1

Filters

A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others.

A filter is a passive filter if it consists of only passive elements R, L, and C.

A lowpass filter passes low frequencies and stops high frequencies.

A highpass filter passes high frequencies and rejects low .

A bandpass filter passes frequencies within a frequency band and blocks or attenuates frequencies outside the band.

A bandstop (band reject) filter passes frequencies outside a frequency band and blocks or attenuates frequencies within the band.

Passive Filters

2

Filter ResponsesIdeal Filter Classification

fLow pass

filterf

High pass filter

fBand pass

filter fBand reject

filter

3

Passive 1st order Lowpass Filters

Plot |H(j)| versus :

Ideal

Actual

1( )1

R R LH jR j L j R L j L R

max

1( ) ( )2cH j H j

1( )1LPF

c

H jj

RL circuit

cRL

4

Passive 1st order Lowpass Filters

Plot |H(j)| versus :

1 ( ) 1( )1 ( ) 1

j CH jR j C j RC

1c RC

1( )

1LPFc

H jj

RC circuit

Ideal

Actual

max

1( ) ( )2cH j H j

5

Passive 1st order Highpass Filters

LR

c

( )1 1

c

c

jj L j L RH jR j L j L R j

RCc1

( )1 ( ) 1 1

c

c

jR j RCH jR j C j RC j

6

Example Effects of Loading on Filter Performance

( ) ( )/

, 1

HPF unloaded

c

sL KsH sR sL s R LR KL

( )

( )

( ) ( ) ( )

, ||

( )

,

1;

L

L

L

L

LL

LL

eqHPF loaded eq L

eq

sLRsL R L

HPF loaded sLRL LsL R

RR R

RRR RL

L

L

c loaded c unloaded c unloaded

ZH Z sL R

R Z

sLRHR sL R sLRR

s Kss K R Ls

RKR RK

7

Example cont`d

For RL = R

Let R=RL=500 and L=5.3mH

Then fc (unloaded)=15kHz

and fc (loaded)= fc’=7.5 kHz

Effects of Loading on Filter Performance

Changes the cutoff frequency

Changes the gain in the passband

8

Passive 2nd order Lowpass Filter

nd

2

2

2

2 2 2 order

1 1( )1 1

(1 )( ) (1 )

( )oLPF

o

sCH sR sL sC sRC s LC

LCs R L s LC

H ss s

1o LC

9

Passive Bandpass Filters

)()1()(

)(11

)(

222

2

sHsss

LCsLRssLR

LCsRsCRsC

sCsLRRsH

BPFo

1o LCB R L

Series RLC circuit

10

Passive Bandpass FiltersParallel RLC circuit

12

1

2 2

( ) , ||11

(1 )( )( 1 ) (1 ) (1 )

( )

eqeq

eq

L CsL sC

L CsL sC

BPFo

Z L CH s Z sL sCR Z sL sC

L C RC sH sR R sL sC L C s RC s LC

s H ss s

11

o LCB RC

11

Passive Bandpass FiltersFrequency response characteristics:Only 2 of the 5 characteristics of a bandpass filter can be specified independently. The others are related as follows:

1 2

2 1

o c c

c c

o

BQ

12

Design a bandpass filter with a center frequency of 2 kHz and a bandwidth of 500 Hz. Use a 250 resistor.

Example

1 1 12 500Hz2 500Hz 2 500Hz 250

1.27 F

B CRC R

2 2

1 1 12 2kHz=(2 2kHz) (2 2kHz) 1.27 F4.97 mH

o LCLC

32 10Note: 4500oQ

The Q is so low in order to achieve the specified bandwidth.

13

Passive (Bandstop) Bandreject FiltersSeries RLC circuit

2

2

2 22

2 2 2

1 1( )1 1(1 ) ( )

( ) (1 )o

BRFo

sL sC s LCH sR sL sC RsC s LC

ss LC H ss R L s LC s s

1o LCB R L

14

Passive (Bandstop) Bandreject FiltersParallel RLC circuit

11

o LCB RC

2

21

2 2

2 2

( ) , ||11

( 1 ) (1 )( )( 1 ) (1 ) (1 )

( )

eqeq

L CsL sC

oBRF

o

R L CH s Z sL sCR Z sL sC

R R sL sC s LCH sR R sL sC L C s RC s LC

s H ss s

15

Example 1Determine what type of filter is shown. Calculate the corner or cutoff frequency. Take R = 2 k, L = 2 H, and C = 2 μF.

16

Example 1 cont’d

6

63

1 1Note: rad/s 500 rad/s2 2 10

2 102 10 22

o

p

LC

CQ RL

R = 2 k, L = 2 H, C = 2 μF

17

Example 2For the circuit shown, obtain the transfer function Vo(ω)/Vi(ω). Identify the type of filter the circuit represents and determine the corner frequency. Take R1 = 100 = R2, L = 2 mH.

This is a highpass filter.

18

Example 2 cont’d

19

Example 3If this bandstop filter is to reject a 200-Hz sinusoid while passing other frequencies, calculate the values of L and C.Take R = 150 and the bandwidth as 100 Hz.

0

220

2 100 Hz 200 rad/s

150 0.2387 H200 rad/s

2 2 400 Hz 400 rad/s1 11 2.66μF

400 rad/s 0.2387 H

o

o

B

R RB LL B

f

LC CL

20

Example 4Design this bandpass filter with a lower cutoff frequency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take R = 20 k. Calculate L, C, and Q.

Too large !!

1 2

2 (20.3 20.1) kHz 400 rad/s2 (20.3 20.1) kHzAssuming high : 40.4 krad/s

2 240.4 krad/s 101400 rad/s

o

o

B

Q

QB

20 k 15.916 H400 rad/s

1 1

140.4 krad/s 101 20 k3.9 pF

o o

R RB LL B

Q CRC QR

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Example 4 cont’dTake R = 20 .

3

20 15.916 mH400 rad/s

1 1 3.9 nF40.4 10 rad/s 101 20o

RLB

CQR

1 2

2 (20.3 20.1) kHz 400 rad/s2 (20.3 20.1) kHzAssuming high : 40.4 krad/s

2 240.4 krad/s 101400 rad/s

o

o

B

Q

QB

22

Example 5Use a 500 nF capacitor to design this band-reject filter. The filter has a center frequency of 4 kHz and a quality factor of 50.

22 3 90

3 9

1 1 1 3.166 mH2 4 10 rad/s 500 10 F

50 3.98k2 4 10 rad/s 500 10 F

o

oo

LCLC

QQ RC RC

23

Active Filters

Passive Filter Disadvantages: Maximum gain in the pass band is 1 – no amplification in the pass band!

Adding a load to a passive filter changes the filter’s characteristics, like the cutoff frequency.

Passive band pass and band reject filters require the use of inductors, which are large, costly, and can introduce stray electromagnetic field effects.

No obvious method to make the filter characteristic more ideal.

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Active Filters

Active filters Based on circuits with op amps, which are active devices, since they

require an external power supply ( VCC).

Eliminate all of the disadvantages of passive filters:

Can create a pass band gain > 1;

Can add loads without changing the filter characteristics;

All four types of filters can be created using op amps, resistors, and capacitors – no inductors;

Can cascade simple (first- and second-order) filters to create higher order filters that are more nearly ideal.

25

First-Order Active Lowpass Filter

Find the transfer function:

2

1

2 2

1 2 1 2 1 2

2 1 2

2

1 ||( )( )( )

1

11

fo

i i

c

c

Z sC RV sH sV s Z R

R sC RR sC R R R R R sC

R R R C Ks R C s

2

2 1

1c

RKR C R

26

ExampleDesign an active low pass filter with a gain of -5 and a cutoff frequency of 1 kHz. Use a 10 nF capacitor.

2

2 9

21

1 2 (1000) rad/s

1 15.916k2 (1000)(10 10 )

15,916 3.184k5 5

c R C

R

RR

27

First-Order Active Highpass Filter

Find the transfer function:

CRs

sRRsCR

sCRsCR

RZZ

sVsVsH

i

f

i

o

1

12

1

2

1

2

11

1)()()(

2

1 1

1c

RKR C R

For K = -1

28

Example 1Design an active high pass filter with a gain of -10 and a cutoff frequency of 500 rad/s. Use a 0.1 F capacitor.

1

1

2 1

1 500 rad/s

1 20k(500rad/s)(0.1 F)10 200k

c R C

R

R R

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Example 2Design a first-order highpass op amp filter whose transfer function is given by

Use capacitances of any value. But you only have 10 k resistors available - no inductors!

( ) 4000H s s s

( )( )

( )f

i

Z sH s

Z s The transfer function of an inverting op amp is

Thus,3 3 3

3 733 3

8

3

3

9

1 10 10 10 10 10 10( ) 110 10 4000 4 101 4000 10 1010 10 10 102.5 10

10 10110 10

25 10

H ss

ss s

s

Therefore, ( ) 10k1( ) where 10k and 25nF

f

i

Z s R

Z s R R CsC

30

First-Order Active Bandpass Filter

+ ++

=

31

First-Order Active Bandpass Filter

Gives c2 Gives c1 Gives K

32

First-Order Active Bandpass Filter

2

2 1

22

1 2 1 2

1 22 20

( ) ( ) ( ) ( )

( )

( )

if

BPF LPF HPF Inv

fc

c c i

f i c

c c c c

c c

H s H s H s H sRs

s s RR R s

s sK s

s s

So this cascading technique only produces broadband filters.

33

ExampleDesign an active band pass filter with a passband gain of 2 and cutoff frequencies of 500 Hz and 50 kHz. Use 2.2 nF capacitors.

Lowpass filter:1

2 (50,000Hz)(2.2nF)1.447 k

LR

Highpass filter:1

2 (500Hz)(2.2nF)144.7 k

HR

Gain:10k

5kf

i

R

R

34

First-Order Active Bandstop FilterGives c1

Gives c2

Gives K

HPF

LPFSumming

Amp.

35

First-Order Active Bandstop Filter

2120

2

20

22121

2211

2

21

1

if)(

)(]2[)(

)]()()[()(

cc

cccc

cccif

cc

c

i

f

HPFLPFInvBRF

sssK

ssssRR

ss

sRR

sHsHsHsH

So again, this technique only produces broadband filters.

Find the transfer function:

36

ExampleDesign a notch filter between 20 krad/s and 100 krad/s with a gain of K = 5. You have only 10 k resistors available.

Lowpass filter:1 5nF

20krad/s 10kLC

Highpass filter:1 1nF

100krad/s 10kHC

Gain:

5 10k

5 10k

fi

i

f

RR

RR

37

Example cont’dNote: Due to the fact that the lowpass and highpass filters used in this example go down with only 20 dB/dec, attenuation in the stopband might be limited.

38

Second-Order Op-Amp FiltersBandpass Filter

iv ov

2 3

1 3

2 21 3 1 2 3

1 3 1 3

1 2 ( )

R Rj CR R

R R R R Rj C j CR R R R

H

1 3 0 20 2 0 0

1 2 3 2 1

1 1 1 2, , , ( )2 2 2

R R RQ R C B H KC R R R Q R C R

Exercise: Try to derive this transfer function.

2Advantage: The bandwidth, thus Q factor, is entirely controlled by .R

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Example 2nd-order Bandpass Filter

Design for 3kHz, K=2, Q=10:

40

Bandstop Filter

iv ov

2 2 2

2 2 2

1 ( )

1 2(2 ) ( )

k j R C

j k RC j R C

H

1 2k

2 2 2

2 22 2 2

1

1 2(2 )

k R C

R C k RC

H

1 0 dBRC

H

01

RC

41

Lowpass Filter

2 2 21 (3 ) ( )k

j k RC j R C

H

1 2 1 2Let andC C C R R R

100 20log dB

40db/dec

k k

H

H

1 3k

oviv

42

Scaling• There are two ways of scaling a circuit: magnitude or impedance

scaling, and frequency scaling.

• Both are useful in scaling responses and circuit elements to values within the practical ranges.

• Magnitude scaling leaves the frequency response of a circuit unaltered.

• Frequency scaling shifts the frequency response up or down the frequency spectrum.

43

Magnitude Scaling Magnitude scaling is the process of increasing all impedances in a network by

the same factor, the frequency response remaining unchanged.

The primed variables are the new values and the unprimed variables are the old values. Consider the series or parallel RLC circuit. We now have

R m R

L m L

C m C

Z k ZZ k ZZ k Z

1 1

m

m

m

R k Rj L k j L

kj C j C

m

m

m

R k RL k L

CCk

1 1 1

/o o

m mL C LCk L C k

44

Frequency Scaling Frequency scaling is the process of shifting the frequency response of a network

up or down the frequency axis while leaving the impedance the same.

Again, if we consider the series or parallel RLC circuit, for the resonant frequency

fk

1 1f

f

R R

j k L j L

j Cj k C

/

/f

f

R RL L k

C C k

1 1

/ /f

o f o

f f

kk

L C LCL k C k

fB k B

45

Magnitude and Frequency Scaling If a circuit is scaled in magnitude and frequency at the same time,

then

m

m

f

m f

f f

R k RkL Lk

CCk k

k B k B

46

Example 1Redesign the active low pass filter to use a 200 pF capacitor.

Use magnitude scaling:

1

2

10nF 50200pF

50(3183.8 ) 159.19 k50(15,915.5 ) 795.775k

m

m

CCkCkC

RR

47

Example 2Redesign the active low pass filter, which was designed for 1 kHz, so that the cutoff frequency is 7.5 kHz, using a 250 pF capacitor.

Use frequency scaling first,then magnitude scaling:

1 2 1

7500 7.51000

10nF 5.3(7.5)(250pF)

250 pF; 5.3(3183.8 ) 16.98k ; 5 84.9k

cc f c f

c

mf m f

k k

C CC kk k k C

C R R R

48

Example 3The fourth-order Butterworth lowpass filter is designed such that the cutoff frequency is ωc = 1 rad/s. Scale the circuit for a cutoff frequency of 50 kHz using 10-k resistors.

mm

f

fm f

kR k R L Lk

CC kk k

34' 10 10Magnitude scaling: 10

1mRkR

35' 2 50 10Frequency scaling: 10

1c

fk

49

mH

50

Example 4The third-order Butterworth filter is normalized to ωc = 1. Scale the circuit to a cutoff frequency of 10 kHz. Use 15 nF capacitors.

44

4

9 4

' 2 10 2 101

'

1 10' 315 10 2 10

cf

m f

mf

k

CCk k

CkC k

4

4

4

10' 1 1.061k310 2H' 33.77 mH3 2 10

' 15nF

m

m

f

R k R

kL L

k

C