V sRCV= + + − 12
1/ 1/
i a a o a a
V V V V V V
R sC sC R
−−
= + + 3
o
a
V
V
sR C
−
= 3
1/
ao
VV
sC R
−
=
Mai Linh, PhD
❖Narrow Bandpass: (high Q values) we need an Op-Amp circuit that can produce a transfer
function with complex conjugate poles (in the figure).
@ inverting input
@ node a:
(transfer function V
0 /V
i)
The standard form of the transfer function for a bandpass filter:
Narrow Bandpass & Bandreject Filters2
2
3 1 3
2 1 1
;;
o
eq
A
RCRCRRC
= = = ()
22
o
As
Hs
ss
−
=
++ 12
12
12
||
eq
RR
R R R
RR
==
+ ()
1
2
2
33
21
eq
s
RC
Hs
ss
R C R R C
−
=
++ ( )
1 1 2 1
1 2 /
i a o
V sRC R R V sRCV= + + − 12
1/ 1/
i a a o a a
V V V V V V
R sC sC R
−−
= + + 3
o
a
V
V
sR C
−
= 3
1/
ao
VV
sC R
−
=
Mai Linh, PhD
83
1)Consider the prototype version of the circuit:
2)Finally, use the scaling is used to specify practical values for the circuit
components.
Narrow Bandpass & Bandreject Filters1 / & 1
o
rad s C F== 1
2
2
3
/
/(2)
2
RQA
RQQA
RQ
=
=−
=
Note: A is passband gain |H(j)|, from slide #37
Mai Linh, PhD
1)Consider the prototype version of the circuit:
2)Finally, use the scaling is used to specify practical values for the circuit
components.
Narrow Bandpass & Bandreject Filters1 / & 1
o
rad s C F== 1
2
2
3
/
/(2)
2
RQA
RQQA
RQ
=
=−
=
Note: A is passband gain |H(j)|, from slide #37
Mai Linh, PhD
84
Example: Design a bandpass filter, using the circuit in Fig. below which has a
center frequency of 3000 Hz, a quality factor of 10, and a passband gain of 2.
Use 0.01 F capacitors in your design. Compute the transfer function of your
circuit and sketch a Bode plot of its magnitude response.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Example: Design a bandpass filter, using the circuit in Fig. below which has a
center frequency of 3000 Hz, a quality factor of 10, and a passband gain of 2.
Use 0.01 F capacitors in your design. Compute the transfer function of your
circuit and sketch a Bode plot of its magnitude response.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
85
SolutionIn the prototype circuit, we easily calculate:
Narrow Bandpass & Bandreject Filters
The scaling factor are
The transfer function:
Bode plot of its magnitude response1
2
3
10 / 2 5,
10 / (200 2) 10 /198,
2(10) 20
R
R
R
==
= − =
== 8
6000 10 /
f m f
k k k== 1
2
3
26.5
268.0
106.1
Rk
R
Rk
=
=
= 26
3770
()
1885.0 355 10
s
Hs
ss
−
=
+ +
Mai Linh, PhD
SolutionIn the prototype circuit, we easily calculate:
Narrow Bandpass & Bandreject Filters
The scaling factor are
The transfer function:
Bode plot of its magnitude response1
2
3
10 / 2 5,
10 / (200 2) 10 /198,
2(10) 20
R
R
R
==
= − =
== 8
6000 10 /
f m f
k k k== 1
2
3
26.5
268.0
106.1
Rk
R
Rk
=
=
= 26
3770
()
1885.0 355 10
s
Hs
ss
−
=
+ +
Mai Linh, PhD
86
❖Bandreject filter that combines low-pass and high-pass filter components with a summing
amplifier
A high-Q active bandreject filter aka.
twin-T notch filter
Summing the currents away from the noninverting input
terminal of the top op amp gives
@ node a
@ node b
To solve the set of 3 equations (1), (2), and (3): use Cramer's rule to solve for V
0:
Narrow Bandpass & Bandreject Filters( )( )
2( )
0
ao
a i a o
VV
V V sC V V sC
R
−
− + − + = 2 2 2 (1)
a o i
V sCR V sCR sCRV+ − + = ( )20
b i b o
bo
V V V V
V V sC
RR
−−
+ + − = 2 2 1 2 (2)
b o i
V RCs V RCs V+ − + = ( ) ( )0 1 0 (3)
ob
o a a b o
VV
V V sC sRCV V sRC V
R
−
− + = − − + + = a
b
Mai Linh, PhD
❖Bandreject filter that combines low-pass and high-pass filter components with a summing
amplifier
A high-Q active bandreject filter aka.
twin-T notch filter
Summing the currents away from the noninverting input
terminal of the top op amp gives
@ node a
@ node b
To solve the set of 3 equations (1), (2), and (3): use Cramer's rule to solve for V
0:
Narrow Bandpass & Bandreject Filters( )( )
2( )
0
ao
a i a o
VV
V V sC V V sC
R
−
− + − + = 2 2 2 (1)
a o i
V sCR V sCR sCRV+ − + = ( )20
b i b o
bo
V V V V
V V sC
RR
−−
+ + − = 2 2 1 2 (2)
b o i
V RCs V RCs V+ − + = ( ) ( )0 1 0 (3)
ob
o a a b o
VV
V V sC sRCV V sRC V
R
−
− + = − − + + = a
b
Mai Linh, PhD
87
The standard form for the transfer function of a
bandreject filter
The transfer function:
Equating both Equ.
In this circuit, we have three parameters (R, C, and ) and two design constraints (
0 and β).
Thus one parameter is chosen arbitrarily. Usually, we select capacitor value.
Narrow Bandpass & Bandreject Filters( )
2 2 2
2 2 2
2( 1) 0
0 2( 1)
110
2( 1) 0 ( 2 ) 4 (1 ) 1
0 2( 1) (2 1)
11
i
i
i
o
RCs sCRV
RCs V
R C s VRCs
V
RCs RCs R C s RC s
RCs RCs
RCs RCs
+
+
+−−
==
+ − + + − +
+ − +
− − + 2
22
2
22
1
(s)
4(1 ) 1
o
i
s
V RC
H
V
s
RC R C
+
==
−
++
22
22
(s)
o
o
s
H
ss
+
=
++ 2
22
1 4(1 )
,.
o
R C RC
−
==
Mai Linh, PhD
The standard form for the transfer function of a
bandreject filter
The transfer function:
Equating both Equ.
In this circuit, we have three parameters (R, C, and ) and two design constraints (
0 and β).
Thus one parameter is chosen arbitrarily. Usually, we select capacitor value.
Narrow Bandpass & Bandreject Filters( )
2 2 2
2 2 2
2( 1) 0
0 2( 1)
110
2( 1) 0 ( 2 ) 4 (1 ) 1
0 2( 1) (2 1)
11
i
i
i
o
RCs sCRV
RCs V
R C s VRCs
V
RCs RCs R C s RC s
RCs RCs
RCs RCs
+
+
+−−
==
+ − + + − +
+ − +
− − + 2
22
2
22
1
(s)
4(1 ) 1
o
i
s
V RC
H
V
s
RC R C
+
==
−
++
22
22
(s)
o
o
s
H
ss
+
=
++ 2
22
1 4(1 )
,.
o
R C RC
−
==
Mai Linh, PhD
88
Narrow Bandpass & Bandreject Filters11
, 1 1
44
oo
R
CQ
= = − = −
Example: Design a high-Q active
bandreject filter (based on the circuit in
Fig.) with a center frequency of 5000
rad/s and a bandwidth (β) of 1000
rad/s. Use 1 F capacitors in your
design.
Mai Linh, PhD
Narrow Bandpass & Bandreject Filters11
, 1 1
44
oo
R
CQ
= = − = −
Example: Design a high-Q active
bandreject filter (based on the circuit in
Fig.) with a center frequency of 5000
rad/s and a bandwidth (β) of 1000
rad/s. Use 1 F capacitors in your
design.
Mai Linh, PhD
89
Solution
In the bandreject prototype filter,
0 = 1 rad/s, R = 1 Ω, and C = 1 F.
With Q =
0/β = 5000/1000 = 5, we found that:
R = 1 / (
0C) = 1 / (5000×10
-6
) = 200(Ω);
= 1- β/4
0 =0.95
Thus: we need resistors:
200 (R), 100 (R/2), 190 (R), and
10 [(1 - )R].
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Solution
In the bandreject prototype filter,
0 = 1 rad/s, R = 1 Ω, and C = 1 F.
With Q =
0/β = 5000/1000 = 5, we found that:
R = 1 / (
0C) = 1 / (5000×10
-6
) = 200(Ω);
= 1- β/4
0 =0.95
Thus: we need resistors:
200 (R), 100 (R/2), 190 (R), and
10 [(1 - )R].
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
90
Solution (cont.)
Bode magnitude plot is:
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Solution (cont.)
Bode magnitude plot is:
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
91
Problem 4: Design an active bandpass filter (the figure below) with Q = 8, A = 5,
and
0 = 1000 rad/s. Use 1 F capacitors and specify the values of all resistors.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Problem 4: Design an active bandpass filter (the figure below) with Q = 8, A = 5,
and
0 = 1000 rad/s. Use 1 F capacitors and specify the values of all resistors.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
92
Solution:
The transfer function:
With
From
From
Narrow Bandpass & Bandreject Filters1
22
2 12
2
3 1 2 3
1
()
2
o
s
RC As
Hs
ssRR
ss
R C R R R C
−
==
++ +
++
3
3
22
1000
125 ( / )
8
o
R
R C C
rad s
Q
= =
= = = 6
3
2 10
16 ( )
125 1
Rk
= =
1 6
1
1 1 1
1.6 ( )
5 125 1 10
A R k
RC A C
−
= = = =
()
26 1 2 2
22
6
1 2 3
2
1600
10
1600 16000 10
o
R R R
R R R C
R
−
++
= =
(Based on slide #79)
Mai Linh, PhD
Solution:
The transfer function:
With
From
From
Narrow Bandpass & Bandreject Filters1
22
2 12
2
3 1 2 3
1
()
2
o
s
RC As
Hs
ssRR
ss
R C R R R C
−
==
++ +
++
3
3
22
1000
125 ( / )
8
o
R
R C C
rad s
Q
= =
= = = 6
3
2 10
16 ( )
125 1
Rk
= =
1 6
1
1 1 1
1.6 ( )
5 125 1 10
A R k
RC A C
−
= = = =
()
26 1 2 2
22
6
1 2 3
2
1600
10
1600 16000 10
o
R R R
R R R C
R
−
++
= =
(Based on slide #79)
Mai Linh, PhD
93
Solution:
Solving for R
2:
Narrow Bandpass & Bandreject Filters( )
()
6
2
22 5
2
1600 10
246 16000
256 10
65.04
R
RR
R
+
= =
=
Mai Linh, PhD
Solution:
Solving for R
2:
Narrow Bandpass & Bandreject Filters( )
()
6
2
22 5
2
1600 10
246 16000
256 10
65.04
R
RR
R
+
= =
=
Mai Linh, PhD
94
Problem 5:
Design an active unity-gain bandreject filter with
0 = 1000 rad/s and Q = 4.
Use 2 F capacitors, and specify the values of R and
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Problem 5:
Design an active unity-gain bandreject filter with
0 = 1000 rad/s and Q = 4.
Use 2 F capacitors, and specify the values of R and
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
95
Solution:
Transfer function, center frequency, and bandwidth:
Narrow Bandpass & Bandreject Filters()
( )
()
()
2 2 2
22
22
2
22
1/
41 1
411
;
o
o
o
s R C s
Hs
ss
ss
RC R C
RC RC
+ +
==
++−
++
−
== ()
()
()
()
6
6
11
500
1000 2 10
411000
250 250
4
4 1 250 250 500 2 10 0.25
0.25
1 0.0625; 0.9375
4
o
o
R
C
Q RC
RC
−
−
= = =
−
= = = =
− = = =
− = = =
Mai Linh, PhD
Solution:
Transfer function, center frequency, and bandwidth:
Narrow Bandpass & Bandreject Filters()
( )
()
()
2 2 2
22
22
2
22
1/
41 1
411
;
o
o
o
s R C s
Hs
ss
ss
RC R C
RC RC
+ +
==
++−
++
−
== ()
()
()
()
6
6
11
500
1000 2 10
411000
250 250
4
4 1 250 250 500 2 10 0.25
0.25
1 0.0625; 0.9375
4
o
o
R
C
Q RC
RC
−
−
= = =
−
= = = =
− = = =
− = = =
Mai Linh, PhD
96
Exercise 1
Find the transfer function V
0/V
i for the circuit shown in Fig. if Z
f is the equivalent
impedance of the feedback circuit, Z
i is the equivalent impedance of the input
circuit, and the operational amplifier is ideal.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Exercise 1
Find the transfer function V
0/V
i for the circuit shown in Fig. if Z
f is the equivalent
impedance of the feedback circuit, Z
i is the equivalent impedance of the input
circuit, and the operational amplifier is ideal.
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
97
Exercise 1
Sol.: summing the currents at the inverting input node yields
Narrow Bandpass & Bandreject Filters()
00
0
io
if
oi
fi
fo
ii
VV
ZZ
VV
ZZ
ZV
Hs
VZ
−−
+=
= −
= = −
Mai Linh, PhD
Exercise 1
Sol.: summing the currents at the inverting input node yields
Narrow Bandpass & Bandreject Filters()
00
0
io
if
oi
fi
fo
ii
VV
ZZ
VV
ZZ
ZV
Hs
VZ
−−
+=
= −
= = −
Mai Linh, PhD
98
Exercise 2
a)Use the results of exercise 1 to find the transfer function of the
circuit showing fig.
b)What is the gain of the circuit as → 0?
c)What is the gain of the circuit as → ∞?
d)Do your answers to (b) and (c) make sense in terms of known
circuit behavior?
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
Exercise 2
a)Use the results of exercise 1 to find the transfer function of the
circuit showing fig.
b)What is the gain of the circuit as → 0?
c)What is the gain of the circuit as → ∞?
d)Do your answers to (b) and (c) make sense in terms of known
circuit behavior?
Narrow Bandpass & Bandreject Filters
Mai Linh, PhD
99
Exercise 2 Sol.:
Narrow Bandpass & Bandreject Filters()
( )
( )
22 2
2 2 2 2
2
22
1/
1/ 1
1/
1/
f
R sC R
aZ
R sC R C s
C
s R C
==
++
=
+
Likewise 1
11
1/
1/
i
C
Z
s RC
=
+ ()
( )
( )()
( )
( )
1 1 12 1 1
2 2 1 2 2 2
1/1/ 1/
1/ 1/ 1/
C s R CC s R C
Hs
s R C C C s R C
+−+
= = −
++
Mai Linh, PhD
Exercise 2 Sol.:
Narrow Bandpass & Bandreject Filters()
( )
( )
22 2
2 2 2 2
2
22
1/
1/ 1
1/
1/
f
R sC R
aZ
R sC R C s
C
s R C
==
++
=
+
Likewise 1
11
1/
1/
i
C
Z
s RC
=
+ ()
( )
( )()
( )
( )
1 1 12 1 1
2 2 1 2 2 2
1/1/ 1/
1/ 1/ 1/
C s R CC s R C
Hs
s R C C C s R C
+−+
= = −
++
Mai Linh, PhD
100
Exercise 2Sol. (cont.):
(d) as → 0 the 2 capacitor branches become open, and the circuit reduces to a resistive
inverting amplifier having a gain of –R
2/R
1.
As → ∞ the 2 capacitor branches approach a short circuit and, in this case, we encounter
an indeterminate situation; namely v
n → v
i but v
n = 0 because of the ideal op amp. At the
same time, the gain of the ideal opamp is infinite, so we have the indeterminate form 0 → ∞.
Although = ∞ is indeterminate we can reason that for finite large values of , H(j) will
approach – C
1/C
2 in value. In other words, the circuit approaches a purely capacitive
inverting amplifier with a gain of
Narrow Bandpass & Bandreject Filters()()
( )
( )
111 1 2 2 2
2 2 2 2 1 1 1
1/
1/
j R CC C R C R
b H j
C j R C C R C R
+−
= = − = −
+
()()
11
22
CCj
c H j
C j C
= − = −
( )( )
2 1 1 2
1/ / 1/ /j C j C or C C−−
Mai Linh, PhD
Exercise 2Sol. (cont.):
(d) as → 0 the 2 capacitor branches become open, and the circuit reduces to a resistive
inverting amplifier having a gain of –R
2/R
1.
As → ∞ the 2 capacitor branches approach a short circuit and, in this case, we encounter
an indeterminate situation; namely v
n → v
i but v
n = 0 because of the ideal op amp. At the
same time, the gain of the ideal opamp is infinite, so we have the indeterminate form 0 → ∞.
Although = ∞ is indeterminate we can reason that for finite large values of , H(j) will
approach – C
1/C
2 in value. In other words, the circuit approaches a purely capacitive
inverting amplifier with a gain of
Narrow Bandpass & Bandreject Filters()()
( )
( )
111 1 2 2 2
2 2 2 2 1 1 1
1/
1/
j R CC C R C R
b H j
C j R C C R C R
+−
= = − = −
+
()()
11
22
CCj
c H j
C j C
= − = −
( )( )
2 1 1 2
1/ / 1/ /j C j C or C C−−
Mai Linh, PhD
101
Example of high- & low-pass filters
❑Recording studio monitoring or large public
systems (i.e., concert systems) utilize active
crossovers composed of active filters. Before the
audio signal is fed to a power amplifier, it is split
into two or more frequency bands. The resulting
signals each feed their own power
amplifier/loudspeaker section.
Just an example of commercial electronic Crossover: Behringer Sonic Exciter SX3040 V2
(from amazon.com)
Mai Linh, PhD
Example of high- & low-pass filters
❑Recording studio monitoring or large public
systems (i.e., concert systems) utilize active
crossovers composed of active filters. Before the
audio signal is fed to a power amplifier, it is split
into two or more frequency bands. The resulting
signals each feed their own power
amplifier/loudspeaker section.
Just an example of commercial electronic Crossover: Behringer Sonic Exciter SX3040 V2
(from amazon.com)
Mai Linh, PhD
102
Fig.: Electronic crossover systemHigh-pass
Low-pass
Power
amplifiers
Treble
Bass
A typical two-way system might be crossed at 800 Hz. It means that frequencies above 800
Hz go to a specialized high frequency-transducer, but frequencies below 800 Hz must go to a
specialized low-frequency transducer.
Thus, the crossover network is a combination of an 800 Hz low-pass filter, and an 800 Hz
high-pass filter. The filter order and alignment vary considerably depending on the
application.
A block diagram of this approach is shown in the
Fig. This is a two-way system. Large concert
sound reinforcement systems may break the audio
spectrum into four or five segments. The resulting
system will show lower distortion and higher
output levels than a passively-crossed system.
Example of high- & low-pass filters
Mai Linh, PhD
Fig.: Electronic crossover systemHigh-pass
Low-pass
Power
amplifiers
Treble
Bass
A typical two-way system might be crossed at 800 Hz. It means that frequencies above 800
Hz go to a specialized high frequency-transducer, but frequencies below 800 Hz must go to a
specialized low-frequency transducer.
Thus, the crossover network is a combination of an 800 Hz low-pass filter, and an 800 Hz
high-pass filter. The filter order and alignment vary considerably depending on the
application.
A block diagram of this approach is shown in the
Fig. This is a two-way system. Large concert
sound reinforcement systems may break the audio
spectrum into four or five segments. The resulting
system will show lower distortion and higher
output levels than a passively-crossed system.
Example of high- & low-pass filters
Mai Linh, PhD
103
➢The basic circuit layout of Electronic crossover
system is shown in Fig.
➢To design the crossover system, using the equal
component value (C
1A = C
2A = C
1B = C
2B = 1 F, and
R
1A = R
2A = R
1B = R
2B = 1 ).
➢For second-order Butterworth filters, the damping
factor is found to be 1.414, and the frequency factor
is unity (indicating that f
c and f
3dB are the same).
Note that the design for both halves is almost the
same. Both sections show an f
c of 800 Hz, and a
damping factor of 1.414.
➢With identical characteristics, it follows that the
component values will be the same in both circuits.vin
+
-
R1A
R2A
Rf1
Ra
+
-
R2BR1B
Rf2
Rb
C1B
C2B
C2A
C1A
Treble
Bass
Two-pole low-
pass circuit
Two-pole high-
pass circuit
A
B
Fig.: Basic filter sub-circuits for
crossover of Example
Example of high- & low-pass filters
Mai Linh, PhD
➢The basic circuit layout of Electronic crossover
system is shown in Fig.
➢To design the crossover system, using the equal
component value (C
1A = C
2A = C
1B = C
2B = 1 F, and
R
1A = R
2A = R
1B = R
2B = 1 ).
➢For second-order Butterworth filters, the damping
factor is found to be 1.414, and the frequency factor
is unity (indicating that f
c and f
3dB are the same).
Note that the design for both halves is almost the
same. Both sections show an f
c of 800 Hz, and a
damping factor of 1.414.
➢With identical characteristics, it follows that the
component values will be the same in both circuits.vin
+
-
R1A
R2A
Rf1
Ra
+
-
R2BR1B
Rf2
Rb
C1B
C2B
C2A
C1A
Treble
Bass
Two-pole low-
pass circuit
Two-pole high-
pass circuit
A
B
Fig.: Basic filter sub-circuits for
crossover of Example
Example of high- & low-pass filters
Mai Linh, PhD
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