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ELE205 Electronics 2

Filters-3

Linga Reddy Cenkeramaddi

Department of ICT

UiA, Grimstad

January 2015

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ACTIVE FILTER CIRCUITS

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DISADVANTAGES OF PASSIVE

FILTER CIRCUITS

Passive filter circuits consisting of resistors, inductors, and

capacitors are incapable of amplification, because the output

magnitude does not exceed the input magnitude.

The cutoff frequency and the passband magnitude of passive

filters are altered with the addition of a resistive load at the output

of the filter.

In this section, filters using op amps will be examined. These op

amp circuits overcome the disadvantages of passive filter circuits.

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FIRST-ORDER LOW-PASS FILTER

+

+

+

vovi

R1R2

C

+

+

+

vovi

Zi

Zf

c

csC

i

f

sK

R

R

Z

ZsH

1

12 ||)(

CRR

RK c

21

2 1

4

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PROTOTYPE LOW-PASS FIRST-

ORDER OP AMP FILTER

Design a low-pass first-order filter with R1=1, having apassband gain of 1 and a cutoff frequency of 1 rad/s.

2 1

2

1

1 11

(1)(1)

1( ) 1

c

c

c

R KR

C FR

H s K s s

5

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FIRST-ORDER HIGH-PASS FILTER

+

+

+

vovi

R1R2C 2

1

2

1 1

( )1

1

f

i c

c

Z R sH s K

Z sRsC

RK

R R C

Prototype high-pass filter

with R1=R2=1 and C=1F.The cutoff frequency is 1

rad/s. The magnitude at the

passband is 1.

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EXAMPLE

Figure shows the Bode

magnitude plot of a high-

pass filter. Using the

active high-pass filter

circuit, determine values

of R1 and R2. Use a 0.1F

capacitor.

If a 10 K load resistor isadded to the filter, how

will the magnitude

response change?

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Notice that the gain in the passband is 20dB, therefore, K=10. Also

note the the 3 dB point is 500 Hz. Then, the transfer function for the

high-pass filter is

2

1

1

2

1 1

1 2

10( )

1500

110 500

20 200

Rs

RsH s

ss

R C

R

R R C

R K R K

Because the op amp in the circuit is

ideal, the addition of any load

resistor has no effect on the

behavior of the op amp. Thus, themagnitude response of the high-

pass filter will remain the same

when a load resistor is connected.

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In the design of both passive and active filters, working with element

values such as 1 , 1 H, and 1 F is convenient. After making

computations using convenient values of R, L, and C, the designercan transform the circuit to a realistic one using the process known as

scaling. There are two types of scaling:magnitude andfrequency.

A circuit is scaled in magnitude by multiplying the impedance at a

given frequency by the scale factor km. Thus, the scaled values of

resistor, inductor, and capacitor become

and /m m m

R k R L k L C C k

where the primed values are the scaled ones.

SCALING

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In frequency scaling, we change the circuit parameter so that at the

new frequency, the impedance of each element is the same as it was

at the original frequency. Let kf

denote the frequency scale factor,

then

/ and /f fR R L L k C C k

A circuit can be scaled simultaneously in both magnitude and frequency.

The scaled values in terms of the original values are

1 andmmf f m

kR k R L L C Ck k k

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This circuit has a center frequency of 1 rad/s, a

bandwidth of 1 rad/s, and a quality factor of 1. Use

scaling to compute the values of R and L that yield

a circuit with the same quality factor but with a

center frequency of 500 Hz. Use a 2 F capacitor.

The frequency scaling factor is:

2 (500)

3141.591fk

The magnitude scaling factor is:6

1 1 1159.155

3141.59 2 10m

f

Ck

k C

o

159.155 50.66

1/ 3141.61 rad/s or 500 Hz.

/ 3141.61 rad/s or 500 Hz.

Q / 1

mm

f

o

kR k R L L mH

k

L C

R L

EXAMPLE

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Use the prototype low-pass op amp filter and scaling to compute the

resistor values for a low-pass filter with a gain of 5, a cutoff frequencyof 1000 Hz, and a feedback capacitor of 0.01 F.

8

1 2

/ 2 (1000) /1 6283.185

1 115915.5

(6283.15)(10 )

15915.5

f c c

m

f

m

k

Ck

k C

R R k R

To meet the gain specification, we can adjust one of the resistor

values. But, changing the value of R 2 will change the cutoff

frequency. Therefore, we can adjust the value of R 1

as

R1=R2/5=3183.1 .

EXAMPLE

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OP AMP BANDPASS FILTERS

A bandpass filter consists of three separate components

1. A unity-gain low-pass filter whose cutoff frequency is

c2, the larger of the two cutoff frequencies

2. A unity-gain high-pass filter whose cutoff frequency is

c1, the smaller of the two cutoff frequencies

3. A gain component to provide the desired level of gain in

the passband.

These three components are cascaded in series. The resultingfilter is called abroadband bandpass filter, because the

band of frequencies passed is wide.

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2

2 1

2

2

1 2 1 2

( )

( )

fo c

i c c i

c

c c c c

RV sH sV s s R

K s

s s

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Standard form for the transfer function of a bandpass filter is

2 2BP

o

s

H s s

In order to convert H(s) into the standard form, it is required that

. If this condition holds, 2 1c c

1 2 2( )c c c

Then the transfer function for the bandpass filter becomes

2

2

2 1 2

( ) c

c c c

K sH ss s

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Compute the values of RL and CL to give us the desired cutoff

frequency

2

1

cL LR C

Compute the values of RH and CH to give us the desired cutoff

frequency1

1c

H HR C

2

2

2 2 1

( )( )

( ) ( )

fc oo

o c o c c i

K jH j K

j j R

To compute the values of Ri and Rf, consider the magnitude of thetransfer function at the center frequency o

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Design a bandpass filter to provide an amplification of 2 within the

band of frequencies between 100 and 10000 Hz. Use 0.2 Fcapacitors.

2 6

1 6

1 1(2 )10000 80

2 (10000) (0.2 10 )

1 1(2 )100 7958

2 (100) (0.2 10 )

c L

L L

c H

H H

RR C

RR C

Arbitrarily select Ri=1 k, then Rf=2Ri=2 K

EXAMPLE

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Like the bandpass filters, the bandreject filter consists of three

separate components

The unity-gain low-pass filter has a cutoff frequency ofc1,which is the smaller of the two cutoff frequencies.

The unity-gain high-pass filter has a cutoff frequency ofc2,which is the larger of the two cutoff frequencies.

The gain component provides the desired level of gain in the

passbands.

The most important difference is that these components are

connected in parallel and using a summing amplifier.

OP AMP BAND REJECT FILTERS

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+

+vi

RLRL

+

RH

RH

CH

+

vo+

Ri

Ri

Rf

CL

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1

1 2

2

1 1 2

1 2

( )

2( )( )

f c

i c c

f c c c

i c c

R sH s

R s s

R s sR s s

1 2

1 1

f

c cL L H H i

RK

R C R C R

The magnitude of the transfer function at the center frequency

2

1 1 2

21 2 1 2

1 1

1 2 2

( ) 2 ( )

( ) ( ) ( )( )

2 2

f o c o c c

oi o c c o c c

f fc c

i c c i c

R j j

H j R j j

R R

R R

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All of the filters considered so far are nonideal and have a slow

transition between the stopband and passband. To obtain a sharper

transition, we may connect identical filters in cascade.

For example connecting two first-order low-pass identical filters in

cascade will result in -40 dB/decade slope in the transition region.

Three filters will give -60 dB/decade slope, and four filters shouldhave -80 db/decade slope. For a cascaded of n protoptype low-pass

filters, the transfer function is

1 1 1 1( )1 1 1 1

n

H ss s s s

HIGHER ORDER OP AMP FILTERS

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But, there is a problem with this approach. As the order of the low-

pass is increased, the cutoff frequency changes. As long as we areable to calculate the cutoff frequency of the higher-order filters,

we can use frequency scaling to calculate the component values

that move the cutoff frequency to its specified location. For an nth-

order low-pass filter with n prototype low-pass filters

2/

22

2

1 1( )

( 1) 2

1 1 1 112 21

2 1 2 1

cn n

cn

n

n

cncn

n n

cn cn

H jj

EXAMPLE

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Design a fourth-order low-pass filter with a cutoff frequency of 500

rad/s and a passband gain of 10. Use 1 F capacitors.

44

2 (500)2 1 0.435 rad/s 7222.39

0.435c fk

6

1138.46

7222.39(1 10 )mk

Thus, R=138.46 and C=1 F. To set the passbandgain to 10, choose Rf/Ri=10. For example Rf=1384.6

and Ri =138.46 .

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+

+vi

138.46

1F

138.46

+

138.46

1F

138.46

+

138.46

1F

138.46

+

138.46

1F

138.46

+

138.46

1384.6

+

vo

25

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By cascading identical prototype filters, we can increase the

asymptotic slope in the transition and control the location of the

cutoff frequency. But the gain of the filter is not constant between

zero and the cutoff frequency. Now, consider the magnitude of thetransfer function for a unity-gain low-pass nth order cascade.

2 2

( ) ( )

1( )

( / ) 1

n

cn n

cn

n

cn

n n

cncn

H s s

H j

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BUTTERWORTH FILTERS

A unity-gain Butterworth low-pass filter has a transfer functionwhose magnitude is given by

ncjH

2/1

1)(

1. The cutoff frequency is c for all values of n.

2. If n is large enough, the denominator is always

close to unity when

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Given an equation for the magnitude of the transfer function,

how do we find H(s)? To find H(s), note that if N is a complex

quantity, the |N|2

=NN*

. Then,

nn

nnn

s

sHsH

sjH

s

sHsHjHjHjH

2

222

2

22

2

)1(1

1)()(

)(1

1

)(1

1

1

1)(

since

)()()()()(

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The procedure for finding H(s) for a given n is:

1. Find the roots of the polynomial 1 + (-1)ns2n

= 0

2. Assign the left-half plane roots to H(s) and

the right-half plane roots to H(-s)

3. Combine terms in the denominator of H(s)to form first- and second-order factors

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Find the Butterworth transfer function for n = 2.

For n=2, 1+(-1)2s4 = 0, then s4 = -1 = 1 1800

2

1

2

13151

2

1

2

12251

2

1

2

11351

2

1

2

1451

0

4

0

3

0

2

0

1

jsjs

jsjs

Roots s2 and s3 are in the left-half plane. Thus,

12

1)(

221221

1)(

2

sssH

jsjssH

EXAMPLE

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Normalized Butterworth Polynomials

)1932.1)(12)(1518.0(s6

)1618.1)(1618.01)(s(s5

)1848.1)(1765.0(s4

)11)(s(s3

)12(s21)(s1

222

22

22

2

2

sssss

sss

sss

s

s

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BUTTERWORTH FILTER CIRCUITS

To construct a Butterworth filter circuit, we cascade first- and second-

order op amp circuits using the polynomials given in the table. A fifth-order prototype Butterworth filter is shown in the following figure:

1

1

s 1618.0

12

ss 1618.11

2 ssvi vo

All odd-order Butterworth polynomials include the factor (s+1), so

all odd-order BUtterworth filters must include a subcircuit to

implement this term. Then we need to find a circuit that provides atransfer function of the form

1

1)(

1

2

sbssH

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+

+vi

RR

C1

C2

Va

+

Vo 0

0)(

2

1

R

VVsCV

R

VVsCVV

R

VV

aoo

oaoa

ia

0)1(

)1()2(

2

11

oa

ioa

VsRCV

VVsRCVsRC

211

2

21

2

2

2

21

2

12

1

)(

12

1

CCsCs

CCR

V

VsH

VsRCsCCRV

i

o

io

211

1

11

2

CCCb

33

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Design a fourth-order low-pass filter with a cutoff frequency of 500

Hz and a passband gain of 10. Use as many 1 K resistor as possible.

From table, the fourth-order Butterworth polynomial is

)1848.1)(1765.0(22

ssss

For the first stage: C1=2/0.765=2.61 F, C2=1/2.61=0.38F

For the second stage: C3=2/1.848=1.08 F, C4=1/1.08=0.924F

These values along with 1- resistors will yield a fourth-order

Butterworth filter with a cutoff frequency of 1 rad/s.

EXAMPLE

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A frequency scale factor of kf=3141.6 will move the cutoff

frequency to 500 Hz. A magnitude scale factor km=1000 will permit

the use of 1 k resistors. Then,R=1 k, C1=831 nF, C2=121 nF, C3= 344 nF, C4=294 nF, Rf= 10

k.

+

+vi

RR

C1

C2

+RR

C3

C4

+

Vo

+

Ri

Rf

35

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The Order of a Butterworth Filter

As the order of the Butterworth filter increases, the magnitude

characteristic comes closer to that of an ideal low-pass filter.

Therefore, it is important to determine the smallest value of n that

will meet the filtering specifications.

|H(jw)| Pass

band

Transition band Stopband

A

P

A

S

WP WS

log10w

)1(log10

1

1

log20

)1(log10

11log20

2

10

210

2

10

210

n

s

n

s

s

n

p

n

p

p

A

A

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)(log

)(log

)(log)(log

110

110

110110

10

10

1010

1.0

1.0

p

s

21.021.0

ps

ps

psps

p

s

A

An

n

s

An

p

A

n

n

p

s

sp

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If p is the cutoff frequency, then

)(log

log

1and2log20

10

10

10

ps

s

pp

n

A

For a steep transition region, 110 1.0 sA Thus,

)(log

05.005.0log10

10

10

05.0

ps

s

ss

A

s

An

A

s

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Determine the order of a Butterworth filter that has a cutofffrequency of 1000 Hz and a gain of no more than -50 dB at 6000

Hz. What is the actual gain in dB at 6000 Hz?

21.3)1000/6000(log

)50(05.0

10

n

Because the cutoff frequency is given, and 10-0.1(-50)>>11p

Therefore, we need a fourth-order Butterworth filter. The actualgain at 6000 Hz is

dB25.6261

1log20

810

K

EXAMPLE

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Determine the order of a Butterworth filter whose magnitude is 10 dB

less than the passband magnitude at 500 Hz and at least 60 dB lessthan the passband magnitude at 5000 Hz.

52.2)10(log

)31000(log

105005000

1000110,3110

10

10

)60(1.0)10(1.0

n

ff psps

sp

Thus we need a

third-orderfilter.

Determine the cutoff frequency.

rad/s26.21789

1000

9

10)(110])(1[log10

66

6

10

c

cc

EXAMPLE

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BUTTERWORTH HIGH-PASS FILTERS

To produce the second-order factors in the Butterworth

polynomial, we need a circuit with a transfer function of

1)(

1

2

2

sbs

ssH

+

+

viR

2

R1C C+

Vo

2

212

2

2

12)(

CRR

s

CR

s

s

V

VsH

i

o

Setting C= 1F

212

2

2

12)(

RRsRs

s

V

VsH

i

o

212

1

11

2

RRRb

41

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NARROWBAND BANDPASS AND

BANDREJECT FILTERS

The cascade or parallel component designs from simpler low-passand high-pass filters will result in low-Q filters. Consider the

transfer function

22

22

5.0

2)(

c

cc

c

cc

c

ss

s

s

s

s

s

ssH

2

1,2 22

ococ Q

Thus with discretereal poles, the highest

quality factor

bandpass filter we can

achieve has Q=1/2

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NARROWBAND BANDPASS AND

BANDREJECT FILTERS

43

An active high Q bandpass filter

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