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15Active Filter Circuits

Assessment Problems

AP 15.1 H(s) = (R2/R 1)ss+(1/R1C)

1

R1C = 1rad/s;

R2

R1 = 1 , . C = 1F

R1 = 1,

. R2 = R1 = 1

s

Hprototype(s) =s+1

(1/R1C) 20,000AP 15.2

H(s)=

1

s+(1/R2C) = s + 5000

R1C = 20,000; C = 5 F

1. R1

=

1

(20,000)(5 106) = 10

R2C = 5000

1. R2= (5000)(5 106) = 40

15-1

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15-2 CHAPTER 15. Active Filter Circuits

AP 15.3 c = 2fc = 2 104 = 20,000 rad/s

. kf = 20,000 = 62,831.85

1

C = Ckf km

0.5 106 =kfkm

1

. km =(0.5 106)(62,831.85) = 31.83

AP 15.4 For a 2nd order prototype Butterworth high pass filter

H(s)=

s2

s2 + 2s + 1

For the circuit in Fig. 15.25

H(s) = (s2 + 2R2C

s2)

1)s+

(

R1R2C2

Equate the transfer functions. ForC = 1F,

2

R2C

=

1

2, . R2 = 2=1.414

1

. R1 = = 0.707 R1R2C2 = 1, 2

AP 15.5 Q=8,K =5,o =1000rad/s,C =1F

For the circuit in Fig 15.26

(1 )

sH(s) =

s2+

(

2R3C

) R1C( )R1 + R2s+

R1R2R3C2

= Kss2 + s + o

=

2

R3C ,

2 R3 =

C

= oQ = 1

8

= 125 rad/s

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Problems 15-3

2106

. R3=

1

(125)(1) = 16 k

K= R1C

1 1. R1= KC = 5(125)(1 10

6) = 1.6k

o = R1 + R2

R1R2R3C2

106=

(1600 + R2)

(1600)(R2)(16,000)

(106)2

Solving forR2,

R2 = (1600 +R2)106,

256 105246R2 = 16,000, R2 = 65.04

AP 15.6 o = 1000 rad/s; Q=4;

C =2F

s2 +(1/R2C2)

H(s) = ] ([ 4(1 )

)1

s2 + RC s+R2C2

1 4(1 )

R=

= s2 + os2 + s +

o ;

1 1

o = =RC ; RC

oC =(1000)(2 106) = 500

= o = 250Q = 1 4

4(1 )

RC

= 250

4(1 ) = 250RC = 250(500)(2 106) = 0.25

1=0.254

= 0.0625; =0.9375

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Problems

P 15.1 Summing the currents at the inverting input node yields

0V iZi

Vo

+0V

o =0Zf

i

P 15.2 [a]

Zf = Z i

H(s) = VoVi = Z

i

R2Zf = R2(1/sC2)[R2 + (1/sC2)] = R2C2s + 1

(1/C2)= s+(1/R2C2)

Likewise

Zi=

(1/C1)

s+(1/R1C1)

. H(s)= (1/C2)[s+(1/R1C1)]

[s + (1/R2C2)](1/C1)

=C

1C2

[s + (1/R1C1)]

[s + (1/R2C2)]

]

[b][j + (1/R1C1)

H(j) = C1C2 j + (1/R2C2)

)

H(j0) =C1C2

( R2C2

= R2R1C1 R1)

[c](j

H(j) = C1C2 j

= C1

C2[d] As 0 the two capacitor branches become open and the circuit reduces to a

resistive inverting amplifier having a gain ofR2/R1.As the two capacitor branches approach a short circuit and in this casewe encounter an indeterminate situation; namely vn vi but vn = 0 becauseof the ideal op amp. At the same time the gain of the ideal op amp is infinite so

we have the indeterminate form 0 . Although = is indeterminate we

can reason that for finite large values of H(j) will approach C1/C2 invalue. In other words, the circuit approaches a purely capacitive inverting

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f

amplifier with a gain of(1/jC2)/(1/jC1) orC1/C2.

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P 15.3 [a] Zf =

Problems 15-5

(1/C2)

s+(1/R2C2)

1 1Zi = R1+ sC1 =Rs

[s + (1/R1C1)]

(1/C2) sH(s) =

[s + (1/R2C2)] R1[s + (1/R1C1)]1 s

=R1C2 [s + (1/R1C1)][s + (1/R2C2)]

1 j[b] H(j) =

R1C2 (j+

H(j0) = 0

[c] H(j) = 0

1R1C1

)1) (j +

R2C2

[d]

P 15.4 [a]

As 0 the capacitorC1 disconnects vi from the circuit. Thereforevo = vn = 0.As the capacitor short circuits the feedback network, thus Zf = 0 andtherefore vo = 0.

K =10(10/20) =3.16=R2

R1

1 1R2= cC = (2)(10

3)(750 109) = 212.21

P 15.5

R1 = R2K =

23

[b]

1[a] R1 =

= 67.11

16

1

cC =(2)(8 103)(3.9 109) = 5.10k

K =10(14/20) =5.01= R2R1

. R2 = 5.01R1 = 25.57 k

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L

= km

R


[b]

P 15.6 For the RC circuit

(1/RC)H(s) = Vo

Vi=

R = kmR;

s+(1/RC)

C = Ckmkf

1

. RC = kmR Ckmkf = k1f

RC

1

RC = kf

(1/RC) f

=kf

H(s)

=

H(s)=

s+(1/RC) = s+ kf

1

(s/kf) + 1

For the RL circuit

H(s) = V0Vi = s + (R/L)

R

= kmR; L

= km

kf L

)R

km

kf L=kf

(R

L

= kf

H(s)=

H

(s)=

(R/L) f

s+(R/L) = s+ kf

1

(s/kf) + 1

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Problems 15-7

P 15.7 For the RC circuit

sH(s) = Vo

Vi =s+(1/RC)

R = kmR; C = Ckmkf

1. RC = RC

kf

s

= 1kf ; RC = kf

s (s/kf)

H(s)= s+(1/R

C) = s+kf = (s/kf) + 1

For the RL circuit

H(s)=

s

s+(R/L)

R = kmR;

R (R )

L = kmLkf

= kfL =kf

H(s)=

P 15.8 H(s)=

L

s (s/kf )s+(R/L) = s+s kf = (s/kf )

+ 1

(R/L)s s

s2 + (R/L)s + (1/LC) = s2s +

o

For the prototype circuit o = 1 and = o/Q = 1/Q.For the scaled circuit

(R

/L

)sH(s)= s

2 + (R/L)s + (1/LC)

where R = kmR; L = km

)

fkm

(R

. R

1

kmkf L =kf

f

km

L = kf

LC = kkmC =kf L

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kfL;and C

= k C

L = kmR

kf

= kf

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1


Q = o= kf o=Q kf

therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit.Also note = kf.

(

kfQ

)s. H(s)= (kf

s2 + Q)s+kf

( )( )1 sQ kf

H(s) = [((s skf

P 15.9 [a] L = 1

H;

R= 1

)2 +1)+1

]

Q kf

C =1F

Q =20 = 0.05

[b] kf = o km = R= 5000 = 100,000o = 40,000; R 0.05

Thus,

R = kmR = (0.05)(100,000) = 5k

[c]

L = kmkfL = 40,000 (1) = 2.5 H

1C = C

kmkf = (40,000)(100,000) =250 pF

P 15.10 [a] Since o = 1/LC and o = 1 rad/s,

C = 1L = Q

(R/L)s[b] H(s)= s2 + (R/L)s +

(1/LC)

(1/Q)sH(s)= s

2 + (1/Q)s + 1

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[c] In the prototype circuit

R=1; L=16H;

. km = R

R=10,000;

Thus

R = kmR = 10k

L = km

Problems 15-9

C = 1L =0.0625

F

kf = o

o = 25,000

kfL = 25,000 (16) = 6.4H

C = Ckmkf

=

[d]

0.0625

(10,000)(25,000) = 250 pF

( )1 s16 25,000

[e] H(s) = (s

25,000

()2 +1

16

s )+1

25,000

1562.5s

H

(s) = s2 + 1562.5s + 625 106

P 15.11 [a] Using the first prototype

o = 1 rad/s;

km = RR = 4

25

Thus,

C =1F; L=1H; R=25

= 1600; kf = oo = 50,000

1600R = kmR =40k;

C = C

L = kmkf L = 50,000 (1) = 32

mH;

1k

mkf =(1600)(50,000) = 12.5 nF

Using the second prototype

o = 1 rad/s; C =25F

L= 1 R=125 = 40

mH;

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km = R kf = oR=40,000;

Thus,

R

= kmR =40k;

C = C

o = 50,000

L = km

kf L = 50,000 (0.04) = 32mH;

25

kmkf = (40,000)(50,000) = 12.5 nF

[b]

P 15.12 For the scaled circuit

( )s2 + 1

H(s) = (s2 + R

L

LC

))s+

(L1

C

L = kmkf L;

1

f2

C = Ckmkf

LC = LC;

)

R = kmR

(R

. RL = kf L

It follows then that

( )kf

s2 + LC

H

(s) = (s2 + R)kfs +

k

f

L(

= [(s

s

kf

LC)

)2 +( 1

LC)( )]

skf)2 +

(R

)+

(1

L kf LC= H(s)|s=s/k

f

P15.13 For the circuit in Fig. 15.31

( )s2 + 1

H(s) = LC( )s2 + s 1

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RC + LC

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Q

It follows that

s2

+

Problems 15-11

1LC

H(s)= s

2

+R

sC+

1

LC

kmwhere R = kmR; L=

C = Ckmkf

1 2

kf L;

1

f

L

C

= LC

f

RC = RC

( )kf

s2 + LCH(s) = (

kfs2 + )s+kf

= (s

RC(

s

kf

LC1

)2 +LC)(s

kf)2 +( 1 )+ 1RC kf LC

= H(s)|s=s/kf

P 15.14 [a] For the circuit in Fig. P15.14(a)

s+1

s s2 +1H(s) = Vo = (

Vi = 1 s2 +Q+s+s

For the circuit in Fig. P15.14(b)

H(s) = Vo s

1Q

)s+ 1

Vi = 1+Qs +Qs

= Q(s2+ 1)

Qs2 + s + Q

H(s)=

s2 + 1(

s2 + 1Q

)s+1

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[b] kf oo =

104;

Q=8;

Replace s with s/kf.(

s104

H(s) = (s

)2 + 1( s

104 )2 + 1

)+1

8 104

= s2 + 108s2 + 1250s + 108

P 15.15 For prototype circuit (a):

QH(s) = Vo = Q

Vi=

Q+ 1s+ 1s Q+ s2s+1

s2 + 1

H(s) = Q(s2+ 1)

Q(s2 + 1) + s=

For prototype circuit (b):

1

(s2 +

1Q

)s+1

H(s) = VoVi = 1+(s2/Q) +1)

= s(+)1

s2 + 1 s+1Q

P 15.16 From the solution to Problem 14.15, o =the two scale factors:

kf = o = 4

100 krad/s and = 12.5 krad/s. Compute

o = 2 100 103

C 1 10 109

km = 1kfC =42.5 109 =

Thus,

R

= kmR = 8000 = 2546.48 L

= k

m

kf L

=

Calculate the cutoff frequencies:

c1 = kf c1 = 4(93.95 103) = 1180.6

krad/s c2 = kf c2 = 4(106.45 103) =

1337.7 krad/s To check, calculate the bandwidth:

=

c2

c1

= 157.1 krad/s = 4 (Checks!)

4 (10 103) = 253.303 H

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103

Problems 15-13

P 15.17 From the solution to Problem 14.24, o = 106 rad/s and = 2(10.61) krad/s.Calculate the scale factors:

kf = o = 0.05o = 50106

km = kfL

L

Thus,

= 0.05(200 106)

50 106= 0.2

20 109R = kmR = (0.2)(750) = 150

Calculate the bandwidth:

= kf = (0.05)[2(10.61

103)] =

To check, calculate the quality factor:

106

C = C

kmkf

3333 rad/s

= = 2 F(0.2)(0.05)

Q= o = 2(10.61 103) = 15

Q

= o

= 50 103

3333

= 15 (Checks)

1P 15.18 [a] km = R

R = 11

L = km

= 1000; kf = CkmC=

(1000)(200 109) = 5000

kf (L) = 5000 (1) = 200mH

[b]V 10/s

1000

( 1

V+ V

0.2s + 1000 + (5 106/s) = 0

s ) 1

V 1000 + s + 1000s + 5 106

10(s + 5000)

=100s

5(s + 5000)

V= 2s

2 + 10,000s + 25 106 =

s2 + 5000s + 12.5 106

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K


25(s + 5000)Io = V

0.2s =s(s2 + 5000s + 12.5 106)

K2 2= K1+

s s+2500j2500+s+2500+j2500

K1 = 0.01; K2 = 0.005

io(t) = 10 10e2500t cos 2500t mA

Since km = 1000 and the source voltage didnt change, the amplitude of thecurrent is reduced by a factor of1000. Since kf =multiplied by 5000.

P 15.19 km = R kf = o

5000 the coefficients oft are

R = 5500 =100; o = 5000

4103

C = Ckmkf = (100)(5000) = 8 nF

50 5 k; 700 70 k

L = kmkfL = 5 000 (20) = 0.4

H

0.05v 0.05100 v = 5 104v

The original expression for the current:

io(t) = 1728 + 2880e20t cos(15t + 126.87) mA

The frequency components will be multiplied by kf = 5000:

20 20(5000) = 105; 15 15(5000) = 75,000

The magnitudes will be reduced by km = 100:

1728 1728/100 = 17.28; 2880 2880/100 = 28.80

The expression for the current in the scaled circuit is thus,

io(t) = 17.28 + 28.80e105t cos(75,000t + 126.87)

mA

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P 15.20 [a] From Eq 15.1 we have

H(s) = Kc

s+c

where K = R2

R1 , c=

. H(s)= Kc

s+c

Problems 15-15

1

R2C

1

where K = R2

R1c =

R2C

By hypothesis R1 = kmR1; R2 = kmR2,

and C = C

kfkm . It follows thatK =Kandc =kfc,therefore

KcH(s) = Kkf

c (

[b] H(s) =

s+kfc = skf

K

(s + 1)K

)+c

[c] H(s) = (s

)

+1 = s+kfkfP 15.21 [a] From Eq. 15.4

H(s) = Kss+c where

K = R1 and

c=

1

R1C

. H(s)= Kss+c where

K= R1

and c=

1R1C

By hypothesis

R1 = kmR1; R2 = kmR2; C

=

It follows that

K =Kandc =kfc

Ks K(s/kf )

C

kmkf

. H(s)=(s+kfc =

skf

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c

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Ks[b]

H(s)=

[c]

(s + 1)Ks

( ) =s +1 (s + kf )

kf

P 15.22 [a] Hhp=

ss+1

;

kf = o = 2000 = 100 1

s

. Hhp

=1

s+2000

. RH

=

11.59 k

RH CH = 2000; (2000)(0.1 106) =

1

Hlp= s+1; kf = o

= 10,000 = 500 1

10,000. lp=

1

s+10,000

. RL

=

1= 318.3

RLCL = 10,000;

s

10,000

(10,000)(0.1 106)

[b] H(s)=

=

s+2000 s+10,000

10,000s

(s + 2000)(s + 10,000)

[c] o = c1c1 =(2000)(10,000) = 100020

rad/s

H

(jo)=

(10,000)(j1000 20) (2000 + j1000 20)(10,000 + j1000 20)

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H(s) = K(s/kf)

H

= (2 + j 20)(10 + j20)

= 0.8333/0

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Problems 15-17

[d] G = 20 log10(0.8333) = 1.58 dB

[e]

P 15.23 [a] For the high-pass section:

kf = o = 8000 = 400 1

sH(s)=

s+8000

1R1 = 3.98 k R2 = 3.98 k

R1(10 109) = 8000;

For the low-pass section:

kf = o = 40012) =800

H(s)=

800

s+800

1R2 = 39.8 k R1 = 39.8 k

R2(10 109) = 800;

0 dB gain corresponds to K = 1. In the summing amplifier we are free tochoose Rf and Ri so long as Rf /Ri = 1. To keep from having many differentresistance values in the circuit we opt forRf = Ri = 39.8 k.

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[b]

s 800[c] H(s)= s+8000 + s+800

= s2 + 1600s + 64 105 2

(s + 800)(s + 8000)

[d] o = (8000)(800) = 80010

10)2 + 1600(j800 10) + 64 1052

H(j800 10) = (800 (800 +j800

j128 104 102

10)(8000 + j800 10)

= (800)2(1 + j 10)(10 + j10)

j2 10

= (1 + j 10)(10 + j 10)

= 0.1818/0

[e] G = 20 log10 0.1818 = 14.81 dB

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[f]

P 15.24 [a] H(s) =

Problems 15-19

(1/sC) (1/RC)

R+(1/sC) =

s+(1/RC)

(1/RC)H(j)= j + (1/RC)

(1/RC)|H(j)| =

2 + (1/RC)2

(1/RC)2

|H(j)|2 =2 + (1/RC)2

[b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then

Va Vi

R1

+ sCVa + VaR2 + sL = 0

Solving forVa yields

Va

=

But

(R2 + sL)Vi

R1LCs2 + (R1R2C + L)s + (R1 +

R2)

Vo = sLVaR2 + sL

Therefore

Vo=

sLVi

R1LCs2 + (L + R1R2C)s + (R1 +

R2)

sLH(s)= R1LCs2 + (L + R1R2C)s + (R1 +

R2)

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jLH(j)= [(R1 + R2) R1LC

2] + j(L + R1R2C)

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L|H(j)| =

[R1 + R2 R1LC2]2 + 2(L + R1R2C)2

2L2|H(j)|2 =

(R1 + R2 R1LC2)2 + 2(L + R1R2C)2

2L2=

R1L2C24 + (L2 + R1R2C2 2R1LC + 2R1R2LC)2 + (R1+ R

[c] Let Va be the voltage across R2 positive at the upper terminal. Then

Va

ViR1

+ VaR2 + VasC + VasC = 0

(0 Va)sC + (0 Va)sC + 0 Vo

R3

R2Vi

=0

. Va=

2R1R2Cs + R1 + R2

and Va = Vo

2R3Cs

It follows directly that

H(s) = VoVi

=

H(j) =

2R2R3Cs

2R1R2Cs + (R1 + R2)

2R2R3C(j)(R1 + R2) + j(2R1R2C)

|H(j)|=

2R2R3C

(R1 + R2)2 + 24R1R2C2

4R2R3C22

|H(j)|2 =

(R1 + R2)2

+ 4R1

R2

C2

2

P 15.25 o = 2fo = 400 rad/s

=2(1000)=2000rad/s

. c2 c1

=2000

c1c2=

o= 400

Solve for the cutoff frequencies:

c1c2= 16 10

4

2

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c2 =16

Problems 15-21

c 1

16

104

2

c 1

c1 =2000

orc1+ 2000c1 16104

2=0

c1 =1000 1062 + 0.16 1062

c1 =1000(1 1.16) = 242.01 rad/s

. c2 =2000+242.01=6525.19

rad/s

Thus,fc1 = 38.52 Hz and fc2 = 1038.52 Hz

Check: = fc2 fc1 = 1000Hz

1c2 =

RLCL = 6525.19

1RL =

(6525.19)(5 106) = 30.65

1c1 =

RH CH = 242.01

1RH =

(242.01)(5 106) = 826.43

P 15.26 o = 1000 rad/s; GAIN = 6

=4000rad/s; C =0.2F

=c2 c1 =4000

o = c1c2

= 1000

Solve for the cutoff frequencies:

. c1+ 4000c1 106

=0

c1 =20001000 5=236.07rad/s

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c2 =4000+c1=4236.07rad/s

Check: = c2 c1 = 4000 rad/s

c1=

1

RLCL

1

. RL=

1

(0.2 106)(236.07) = 21.81k

RH CH = 4236.07

RH =

Rf

1(0.2 106)(4236.07) = 1.18k

Ri = 6

IfRi = 1 k Rf = 6Ri = 6 k

1P 15.27 [a] y=20log10

1+2

n

= 10 log10(1 + 2n)

From the laws of logarithms we have

y=

( 10)

ln 10

ln(1 + 2n)

Thus

dy

d

=

( 10 ) 2n2n1

ln 10 (1 + 2n)

x=log10= lnln 10. ln = x ln 10

1 d

dx = ln 10, dx = ln 10)

dy ( dy )(d

dx = d dx

at = c = 1

rad/s

dy

= 20n2n1+2n dB/decade

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dx = 10n .

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[b] y=20log10

= 10n

1[ 1+2]n

Problems 15-23

= 10n log10(1 + 2)

ln 10 ln(1 +

2

)dy 10n

d = ln

10

As before

d

( )1

1+2

2=

dy

20n

(ln 10)(1 + 2)

20n2

dx = (ln 10); dx = (1 + 2)

At the cornerc = 21/n 1 . c = 21/n 1

dy 20n[21/n 1]dB/decade.

dx = 21/n

[c] For the Butterworth Filter For the cascade of identical sections

n dy/dx (dB/decade) n dy/dx (dB/decade)

1 10 1 10

2 20 2 11.72

3 30 3 12.38

4 40 4 12.73 13.86

[d] It is apparent from the calculations in part (c) that as n increases the amplitudecharacteristic at the cutoff frequency decreases at a much faster rate for the

Butterworth filter.

Hence the transition region of the Butterworth filter will be much narrower

than that of the cascaded sections.

(0.05)(30)=P 15.28 [a]

n

=

log10(7000/2000

)

2.76

. n=3

[b]

P 15.29 [a]

1Gain = 20 log10 = 32.65 dB

1+(7000/2000)6 For the scaled circuit

1/(R)2C 1C2H(s) =

s2+

where

2RC1 s

1+

(R)2C1C2

R = kmR; C1 =C1/k

fk

m; C2

=C2/k

fk

m

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s


It follows that

1

(R)2C1C2 = R

2Cf1C

2

2

RC

= 2kf1 RC1

kf/R2C1C2. H(s)=

=

s2 +2kfRC1 s +R2Cf1C2

1/R2C1C2( (

s 2 s 1kf)2 +

)+

RC1 kf R2C1C2

P 15.30 [a] H(s) = 1

(s + 1)(s2 + s + 1)

[b] fc = 2000Hz;

H(s) =(skf

=

c = 4000 rad/s; kf = 4000

1

+ 1)[( s + 1]kf )2 + kfkf

(s + kf )(s2 + kf s + kf )

(4000)3=

(s + 4000)[s2 + 4000s + (4000)2]

64[c]

H(j14,000)=

(4 + j14)(180 + j56)

= 0.02332/ 236.77

Gain = 20 log10(0.02332) = 32.65 dB

P 15.31 [a] In the first-order circuit R = 1 and C = 1 F.

km = R = 1000; kf = o =4000

R = 1 1R = kmR =

1000;

o = 2( 1

1C = C

kmkf = (1000)(4000)

= 79.58 nF

In the second-order circuit R = 1 , 2/C1 = 1 so C1 = 2 F,and

C2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit

R = kmR = 1000;2

C1 = kC1

C2 = kC2

mkf= (1000)(4000) = 159.15

nF

0.5mkf

=(1000)(4000) = 39.79nF

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[b]

P 15.32 [a] n= (0.05)(48)log10(2000/500) =

3.99

Problems 15-25

. n=4

From Table 15.1 the transfer function of the first section is

H1(s)=

s2

s2 + 0.765s + 1

For the prototype circuit

2

R2 = 0.765; R2 = 2.61 ; R1 = 1R2 = 0.383

The transfer function of the second section is

H2(s)= s

2

s2 + 1.848s + 1

For the prototype circuit

2

R2 = 1.848; R2 = 1.082 ; R1 = 1R2 = 0.9240

The scaling factors are:

kf = oo = 2( 1

C = Ckmkf

= 4000

110 109 =4000km

1

km=

4000(10 109) = 7957.75

Therefore in the first section

R1 = kmR1 = 3.04k; R2 = kmR2 = 20.80k

In the second section

R1 = kmR1 = 7.35k; R2 = kmR2 = 8.61k

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[b]

P 15.33 n=5:1+(1)5s10= 0; s10 = 1

s10 = 1/0 + 36k

k sk+1

0 1/0

1 1/36

2 1/72

3 1/108

4 1/144

5 1/180

6 1/216

7 1/252

8 1/288

9 1/324

Group by conjugate pairs to form denominator polynomial.

(s + 1)[s (cos 108 + j sin 108)][s (cos 252 + j sin

252)]

[s (cos 144 + j sin 144)][s (cos 216 + j sin

216)] = (s + 1)(s + 0.309 j0.951)(s + 0.309 +

j0.951)

(s + 0.809 j0.588)(s + 0.809 + j0.588)

which reduces to

(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)

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Problems 15-27

n=6:1+(1)6s12= 0 s12 = 1

s12 = 1/15 +36k

k sk+10 1/15

1 1/45

2 1/75

3 1/105

4 1/135

5 1/165

6 1/195

7 1/225

8 1/255

9 1/285

101/315

111/345

Grouping by conjugate pairs yields

(s + 0.2588 j0.9659)(s + 0.2588 + j0.9659)

(s + 0.7071 j0.7071)(s + 0.7071 + j0.7071)

(s + 0.9659 j0.2588)(s + 0.9659 + j0.2588)

or(s2 + 0.518s + 1)(s2 + 1.414s + 1)(s2 + 1.932s + 1)

s2

P 15.34 H(s) =s2

+

H(s) =

2kmR2(C/kmkf

)s

s2

1

+kmR1kmR2(C2/kmkf )

s2 +2kfR2C s +R1Rf2C2

(s/kf)2= (

2 s 1

(s/kf)2+

R2C kf )+R1R2C

2

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P 15.35 [a] n= (0.05)(48) = 3.99 n=4log10(32/8)

From Table 15.1 the transfer function is

H(s)

=

1

(s2 + 0.765s + 1)(s2 + 1.848s + 1)The capacitor values for the first stage prototype circuit are

2 C1 = 0.765

C2 = 1C1 = 0.38

F

C1 = 2.61 F

The values for the second stage prototype circuit are

2

C1 = 1.848

C2 = 1C1 = 0.92

F

C1 = 1.08 F

The scaling factors are

km = RR

=1000;

kf = oo = 16,000

Therefore the scaled values for the components in the first stage are

R1 = R2 = R = 1000

C1=

C2=

2.61

(16,000)(1000) = 52.01nF

0.38

(16,000)(1000) = 7.61nF

The scaled values for the second stage areR1 = R2

= R = 1000

C1=

C2=

1.08

(16,000)(1000) = 21.53nF

0.92

(16,000)(1000) = 18.38nF

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Problems 15-29

[b]

P 15.36 [a] The cascade connection is a bandpass filter.

[b] The cutoff frequencies ar2 kHz and 8 kHz.

The center frequency is (2)(8) = 4 kHz.The Q is 4/(8 2) = 2/3 = 0.67

[c] For the high pass section kf = 4000. The prototype transfer function is

s4Hhp(s)=

(s2 + 0.765s + 1)(s2 + 1.848s

+ 1)

(s/4000)4H hp(s)=

[(s/4000)2 + 0.765(s/4000) +

1] 1[(s/4000)2 + 1.848(s/4000) + 1]

=s4

(s2 + 3060s + 16 1062)(s2 + 7392s + 16 1062)

For the low pass section kf = 16,000

Hlp(s)=

1

(s2 + 0.765s + 1)(s2 + 1.848s

+ 1)1

Hlp(s) =

[(s/16,000)2 + 0.765(s/16,000) +

1]

1[(s/16,000)2 + 1.848(s/16,000) + 1]

=(16,000)4

([s2 + 12,240s + (16,000)2)][s2 + 29,568s +

(16,000)2]

The cascaded transfer function is

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H(s) = Hhp(s) lp(s)

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For convenience let

D1 = s2 + 3060s + 16 1062

D2 = s2 + 7392s + 16 1062

D3 = s2 + 12,240s + 256

1062

D4 = s2 + 29,568s + 256

1062

Then

H(s) = 65,536 10124s4

D1D2D3D4

[d] o = 2(4000) = 8000 rad/s

s=j8000

s4 = 4096 10124

D1 = (16 1062 64 1062) + j(8000)

(3060)

= 1062(48 + j24.48) =

1062(53.88/152.98) D2 = (16 1062 64

1062) + j(8000)(7392)

= 1062(48 + j59.136) =

1062(76.16/129.07) D1 = (256 1062 64

1062) + j(8000)(12,240)

= 1062(192 + j97.92) =

1062(215.53/27.02) D1 = (256 1062 64

1062) + j(8000)(29,568)

= 1062(192 + j236.544) =

1062(304.66/50.93)

(65,536)(4096)8 1024H(jo)= (8 1024)[(53.88)(76.16)(215.53)(304.66)/360]

= 0.996/ 360 = 0.996/0

P

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7[a] F

r

o

m

the statement of the problem, K = 10 ( = 20 dB). Therefore for theprototype bandpass circuit

R1 = Q

K = 10

=1.6 R2 = Q2Q2 K = 502

R3 = 2Q = 32

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Problems 15-31


kf = o

o = 2(6400) = 12,8001

km = CCkf =(20 109)(12,800) = 1243.40

Therefore,

R1 = kmR1 = (1.6)(1243.40) = 1.99k

R2 = kmR2 = (16/502)(1243.40) =

39.63

R3 = kmR3 = 32(1243.40) = 39.79k

[b]

P 15.38 From Eq 15.58 we can write

( )( )(2 R3C 1

)sH(s)=

or

R3C2

s2 +R3Cs

2 R1C+ R1+R2R1R2R3C

2

H(s)=

s2

+

( )( )R3 22R1 R3C s2

+ R1+R2R3C s R1R2R3C2

Therefore

2 o R1 + R2R3C = = Q; R1R2R3C2 = o;

and K = R32R1

By hypothesis C = 1 F and o = 1 rad/s

2R3 = Q

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3 = 2Q

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R1 = R32K = K

R1 + R2R1R2R3 = 1

Q (Q )

K +R2= K

R2 = Q

(2Q)R2

2Q2 K

P 15.39 [a] First we will design a unity gain filter and then provide the passband gain with

an inverting amplifier. For the high pass section the cut-off frequency is 500

Hz. The order of the Butterworth is

n= (0.05)(20)log10(500/200) = 2.51

. n=3

s3Hhp(s)= (s + 1)(s2 + s + 1)

For the prototype first-order section

R1 = R2 = 1 , C = 1 F

For the prototype second-order section

R1 = 0.5 , R2 = 2 , C = 1 F


kf = oo = 2(500) = 1000

km = CCkf

=

1 106

(15 109)(1000) =

15

In the scaled first-order section

R1 = R2 = kmR1 = 10615 (1) = 21.22

k

C =15nF

In the scaled second-order section

R1 = 0.5km = 10.61k

R2 = 2km = 42.44k

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Problems 15-33

C =15nF

For the low-pass section the cut-off frequency is 4500 Hz. The order of theButterworth filter is

(0.05)(20)

n= log10(11,250/4500) =2.51;

1

. n=3

Hlp(s)=

(s + 1)(s2 + s + 1)

For the prototype first-order section

R1 = R2 = 1 , C = 1 F

For the prototype second-order section

R1 = R2 = 1 ; C1 = 2 F; C2 = 0.5 FThe low-pass scaling factors are

km = R kf = oR =104

; o = (4500)(2) = 9000

For the scaled first-order section

R1 = R2 = 10k; 1C = C

kfkm = (9000)(104)=

3.54 nF

For the scaled second-order sectionR1 = R2 = 10k

C1 = kfk =

m

C2 = kfk =

m

2

(9000)(104) = 7.07nF

0.5

(9000)(104) = 1.77nF

GAIN AMPLIFIER

20 log10 K = 20 dB, . K =10

Since we are using 10 k resistors in the low-pass stage, we will use Rf= 100 k and Ri = 10 k in the inverting amplifier stage.

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H


[b]

P 15.40 [a] Unscaled high-pass stage

s3Hhp(s)= (s + 1)(s2 + s + 1)

The frequency scaling factor is kf = (o/o) = 1000. Therefore thescaled transfer function is

H hp(s) = (s

1000

+

=

(s/1000)3) [( ]

s1 )2 + s1000 1000 + 1s3

(s + 1000)[s2 + 1000s + 1062]

Unscaled low-pass stage

Hlp(s)=

1

(s + 1)(s2 + s + 1)

The frequency scaling factor is kf = (o/o) = 9000. Therefore thescaled transfer function is

Hlp(s) =

(s

9000+

=

1) [(

s1 )2 +

( s )+1]

9000 9000(9000)3

(s + 9000)(s2 + 9000s + 81 1062)

Thus the transfer function for the filter is

H(s) = 10Hhp(s) lp(s) =7

D1D2D3D4

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Problems 15-35

where

D1 = s +

1000

D2 = s +

9000

D3 = s2 + 1000s + 1062

D4 = s2 + 9000s + 81 1062

[b] At f = 200 Hz =400rad/s

D1(j400) = 400(2.5 + j1)

D2(j400) = 400(22.5 + j1)

D3(j400) = 4 1052(2.1 +

j1.0)

D4(j400) = 4 1052(202.1 +

j9)

Therefore

D1D2D3D4(j400) =

25661014(28,534.82/52.36)

(7293 1010)(64 1063)H(j400)= 256

6 1014(28,534.82/52.36)

= 0.639/ 52.36

20 log10 |H(j400)| = 20 log10(0.639) = 3.89 dB

At f = 1500 Hz, =3000rad/s

Then

D1(j3000) = 1000(1 + j3)

D2(j3000) = 3000(3 + j1)

D3(j3000) = 1062(8 + j3)

D4(j3000) = 9 1062(8 +

j3)

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H(j3000) = (729 3 1010)(27 109 327 10186(730/270 )

= 9.99/90

20 log10 |H(j3000)| = 19.99 dB

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[c] From the transfer function the gain is down 19.99 + 3.89 or23.88 dB at 200 Hz.Because the upper cut-off frequency is nine times the lower cut-off frequency

we would expect the high-pass stage of the filter to predict the loss in gain at 200

Hz. For a 3nd order Butterworth

1

GAIN = 20 log10 = 23.89 dB.1+(500/200)6

1500 Hz is in the passband for this bandpass filter, and is in fact the centerfrequency. Hence we expect the gain at 1500 Hz to equal, or nearly equal,20 dB as specified in Problem 15.39. Thus our scaled transfer functionconfirms that the filter meets the specifications.

P 15.41 [a] From Table 15.1

Hlp(s) =(s2 + 0.518s + 1)

(s2 +

1

2s + 1)(s2

+ 1.932s +1)

1Hhp(s) = (

1s2+

Hhp(s) =

(1

0.518 )+1)(1s s2+

s6

( (1 1

2 )+1)(

1)+1

)

s s2 + 1.932 s

(s2 + 0.518s + 1)(s2

+

P

15.42 [a] kf = 25,000(s/25,000)6

2s + 1)(s2 + 1.932s + 1)

H hp(s)= [(s/25,000)

2 + 0.518(s/25,000)

+ 1]

1

=

[(s/25,000)2 + 1.414s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 +

s6

(s2 + 12,950s + 625 106)(s2 + 35,350s + 625 106)

1(s2 + 48,300s + 625 106)

(25,000)6[b]

H(j25,000)=

=

[12,950(j25,000)][35,350(j25,000)]

[48,300(j25,000)]

(25,000)3

(12,950)(25,350)(48,300)j3

= 0.7067/ 90

20 log10 |H(j25,000)| = 3.02 dB

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G1 + G2 + G3

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b1 = KG2 + G2 + boC

1

G2

b1 = G2(1 + K) +boC1

G2

Solving this quadratic equation forG2 we get

b1 boC14(1 + K)G2 = b1

2(1 + K) 4(1 + K)2

b1=

b1 4bo(1 +K)C12(1 + K)

ForG2 to be realizable

b1

C1 0.5we have amplification, and if < 0.5 we have attenuation.Also note the amplification an attenuation are symmetric about = 0.5. i.e.

1|H(j)|=0.6= |H(j)|=0.4

Yes, the circuit can be used as a treble volume control because

The circuit has no effect on low frequency signals

Depending on the circuit can either amplify ( > 0.5) or attenuate ( < 0.5)signals in the treble range

The amplification (boost) and attenuation (cut) are symmetric around = 0.5.When = 0.5 the circuit has no effect on signals in the treble frequency range.

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P 15.59 [a] |H(j)|=1 = Ro(R4+ R 3)

R3(R4 + Ro)=

Problems 15-55

= 9.99(5.9)(565.9)

[b]

[c]

maximum boost = 20 log10 9.99 = 19.99 dB

|H(j)|=0 = R3(R4 +R 3)

Ro(R4 + Ro)

maximum cut = 21.93 dB

R4 = 500 k; Ro = R1 + R3 + 2R5 = 65.9 k

. R4 = 7.59Ro

Yes, R4 is significantly greater than Ro.

(2R3 + R4) + j Ro[d] |H(j/R3C2)|=1=

R3 (R4 + R3)

(2R3 + R4) + j(R4 + Ro)

511.8 + j 65.9

=

(505.9) 5.9

511.8 + j565.9

= 7.44

20 log10 |H(j/R3C2)|=1 = 20 log10 7.44 = 17.43 dB

[e] When = 0

|H(j/R3C2)|=0=

(2R3 + R4) + j(R4 + Ro)

(2R3 + R4) + j RoR3 (R4

+ R3)

Note this is the reciprocal of |H(j/R3C2)|=1.

20 log10 |H(j/R3C2)|+0 = 17.43 dB

[f] The frequency 1/R3C2 is very nearly where the gain is 3 dB off from its

maximum boost or cut. Therefore for frequencies higher than 1/R3C2 thecircuit designer knows that gain or cut will be within 3 dB of the maximum.

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15-56

P 15.60

CHAPTER 15. Active Filter Circuits

|H(j)| = [(1 )R4+ R

o][R4+

R 3]

[(1 R4 + R3][R4 +Ro]

]

]

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16Fourier Series

Assessment Problems

AP 16.1 av =1

T

2T /3 T

Vm dt + 10 T 2T /3

[ 2T /3

(

Vm

3

)

dt = 79 Vm

= 7 V

) ]

ak = 2T 0

TVm cos k0t dt +

2T /3

) ) )

( Vm

3

cos k0t dt

)

=( 4Vm

3k0T

( 4k (6 ( 4ksin = sin

3 k 3[

bk = 2T

2T /3

Vm sin k0t dt+0

T( Vm

2T /3 3

) ]

sin k0t dt

=)[

( 4Vm( 4k

1cos3k0T 3

)] )[ )](6 ( 4k

= 1cosk 3

AP 16.2 [a] av = 7 = 21.99 V

[b] a1 = 5.196a2 =2.598

b1 = 9 b2 = 4.5)

a3 = 0 a4 = 1.299a5 = 1.039

b3 = 0 b4 = 2.25 b5= 1.8[c] 0=

( 2

T

= 50 rad/s

[d] f3 = 3f0 = 23.87 Hz

[e] v(t) = 21.99 5.2 cos 50t + 9 sin 50t + 2.6 cos 100t + 4.5 sin 100t

1.3 cos 200t + 2.25 sin 200t + 1.04 cos 250t + 1.8 sin 250t + V

AP 16.3 Odd function with both half- and quarter-wave symmetry.

(

6Vm)vg (t) =T

t, 0tT/6; 16-1

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= , ak = or a

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