Analogue Filter Design
Module: SE2A2 Signals and Telecoms
Lecturer: James Grimbleby
y
URL: http://www.personal.rdg.ac.uk/~stsgrimb/
email: jj.b.grimbleby
g
y reading.ac.uk
g
Number of Lectures: 5
Reference text:
Design with Operational Amplifiers and Analog Integrated
Circuits (3rd edition)
Sergio Franco
McGraw-Hill 2003 ISBN 0071207031
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 1
Analogue Filter Design
Reference text:
Design with Operational
Amplifiers and Analog
Integrated Circuits (3rd ed)
Sergio Franco
McGraw-Hill 2003
ISBN 0071207031
Approx
pp o £
£43
3
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 2
Analogue Filter Design Syllabus
This course of lectures deals with the design of passive and
active analogue filters
Th topics
The
t i that
th t will
ill b
be covered
d iinclude:
l d
q
y
filter approximations
pp
Frequency-domain
Filter transformations
Passive equally-terminated
q
y
ladder filters
Passive ladder filters from filter design tables
Active filters – cascade synthesis
Component value sensitivity
Copying methods
Generalised immittance converter (GIC)
Inductor simulation using GICs
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 3
Analogue Filter Design Prerequities
You should be familiar with the following topics:
SE1EA5: Electronic Circuits
Circuit analysis using Kirchhoff’s Laws
Thévenin and Norton's theorems
The Superposition Theorem
Semiconductor devices
Complex impedances
Frequency response function – gain and phase
The infinite-gain approximation
SE1EC5: Engineering Mathematics
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 4
Analogue Filter Design
Filters are normally used to modify the frequency spectrum of
a signal and are therefore specified in the frequency domain:
Gain
(dB)
Max pass-band gain
Min pass-band
pass band gain
Max stop-band gain
Max
p
stopband
gain
g
log(freq)
og( eq)
Stop-band
g
edge
James Grimbleby
Pass-band
edges
g
Stop-band
edge
g
School of Systems Engineering - Electronic Engineering
Slide 5
Analogue Filter Design
The design procedure for analogue filters has two distinct
stages
In the first stage a frequency response function H(jω) is
derived which meets the specification
In the
e seco
second
d sstage
age a
an e
electronic
ec o c ccircuit
cu is
s des
designed
g ed to
o
generate the required frequency response function
Filter circuits can be constructed entirely from passive
components or can contain active elements such as
operational amplifiers
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 6
Frequency-Domain
Frequency
Domain Filters
The design of frequency-domain
frequency domain filters usually starts by
deriving a low-pass prototype filter normalised to a cut-off
frequency
q
y of 1 rad/s
This p
prototype
yp filter is then transformed to the required
q
type
yp
and cut-off frequency:
Low-pass
Low
pass
High-pass
Band-pass
Band-stop
This procedure results in a frequency response function
which meets the specification
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 7
Frequency-Domain
Frequency
Domain Filters
An ideal normalised sharp
p cut-off low-pass
p
filter has a
frequency response:
Gain
|H(jω)|=1 0
|H(jω)|=1.0
00
0.0
0.0
Angular
ω =1.0 frequency
Unfortunately
Unfort
natel such
s ch a filter is not realisable and it is
necessary to use approximations to the ideal response
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 8
Frequency-Domain
Frequency
Domain Filters
The frequency response function of a circuit containing no
distributed elements is a rational function of jω :
H ( jω ) =
a0 + a1( jω ) + a2 ( jω )2 + .. + an ( jω )n
b0 + b1( jω ) + b2 ( jω )2 + .. + bn ( jω )n
The first stage in the design is to choose suitable values for
the order n and the coefficients a0..an and b0..bn
Five different approximations will be considered: Butterworth
Butterworth,
Chebychev, Inverse Chebychev, Elliptic and Bessel
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 9
Butterworth Approximation
The Butterworth
B tter orth appro
approximation
imation gain is ma
maximally
imall flat
That is to say the gain in the pass-band
pass band (below ω=1) is as flat
as possible
The gain falls off monotonically in both pass-band (below
ω=1) and stop-band (above ω=1)
The gain of a Butterworth filter of order n is give by:
H ( jω ) =
James Grimbleby
1
1 + ω 2n
School of Systems Engineering - Electronic Engineering
Slide 10
Butterworth Approximation
Gain
(dB)
0
-3
gs
ω=1 ω= ωs
James Grimbleby
School of Systems Engineering - Electronic Engineering
log(freq)
Slide 11
Butterworth Approximation
The poles pk of a Butterworth frequency response function
are given by:
pk = xk + jy k
where:
(2k − 1)
xk = − sin
π
2n
k = 1, 2, K, n
(2k − 1)
y k = cos
π
2n
Values of k from 1 to n are substituted into this formula
giving the n poles
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 12
Butterworth Approximation
The poles of the Butterworth
B tter orth response p1, p2, ... , pn are
then combined to give the Butterworth frequency response
function:
1
H ( jω ) =
( jω − p1)( jω − p2 )..( jω − pn )
The Butterworth approximation is an all
all-pole
pole response
That is, the numerator consists simply of a constant and
there are no zeros
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 13
Butterworth Approximation
3rd-order Butterworth approximation:
(2k − 1)
( 2k − 1)
pk = − sin
π
π + j cos
6
6
k
1
2
3
James Grimbleby
pk
– 0.5 +j 0.866
– 1.0
1 0 +j 0.0
00
– 0.5 – j 0.866
School of Systems Engineering - Electronic Engineering
Slide 14
Butterworth Approximation
Poles of H(jω):
(j )
– 0.5
+ j 0.866
– 0.5
– j 0.866
– 1.0
+ j 0.0
Combining the poles:
1
H ( jω ) =
( jω + 0.5 − j0.866)
j0 866)( jω + 0.5 + j0.866)
j0 866)( jω + 1.0)
1
=
(( jω )2 + jω + 1.0)( jω + 1.0)
1
=
( jω )3 + 2.0( jω )2 + 2.0jω + 1.0
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 15
Butterworth Approximation
What order n is required to meet specification?
Gain (dB)
0
-3
gs
ω=1 ω= ωs
James Grimbleby
School of Systems Engineering - Electronic Engineering
log(freq)
Slide 16
Butterworth Approximation
The gain g of a Butterworth filter, expressed in dB, is given
b
by:
g = 20 log10 H ( jω )
= 20 log
l 10 (1 + ω 2n )-1/2
= −10 log10 (1 + ω 2n )
The stop-band edge is at ωs, and the stop-band gain is
required to be less than gs dB:
g s ≥ −10 log
l 10 (1 + ωs 2n ) ≈ −10 log
l 10 ωs 2n
g s ≥ −20n log10 ωs
− gs
n≥
20 log10 ωs
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 17
Butterworth Approximation
Example:
p
Pass-band gain:
Stop-band gain:
Pass-band edge:
Stop-band edge:
gp ≥ -3 dB
gs ≤ -40 dB
ωp = 1.0
ωs = 1.5
Using the formula for the filter order:
− gs
40.0
n≥
=
20 log10 ωs 20 log10 1.5
≥ 11.36
Thus a Butterworth approximation of order 12 is required to
meet the specification
p
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 18
Butterworth Approximation
Gain
n (dB)
Response of 12th-order
12th order Butterworth approximation:
0 01
0.01
James Grimbleby
gp ≥ -3 dB
gs ≤ -40 dB
ωp = 1.0
ωs = 1.5
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
10
1.0
Slide 19
Chebychev Approximation
The Chebychev
Cheb che approximation
appro imation gain oscillates bet
between
een 0
dB and gp dB in the pass-band (below ω=1)
In the stop-band (above ω=1) the gain falls off
monotonically
The Chebychev approximation is not a single
approximation for each n, but a group of approximations
with different values of the pass-band
pass band ripple gp
Like the Butterworth approximation, the Chebychev
approximation is an all-pole response
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 20
Chebychev Approximation
Gain
(dB)
0
gp
gs
ω=1 ω= ωs
James Grimbleby
l (f )
log(freq)
School of Systems Engineering - Electronic Engineering
Slide 25
Chebychev Approximation
Gain
n (dB)
Response of 6th
6th-order
order Chebychev approximation:
0.01
James Grimbleby
gp ≥ -3 dB
gs ≤ -40 dB
ωp = 1.0
ωs = 1.5
15
Frequency
q
y ((Hz))
School of Systems Engineering - Electronic Engineering
1.0
Slide 26
Inverse Chebychev Approx
The Inverse Chebychev approximation gain falls off
monotonically in the pass-band (below ω=1)
In the stop-band (above ω=1) the gain at first falls to -∞ dB,
and
d th
then oscillates
ill t b
between
t
-∞ dB and
d gs dB
The Inverse Chebychev approximation is not a single
approximation for each n, but a group of approximations with
different values of the stop-band
stop band ripple gs
The Inverse Chebychev approximation has imaginary zeros
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 27
Inverse Chebychev Approx
Gain
(dB)
0
gp
gs
ω=1 ω= ωs
James Grimbleby
log(freq)
School of Systems Engineering - Electronic Engineering
Slide 28
Inverse Chebychev Approx
Gain (dB)
Response of 6th-order Inverse Chebychev approximation:
0 01
0.01
James Grimbleby
gp ≥ -3 dB
gs ≤ -40 dB
ωp = 1.0
ωs = 1.5
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
10
1.0
Slide 29
Elliptic Approximation
The elliptic response gain oscillates between 0 and gp dB in
the pass
pass-band
band (below ω=1)
The gain in the stop-band
stop band (above ω=1) oscillates between
gs dB and -∞ dB
The elliptic approximation has imaginary zeros
For a given filter specification the elliptic approximation gives
the lowest order frequency response function
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 30
Elliptic Approximation
Gain
(dB)
0
gp
gs
ω=1 ω= ωs
James Grimbleby
log(freq)
School of Systems Engineering - Electronic Engineering
Slide 31
Elliptic Approximation
Gain (dB)
Response of 4th
4th-order
order elliptic approximation:
0 01
0.01
James Grimbleby
gp ≥ -3 dB
gs ≤ -40 dB
ωp = 1.0
ωs = 1.5
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
10
1.0
Slide 32
Time-Domain
Time
Domain Response
Unit-step
Unit
step response of 6th
6th-order
order Chebychev:
g(t))
1.0
0.0
James Grimbleby
1 0s
1.0s
Time
3 0s
3.0s
School of Systems Engineering - Electronic Engineering
Slide 33
Bessel Approximation
The Bessel appro
approximation
imation is used
sed where
here a freq
frequency-domain
enc domain
filter is required which also has a good time-domain
behaviour
All filters generate a frequency
frequency-dependent
dependent phase shift
In a Bessel filter the phase varies approximately linearly with
frequency and the different frequency components are
delayed by the same amount
A time
time-domain
domain waveform is therefore delayed, but is not
seriously distorted
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 34
Bessel Approximation
The Bessel approximation has an all-pole frequency
response function:
a0
H ( jω ) =
b0 + b1( jω ) + b2 ( jω )2 + .. + bn ( jω )n
I the
In
th high-frequency
hi h f
limit
li it ω→∞:
H ( jω ) ≈
a0
bn ω n
The coefficients are related to Bessel functions:
a0 = b0
bi =
James Grimbleby
(2n − i )!
2n −i i ! (n − i )!
School of Systems Engineering - Electronic Engineering
Slide 35
Bessel Approximation
Gain ((dB)
G
Response of 8th
8th-order
order Bessel approximation:
0 01
0.01
James Grimbleby
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
10
1.0
Slide 36
Bessel Approximation
Unit-step
Unit
step response of 8th
8th-order
order Bessel:
g(t)
10
1.0
0.0
James Grimbleby
1 0s
1.0s
Time
3 0s
3.0s
School of Systems Engineering - Electronic Engineering
Slide 37
Low Pass to Low
Low-Pass
Low-Pass
Pass
Transformation
This transformation shifts the cut-off frequency to ω0:
jω
ω=1.0
James Grimbleby
→
jω
ω0
ω= ω0
School of Systems Engineering - Electronic Engineering
Slide 38
Low Pass to Low
Low-Pass
Low-Pass
Pass
Transformation
Design example: A Chebychev low-pass filter is required with
th ffollowing
the
ll i specification:
ifi ti
Pass-band
P
b d gain:
i
Stop-band gain:
Pass band edge:
Pass-band
Stop-band edge:
gp ≥ -3
3 dB
gs ≤ -50 dB
fp = 1000 Hz
fs = 2000 Hz
This corresponds to a normalised low-pass filter with ωp=1.0,
ωs=2.0
=2 0
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 39
Low Pass to Low
Low-Pass
Low-Pass
Pass
Transformation
Normalised low-pass filter:
Pass-band
P
b d gain:
i
Stop-band gain:
P
Pass-band
b d edge:
d
Stop-band edge:
gp ≥ -3
3 dB
gs ≤ -50 dB
ωp=1.0
10
ωs=2.0
This specification can be met by a 5th-order Chebychev
approximation with 3 dB pass
pass-band
band ripple:
H ( jω ) =
0.06265
( jω )5 + 0.5745( jω )4 + 1.415( jω )3
+ 0.5489( jω )2 + 0.4080( jω ) + 0.06265
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 40
Low Pass to Low
Low-Pass
Low-Pass
Pass
Transformation
Applying the low-pass to low-pass transformation with
ω0=2π×1000=6283:
0.06265
H ( jω ) =
5
4
3
j
j
j
ω
ω
ω
⎧
⎫
⎧
⎫
⎧
⎫
+
0
.
5745
+
1
.
415
⎨
⎬
⎨
⎬
⎨
⎬
⎩ 6283 ⎭
⎩ 6283 ⎭
⎩ 6283 ⎭
2
=
⎧ jω ⎫
⎧ jω ⎫
+ 0.5489 ⎨
⎬ + 0.4080 ⎨
⎬ + 0.06265
⎩ 6283 ⎭
⎩ 6283 ⎭
0.06265
1.021× 10 −19 ( jω )5 + 3.686 × 10 −16 ( jω )4 + 5.705 × 10 −12 ( jω )3
+ 1.390 × 10 −8 ( jω )2 + 6.493 × 10 −5 ( jω ) + 0.06265
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 41
Low Pass to Low
Low-Pass
Low-Pass
Pass
Transformation
Gain ((dB)
G
5th-order Chebychev low-pass with cut-off frequency 1000 Hz :
Frequency (Hz)
James Grimbleby
1000
School of Systems Engineering - Electronic Engineering
Slide 42
Low Pass to High
Low-Pass
High-Pass
Pass
Transformation
This transformation converts to a high-pass response and
shifts the cut-off frequency to ω0:
jω
ω=1.0
James Grimbleby
→
ω0
jω
ω= ω0
School of Systems Engineering - Electronic Engineering
Slide 43
Low Pass to High
Low-Pass
High-Pass
Pass
Transformation
Design example: A Chebychev high-pass filter is required
with the following specification:
Pass-band
Pass
band gain:
Stop-band gain:
Pass-band
Pass
band edge:
Stop-band edge:
gp ≥ -3
3 dB
gs ≤ -25 dB
fp = 2000 Hz
fs = 1000 Hz
This corresponds to a normalised low-pass filter with
ωp=1.0,
1.0, ωs=2.0
2.0
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 44
Low Pass to High
Low-Pass
High-Pass
Pass
Transformation
Normalised low-pass filter:
Pass-band gain:
Stop-band gain:
Pass-band edge:
Stop-band edge:
gp ≥ -3 dB
gs ≤ -25 dB
ωp=1.0
ωs=2.0
This specification can be met by a 3rd-order Chebychev
approximation
i ti with
ith 3 dB pass-band
b d ripple:
i l
H ( jω ) =
James Grimbleby
0.2506
( jω )3 + 0.5972( jω )2 + 0.9284( jω ) + 0.2506
School of Systems Engineering - Electronic Engineering
Slide 45
Low Pass to High
Low-Pass
High-Pass
Pass
Transformation
H ( jω ) =
0.2506
( jω )3 + 0.5972( jω )2 + 0.9284( jω ) + 0.2506
Applying the low-pass to high-pass transformation with
ω0=2π×2000:
0.2506
H ( jω ) =
3
2
⎧12566 ⎫
⎧12566 ⎫
⎧12566 ⎫
⎨
⎬ + 0.5972⎨
⎬ + 0.9284 ⎨
⎬ + 0.2506
⎩ jω ⎭
⎩ jω ⎭
⎩ jω ⎭
=
0.2506( jω )3
1.984 × 1012 + 9.431× 107 ( jω ) + 1.167 × 10 4 ( jω )2 + 0.2506( jω )3
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 46
Low Pass to High
Low-Pass
High-Pass
Pass
Transformation
Gain
n (dB))
3rd-order Chebychev high-pass with cut-off frequency 2000 Hz :
1000
James Grimbleby
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
Slide 47
Low Pass to Band
Low-Pass
Band-Pass
Pass
Transformation
This transformation converts to a band-pass response
centred on ω0 with relative bandwidth k2:
jω
→
ω=1 0
ω=1.0
James Grimbleby
⎧ω0 jω ⎫
+
⎨
⎬
⎩ jω ω0 ⎭
k
ω= ω0
School of Systems Engineering - Electronic Engineering
Slide 48
Low Pass to Band
Low-Pass
Band-Pass
Pass
Transformation
Gain ((dB)
G
10th-order Chebychev band-pass centred on f=1000 Hz
with relative bandwidth k2=4:
James Grimbleby
1000
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
Slide 49
Low Pass to Band
Low-Pass
Band-Stop
Stop
Transformation
This transformation converts to a band-stop response
centred on ω0 with relative bandwidth k2:
jω
→
ω=1
ω
1.0
0
James Grimbleby
k
⎧ω0 jω ⎫
+
⎨
⎬
⎩ jω ω0 ⎭
ω= ω0
ω
School of Systems Engineering - Electronic Engineering
Slide 50
Low Pass to Band
Low-Pass
Band-Stop
Stop
Transformation
Gain (dB)
10th-order Chebychev band-stop centred on f=1000 Hz
with relative bandwidth k2=4:
James Grimbleby
1000
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
Slide 51
Passive Filter Realisation
Passive filters are usually realised as equally-terminated
ladder filters
This type
yp of filter has a resistor in series with the input
p and a
resistor of nominally the same value in parallel with the
output; all other components are reactive (that is inductors or
capacitors)
Passive equally-terminated ladder filters are not normally
used at frequencies below about 10 kHz because they
contain inductors
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 52
Passive Filter Realisation
Equally-terminated low-pass all-pole ladder filter:
R
Input
R Output
This type of filter is suitable for implementing all-pole
designs such as Butterworth
Butterworth, Chebychev and Bessel
Order = number of capacitors + number of inductors
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 53
Passive Filter Realisation
Alternative equally-terminated low-pass all-pole ladder filter:
R
Input
R Output
This type of filter is suitable for implementing all-pole designs
such as Butterworth
Butterworth, Chebychev and Bessel
Order = number of capacitors + number of inductors
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 54
Passive Filter Realisation
Equally terminated low-pass
Equally-terminated
low pass ladder filter with imaginary
zeros in response:
R
Input
R Output
This type of filter is suitable for implementing an Inverse
Chebychev or Elliptic response
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 55
Passive Filter Realisation
Alternative equally-terminated
equally terminated low-pass
low pass ladder filter with
imaginary zeros in response:
R
Input
R
p
Output
This type of filter is suitable for implementing an Inverse
Chebychev or Elliptic response
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 56
Passive Filter Realisation
Low pass filters can be changed to high-pass
Low-pass
high pass filters by
replacing the inductors by capacitors, and the capacitors
by
y inductors:
R
Input
p
R Output
p
This type of filter is suitable for implementing all-pole
designs such as Butterworth, Chebychev and Bessel
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 57
Passive Filter Realisation
Low-pass filters can be changed to band-pass filters by
replacing the inductors and capacitors by LC combinations
L
C
L=
1
ω02C
C
L
L
C
C=
1
ω02L
where ω0 is the geometric centre frequency of the
passband
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 58
Passive Filter Realisation
Low-pass filters can be changed to band-pass filters by
replacing the inductors and capacitors by LC combinations
R
Input
R
Output
This type of filter is suitable for implementing all-pole
designs such as Butterworth, Chebychev and Bessel
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 59
Component Value Determination
Suitable component values can be determined by the
following procedure:
1.
Select a suitable filter circuit
2.
Obtain its frequency-response function in symbolic
form
3.
Equate coefficients of the symbolic frequencyresponse function and the required response
4.
Solve the simultaneous non-linear equations to
obtain the component values
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 60
Component Value Determination
A simpler procedure is to use filter design tables
Normalised low-pass
low pass Butterworth
(R = 1.0 Ω, ω0 = 1.0 rad/s):
n
2
3
4
5
6
7
C1
1.414
1.000
0.765
0.618
0.518
0.445
James Grimbleby
L2
1.414
2.000
1.848
1.618
1.414
1.247
C3
1.000
1.848
2.000
1.932
1.802
L4
C5
L6
C7
0.765
1.618 0.618
1.932 1.414 0.518
2.000 1.802 1.247 0.445
School of Systems Engineering - Electronic Engineering
Slide 61
Component Value Determination
Design example: a 5th
5th-order
order low-pass
low pass Butterworth filter with
cut-off (-3 dB) frequency 2 kHz
Normalised 5th-order Butterworth low-pass filter:
1Ω
Ω
1.618H
1.618H
Input
1Ω
0.618F
James Grimbleby
2.000F
Output
0.618F
School of Systems Engineering - Electronic Engineering
Slide 62
Component Value Determination
Scale impedances: multiply resistor and inductor values by
k, divide capacitor values by k
Choose k = 10000
10kΩ
Ω
16.18kH 16.18kH
10kΩ
Ω
Input
Output
61.8μF
James Grimbleby
200μF
61.8μF
School of Systems Engineering - Electronic Engineering
Slide 63
Component Value Determination
Scale
S
l ffrequency: divide
di id iinductor
d t and
d capacitor
it values
l
b
by k
where k = 2π×2000 = 1.257×104
10kΩ
1.288H
1.288H
10kΩ
Input
Output
4 918 F
4.918nF
James Grimbleby
15 92 F
15.92nF
4 918 F
4.918nF
School of Systems Engineering - Electronic Engineering
Slide 64
Component Value Determination
Gain (dB)
Response of 5th
5th-order
order low-pass
low pass Butterworth filter:
100
James Grimbleby
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
10000
Slide 65
Component Value Determination
Normalised 1 dB ripple low-pass Chebychev
(R = 1.0
1 0 Ω,
Ω ω0 = 1.0
1 0 rad/s):
n
2
3
4
5
6
7
James Grimbleby
C1
0.572
2.216
0.653
2.207
0.679
2.204
L2
3.132
1.088
4.411
1.128
3.873
1.131
C3
2.216
0.814
3.103
0.771
3.147
L4
C5
L6
C7
2.535
1.128 2.207
4.711 0.969 2.406
1.194 3.147 1.131 2.204
School of Systems Engineering - Electronic Engineering
Slide 66
Component Value Determination
Design example: a 3rd-order
3rd order high-pass
high pass Chebychev filter with
1 dB ripple and a cut-off (-3 dB) frequency 5 kHz
Normalised 3rd-order Chebychev high-pass filter:
1Ω
Ω
Input
p
0.451H
1
1.088
0 919F
0.919F
0.451H
1Ω
Output
p
1
2.216
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 67
Component Value Determination
Scale impedances: multiply resistor and inductor values by k,
divide capacitor values by k
Choose k = 1000
1000Ω
Ω
Input
p
James Grimbleby
451H
919μF
451H
1000Ω
Ω
Output
p
School of Systems Engineering - Electronic Engineering
Slide 68
Component Value Determination
Scale
S
l ffrequency: divide
di id iinductor
d t and
d capacitor
it values
l
b
by k
where k = 2π×5000 = 3.142×105
1000Ω
Ω
29 25nF
29.25nF
Input
p
1000Ω
Ω
Output
p
14.36mH 14.36mH
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 69
Component Value Determination
Gain (dB)
Response of 3rd-order
3rd order high-pass
high pass Chebychev filter:
1k
James Grimbleby
Frequency (Hz)
School of Systems Engineering - Electronic Engineering
100k
Slide 70
Alternative Configuration
Design example: a 3rd-order
3rd order high-pass
high pass Chebychev filter with
1 dB ripple and a cut-off (-3 dB) frequency 5 kHz
Normalised 3rd-order Chebychev high-pass filter:
1
2.216
0 451F
0.451F
0 451F
0.451F
1Ω
Input
p
0.919H
0
9 9
1Ω
Output
p
1
1.088
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 71
Alternative Configuration
Scale impedances: multiply resistor and inductor values by k,
divide capacitor values by k
Choose k = 1000
1000Ω
Input
p
James Grimbleby
451μF
919H
9
9
451μF 1000Ω
Output
p
School of Systems Engineering - Electronic Engineering
Slide 72
Alternative Configuration
Scale
S
l ffrequency: divide
di id iinductor
d t and
d capacitor
it values
l
b
by k
where k = 2π×5000 = 3.142×105
1000Ω 14.36nF 14.36nF 1000Ω
Input
James Grimbleby
29.25mH
Output
School of Systems Engineering - Electronic Engineering
Slide 73
Component Value Sensitivity
Gain (dB)
5th-order 3dB ripple low-pass Chebychev equallyterminated ladder filter with cut-off frequency 1000 Hz:
±10% variation of
components values
Frequency (Hz)
James Grimbleby
1000
School of Systems Engineering - Electronic Engineering
Slide 74
Active Filters: Cascade Synthesis
The frequency response function is first split into secondorder factors
H ( jω ) =
a0 + a1( jω ) + a2 ( jω )2 + .. + an ( jω )n
b0 + b1( jω ) + b2 ( jω )2 + .. + bn ( jω )n
a10 + a11( jω ) + a12 ( jω )2 a20 + a21( jω ) + a22 ( jω )2
=
×
× ...
b10 + b11( jω ) + b12 ( jω )2 b20 + b21( jω ) + b22 ( jω )2
= H1( jω ) × H 2 ( jω ) × ...
Each factor is implemented using
g a second-order active filter
These filters are then connected in cascade
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 75
Active Filters: Cascade Synthesis
If the response to be implemented is derived from an allpole approximation then the second-order factors will be of
simple
i l llow-pass (Hlp),
) b
band-pass
d
(Hbp) or hi
high-pass
h
(Hhp)
form:
a0
H lp ( jω ) =
b0 + b1( jω ) + b2 ( jω )2
H bp ( jω ) =
H hp
h ( jω ) =
James Grimbleby
a1( jω )
b0 + b1( jω ) + b2 ( jω )2
a2 ( jω )2
b0 + b1( jω ) + b2 ( jω )2
School of Systems Engineering - Electronic Engineering
Slide 76
Sallen-Key
Sallen
Key Low-Pass
Low Pass Filter
C1
R1
R2
Output
Input
C2
H lpp ( jω ) =
1
1 + 2T2 ( jω ) + T1T2 ( jω )2
where : R1 = R2
James Grimbleby
T1 = R1C1 T2 = R2C2
School of Systems Engineering - Electronic Engineering
Slide 77
Resonance Frequency and Q
Q-factor
factor
Sallen-Key:
H lp ( jω ) =
Standard response: H lp ( jω ) =
Thus:
or:
James Grimbleby
1
ω02
= T1T2
1
ω0 =
T1T2
1
1 + 2T2 ( jω ) + T1T2 ( jω )2
1
j ω ( j ω )2
1+
+
ω0Q
ω02
1
= 2T2
ω0Q
1
1 T1
Q=
=
2ω0T2 2 T2
School of Systems Engineering - Electronic Engineering
Slide 78
Sallen-Key
Sallen
Key Low-Pass
Low Pass Filter
Gain(dB)
20 dB
Q=2
0 dB
Q = 10
1
Q=
2
-20 dB
-12
12 dB / octave
-40 dB
10
James Grimbleby
100
1000
10000
Frequency (rad/s)
100000
School of Systems Engineering - Electronic Engineering
Slide 79
Sallen-Key
Sallen
Key High-Pass
High Pass Filter
R1
C1
Input
C2
R2
H hp ( jω ) =
T1T2 ( jω )2
1 + 2T1( jω ) + T1T2 ( jω )2
where : C1 = C2
James Grimbleby
O t t
Output
T1 = R1C1 T2 = R2C2
School of Systems Engineering - Electronic Engineering
Slide 80
Sallen-Key
Sallen
Key High-Pass
High Pass Filter
Gain(dB)
20 dB
Q=2
Q = 10
0 dB
1
Q=
2
-20 dB
12 dB / octave
-40 dB
10
James Grimbleby
100
1000
10000
Frequency (rad/s)
100000
School of Systems Engineering - Electronic Engineering
Slide 81
Rauch Band
Band-Pass
Pass Filter
R1
C2
C1
R2
Input
Output
H bp ( jω ) =
1 + 2T2 ( jω ) + T1T2 ( jω )2
where : R1 = R2
James Grimbleby
− T1( jω )
T1 = R1C1 T2 = R2C2
School of Systems Engineering - Electronic Engineering
Slide 82
Rauch Band
Band-Pass
Pass Filter
Gain(dB)
Q = 10
40 dB
20 dB
Q=5
Q=2
0 dB
10
James Grimbleby
100
1000
10000
Frequency (rad/s)
100000
School of Systems Engineering - Electronic Engineering
Slide 83
All-Pole
All
Pole Cascade Synthesis
Design example: 5th-order Chebychev low-pass filter has
numerator and denominator coefficients:
a0 = 0.06265
a1 = 0.0
00
a2 = 0.0
a3 = 0.0
00
a4 = 0.0
a5 = 0.0
00
b0 = 0.06265
b1 = 6.493×10
6 493 10-55
b2 = 1.390×10-8
b3 = 5
5.705×10
705×10-12
b4 = 3.686×10-16
19
b5 = 1
1.021×10
021×10-19
The frequency response is factored into second order
sections by finding the poles (there are no zeros)
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 84
All-Pole
All
Pole Cascade Synthesis
The poles are the roots of the equation obtained by setting
the denominator polynomial to zero:
-9.011×10+2 + j3.751×10+3
-9.011×10
9 011 10+2 - j3.751×10
j3 751 10+3
-3.463×10+2 + j6.070×10+3
-3.463×10
3 463 10+2 - j6.070×10
j6 070 10+3
-1.115×10+3 + j0.0
The frequency response is of 5th-order so that the factors
are two conjugate pairs of complex roots and a single real
root
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 85
All-Pole
All
Pole Cascade Synthesis
Conjugate
j g
p
pole p
pairs are combined:
-9.011×10+2 + j3.751×10+3
-9.011×10
9 011×10+22 - j3.751×10
j3 751×10+33
D( jω ) = (jω + 9.011
9 011× 10 2 − j3.751× 103 ) ×
(jω + 9.011× 10 2 + j3.751× 103 )
= ( jω )2
+ (9.011× 10 2 + 9.011× 10 2 )( jω )
2 2
3
+ (9.011
9 011× 10 ) + (3.751× 10 )
2
= ( jω )2 + 1.802 × 10 +3 ( jω ) + 1.488 × 10 +7
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 86
All-Pole
All
Pole Cascade Synthesis
The denominator of the 1st-order section:
D( jω ) = ( jω − p ) = jω + 1.115 × 10 +3
Dividing through by 1.115×10+3:
D( jω ) = 1 + 8.966 × 10 − 4 ( jω )
Frequency
eque cy response
espo se function
u c o o
of 1st-order
s o de filter:
e
1
H ( jω ) =
1 + RC( jω )
Thus:
RC = 8.966 × 10 − 4
Let R=10 kΩ; then C=89.66 nF.
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 87
All-Pole
All
Pole Cascade Synthesis
Denominator of the 1st 2nd-order section:
D( jω ) = ( jω )2 + 1.802 × 10 +3 ( jω ) + 1.488 × 10 +7
Di idi th
Dividing
through
hb
by 1
1.488×10
488 10+7:
D( jω ) = 1 + 1.211× 10 − 4 ( jω ) + 6.718 × 10 −8 ( jω )2
Frequency response function of 2nd-order filter:
1
H lp ( jω ) =
1 + 2T2 ( jω ) + T1T2 ( jω )2
2T2 = 1.211× 10 − 4
→ T2 = 6.054 × 10 −5
T1T2 = 6.718 × 10 −8
→ T1 = 1.110 × 10 −3
Thus:
Let R1=R
R2=10
10 kΩ; C1=111
111.0
0 nF,
nF C2=6
6.054nF
054nF
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 88
All-Pole
All
Pole Cascade Synthesis
Denominator of the 1st 2nd-order section:
D( jω ) = ( jω )2 + 6.926 × 10 + 2 ( jω ) + 3.696 × 10 +7
Di idi th
Dividing
through
hb
by 3
3.696×10
696 10+7:
D( jω ) = 1 + 1.874 × 10 −5 ( jω ) + 2.706 × 10 −8 ( jω )2
Frequency response function of 2nd-order filter:
1
H lp ( jω ) =
1 + 2T2 ( jω ) + T1T2 ( jω )2
2T2 = 1.874 × 10 −5
→ T2 = 9.369 × 10 −6
T1T2 = 2.706 × 10 −8
→ T1 = 2.888 × 10 −3
Thus:
Let R1=R
R2=10
10 kΩ; C1=288
288.8
8 nF,
nF C2=0
0.9369nF
9369nF
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 89
All-Pole
All
Pole Cascade Synthesis
Complete cascade synthesis of Chebychev active filter:
111.0nF
10kΩ
10kΩ
288.8nF
10kΩ
10kΩ
10kΩ
89 66nF
89.66nF
6 054nF
6.054nF
0 9369nF
0.9369nF
Input
James Grimbleby
Output
School of Systems Engineering - Electronic Engineering
Slide 90
General Cascade Synthesis
If the response is not deri
derived
ed from an all
all-pole
pole appro
approximation
imation
then the second-order factors will be of general form:
H ( jω ) =
a0 + a1( jω ) + a2 ( jω )2
b0 + b1( jω ) + b2 ( jω )2
Rauch or Sallen-Key second-order filters are unsuitable and
a more complex filter configuration must be used
The ring
ring-of-three
of three filter (aka the bi-quad
bi quad or state
state-variable
variable
filter) can be used
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 91
Component Value Sensitivity
Gain ((dB)
G
5th-order Chebychev low-pass implemented as a cascade
of active Sallen-Key sections:
Frequency (Hz)
James Grimbleby
1000
School of Systems Engineering - Electronic Engineering
Slide 92
Copying Methods
Copying methods are ways of designing active filters with
the same low sensitivity properties of passive equallyterminated ladder filters
The starting point for all copying methods is a prototype
passive filter
This is then copied in some way which preserves the
properties
p
of the p
passive filters but which
desirable p
eliminates the inductors
The copying method that will be described here is based on
inductor simulation
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 93
Positive Immittance Converters
V1 I1
V3 Z4
V1 = V3 + I1Z 4
Z3
V1
Z2
V2
Z3
Z2
V1 = V3
+ V2
Z 2 + Z3
Z 2 + Z3
Z0
V1 = V2
Z0 + Z1
Z1
V1
Z0
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 94
Positive Immittance Converters
V1 = V3 + I1Z 4
Z3
Z2
V1 = V3
+ V2
Z 2 + Z3
Z 2 + Z3
Z0
V1 = V2
Z0 + Z1
Substitute to remove V2 and V3:
V1(Z2 + Z3 ) = V3Z2 + V2Z3
Z3 (Z1 + Z0 )
= V3Z2 + V1
Z0
Z3 (Z1 + Z0 )
= Z2 (V1 − I1Z 4 ) + V1
Z0
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 95
Positive Immittance Converters
Multiply both sides by Z0:
V1(Z0Z2 + Z0Z3 ) = V1Z0Z2 − I1Z0Z2Z 4
+ V1(Z1Z3 + Z0 Z3 )
Remove cancelling terms:
V1Z1Z3 = I1Z0Z2Z 4
The impedance Zi of the PIC is given by:
V1 Z0Z2Z 4
Zi = =
I1
Z1Z3
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 96
Inductor Simulation
Let Z0, Z1 ,Z2,and Z4 be resistors of value
al e R,
R and Z3 be a
capacitor of value C:
or:
Z0 Z 2 Z 4
R3
=
= jωCR 2
Zi =
Z1Z3
R / j ωC
Zi = jωL
where L = CR 2
The PIC therefore simulates a grounded inductor of value
CR2
By correct choice of R and C the PIC can be made to
y required
q
inductance.
simulate any
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 97
Copying Methods
5th order Cheb
5th-order
Chebychev
che high
high-pass
pass passive
passi e filter
filter:
9 134nF
10kΩ 9.134nF
Input
4.178H
7 015nF
7.015nF
9 134nF
9.134nF
4.178H
10kΩ
Output
Let R
R=10
10 kΩ; then:
CR 2 = 4.178 H
James Grimbleby
→
C = 4.178 × 10 −8 F
School of Systems Engineering - Electronic Engineering
Slide 98
Copying Methods
5th-order Chebychev high-pass active filter:
10kΩ 9.134nF
7.015nF
9.134nF
10kΩ
41.78nF
41.78nF
James Grimbleby
10k
10kΩ
10kΩ
10k
10kΩ
10kΩ
10kΩ
10kΩ
Ou
utput
Inp
put
10kΩ
10kΩ
School of Systems Engineering - Electronic Engineering
Slide 99
Copying Methods
3th-order elliptic high-pass passive filter:
1kΩ
382nF
395nF
815nF
Input
1kΩ
Output
2.29H
Let R=1 kΩ; then:
CR 2 = 2.29H
James Grimbleby
→
C = 2.29 × 10 − 6 F
School of Systems Engineering - Electronic Engineering
Slide 100
Copying Methods
3th-order elliptic
high-pass
active filter:
1kΩ 382nF
815nF 395nF
1kΩ
2.29μF
Input
Output
1k
1kΩ
1kΩ
1kΩ
1kΩ
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 101
Analogue Filter Design
© J. B. Grimbleby, 19 February 2009
James Grimbleby
School of Systems Engineering - Electronic Engineering
Slide 110