chapter 16 - filter circuits

8/9/2019 Chapter 16 - Filter Circuits

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Chapter 16: Filter Circuits

Exercises

Ex. 16.3-1

1( )

11 1250

( )1250 12501

1250

H sn s

s H s H

n s s

=+ = = = + +

ProblemsSection 16.3: Filters

P16.3-1Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having acutoff frequency equal to 1 rad/s.

2

1( )

( 1)( 1)n

H ss s s

=+ + +

Frequency scaling so that = 2 100=628 rad/s:c

3

L 2 2 22

1 628 247673152( )

( 628)( 628 628 ) ( 628)( 628 394384)1 1628 628 628

H ss s s s s ss s s

= = =+ + + + + +

+ + +

P16.3-2Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having acutoff frequency equal to 1 rad/s and a dc gain equal to 1.

2

1( )

( 1)( 1 H s

n s s s=

)+ + +

Multiplying by 5 to change the dc gain to 5 and frequency scaling to change the cutoff frequency to c = 100 rad/s:

3

L 2 2 22

5 5 100 5000000( )

( 100)( 100 100 ) ( 100)( 100 10000)1 1

100 100 100

H ss s s s s ss s s

= = =+ + + + + + + + +

1


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P16.3-3Use Table 16-3.2 to obtain the transfer function of a third-order Butterworth high-pass filterhaving a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

3

2

5( )

( 1)( 1

s H s

ns s s

=)+ + +

Frequency scaling to change the cutoff frequency to 100 rad/s:c =

3

3 3

H 2 2 22

55 5100( )

( 100)( 100 100 ) ( 100)( 100 10000)1 1

100 100 100

s

s s H s

s s s s s ss s s

= = =

+ + + + + + + + +

P16.3-4

Use Table 16-3.2 to obtain the transfer function of a fourth-order Butterworth high-pass filterhaving a cutoff frequency equal to 1 rad/s and a dc gain equal to 5.

( ) ( )( )4

2 2

5

0.765 1 1.848 1n

s H s

s s s s=

+ + + +

Frequency scaling can be used to adjust the cutoff frequency 500 hertz = 3142 rad/s:

( ) ( )( )

4

4

H 2 2 2 2 2

553142

2403.6 3142 5806.4 3142

0.765 1 1.848 13142 3142 3142 3142

s

s H s

s s s ss s s s

= = + + + + + + + +

2

P16.3-5First, obtain the transfer function of a second-order Butterworth low-pass filter having a dc gainequal to 2 and a cutoff frequency equal to 2000 rad/s:

( )L 2 22 8000000

2828 40000001.414 1

2000 2000

H ss ss s

= =+ + + +

Next, obtain the transfer function of a second-order Butterworth high-pass filter having apassband gain equal to 2 and a cutoff frequency equal to 100 rad/s:

2


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P16.3-8

( )( )

( )

2 2

N 222 2

250 24 62500

14 4 250 250 625002501

ss

H ss ss s

+

= = + + + +

P16.3-9

( )( )

2

2 4

L 222 2

250 4 2504

250 250 625002501

H ss ss s

= =

+ + + +

P16.3-10

( )( )

2

2 4

H 222 2

44

250 250 625002501

s s H s

s ss s

= =

+ + + +

4


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Section 16.4: Second-Order Filters

P16.4-1

The transfer function is

( ) ( )( )

2o

22s

2

1

11

1 11

1

s LC s

s L ss L

V s s LC s s LC R C H s

s L sV s s LCR s L Rs L R sC s s LC R C L C

Rs L

C s

++= = = = =

+ + + +++

+

+

so

2 00 01 11 , and C K Q LC RC Q L

= = = = = RC R

20

1Pick 1 F. Then 1 H and 1000

LC L R Q

C C

= = = = =

P16.4-2


0

2

1

( )( )1( )s

I s LC T ss I s s

RC LC

= =+ +

so

2 00 0

1 11 , and

C k Q

LC RC Q L

= = = = = RC R

20

1Pick 1 F then 25 H and 3535

LC L R Q

C C

= = = = =

1


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P16.4-3


21

22 2

1

1

( )1 1

2

R RC T s

Rs s R C R R C

=

+ + +

Pick 0.01 FC = , then

0

01

1

12000 50000 50

12 8333 8.33

2

R k RC

R R R k

Q RC R Q

= = = =

= + = = =

P16.4-4

1 2 3Pick 0.02 F. Then 40 k , 400 k and R =3.252 k .C R R = = =

P16.4-5

1 2Pick 1 FC C C = = = . Then6

01 2

10

R R =

and1 10

2 21 2

1

R RQ R

R C Q R Q = = =

6

2 1 110

In this case, since 1, we have and 1000 1 k 1000

Q R R R= = = = =

2


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P16.4-6

The node equations are

( ) ( )

( ) ( ) ( ) ( )( )

20 a

22

0 a1 a i

1

1

0

RV s V s

R C s

V s V sC s V s V s

R

=+

=

Doing a little algebra

( ) ( ) ( )o a 1 11 1

1V sV s s C s C V s

R R

+ =

i

Substituting for ( )aV s

( ) ( ) ( )o 2 2 1 1o 11 2 2 1

1 1V s R C s R C sV s s C V s R R C s R

+ + + =

i

( )( )

( )( )( )

21 1 2 2o 1 2 1 2

2i 1 2 1 2 1 12 2 2 2 1 1 11 1

C s R R C sV s R R C C s

V s R R C C s R C s R C s R C s R C s= =

+ + + + +

The transfer function is:

( ) ( )( )

20

2i

2 2 1 2 1 2

1V s s

H ssV s s

R C R R C C

= =+ +

0 2 21 2 1 20

2 11 2

1 1Pick 1 F. Then and

RC C C Q R Q R

R C Q RC R R

= = = = = = = .

1 2 0

1In this case and 1000 R R R R

C R = = = = .

3


8/26

P16.4-7

( ) 02i

1 1( )

1 1( )V s C s LC

T s RV s L s R s s

C s L LC

= = =+ + + +

2 6When 25 , 10 H and 4 10 F, then R L C = = = the transfer function is

( )6

2 6

25 102500 25 10

H ss s

=+ +

so6

old 25 10 5000 = = and

new

f old

2500.055000k

= = =

The scaled circuit is

P16.4-8

The transfer function of this circuit is

4


9/26

( ) ( )( ) ( )( )

2

2 2 2 1 1 20

i 1 1 2 2 21

11 1 2 2 1 2 1 2

11

1 1 1 1 1 1

Rs

R C s R C s R C V s H s

V s R C s R C s R s sC s R C R C R R C C

+= = = =

+ ++ + + +

( )m

1 1 2 2

100 F 500 FPick 1000 so that the scaled capacitances will be 0.1 F and 0.5 F.

1000 1000

Before scaling 20 , 100 F, = 10 and 500 F

k

R C R C

= =

= = =

=

( ) 2 5-100700 10

s H s

s s=

+ +

( )1 1 2 2After scaling 20000 20 k , 0.1 F, 10000 10 k , 0.5 F R C R C = = = = = =

( ) 2 5100

700 10s

H ss s

=+ +

P16.4-9

This is the frequency response of a bandpass filter, so

( )

0

02 20

k sQ H s

s sQ

=+ +

From peak of the frequency response

6 60 2 10 10 62.8 10 rad/s and =10 dB 3.16k = = =

Next6 6 6 60 BW (10.1 10 9.9 10 ) 2 (0.2 10 )2 1.26 10 rad/s

Q

= = = =

So the transfer function is

6 6

2 6 2 12 2 6

3.16(1.26)10 (3.98)10( )

(1.26)10 62.8 10 (1.26)10 3.944 10s s

H ss s s s

= =+ + + + 15

5


10/26

P16.4-10

(a)( ) ( )( )

2220

1 1 2s 1 1

22

1

= where and1

R j C j

RC R

j C

= = =

+

ZVH Z Z

V Z

( ) 2 1 1 21 1 2 2

1 2

1 1= where ,

1+ 1+

j R C

R C R j j

= =

HC

(b) 1 21 1 2 2

1 1,

R C R = =

C

(c)

( )( )

2 1 22 11 2 1

1

11

=

0+ 1+0

R C R j R C R C R j

= =

H =

P16.4-11

Voltage division:

( ) ( ) ( ) ( )0 1 1

1

s , (1 )1

C sV s V V s s RC V s

RC s

= = ++

0

KCL:( )1 s 1 0 1 0+ + 0

V V V V V V nC s

m R R

=

Combining these equations gives:

2 20

1 s

V s C V s C s n R C

m R m m R

+ + + =

6


11/26

Therefore( )( ) ( )

022 2 2

0 0

V 1 1( ) =

V 1 1s1

s m R C n m R C s j

Q

= =+ + +

+

H

01where and

1m nQ

mmn RC = = +

P16.4-12

22

2 2 2 1 20

1 1S 21

11 1 1 2 2 1 2 1 2

1 11( )

( ) = 1 1( ) 1 1 1

R R s

C s R C s R C V s H s R C sV s R s sC s C s R C R C R R C C

+

= = =+ + + + +

Substituting the element values ,1 100 R = 2 200 R = , 1 0.02 FC = and wedetermine the center frequency and bandwidth of the filter to be

2 50 pFC =

( )

( )

01 2 1 2

0

1 1 2 2

170.7 krad sec= 2 11.25 kHz

1 1150 k rad =2 23.9 kHz

R R C C

BW sQ R C R C

= =

= = + =

7


12/26

P16.4-13

( ) a 01 a s1 2 20

2a s2 0

1 1 1 2 1 22

1= 0 s( )

( )=1 1( ) ss 0

V V C s V V R R C V s H s

V V s sC V R C R R C C R

+

=

+ + =

Comparing this transfer function to the standard form of the transfer function of a second orderbandpass filter and substituting the element values 1 1 k R = , 2 100 R = , 1 1 FC = and

2 0.1 FC = gives:

40

1 2 1 2

3

1 1

0

110 rad sec

1BW = 10 rad/sec

10BW

R R C C

R C

Q

= =

=

= =

8


13/26

P16.4-14

Node equations:

( )( ) ( ) ( )

c s c 0c

s cs 0

( ) ( ) ( ) ( )( ) 0

( ) 0

V s V s V s V sa C sV s

R R

V s V sC s V s V sa R

+ + =

+ =

Solving these equations yields the transfer function:

( )( )

( )

( )

22

0

2s2

2 1 1

( )2 1 1

s a sa R C V s R C

H sV s s s

a R C R C

+ + +

= = + +

We require 51

10 RC

= . Pick 0.01 F then 1000C R = = . Next at 0s j =

( )2

0

2

12 2

a aa

a

+= = +H

The specifications require

( )2

0201 1 202a

a = = + =H

9


14/26

P16.4-15

Node equations:

( )

a a 0

2 1

a ba

b s b ab 0

=0

0

+ +

V V V

R RV V

C sV R

V V V V C s V V 0

R R

+

+ =

=

Solving the node equation yields:

( )( )

12 2

20

s 122 2

2

11

1 12

R

R R C V s

V s Rs s

R R C R C

+

= + +

( ) ( )0 3 91 1

41.67 k rad sec1.2 10 20 10 RC

= = =

10


15/26

Section 16.5: High-Order Filters

P16.5-1This filter is designed as a cascade connection of a Sallen-Key low-pass filter designed asdescribed in Table 16.4-2 and a first-order low-pass filter designed as described in Table 16.5-2.

Sallen-Key Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_sklp.mcd)

A 2=Calculate the dc gain.

R A 1( ) 1.592 10 4=R 1.592 10 4=A 31

Q:=R

1

C 0:=Calculate resistance values:

C 0.1 106

:=Pick a convenient value for the capacitance:

Q 1=0 628=Q0b

:=0 a:=Determine the Filter Specifications:

b 628:=a 6282

:=Enter the transfer function coefficitents:

c----------------- .s^2 + bs + a

The transfer function is of the form H(s) =

1


16/26

First-Order Low-Pass Filter:

MathCad Spreadsheet (p16_5_1_1stlp.mcd)

The transfer function is of the form H(s) =-k

-------- .s + p

Enter the transfer function coefficitents: p 628:= k 0.5p:=

Pick a convenient value for the capacitance: C 0.1 10 6

:=

Calculate resistance values: R21

C:= R1

1

C k := R1 3.185 10 4= R2 1.592 10 4=

P16.5-2This filter is designed as a cascade connection of a Sallen-key high-pass filter, designed asdescribed in Table 16.4-2, and a first-order high-pass filter, designed as described in Table 16.5-2.

The passband gain of the Sallen key stage is 2 and the passband gain of the first-order stage is2.5 So the overall passband gain is 2 2.5 = 5

Sallen-Key High-Pass Filter:

2


17/26

MathCad Spreadsheet (p16_5_2_skhp.mcd)

A 2=Calculate the passband gain.

R A 1( ) 1 105=R 1 10 5=A 31

Q:=R

1


C 0.1 106


Q 1=0 100=Q0b


b 100:=a 10000:=Enter the transfer function coefficitents:

A s^2----------------- .s^2 + bs + a


First-Order High-Pass Filter:

MathCad Spreadsheet (p16_5_2_1sthp.mcd)

The transfer function is of the form H(s) =-ks

-------- .s + p

Enter the transfer function coefficitents: p 100:= k 2.5:=

Pick a convenient value for the capacitance: C 0.1 106

:=

Calculate resistance values: R11

C p:= R2 k R1:= R1 1 10 5= R2 2.5 10 5=

3


18/26

P16.5-3This filter is designed as a cascade connection of a Sallen-key low-pass filter, a Sallen-key high-pass filter and an inverting amplifier.



A 1.586=Calculate the dc gain.

R A 1( ) 2.93 10 3=R 5 10 3=A 31

Q:=R

1


C 0.1 106


Q 0.707=0 2 103=Q

0

b



c----------------- .s^2 + bs + a


4


19/26



A 1.586=Calculate the passband gain.

R A 1( ) 5.86 10 4=R 1 10 5=A 31

Q

:=R1


C 0.1 106


Q 0.707=0 100=Q0b


b 141.4:=a 10000:=Enter the transfer function coefficitents:

c s^2----------------- .s^2 + bs + a


Amplifier: The required passband gain is61.6 10

4.00141.4 2828

= . An amplifier with a gain equal to

4.01.59

2.515= is needed to achieve the specified gain.

5


20/26

P16.5-4This filter is designed as the cascade connection of two identical Sallen-key bandpass filters:

Sallen-Key BandPass Filter:

MathCad Spreadsheet (p16_5_4_skbp.mcd)

A Q 2=Calculate the pass-band gain.

R A 1( ) 4 104=2 R 8 104=R 4 10 4=

A 31

Q:=R

1


C 0.1 106


Q 1=0 250=Q0b



cs----------------- .s^2 + bs + a


6


21/26

P16.5-5This filter is designed using this structure:



A 1.586=Calculate the dc gain.

R A 1( ) 5.86 10 4=R 1 10 5=A 31

Q:=R

1


C 0.1 106


Q 0.707=0 100=Q

0b


b 141.4:=a 10000:=Enter the transfer function coefficitents:

c----------------- .s^2 + bs + a


7


22/26



A 1.586=Calculate the passband gain.

R A 1( ) 2.93 10 3=R 5 10 3=A 31

Q

:=R1


C 0.1 106


Q 0.707=0 2 103=Q

0b



c s^2----------------- .s^2 + bs + a


Amplifier: The required gain is 2, but both Sallen-Key filters have passband gains equal to 1.586.

The amplifier has a gain of 2

1.261.586

= to make the passband gain of the entire filter equal to 2.

8


23/26


24/26

P16.5-7

(a) Voltage division gives: ( )( )( )

1 1 1a

s 11

1 1

V s R R C s H s

V s R C s RC s

= = =++

(b) Voltage division gives: ( ) ( )( )2

b1 2

V s L s H sV s R L s

= = +

(c) Voltage division gives:

Doing some algebra:

( ) ( )( )

( )( )

1 22c

s 21 2

||

1||

R R L sV s L s H s

V s R L s R R L sC s

+= =

++ +

( ) ( )( )

( )( )

( )

( )

( )( )

( )

1 2

1 22c

s 21 2

1 22

1 2 12

1 2 1 1 2 2

1 2

221 1 2 1 2

21

21 1 2 1 2

1

R R L s

R R L sV s L s H s

V s R L s R R L s

C sR R L s R R C s R L C s L s

R R C s R LC s R R L s R L s

R C s R L s L s R L s R LC s R R C L s R R

R L C s

R LC s R R C L s R R

+

+ += =

+ ++

+ +

+=

+ + + + +

+=

++ + + +

=+ + + +

(d) ( ) ( ) ( )c a b H s H s H s because the 2 , R L s voltage divider loads the 11 , RC svoltage

divider.

10


25/26

P16.5-8

( ) 100 20

1 1 1 1200 20,000 20 4000

20001 1 1 1

20 200 4000 20,000

H ss s s s

s s s s

=

+ + + +

= + + + +

P16.5-9(a) The transfer function of each stage is

( )

2

2 2

2 22 1

i1 1 1

1

1 1||

1

1

RC s

R R R R

R C s RC s C s H s

2 R R R R

+ += = = =

+ C s

The specification that the dc gain is 0 dB = 1 requires 2 1 R R= .

The specification of a break frequency of 1000 rad/s requires2

11000

R C = .

Pick 0.1 FC = . Then 2 10 k R = so 1 10 k R = .

(b)( ) ( )

2

2

1 1 1 110, 000 40.1 dB

1011 101 11000 1000

j j

= = = = + + +

H H

11


26/26

Section 16.7 How Can We Check?

P16.7-1

0

0

100

10000 100 rad and 25 4 525s Q

Q

= = = = =

This filter does not satisfy the specifications.

P16.7-2

00

100 7510000 100 rad , 25 4 and 3

25 25s Q k

Q

= = = = = = =

This filter does satisfy the specifications.

P16.7-3

00

20 600400 20 rad s, 25 0.8 and 1.5

25 400Q k

Q

= = = = = = =


P16.7-4

00

25 750625 25 rad s , 62.5 0.4 and 1.2

62.5 625Q k

Q

= = = = = = =


P16.7-5

00

12144 12 rad/s and 30 0.4

30Q

Q

= = = = =

This filter does not satisfy the specifications.

chapter 16 - filter circuits

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