sintesis filtros
TRANSCRIPT
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Lecturer: James Grimbleb
URL: http://www.personal.rdg.ac.uk/~stsgrimb/
email: .b. rimbleb readin .ac.uk
Number of Lectures: 5
Reference text:
Circuits (3rd edition)
McGraw-Hill 2003 ISBN 0071207031
School of Systems Engineering - Electronic Engineering Slide 1James Grimbleby
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Design with Operational
Integrated Circuits (3rd ed)
McGraw-Hill 2003
A rox £43
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This course of lectures deals with the design of passive andactive analogue filters
e op cs a w e covere nc u e:
Fre uenc -domain filter a roximations
Filter transformations
Passive e uall -terminated ladder filtersPassive ladder filters from filter design tables
Active filters – cascade synthesis
Component value sensitivityCopying methods
Generalised immittance converter (GIC)
Inductor simulation using GICs
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You should be familiar with the following topics:
Circuit analysis using Kirchhoff’s Laws
Thévenin and Norton's theorems
The Superposition Theorem
Semiconductor devicesComplex impedances
Frequency response function – gain and phase
The infinite-gain approximation
School of Systems Engineering - Electronic Engineering Slide 4James Grimbleby
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a signal and are therefore specified in the frequency domain:
Gain(dB)
Max pass-band gain
-Max
sto -Max stop-band gain
band
gain
lo fre
Stop-band
ed e
Stop-band
ed e
Pass-band
ed es
School of Systems Engineering - Electronic Engineering Slide 5James Grimbleby
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stages
In the first stage a frequency response function H (jω) is
In the second sta e an electronic circuit is desi ned togenerate the required frequency response function
Filter circuits can be constructed entirely from passivecomponents or can contain active elements such as
operational amplifiers
School of Systems Engineering - Electronic Engineering Slide 6James Grimbleby
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-
-
deriving a low-pass prototype filter normalised to a cut-off
fre uenc of 1 rad/s
This rotot e filter is then transformed to the re uired t e
and cut-off frequency:
-
High-pass
Band-stop
This procedure results in a frequency response function
which meets the specification
School of Systems Engineering - Electronic Engineering Slide 7James Grimbleby
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-
An ideal normalised sharp cut-off low-pass filter has a
frequency response:
Gain
.
Angular
ω =1.0 frequency
.
0.0
n or una e y suc a er s no rea sa e an s
necessary to use approximations to the ideal response
School of Systems Engineering - Electronic Engineering Slide 8James Grimbleby
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-
distributed elements is a rational function of jω :
nnaaaa ) j(..) j() j( 2
210 ωωω ++++n
nbbbb ) j(..) j() j( 2210 ωωω ++++
The first stage in the design is to choose suitable values for
the order n and the coefficients a0..an and b0..bn
,
Chebychev, Inverse Chebychev, Elliptic and Bessel
School of Systems Engineering - Electronic Engineering Slide 9James Grimbleby
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e u erwor approx ma on ga n s max ma y a
-as possible
The gain falls off monotonically in both pass-band (below
= - =
nH
21) j(
ω
ω
+=
School of Systems Engineering - Electronic Engineering Slide 10James Grimbleby
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(dB)
0-3
g s
log(freq)ω=1 ω= ωs
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The poles pk of a Butterworth frequency response function
k k k y x p j+=
where:
k k )12()12( −−−
nk
nn
,,2,1
22
K=
Values of k from 1 to n are substituted into this formula
giving the n poles
School of Systems Engineering - Electronic Engineering Slide 12James Grimbleby
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e po es o e u erwor response p1, p2 , ... , pn are
then combined to give the Butterworth frequency response
) j().. j() j(
) j(
21 n p p p
H
−−−
=ωωω
ω
-
there are no zeros
School of Systems Engineering - Electronic Engineering Slide 13James Grimbleby
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Poles of H (jω):
– 0.5 + j 0.866 – 0.5 – j 0.866 – 1.0 + j 0.0
Combining the poles:
1) j( =ωH
1
.....
=
+++−+ ωωω
)0.1 j(1.0) j) j((2
+++ ωωω
1.02.0j) j(0.2) j( 23 +++=
ωωω
School of Systems Engineering - Electronic Engineering Slide 15James Grimbleby
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What order n is required to meet specification?
Gain (dB)
0-
g s
log(freq)ω
=1 ω
=ω
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The gain g of a Butterworth filter, expressed in dB, is giveny:
) j(log20 10 H g ω=
og2
-10
n
ω
−=+=
The stop-band edge is at ωs, and the stop-band gain isrequired to be less than g s dB:
sssss
ng
g
ω
ωω
101010
log20
ogog
−≥
−≈+−≥
sg n
−≥
School of Systems Engineering - Electronic Engineering Slide 17James Grimbleby
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-
( d B )
≥ -3 dB
G a i
g s ≤ -40 dB
ω = 1.0
ωs = 1.5
School of Systems Engineering - Electronic Engineering Slide 19James Grimbleby
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e e yc ev approx ma on ga n osc a es e ween
dB and g p dB in the pass-band (below ω=1)
In the stop-band (above ω=1) the gain falls off
approximation for each n, but a group of approximations
- p
approximation is an all-pole response
School of Systems Engineering - Electronic Engineering Slide 20James Grimbleby
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(dB)
0g p
g s
log(freq)ω=1 ω= ωs
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Res onse of 6th-order Cheb chev a roximation:
(
d B )
g p ≥ -3 dB
G a i n g s ≤ -40 dB
ω p = 1.0
ω
s
= .
Fre uenc Hz 1.00.01School of Systems Engineering - Electronic Engineering Slide 26James Grimbleby
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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,
an en osc a es e ween -∞ an g s
approximation for each n, but a group of approximations with
- s
School of Systems Engineering - Electronic Engineering Slide 27James Grimbleby
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( d B )
≥ -3 dB
G a i n g s ≤ -40 dB
ω = 1.0
ωs = 1.5
School of Systems Engineering - Electronic Engineering Slide 29James Grimbleby
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The elliptic response gain oscillates between 0 and g p dB in
-
-
g s dB and -∞ dB
The elliptic approximation has imaginary zeros
For a given filter specification the elliptic approximation givesthe lowest order fre uenc res onse function
School of Systems Engineering - Electronic Engineering Slide 30James Grimbleby
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a n
(dB)
0g
g s
log(freq)ω=1 ω= ω
s
School of Systems Engineering - Electronic Engineering Slide 31James Grimbleby
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-
( d B )
≥ -3 dB
G a i n g s ≤ -40 dB
ω = 1.0
ωs = 1.5
School of Systems Engineering - Electronic Engineering Slide 32James Grimbleby
..
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-
- -
1.0
g ( t
0.0
School of Systems Engineering - Electronic Engineering Slide 33James Grimbleby
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e esse approx ma on s use w ere a requency- oma n
filter is required which also has a good time-domain
-
frequency and the different frequency components are
-
seriously distorted
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The Bessel approximation has an all-pole frequencyresponse function:
a0
nnbbbb ) j(..) j() j( 2210 ωωω ++++
n e g - requency m ω→∞:
an
nb ω
≈
00 ba =
)!2( i nb
i ni
−
−=
−
School of Systems Engineering - Electronic Engineering Slide 35James Grimbleby
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-
d
B )
G a i n
School of Systems Engineering - Electronic Engineering Slide 36James Grimbleby
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- -
.
t ) g (
0.0
School of Systems Engineering - Electronic Engineering Slide 37James Grimbleby
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- -
TransformationThis transformation shifts the cut-off frequency to ω0:
j jω
ωω →
ω= ω0 ω=1.0
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- -
Transformation
Design example: A Chebychev low-pass filter is required with
e o ow ng spec ca on:
ass- an ga n: g p -
Stop-band gain: g s ≤ -50 dB
- p
Stop-band edge: f s = 2000 Hz
This corresponds to a normalised low-pass filter with ωp=1.0,s .
School of Systems Engineering - Electronic Engineering Slide 39James Grimbleby
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- -
TransformationNormalised low-pass filter:
ass- an ga n: g p ≥ -Stop-band gain: g s ≤ -50 dB
ass- an e ge: ωp= .
Stop-band edge: ωs=2.0
This specification can be met by a 5th-order Chebychev
62650.0=
-
) j(415.1) j(5745.0) j(
2
345 ++ ωωω
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- -
Transformation Applying the low-pass to low-pass transformation with
ω0=2π×1000=6283:
j j.) j( 345
⎫⎧⎫⎧⎫⎧=
ωωω
ωH
6283.
6283.
6283
2
⎭⎩⎭⎩⎭⎩
06265.06283
4080.06283
5489.0 +⎭⎬
⎩⎨+
⎭⎬
⎩⎨+
ωω
10705.510686.310021.1
62650.0312416519 ×+×+×= −−−
ωωω
06265.0) j(10493.6) j(10390.1 528 +×+×+ −−ωω
School of Systems Engineering - Electronic Engineering Slide 41James Grimbleby
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- -
Transformation5th-order Chebychev low-pass with cut-off frequency 1000 Hz :
d
B )
a i n
School of Systems Engineering - Electronic Engineering Slide 42James Grimbleby
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- -
TransformationThis transformation converts to a high-pass response and
shifts the cut-off frequency to ω0:
ωω j 0→
ω= ω0ω=1.0
School of Systems Engineering - Electronic Engineering Slide 43James Grimbleby
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- -
TransformationDesign example: A Chebychev high-pass filter is required
- p -
Stop-band gain: g s ≤ -25 dB
- =p Stop-band edge: f s = 1000 Hz
This corresponds to a normalised low-pass filter withω =1.0 ω =2.0
School of Systems Engineering - Electronic Engineering Slide 44James Grimbleby
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- -
TransformationNormalised low-pass filter:
Pass-band gain: g p ≥ -3 dBStop-band gain: g s ≤ -25 dB
Pass-band edge: ωp=1.0
Stop-band edge: ω
s=2.0
This specification can be met by a 3rd-order Chebychev
approx ma on w pass- an r pp e:
2506.0) j(9284.0) j(5972.0) j(
.) j(
23 +++=
ωωω
ωH
School of Systems Engineering - Electronic Engineering Slide 45James Grimbleby
- -
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- -
Transformation
2506.0) j(9284.0)(5972.0) j(
2506.0) j(
23 +++=
ωωω
ωH
Applying the low-pass to high-pass transformation with
ω0=2π×2000:2506.0
ω =H
2506.012566
9284.012566
5972.012566
ωωω
+⎬⎫
⎨⎧
+⎬⎫
⎨⎧
+⎬⎫
⎨⎧
324712 2506.010167.110431.910984.1
)(2506.0
ωωω
ω
+×+×+×=
School of Systems Engineering - Electronic Engineering Slide 46James Grimbleby
- -
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- -
Transformation3rd-order Chebychev high-pass with cut-off frequency 2000 Hz :
( d B
G a i
School of Systems Engineering - Electronic Engineering Slide 47James Grimbleby
- -
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TransformationThis transformation converts to a band-pass response
centred on ω0 with relative bandwidth k 2 :
⎬⎫⎨⎧ +0 jωω
k → 0 jω
School of Systems Engineering - Electronic Engineering Slide 48James Grimbleby
0.
- -
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Transformation
with relative bandwidth k 2=4:
d
B )
a i n
School of Systems Engineering - Electronic Engineering Slide 49James GrimblebyFrequency (Hz)1000
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Transformation10th-order Chebychev band-stop centred on f =1000 Hz
with relative bandwidth k 2=4:
)
n ( d B
G
a i
School of Systems Engineering - Electronic Engineering Slide 51James GrimblebyFrequency (Hz)1000
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Passive filters are usually realised as equally-terminated
ladder filters
This type of filter has a resistor in series with the input and a
resistor of nominally the same value in parallel with the
output; all other components are reactive (that is inductors orcapacitors)
Passive equally-terminated ladder filters are not normally
used at frequencies below about 10 kHz because they
contain inductors
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Equally-terminated low-pass all-pole ladder filter:
R
This type of filter is suitable for implementing all-pole,
=School of Systems Engineering - Electronic Engineering Slide 53James Grimbleby
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Alternative equally-terminated low-pass all-pole ladder filter:
R
This type of filter is suitable for implementing all-pole designs,
=School of Systems Engineering - Electronic Engineering Slide 54James Grimbleby
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- -zeros in response:
R
R Input Output
This type of filter is suitable for implementing an Inverse
Chebychev or Elliptic response
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- -imaginary zeros in response:
R Input
Out ut
This type of filter is suitable for implementing an Inverse
Chebychev or Elliptic response
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- -replacing the inductors by capacitors, and the capacitors
b inductors:
R
R In ut Out ut
This type of filter is suitable for implementing all-pole
designs such as Butterworth, Chebychev and Bessel
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- -
replacing the inductors and capacitors by LC combinations
C L
1=
C
0ω
C LL 1
L2
0ω
=
where ω0 is the geometric centre frequency of the
passband
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replacing the inductors and capacitors by LC combinations
R
R Input Output
This type of filter is suitable for implementing all-pole
designs such as Butterworth, Chebychev and Bessel
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following procedure:
1. Select a suitable filter circuit
2. Obtain its frequency-response function in symbolic
form
3. Equate coefficients of the symbolic frequency-
response function and the required response
4. Solve the simultaneous non-linear equations to
obtain the component values
School of Systems Engineering - Electronic Engineering Slide 60James Grimbleby
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-
(R = 1.0 Ω, ω0 = 1.0 rad/s):
n C1 L2 C3 L4 C5 L6 C7
2 1.414 1.4143 1.000 2.000 1.000
4 0.765 1.848 1.848 0.765
5 0.618 1.618 2.000 1.618 0.618
6 0.518 1.414 1.932 1.932 1.414 0.518
7 0.445 1.247 1.802 2.000 1.802 1.247 0.445
School of Systems Engineering - Electronic Engineering Slide 61James Grimbleby
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- -cut-off (-3 dB) frequency 2 kHz
Normalised 5th-order Butterworth low-pass filter:
1Ω 1.618H 1.618H
0.618F 0.618F2.000F
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k , divide capacitor values by k
Choose k = 10000
10kΩ 10kΩ16.18kH 16.18kH
61.8μF 61.8μF200μF
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ca e requency: v e n uc or an capac or va ues y
where k = 2π×2000 = 1.257×104
10kΩ 10kΩ1.288H 1.288H
Input Output
. n . n . n
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- -
( d B )
G a i n
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Normalised 1 dB ripple low-pass Chebychev
= =. , .
2 0.572 3.132
3 2.216 1.088 2.2164 0.653 4.411 0.814 2.535
5 2.207 1.128 3.103 1.128 2.207
6 0.679 3.873 0.771 4.711 0.969 2.406
7 2.204 1.131 3.147 1.194 3.147 1.131 2.204
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- -1 dB ripple and a cut-off (-3 dB) frequency 5 kHz
Normalised 3rd-order Chebychev high-pass filter:1
1Ω .088.1
1ΩIn ut Out ut0.451H 0.451H
216.2
1
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divide capacitor values by k
Choose k = 1000
1000Ω 1000Ω
In ut Out ut451H 451H
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ca e requency: v e n uc or an capac or va ues y
where k = 2π×5000 = 3.142×105
1000Ω . 1000Ω
In ut Out ut
14.36mH 14.36mH
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- -
( d B )
G a i n
School of Systems Engineering - Electronic Engineering Slide 70James Grimbleby
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- -1 dB ripple and a cut-off (-3 dB) frequency 5 kHz
Normalised 3rd-order Chebychev high-pass filter:1
.216.2
.
1Ω
1Ω0.919HIn ut Out ut
088.1
1
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divide capacitor values by k
Choose k = 1000
919HIn ut Out ut
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ca e requency: v e n uc or an capac or va ues y
where k = 2π×5000 = 3.142×105
1000Ω 14.36nF 1000Ω14.36nF
29.25mHInput Output
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5th-order 3dB ripple low-pass Chebychev equally-terminated ladder filter with cut-off frequency 1000 Hz:
B )
a i n ( d
±10% variation of
G
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order factors
) j(..) j() j() j( 2210 ++++= ωωω
ω aaaaH n
n
n
) j() j() j() j( 2222120
2121110
×
++
×
++
=
ωωωω aaaaaa
) j() j() j() j( 222120121110
=
++++ ωωωω bbbbbb
Each factor is implemented using a second-order active filter
These filters are then connected in cascade
School of Systems Engineering - Electronic Engineering Slide 75James Grimbleby
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If the response to be implemented is derived from an all-
pole approximation then the second-order factors will be of
s mp e ow-pass lp , an -pass bp or g -pass hp
form:
2210
0
) j() j() j(
ωω
ω
bbb
aH lp
++=
21 ) j(
) j( ω
ωa
H bp =
22 ) j( ωa2
210 ) j() j( ωω bbb p
++
School of Systems Engineering - Electronic Engineering Slide 76James Grimbleby
- -
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C 1
R 1 R 2
OutputInputC 2
1) j(H l =ω
22211121
212
:where C R T C R T R R ===
++ ωω
School of Systems Engineering - Electronic Engineering Slide 77James Grimbleby
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2212 ) j() j(21
) j(ωω
ω
T T T H lp
++=Sallen-Key:
Standard response:2
1) j( ω =H lp
200
1ωω
++Q
Thus: 221 211
T T T ==00ω
or:2
1
20210
22 T T Q
T T ===
ω
ω
School of Systems Engineering - Electronic Engineering Slide 78James Grimbleby
- -
Gain(dB)
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Gain(dB)
20 dB
2=Q 10=Q
0 dB
2
1=Q
-20 dB-
1000 10000 10000010010-40 dB
School of Systems Engineering - Electronic Engineering Slide 79James Grimbleby
- -
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R
C 1 C 2
u puInputR 2
221 ) j() j( T T H = ωω
211
:where
) j() j(21
C R T C R T C C
T T T
===
++ ωω
School of Systems Engineering - Electronic Engineering Slide 80James Grimbleby
- -
Gain(dB)
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Gain(dB)
20 dB
2=Q 10=Q
0 dB
1=Q
-20 dB
1000 10000 10000010010-40 dB
School of Systems Engineering - Electronic Engineering Slide 81James Grimbleby
-
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R C 2
R 2 C
1
OutputInput
21 )() j( T H bp −= ω
ω
22211121
212
:where C R T C R T R R ===
School of Systems Engineering - Electronic Engineering Slide 82James Grimbleby
-
Gain(dB)
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( )
40 dB 10=Q
20 dB 5=Q
2=Q
0 dB
1000 10000 10000010010
School of Systems Engineering - Electronic Engineering Slide 83James Grimbleby
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Design example: 5th-order Chebychev low-pass filter has
numerator and denominator coefficients:
a0 = 0.06265 b0 = 0.06265a1 = . 1 = . × -
a2 = 0.0 b2 = 1.390×10-8
-3 . 3 . ×a4 = 0.0 b4 = 3.686×10-16
-5 . 5 .
sections by finding the poles (there are no zeros)
School of Systems Engineering - Electronic Engineering Slide 84James Grimbleby
-
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The poles are the roots of the equation obtained by setting
the denominator polynomial to zero:
-9.011×10+2 + j3.751×10+3
- . × - . ×-3.463×10+2 + j6.070×10+3
- . × - . ×-1.115×10+3 + j0.0
The frequency response is of 5th-order so that the factors
root
School of Systems Engineering - Electronic Engineering Slide 85James Grimbleby
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Conjugate pole pairs are combined:-9.011×10+2 + j3.751×10+3
+2 +3- . - .
32
32 )10751.3 j109.011(j
..
×+×+
−
ω
2) j(= ω
322
) j)(109.011109.011(
2
×+×+ ω
732 10488.1) j(10802.1) j(
..
++ ×+×+= ωω
School of Systems Engineering - Electronic Engineering Slide 86James Grimbleby
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310115.1 j) j() j( +×+=−= ωωω pD
Dividing through by 1.115×10+3:
Fre uenc res onse function of 1st-order filter:
) j(10966.81) j( ωω −×+=D
1) j( ωH =
Thus: 4×
Let R =10 kΩ; then C =89.66 nF.
. ×
School of Systems Engineering - Electronic Engineering Slide 87James Grimbleby
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Denominator of the 1st 2nd-order section:732 10488.1) j(10802.1) j() j( ++ ×+×+= ωωωD
v ng roug y . × :
284 −−=
Frequency response function of 2nd-order filter:
..
2212 ) j() j(21
) j(ωω
ω
T T T H lp
++=
Thus:
5
2
4
2 10054.610211.12
−−
×=→×= T T
= = = =
121 10110.110718.6 −− ×=→×= T T T
School of Systems Engineering - Electronic Engineering Slide 88James Grimbleby
. , .
-
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Denominator of the 1st 2nd-order section:722 10696.3) j(10926.6) j() j( ++ ×+×+= ωωωD
v ng roug y . × :
285 −−=
Frequency response function of 2nd-order filter:
..
2212 ) j() j(21
) j(ωω
ω
T T T H lp
++=
Thus:
6
2
5
2 10369.910874.12
−−
×=→×= T T
= = = =
121 10888.210706.2 −− ×=→×= T T T
School of Systems Engineering - Electronic Engineering Slide 89James Grimbleby
. , .
-
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Complete cascade synthesis of Chebychev active filter:
111.0nF 288.8nF
10kΩ 10kΩ 10kΩ 10kΩ
. . .
Input Output
School of Systems Engineering - Electronic Engineering Slide 90James Grimbleby
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e response s no er ve rom an a -po e approx ma on
then the second-order factors will be of general form:
2210 ) j() j( ωωω
aaaH ++=
210 ) j() j( ωω bbb ++
Rauch or Sallen-Key second-order filters are unsuitable and
- - - -
filter) can be used
School of Systems Engineering - Electronic Engineering Slide 91James Grimbleby
5th-order Chebychev low-pass implemented as a cascade
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of active Sallen-Key sections:
d B )
a i n
School of Systems Engineering - Electronic Engineering Slide 92James Grimbleby
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the same low sensitivity properties of passive equally-
passive filter
This is then copied in some way which preserves the
desirable ro erties of the assive filters but which
eliminates the inductors
The copying method that will be described here is based on
inductor simulation
School of Systems Engineering - Electronic Engineering Slide 93James Grimbleby
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1
3
4131 Z I V V +=3
V 1 322
3231
Z Z V
Z Z V V
++
+=
2 V 210
021
Z Z V V
+=
Z 1
V Z 0
School of Systems Engineering - Electronic Engineering Slide 94James Grimbleby
4131 Z I V V +=
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41313
22
31Z Z
Z V
Z Z
Z V V
++
+=
021
Z V V =
Substitute to remove V and V :
10
3223321 )( Z V Z V Z Z V +=+
013
123
)(
Z
Z Z Z
V Z V
+
+=
01314112
)()(
Z Z Z V Z I V Z
++−=
School of Systems Engineering - Electronic Engineering Slide 95James Grimbleby0
0
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0
−=
)( 30311 Z Z Z Z V ++
Remove cancelling terms:
4201311 Z Z Z I Z Z V =
i
311 Z Z I Z i ==
School of Systems Engineering - Electronic Engineering Slide 96James Grimbleby
e an e res s ors o va ue an e a
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e 0, 1 , 2,an 4 e res s ors o va ue , an 3 e acapacitor of value C :
23
420 j CR R Z Z Z Z i ω===
or:2
i ==
CR 2
By correct choice of R and C the PIC can be made to
simulate an re uired inductance.
School of Systems Engineering - Electronic Engineering Slide 97James Grimbleby
or er e yc ev g pass pass ve er:
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-or er e yc ev g -pass pass ve er:
. . .
Input Output4.178H 4.178H 10kΩ
=
F10178.4H178.4 82 ×C CR
School of Systems Engineering - Electronic Engineering Slide 98James Grimbleby
5th-order Chebychev high-pass active filter:
9 134 F 9 134 F7 015 F
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9.134nF 9.134nF7.015nF10kΩ
10kΩ 10kΩ
u t O
41.78nF41.78nF
I n t p
u t
Ω
Ω10kΩ
School of Systems Engineering - Electronic Engineering Slide 99James Grimbleby
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395nF382nF
Input Output815nF
1kΩ
2.29H
Let R =1 kΩ; then:62
×
School of Systems Engineering - Electronic Engineering Slide 100James Grimbleby
×
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high-pass 815nF 395nF382nF1kΩ
1kΩ
2.29μF
Ω
1kΩ1kΩ
School of Systems Engineering - Electronic Engineering Slide 101James Grimbleby