5 Filter Part 2 (Lecture 19, 20, 21, 22).pdf5 Filter Part 2 (Lecture 19, 20, 21, 22).pdf5 Filter Part 2 (Lecture 19, 20, 21, 22).pdf
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Slides: 62 pages
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Filters
Basic Realization of Second Order Filter
Second Order LCR Resonator
Use of this resonator to derive circuit realizations for the
various second-order filter functions
The Resonator Natural Modes
Thenaturalmodesoftheparallelresonancecircuitcanbe
determinedbyapplyinganexcitationthatdoesnotchange
thenaturalstructureofthecircuit.
Resonatorisexcitedwithacurrentsource‘I’connectedin
parallel
Circuitisconcernedwithanindependentidealcurrent
sourceisequivalenttoanopencircuit
Findthepolesofanyresponsefunction
Second Order LCR Resonator
The Resonator Natural Modes
V
oacrosstheresonatorastheresponseandthusobtainthe
responsefunctionV
o/I=Z
Z is the impedance of the parallel resonance circuit
In terms of the admittance Y
Second Order LCR Resonator
Alternative way of exciting LCR resonator
NodexofinductorLhasbeendisconnectedfrom
groundandconnectedtoanidealvoltagesourceV
i
Idealindependentvoltagesourceisequivalenttoa
shortcircuit
Notalterthenaturalstructureoftheresonator
SelectVoastheresponsevariableandfindthetransfer
functionV
0/V
i
Second Order LCR Resonator
Realization of Transmission Zeros
TransferfunctionV
0/V
iwouldbeaccordingtothefilter
specification
Nodeslabeledx,y,orzcanbedisconnectedfromground
andconnectedtoV
iwithoutalteringthecircuit’snatural
modes
Circuittakestheformofavoltagedivider
Realization of the Low-Pass Function
Node x is disconnected from ground and connected to Vi
sLbecomes infinite at s = ∞
shunt impedance becomes zero at s = ∞
Realization of the High-Pass Function
Node y is disconnected from ground and connected to V
i
Series capacitor introduces a transmission zero at s = 0 (dc)
Shunt inductor introduces another transmission zero at s = 0
(dc)
ω
0and Q are the natural mode parameters
a
2is the high-frequency transmission
Value of a
2can be determined from the circuit by observing
that as s approaches ∞
S
2
Realization of the Band-Pass Function
Onezeroats=0isrealizedbytheshuntinductor,andone
zeroats=∞isrealizedbytheshuntcapacitor
Centerfrequencyω
0theparallelLC-tunedcircuitexhibits
aninfiniteimpedanceandthusnocurrentflowsinthe
circuit
At,ω=ω
0,V
o=V
i;Center-frequencygainofthebandpass
filterisunity
Realization of the Notch Functions
Disconnecting both nodes x and y from ground and connecting
them together to V
i
The impedance of the LC circuit becomes infinite
at ω= ω
0= 1/√(LC)
Realization of the Notch Functions
To obtain a notch-filter realization in which the notch
frequency ω
nis arbitrarily placed relative to ω
0
Simple notch-filter when ω
n= ω
0
(ω
n)
2
= 1/L
1C
1
(ω
0)
2
= 1/(L
2││L
1) (C
2+C
1)
Realization of the All-Pass Function
All-passrealizationwithaflat
gainof0.5
Disadvantageoflackinga
common groundterminal
betweentheinputandthe
output
Inductance Simulation
Inductor Replacement
Antoniuoinductance simulation
circuit
Second-Order Active Filters Based
on Inductor Replacement
The Op Amp–RC Resonator
Thenodefromwhereoutput
istakenisnotveryconvenient
totakeanoutput.Iftheoutput
loadishighthenthefilter
characteristicswillbechanged
accordingtotheload.
So, it is
convenient
to connect
a buffer
amplifier
to the
output of
the filter
Second Order Filter by Op-amp RC
Resonator
Second Order Filter by Op-amp RC
Resonator
Second Order Filter by Op-amp RC
Resonator
Second Order Filter by Op-amp RC
Resonator
X
X
Second Order Filter by Op-amp RC
Resonator
Problem 4
Identifythetypeoffilter.
Alsofindω
0andQ
Second Order Filer by Op-amp RC
Resonator
Sallen-Keytopology
MultipleFeedbacktopology
Twointegratorlooptopology
Sallen-Key Band Pass Filter
Gain (K)=1+(R
B/R
A)
Centre frequency (f
0)=1/(2πRC)
Q-factor: 1/(3-K)
Gain at f
0=K/(3-K)
Sallen-Key Band Reject Filter
Gain (K)=1+(R
B/R
A)
Centre frequency (f
0)=1/(2πRC)
Pass-band gain=K
Q-factor: 1/[2(2-K)]
Problem 5
Design a 2
nd
order unity gain Butterworth Sallen-Key
Low pass filter at ω
0= 1 kHz
Cascading of two filter stages
(Butterworth)
Ref: Operational Amplifiers … by David A Bell
For 1.5 dbf
c1 = f
c/0.65 For 1.5 dB f
c2 = f
c/0.8
1.5=20log1+
??????
??????
??????
??????1
2
1
2
1.5=20log1+
??????
??????
??????
??????2
4
1
2
fc : Cut-off frequency of 3
rd
order filter
f
c1and f
c2are the individual cut-off frequency of 1
st
and 2
nd
order stages.
Second-Order Active Filters Based on the
Two-Integrator-Loop Topology
Anotherfamilyofopamp–RCcircuitsthat
realizesecond-orderfilterfunctions
Twointegratorsconnectedincascadeinan
overallfeedbackloop
Two-integrator-loopbiquadraticcircuitor
biquadcommonlydesignedwithsecond-
orderhigh-passtransferfunction
Derivation of the Two-Integrator-Loop Biquad
Derivation of the Two-Integrator-Loop Biquad
Derivation of the Two-Integrator-Loop Biquad
Two-integrator-loopbiquadrealizesthethreebasicsecond-
orderfilteringfunctions,LP,BP,andHP,simultaneously.
Thisversatilityhasmadethecircuitverypopularandhasgiven
itthenameuniversalactivefilter.
Circuit Implementation of Two-Integrator-
Loop Biquad
ReplaceeachintegratorwithaMillerintegratorcircuit
having
Replacethesummerblockwithanop-ampsumming
circuitthatiscapableofassigningbothpositiveand
negativeweightstoitsinputs.
Theresultingcircuit,knownastheKerwin–Huelsman–
NewcomborKHNbiquad
Selectsuitablypracticalvaluesforthecomponentsof
ω0,Q,andK
Two Integrator loop topology
(Practical Circuit)
Two Integrator loop topology
(Practical Circuit)
Two Integrator loop topology
(Practical Circuit)
Two Integrator loop topology
(Practical Circuit)
Two Integrator loop topology (Practical Circuit)
Realization of All Filter (Practical
Circuit)
All Pass Filter
Realization of Notch Filter (Practical
Circuit)
All Pass Filter
A notch is obtained by selecting R
B= ∞ and