a second using ideal components all the biquad topologies ...ece.tamu.edu/~s-sanchez/622 lecture 3...

• 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

QQ

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