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• A second order filter is also known as a Biquad. • A second – order filter consists of a two integrator loop of one lossless and one
lossy integrator• Using ideal components all the biquad topologies have the same transfer
function.• Biquad with real components are topology dependent .
We will cover the following material:- Biquad topologies- Voltage scaling to equalize signal at all filter nodes.- State variable Biquad derivation.- Tow-Thomas Biquad Design Procedure- Fully Balanced topologies using single ended Op Amps- Non-ideal integrators and Biquads.- Mason Rule to obtain transfer functions by inspection
622 Active Filters ( Edgar Sánchez-Sinencio)
1
TWO-INTEGRATOR LOOP FAMILIES
2o
1o1
QK
o
2o
1o
QKs
1
Family 1a
Two-integrator loops consists of one lossless and one lossy integrator, furthermore one is positive and the other is negative.
Why do we consider different filter topologies to obtain the same (ideal) transfer function ?
2
2o
1o1
QK
2o
1
Q
o
Ks
Family 1b
2
ooQ
ss
K1)s(D 21
1S;sKs)s(Ds BWKooQ
22Q21
1S20
1o
All loops must be negative.
3
s1
2o
1o1
s1
1QK2QK
2o
1o
2QKs
1
1QKs
1
Two-Integrator Loop Family 2, ideal to avoid CMF in Fully Differential Structures
• Lossy integrators are easyto implement
Topology using two lossy integrators
4
2
2
ooQQ
s
KK
ss
K
s
K1)s(D 212121
)KK(s)KK(s)s(Ds212121 QQooQQ
22
2121 QQoo2o KK
21 QQo KK
QBW
1
221
1
1Q
1
1Q
Q
QQQ
QBWK
QBWK
K
K1
1
KK
KS;
BW
K1S
21
212121
1
2
2o
1o
oo
QQQQoo
oo KK
1
1
KKS
Low sensitivity structure:
5
s1
2o
1o1
s1
s
K1oQ
Family 3 Self-Loop
s
K
s1)s(D 121 oQ
2
oo
211 oooQ22 sKs)s(Ds
Design Equations: PICK,BW,Given o
22 oQ andK 21 QQ KBWK
2221 QQoo2o K)KBW(
2221 QQ2ooo K)BWK(
A self loop can be implemented by adding a resistor to one (lossless) integrator.
Self-Loop two integrator loop
6
SCALING for Active-RC, MOSFET-C and SC implementations
o
KZ
1oZ
1Z
2oV
2oZ
QZ
3oV
1oV
1CZ
'1Z
11 oo VV k
/CC/SC
1
SCZZ 11
11CC 11
kk
kk
'1
'1
'1
'1
'1 RRRZZ k k k
Note that the voltages V02 and V03 were not modified
Modify all the impedances connected to the output under consideration.
An example.
7
VOLTAGE SWING SCALING
A good filter design approach requires to have a proper voltage swingat a frequency range for all internal and output nodes of the filter.
Let us consider the different types of filter implementations.
o o
iV
KR
1oR
1C
1oV
'1R
1R
2oV2oR
QR
2C
3oV
s1
1o CR
1
1
2o CR
1
2
1
s1
3oV
oo
inV2oV
o
1oV
o
2QCR
1
'1
1
21oo2Q
22
R
R
CCRR
1
CR
1ss)s(Ds
21
21oo2Q
2
2Q1k
2
2Q1K
in
o1
CCRR
1
CR
1ss
CR
1s
CR
1
)s(D
ssCR
11
sCR
1
V
V)s(H
21
1
1KCR
1
8
)s(HV
V)s(H 1
in
o2
2
21oo2Q
2
2o1K
in
o3
CCRR
1
CR
SS
CRCR
1
V
V)s(H
21
23
Let us consider that we want
K)j(H o3
2o1K2ooo1K
Q
2Q
o
2o1Ko3
CRCR
QK
RCR
R
CR
CRCR
1
)j(H
22
2
1K0 CR
QK
2o1oo
CR
1
CR
1
1
'1
1
21oo
2o
oo1
QK
R
R
CCRR
1;
RKC
RRThus
212
9
o
2Qo
1K
2Q
2Q
o
2Qo
1Ko1
j
CR
1j
CR
CR
CRj
CR
1j
CR
1
)j(H
EXAMPLENUMERICAL
2K4
3Q1o
21K
Q
2
2Q
20
o1K
2Qo1
Q
11
CR
R
CR
1
CR
CR)j(H
10
THEN
1CR
1
CR
1;
43
1
QCR
1o
2o1o
o
2Q 1
5R
1RR
1CC
Q
oo
21
21
5.22
5RK
1025
11
15.2
15)j(H o1
VERIFYTO
10
51
2)j(H o3
11
We want to make
)j(H)j(H o3oi
11 oo V5
1V
5
R&
5SC
1R&
SC
1 1
1
11
1
5
RR
C5C
'1'
1
11
Before scaling
12
1
3
1 oo1o
o
1Kio VV;
CSR
V)
CSR
1(VV
After
.VV;5
1)
CSR
V(
5
1)
CSR
1(VV
12
1
3
1 oo1o
o
1Kio
This usually can be done when enough degrees of freedom exist.
Thus we have modified and without changing1oV
2oV3oV
12
STATE-VARIABLE FILTER ARCHITECTURES
• Derived from state-variable techniques to solve differential
equations. The objective is to render an expression that
can be implemented using integrator building blocks
• Example
)1()(
)()(
20
2
sQ
s
Ks
sV
sVsH
oi
o
Where o is the cut-off frequency, (o/Q) is the bandwidth, Qis the quality (selectivity) factor.
Let us rewrite the above equation by multiplying
)2(
)(1
1
)(
)(
)(2
sXssQ
sXs
K
sV
sV
ooi
o
13
Both numerator and denominator by X(s)/s2. Thus (2) can be split into
)3()(
)(
)3()()(
)()(2
2
bs
sXKsV
as
sX
s
sX
QsVsX
o
oo
i
Observe that 1/s is an integral operator. Thus we implement (3) by using integral building block.
14
Remember that each integrator has its time-constant . Wecan associate more than one time-constant with each integrator.
SVi
o
K
Qo /
X(s)
o
o
K1
I
2
V02 (s)
Vo(s)
Also observe that V02 correspond to a low-pass, Vo to a bandpassand X(s) to a high pass. K1 could be 1 or any other suitable value
15
This two-integrator biquad consists of
– One lossless integrator (I2)
– One Lossy integrator (I1)
– Two negative closed loops: One determines the centerfrequency, the other the bandwidth (or Q).
– One of two integrators must be positive, the other negative.
–The lossy integrator must have a negative feedback.
16
Second-Order Filter Structures
V1
V2 V3V4
-K
1/s 1/s
I1
o I2
-o/Q
-o
221
4)(
oo
o
sQ
s
K
V
VsH
17
Zero implementation by addition of outputs technique
Vo
S
1
z
z
Q
Z
V2
-K
Ss
1
s
1o
-o/Q
-o
22
2
1 )/(
))/(()(
oo
ozzo
sQs
KsQs
V
VsH
V1
18
Feed-Forward Zeros Implementation
K
z
z
QK
ViVo
ZK
S 1/s S1/s
I1I2
o/Qo
220
2
1 )/(
)/()(
ooo
zzzo
sQs
sQsK
V
VsH
o
o
S
Why H(s) is not correct?
19
s/1
1oK
2oK
QK
o
LPK
o1V
s/1
K2oV
1oV
BPPK
BPNK
2o1oQ
QBPNBPP2o2
2oLP
1
o2
KKK
s
K1
)s/K1)(s/Ks/KK(s/KKK
)s(V
)s(V)s(H R
2o1oQ2
2oLPBPNBPP2oQ2
KKKsKs
KK)KKK)(Ks()s(H
)K(KKKsKs
K/KKKs)K/KKK(s
)s(V
)s(V)s(H BPP
2o1oQ2
BPP1oBPNBPPLPQ2
1
1o1
Tow-Thomas Biquadwith Feedforward Zero implementations
s2
Tow-Thomas Biquad
20
Feedforward Tow-Thomas (TT) Biquad Circuit
R1
C1
R4
+ +
+
A1 A2 A3
R6
V1
R5
R7 R2
R8
R3
C2
1
2 3
411 5
6
Observe that the regular TT Biquad does not implement a highpass output
21
Description of the Parameters for the Tow-Thomas Filter
109237
8
91
2
109537
6
7
6
9491
2
6
8
CCRR
1
R
R
CR
1ss
CCRR
1
R
R
R
R
CR
1
CR
1s
R
RsT
General Transfer Function
where
74
6
110753
96z
10732
981p
109753
62z
109327
82p
RR
R
R
1
CRRR
CRQ,
CRRR
CRRQ
CCRRR
R,
CCRRR
R
and
64
5
2LP
65
74
81BP
5
6
741
6
8HP
R,Rfor,R
RH
R,Rfor,RR
RRH
R,R
RRRfor,
R
RH
For the bandstop (notch)
8
265
6
741
6
8notch
R
RRR,
R
RRRfor,
R
RH
22
Design Equations for the Tow-Thomas Filter
Let
CCC
RRR
RaR
RR
21
'87
32
2
3
'4
6
1'
5
6z
1p
'5
6p
RR
R
R
1
RRR
RQ,
aR
RQ
RRR
R
C
1,
aRC
1
'
62
5
6
'4
1
6
'
notch
64
5
2
LP
65
4
1HP
5HP4
6
'4
1
6
'
HP
R
RRaR,
R
RRRfor,
R
RH
R,Rfor,R
RaH
R,Rfor,R
RH
R,HRR
RRRfor,
R
RH
23
Tow - Thomas Biquad
** Description of the passive components
r1 2 3 1596698
r2 11 5 100000
r3 6 2 100000
r4 2 1 1596698
r7 3 4 100000
r8 4 11 100000
c1 2 3 9.7491D-11
c2 5 6 9.7491D-11
* Description of Op Amps
E1 3 0 0 2 2D5
E2 11 0 0 4 2D5
E3 6 0 0 5 2D5
*
VIN 1 0 AC 1
*
.AC LIN 100 6000 20000
.PLOT AC VDB(11) VP(11)
.PROBE
.END
PSPICE Input file of Tow Thomas Filter
24
Tow - Thomas Biquad
** Description of the passive components
r1 2 3 1596698
r2 11 5 100000
r3 6 2 100000
r4 2 1 1596698
r7 3 4 100000
r8 4 11 100000
c1 2 3 9.7491D-11
c2 5 6 9.7491D-11
* Description of Op Amps
E1 3 0 0 2 2D5
E2 11 0 0 4 2D5
E3 6 0 0 5 2D5
*
VIN 1 0 AC 1
*
.AC LIN 100 6000 20000
.PLOT AC VDB(11) VP(11)
.PROBE
.END
25
KHN Fully-Differential Version
-+-+ +
--+
+
-+-
XR 03R
3R QR
01R
1C
1C3R
2R
2R
QR
03R
XR
02R
02R
2C
2C
iV
iV
LPV
LPV
26
Kerwin-Huelsman-Newcomb Biquad Circuit
R6
R4
+ +
+
A1 A2 A3V1
R5 R1 R2
R3
C8
1
2 3
411 5
6
V3
V2
R9
C7
V4
12
27
Simple Design Equations for the KHN Filter
Let
CCC
RRR
RRbRR
87
'21
6542
bb
RRRRQ
CbR
p
p
293
/11
//1
1
'
,
933
2
5
3
||
,||
//1
/11||
RR
LP
RR
BP
HP
H
H
RRRR
bH
28
KHN Biquad Design Procedure
A simple design procedure is described as follows:
1. Assume that wp, Qp, and H=K are the design specifications.
2. Select convenient values* or R’, C, and R to determine thevalues of R1, R2, C7, C8, R4, and R6.
3. Calculate the following element values:
pCRb
'
1
For the HP case:
1/119
3
2
HPp HbbQ
R
HPp
R
HbQ
RR
* A rule of thumb for choosing R’ and R is to make them proportional to 10fp , whenQp > 10, or else make R’ and R proportional to fp. Then C should be made proportional to 1/R’ p .
29
For the BP case:
BPp
BP
HbbQ
RR
H
RR
1/11 29
3
For the LP case:
LPpp
HQbR
HQbQb
bRR
RLPp
29
3
1
30
Example Design a bandpass modified KHN filter having a gainof HBP =3, QP = 20, and o = 2p x 103r/s.
Procedure
1. Let us choose R=10 kW, C=0/01 mF, and R’=11.254 kW; that isR1= R2 = 11.254kW and R4 = R5 = 10kW.
2. Since the values of R1 and R2 are 11.254kW, then
414207.1'1
pCRb
31
3. The expression for ;26
b
RR then W kR 56
4. For this case
W kH
RR
BP
33.33
and
W
260
1/11 29
BPp HbbQ
RR
Exercise.- Propose component value condition to make this structure Fully Balance Fully Symmetric Structure
32
KERWIN-HUELSMAN-NEWCOMB Biquad Circuit with f0=1KHz, Q=20, Peak
value gain =3
** Description of the passive components
r1 4 3 11.254K
r2 11 5 11.254K
r3 1 12 3.3K
* varying r4 values and parameters with the .step statement
.param R=1
r4 1 11 {R}
r5 2 6 10K
r6 2 3 5K
r9 1 0 260
c7 4 11 0.01U
c8 5 6 0.01U
* Description of Op Amps
E1 3 0 1 2 2D5
E2 11 0 0 4 2D5
E3 6 0 0 5 2D5
*
VIN 12 0 AC 1
*
.AC LIN 100 500 2K
.step DEC param R 300 30K 10
.PLOT AC VDB(11) VP(11)
.PROBE
.END 33
34
35
Second-Order Filter Characteristicsand Mason Rule to obtain Transfer Functions
• Second-Order Transfer function characteristics.• How can you use this information for better design and tuning
• Time domain measurements and characterization
Analog and Mixed-Signal Center.
Texas A&M University
Edgar Sánchez-Sinencio
• Mason Rule is a good tool, borrow from control theory, to easily obtain transfer functions from a signal flow graph
622 Active Filters by Edgar Sánchez-Sinencio
PROPERTIES of SECOND-ORDER SYSTEMS
AND MASON’S RULE to determine transfer functions by
inspection
• Mathematical definitions and properties of Second-OrderSystems.
• Building block second-order system architectures and properties.
• Mason’s Rule to obtain easily transfer functions and to facilitate the generation of new architectures.
SECOND-ORDER FILTER TYPES
Second-order blocks are important building blocks since with a combination of themallows the implementation of higher-order filters. The general order transfer function inthe s-plane has the form:
2p
o2
322
1
Q
ss
KsKsK)s(H
Particular conventional cases are:
Lowpass i.e.,
Bandpass i.e.,
Highpass i.e.,
(Notch) Band-Elimination i.e.,
Allpass i.e.,
0KK 21
0KK 31
0KK 32
0K2
2o3
o21 Kand
QK,1K
One interesting case used for amplitude equalization is the “equalizer” sometimes referred
to as Bump (DIP) Equalizer. In this case, and .
Specific structures have different properties. Some structures have enough degrees of
freedom to allow them to change independently , Q (or BW) and a particular gain
where is a particular frequency, i.e., for the LP, BP and HP cases.
Furthermore, some structures have the property to have constant Q or BW while varying
. We will illustrate later, by examples, some of the structures with such properties.
QkKK1K o
22o31
o )(p
H
p ,,0 op
of
Properties of Second-Order Systems in the time domain
2o
2222o
o2 s2s)s(sQ
s
where2o
o
Q4
11,
Q2
)t(sineKK)t(v t21os
)s(N
2)s(
inv
t
1
))s(N(fK,K 21
1t1
2to
p
N2t2
CYCLESN
e
A
AHow to determine the pole location from a step response?
Sinusoidal steady-state response1
ttsin
)s(N
2o
o2
Qss
)](tsin[)(M
o
)(M
maxM
2
Mmax 10Q
QBW o
2o222
ojs
2o
o2
Q)(
)j(N
Qss
)s(N)(M
Only two measurements are necessary to fix the position of the complex poles. Themeasurement of the frequency of peaking determines the magnitude of the poles,and the measurement of the 3-dB bandwidth determines
,o.Q/o
Second-Order Low-Pass Networks
2o
o2
Qss
H)s(T
Since H is merely a magnitude scale factor, let .H 2o
1Q
1ss
1
Qss
)s(T
o
2
o
2o
o2
2o
1. For Therefore, low frequencies are passed.
2. For Therefore, high frequencies are attenuated.
3. For the magnitude peaks at
The peak occurs at a frequency lower than For Q>5, the frequency of
peaking practically equals (within 1%).
4. For Q>5, the 3-dB bandwidth is practically equal to
5. At and the phase is
6. At the phase is
7. For Q>5, the phase undergoes a rapid shift of radians about
.1)j(T,1/ o
.)/()j(T,1/ 2oo
,2/1Q 2o Q2
11
.o
o
.s/radQ/o
Q)j(T,1/ oo .2/p
.p,1/ o
p .o
H1 w( )
H2 w( )
H3 w( )
H4 w( )
H5 w( )
H6 w( )
H7 w( )
w
0 0.5 1 1.5 20
2
4
6
8
10
arg H1 w( )( )
arg H2 w( )( )
arg H3 w( )( )
arg H4 w( )( )
arg H5 w( )( )
arg H6 w( )( )
arg H7 w( )( )
w
0 0.5 1 1.5 23.5
3
2.5
2
1.5
1
0.5
0
10Q
o
o
2o
o2
2o
Qss
Magnitude and phase of
2
p
p
Magnitude
Phase
1
2
3
5
1235
The Second-Order Band-Pass Function
2o
o2
Qss
Hs)s(T
1Q
1ss
s
Q
1
Qss
sQ
)s(T
o
2
o
o
2o
o2
o
The second-order band-pass function has one zero at the origin and anotherat infinity:
To normalize the peak value of the magnitude function to unity, let :)Q/(H o
H7 w( )
jw
Q
.
w2
jw
Q
. 1
H1 w( )
H2 w( )
H3 w( )
H4 w( )
H5 w( )
H6 w( )
H7 w( )
w
0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
arg H1 w( )( )
arg H2 w( )( )
arg H3 w( )( )
arg H4 w( )( )
arg H5 w( )( )
arg H6 w( )( )
arg H7 w( )( )
w
0 0.5 1 1.5 22
1.5
1
0.5
0
0.5
1
1.5
2
Magnitude
Phase
5.0Q
1
2
3
5
10Q
o
o
10Q
5.0Q
1235
2o
o2
o
Qss
sQ
Magnitudeand phase of
1t
1
and
2t
2N
p
Re
Im
oj
Im
Q2
o
Re Re
Im
Re
Im
o
o
POLE-ZERO LOCI
Constant Q
Constant BW
Constant
Pole locations and properties
In practical implementations besides the general specifications
other particular specifications are imposed which
are application dependent. Among them are silicon area, dynamic
range, power supply rejection ratio, power consumption, tolerance,
accuracy, and sensitivity. This last parameter is often used as a
figure of merit. i.e.,
The above definition is usually referred as normalized sensitivity
due to the (x/p) factor. x and p are the variable of the network
(i.e. R, C, ) and the variable under consideration
p
x
x
pSp
x
))(H,Q,( po
))(H,Q,.,e.i( po
mg
)1(
The General Input-Output Gain Mason Formula
We can reduce complicated block diagrams to canonical form, from which the control ratio is easily written:
It is possible to simplify signal flow graphs in a manner similar to that of block diagram reduction. But it is also possible,
and much less time-consuming, to write down the input-output relationship by inspection from the original signal flow graph.
This can be accomplished using the formula presented below. This formula can also be applied directly to block diagrams,
but the signal flow graph representation is easier to read - especially when the block diagram is very complicated.
Signal Flow Graphs
Let us denote the ratio of the input variable to the output variable by T. For linear feedback control systems, T=Vout/Vin.
For the general signal flow graph presented in preceding paragraphs Vout is the output and Vin is the input.
The general formula for and signal flow graph is
where = the ith forward path gain
= jth possible product of k non-touching loop gains
= 1 - (sum of all loop gains) + (sum of all gain-products of 2 non-touching loops) - (sum of all gain-
products of 3 non-touching loops + …
= evaluated with all loops touching eliminated.
Two loops, paths, or a loop and a path are said to be non-touching if they have no nodes in common.
is called the signal flow graph determinant or characteristic function, since is the system characteristic
equation.
GH
G
V
V
in
out
1
i
iiP
T
iP
jkP
k j
jk1k P)1(1
j j
3jj
2j1j PPP1
i iP
0
ExamplesLet us determine the control ratio Vout/Vin and the canonical block diagram of the feedback
control system shown below:
The signal flow graph is
There are two forward paths:
inV
1G 4G
1H
2G
2H
outV
3G
inV
1 41GG 2G 1 1
outV
1H
2H
3G
43124211 GGGP,GGGP
There are three feedback loops:
There are no non-touching loops, and all loops touch both forward paths; then
Therefore the control ratio is
From Equations (8.3) and (8.4), we have
Therefore
The canonical block diagram is there fore given by
The negative summing point sign for the feedback loop is a result of using a positive sign in the GH
formula above.
24313124212114111 HGGGP,HGGGP,HGGP
1,1 21
24312421141
4314212211
HGGGHGGGHGG1
GGGGGGPP
R
CT
24312421141
3241
HGGGHGGGHGG1
)GG(GG
)HHGHG(GGGHand)GG(GGG 12223413241
32
1232
GG
HH)GG(
G
GHH
R
32
1232
GG
HH)GG(
)GG(GG 3241 C
G
H
Draw a signal flow graph for the following resistance network in whichis the voltage across
The five variables are is the input. The four independent
equations derived from Kirchoff’s voltage and current laws are
The signal flow graph can be drawn directly from these equations:
In Laplace transform notation, the signal flow graph is given by
.0)0(v)0(v 32
2v .C1
Example
o o
o o
1v
1R
2C
2
1C 3v
1i 2i
2R
121321 vand;iand,i,v,v,v
t
0
t
02
11
122
11
11 ,dti
C
1dti
C
1v,v
R
1v
R
1i
t
02
233
22
22 dti
C
1v,v
R
1v
R
1i
1R/1 t
01 dt)C/1(
2i1v
1R/1 t
01 dt)C/1( 1
1i 2v
2R/1 t
02 dt)C/1(
2R/1
3v 3v
1sC/1
2I1V
1R/1 1sC/1 1
1I
1R/1
2V
2R/1 2sC/1
2R/1
3V 3V
51
ECEN 622 (ESS)
Example of use of Mason Rule in a 3rd Orders State-Variable (observable) Filter
S S S S1/s 1/s 1/s
Vi
bo b1 b2 b3
Vo
-a2
-a1
-ao
o1
22
3o1
22
33
3o
212
32
21
3o
1asasas
bsbsbsb
s
a
s
a
s
a1
bs
b
s
b
s
b
sH
What is wrong with this application of Mason’s Rule?
52
o21112
23
o211132222
211232323
32
3o21
2112
2112
3222
2211
321o
2
aBBsBasas
bBBsaBbaBbsBBbabBbsbH
s
aBB
s
Ba
s
a1
s
aB
s
a1b
s
a1
s
Bb
s
BBb
s
BBb
sH
S S S S1/s 1/s 1/s
Vi
bob1 b2
b3
Vo
-a2
-a1
-ao
B1 B2
53
A possible implementation of H2 (s) using Active-RC follows
C/b3
Rb3
Rb1
Rbo
C CC
RB1RB2
Vin
Ra2
Ra1
Rao-1
Vo
