active filter (part 2)1 m2-3s active filter (part ii) biquadratic function filters positive feedback...

Active Filter (Part 2) 1

M2-3S Active Filter (Part II)

• Biquadratic function filters

• Positive feedback active filter: VCVS

• Negative feedback filter: IGMF

• Butterworth Response

• Chebyshev Response

Active Filter (Part 2)

22

22

2

2

)(

PP

P

ZZ

Z

sQ

s

sQ

s

Kbass

dcsssH

Realised by:

(I) Positive feedback

(II) Negative feedback

Biquadratic function filters


(I) Low Pass

(II) High Pass

(III) Band Pass

(IV) Band Stop

(V) All Pass

222

11)(

P

P

P sQ

sK

bassKsH

22

2

2

2

)(

P

P

P sQ

s

sK

basss

KsH

222

)(

P

P

P sQ

s

sK

basss

KsH

22

22

2

2

)(

P

P

P

Z

sQ

s

sK

bass

bsKsH

22

22

2

2

)(

P

P

P

Z

Z

Z

sQ

s

sQ

sK

bass

bassKsH

Biquadratic functions


Low-Pass Filter

1

1

2

pQ

ss

sH

0 1 2 3 40

0.5

1

1.5 Qp = 1.5

Qp = 1

Frequency

K = 1, p = 1

Voltage Gain

Qp = 2

1

0 1 2 3 4

-30

-20

-10

0

Frequency

K = 1, p = 1

Qp = 1.5

Qp = 1

Voltage Gain (dB)

Qp = 2

1


High-Pass Filter

12

2

pQ

ss

ssH

0 1 2 3 40

0.5

1

1.5 Qp = 1.5

Qp = 1

Frequency

K = 1, p = 1

Voltage Gain

Qp = 2

1

0 1 2 3 4-20

-15

-10

-5

0

5

Frequency

K = 1, p = 1

Qp = 1.5Qp = 1

Voltage Gain (dB)

Qp = 2

1


Band-Pass Filter

12

pQ

ss

ssH

0 2 4 6 80

0.5

1

1.5 Qp = 1.5

Qp = 0.5

Qp = 0.1

Frequency

K = 1, p = 1

Voltage Gain

Qp = 1

Qp = 2

1

0 2 4 6 8-30

-25

-20

-15

-10

-5

0

5

Frequency

K = 1, p = 1

Qp = 1.5

Qp = 0.5

Qp = 0.1

Voltage Gain (dB)

Qp = 2

1Qp = 1


Band-Stop Filter

1

2

2

22

pQ

ss

ssH

0 2 4 6 80

1

2

3

4

5

Qp = 1.5

Frequency

K = 1, p = 1, z = 2

Vol

tage

Gai

n

Voltage Gain

Qp = 1

Qp = 2

1

0 2 4 6 8-30

-20

-10

0

10

Frequency

K = 1, p = 1, z = 2 Qp = 1.5

Voltage Gain (dB)

Qp = 2

1


Voltage Controlled Votage Source (VCVS) Positive Feedback Active Filter (Sallen-Key)

By KCL at Va: i i ii f a 0

iV VZii a

1

iV

Z Zaa

2 4

where,V VZ

V VZ

VZ Z

i a o a a

1 3 2 4

0

Therefore, we get

VZ

VZ

VZ Z Z Z

i oa

1 3 1 3 2 4

1 1 10

Re-arrange into voltage group gives:

(1)

Z1 Z2

Z3

Z4

K

Vi Vo

Va

if

ii ia

3Z

VVi ao

f


But, V Ki Z KZVZ Zo a

a 44

2 4(2)

Substitute (2) into (1) gives V

ZVZ

V Z Z

KZ Z Z Z Zi o o

1 3

2 4

4 1 3 2 4

1 1 10

or

H

VV

KZZ

KZZZ Z Z

Z Z

o

i

1

3

1 2

3 4 41 21 1

1

In admittance form:

H

K

YY Y

YY

KY YYY

1

1 114

1 2

3

1

3 4

1 2

* This configuration is often used as a low-pass filter, so a specific example will be considered.

(3)

(4)


VCVS Low Pass Filter H s Ks as b

( ) 1

2

In order to obtain the above response, we let:

444

3332211

11

11

sCCjZ

sCCjZRZRZ

R 1 C 3R 2

C 4

Then the transfer function (3) becomes:

224321

2

31214

'

11)(

P

P

P sQ

s

K

CCRRsKCsRRRsC

KsH

(5)


Equating the coefficient from equations (6) and (5), it gives:

42

31

31

42

32

4142314321 1

1

111

CR

CRK

CR

CR

CR

CRQ

CRCRCCRRPP

Now, K=1, equation (5) will then become,

H ssC R R s RR CC

( )

1

1 4 1 22

1 2 3 4

we continue from equation (5),

11)(

312144321

2

KCsRRRsCCCRRsK

sH

43214321

312142

4321

11

1

)(

CCRRCCRR

KCRRRCss

CCRRK

sH


Simplified Design (VCVS filter)

mR nCR

C

22211

1)(

CnmRsmsRCsH

sCCjZ

snCnCjZRZmRZ

11

1

)(

1 4321

Comparing with the low-pass response:

22

1)(

PP

P sQ

sKsH

It gives the following:

1

1

m

mnQ

nmRCpp


Example (VCVS low pass filter)To design a low-pass filter with and

2

1 Hz512 QfO

vin C vo

+

-

mR R

nCLet m = 1

2

1

211

1

1

nn

m

mnQ

P

n = 2

)512(22

1

21

11Hz

RCRCmnRCP

nFC 100Choose

Then kR 2.2~198,2 vin 100nF vo

+

-

2k2 2k2

200nF

What happen if n = 1?


VCVS High Pass Filter

vin

C1

vo

+

-

R3

R4

C2

22

2

2143132414

2

2 '

11

111)(

P

P

P sQ

s

SK

CCRRK

CRCRCRss

KSsH


VCVS Band Pass Filter

vin

R1

vo

+

-

R3

R4

C2

C5

22

52431

31

24

5

3415

2

51'

1111

)(

P

P

P sQ

s

SK

CCRRR

RR

CR

CK

RRRC

ss

CR

sK

sH


Infinite-Gain Multiple-Feedback (IGMF) Negative Feedback Active Filter

V i

Z 1

V x

Z 3

Z 2

Z 4 Z 5

V o

)2( ,at KCLBy

)1( 0 0

4321x

5

3

3

5

Z

VV

Z

V

Z

V

Z

VVV

VZ

ZVV

Z

ZVvv

oxxxxi

oxxoii

substitute (1) into (2) gives VZ

ZZZ

VZZ Z

VVZ

ZZ Z

VVZ

io o

oo

o

1

3

1 5

3

5 2 5

3

4 5 4

(3)

Note: because no current flows into v+, v- terminals of op-amp. Therefore, from KCL at node v- : Vo/Z5+ Vx/Z3 = 0


rearranging equation (3), it gives,

HVV

ZZ

Z Z Z Z Z Z Z

o

i

1

1 1 1 1 1 11 3

5 1 2 3 4 3 4

HVV

YY

Y Y Y Y Y Y Yo

i

1 3

5 1 2 3 4 3 4

Or in admittance form:

Z1 Z2 Z3 Z4 Z5

LP R1 C2 R3 R4 C5

HP C1 R2 C3 C4 R5

BP R1 R2 C3 C4 R5

Filter Value


IGMF Band-Pass Filter

H s Ks

s as b( )

2Band-pass:

To obtain the band-pass response, we let

5544

433

32211 11

11

RZsCCj

ZsCCj

ZRZRZ

R 1

C 3R 2

C 4 R 5H s

sCR

sCC sC CR R R R

( )

3

1

23 4

3 4

5 5 1 2

1 1 1

*This filter prototype has a very low sensitivity to component tolerance when compared with other prototypes.


Simplified design (IGMF filter)

R 1

C

C R 5

22

551

1

21)(

CsR

Cs

RR

R

sC

sH

Comparing with the band-pass response

22

)(

P

P

P sQ

s

sKsH

Its gives,

2

11

5

51

2 CR

1- K

2

1

1QjH

R

RQ

RRCppp


Example (IGMF band pass filter)To design a band-pass filter with and 10 Hz512 QfO

)512(21

51

HzRRCP

251 741,662,9 100 RRnFC

102

1

1

5 R

RQ

P

170,62 4.155 51 RR vin vo+

-

100nF

4.155170,62

100nF

With similar analysis, we can choose the following values:

700,621 and 554,1 10 51 RRnFC


Butterworth Response (Maximally flat)

n

o

jH2

1

1

where n is the order

Normalize to o = 1rad/s

162.1162.01

124.324.524.524.3

185.1177.0

161.241.361.2

11

122

12

1

22

23455

22

2344

2

233

22

1

sssss

ssssssB

ssss

sssssB

sss

ssssB

sssB

ssB

Butterworth polynomials

Butterworth polynomials:

)(

1)(ˆ

jBjH

n

njH

21

1)(ˆ


Butterworth Response


Second order Butterworth responseStarted from the low-pass biquadratic function

22

1)(

P

P

P sQ

sKsH

2

111 QKp

njH

jH

jH

jH

jjH

sssH

222

4

242

222

2

2

1

1

1

1)(

1

1)(

221

1)(

21

1)(

12

1)(

)polynomial butterwothorder (second12

1)(

For


Bode plot (n-th order Butterworth)

dB/.decade20n - :orderth -nFor

dB/decade 60- :order 3rdFor

dB/decade 40- :order 2ndFor

ecade(-20n)dB/d have ldfilter wouh Butterwort The

10 condition, decadeFor

)log(20 )(ˆ

1 suppose

)1log(20)(ˆ

1

1log20)(ˆ

:form dBIn 1

1)(ˆ

x

2

2

2

njH

jH

jH

jH

n

n

n

Butterworth response

-20

-40

-60

-80

-100

x

x

1st order

2nd order3rd order4th order5th order

volta

ge g

ain

(dB

)


Second order Butterworth filter

224321

2

31214

'

11)(

P

P

P sQ

s

K

CCRRsKCsRRRsC

KsH

42

31

31

42

32

41 1

1

CR

CRK

CR

CR

CR

CRQ

P

Setting R1= R2 and C1 = C2

KKKQ

P

3

1

12

1

1111

1

Now K = 1 + RB/ RA

vinC4 vo

+

-

R1 R2C3

RB

RA

A

B

A

B

P

R

R

R

RKQ

2

1

13

1

3

1

For Butterworth response:

2

1P

Q

A

B

P

R

RQ

2

1

2

1

Therefore, we have 414.122 A

B

R

R

We define Damping Factor (DF) as:

414.121

A

B

P R

R

QDF


Damping Factor (DF)

• The value of the damping factor required to produce desire response characteristic depends on the order of the filter.

• The DF is determined by the negative feedback network of the filter circuit.

• Because of its maximally flat response, the Butterworth characteristic is the most widely used.

• We will limit our converge to the Butterworth response to illustrate basic filter concepts.


Values for the Butterworth response 162.1162.01124.324.524.524.3

185.1177.0161.241.361.2

11122

12

1

2223455

222344

2233

22

1

sssssssssssB

sssssssssB

sssssssB

sssB

ssB

Roll-off

dB/decade

1st stage 2nd stage 3rd stage

Order poles DF poles DF poles DF

1 -20 1 optional

2 -40 2 1.414

3 -60 2 1.000 1 1.000

4 -80 2 1.848 2 0.765

5 -100 2 1.000 2 1.618 1 0.618

6 -120 2 1.932 2 1.414 2 0.518


Forth order Butterworth Filter2121 , CCRR 4343 , CCRR

kR

kRR

R

R

R

RDF

B

AB

A

B

A

B

52.1

10152.0152.0

152.0

848.12

kR

kRR

R

R

R

RDF

B

AB

A

B

A

B

17.27

22235.1235.1

235.1

765.02

+

-

R2 8.2 k

C1 0.01 F

+

-

Vout

+15 VR1 8.2 k

RB 1.5 k

RA 10 k

R3 8.2 k R4 8.2 k+15 V

-15 V -15 V

C3 0.01 F

C2

0.01 F C4

0.01 FRB 27 k

RA 22 k

741C 741C


Chebyshev Response (Equal-ripple)

221

1

nCjH

Where determines the ripple and is the Chebyshev cosine polynomial defined as 2

nC


Chebyshev Cosine Polynomials

21

355

244

33

22

1

0

2

52016

188

34

12

1

nnn CCC

C

C

C

C

C

C


Second order Chebychev Response

Example: 0.969dB ripple gives = 0.5, 12 22 C

221

1

nCjH

25.1

1

125.01

1

1

1

24

22

2222

nCH

js

2

1

2

5112Roots:

jsjs

sHsH

21

21

1

2222

25.1

1

)(

24

22/

22

ss

sHsHHfs

969.0)1

1(log20

210

Note:


Roots

Roots of first bracketed term

899.0566.0

2

1

4

5

2

1

2

1

4

5

2

1

2

1

2

21

40

j

j

j

j

s

Roots of second bracketed term

899.0566.0

2

1

4

5

2

1

2

1

4

5

2

1

2

1

2

21

40

j

j

j

j

s

899.0556.0899.0566.0

1

899.0556.0899.0556.0

122 jsjs

．jsjs

sHsH

or 117.1112.0

1

899.0556.0

1

899.0556.0899.0556.0

12222

sssjsjs

sH

))2

sin()2

(cos(2)/(tan22 1 jCeCCeebabja jjabj

Note:

active filter (part 2)1 m2-3s active filter (part ii) biquadratic function filters positive feedback...

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