Electric Circuits, 9th Edition P62 pps

15.5 Narrowband Bandpass and Bandreject Filters The cascade and parallel component designs for synthesizing bandpass and bandreject filters from simpler low-pass and high-pass filters

/ A S S E S S M E N T PROBLEM

Objective 3—Understand how to use cascaded first- and second-order Butterworth filters

15.4 For the circuit in Fig 15.25, find values of i?t Answer: R x = 0.707 Q,, R 2 = 1.41 O

and R 2 that yield a second-order prototype

Butterworth high-pass filter

NOTE: Also try Chapter Problems 15.36,15.38 and 15.39

15.5 Narrowband Bandpass and

Bandreject Filters

The cascade and parallel component designs for synthesizing bandpass and bandreject filters from simpler low-pass and high-pass filters have the

restriction that only broadband, or low-Q, filters will result (The Q, of course, stands for quality factor.) This limitation is due principally to the

fact that the transfer functions for cascaded bandpass and parallel band-reject filters have discrete real poles The synthesis techniques work best for cutoff frequencies that are widely separated and therefore yield the lowest quality factors But the largest quality factor we can achieve with discrete real poles arises when the cutoff frequencies, and thus the pole locations, are the same Consider the transfer function that results:

S + (x)c/\S + CO,

SOOc

s 2 + 2(o c s + at 2

0.5/3.5

Eq 15.50 is in the standard form of the transfer function of a bandpass filter, and thus we can determine the bandwidth and center frequency directly:

From Eqs 15.51 and 15.52 and the definition of Q, we see that

(Jin 00, 1

r

(15.53)

Thus with discrete real poles, the highest quality bandpass filter (or

band-reject filter) we can achieve has Q = 1/2

To build active filters with high quality factor values, we need an op amp circuit that can produce a transfer function with complex conjugate poles Figure 15.26 depicts one such circuit for us to analyze At the invert-ing input of the op amp, we sum the currents to get

K Figure 15.26 • An active high-Q bandpass filter

^ -Vg

X/sC R

Solving for V v

K = -£-• (15.54)

At the node labeled a, we sum the currents to get

Solving for V,-,

v,

v,

-R

= (1

V V

I

+ 2sR x C

- va v,

\/sC 1/sC

+ RJR 2 )V A

-Ri

sR t CV 0 (15.55) Substituting Eq 15.54 into Eq 15.55 and then rearranging, we get an

expression for the transfer function V 0 / V t:

—s

where

H{S) = - — ^ , (15.56)

^2 + TTTP + o R3C RCHR3C2

Rsq = R1]\R2 =

R { + R 2

Since Eq 15.56 is in the standard form of the transfer function for a

bandpass filter, that is,

s~ + ps + a>;

we can equate terms and solve for the values of the resistors, which will

achieve a specified center frequency («„), quality factor (Q), and passband

gain (K):

fi = ^ ; (15.57)

Kf3 = — ; (15.58)

At this point, it is convenient to define the prototype version of the

circuit in Fig 15.25 as a circuit in which o> 0 = 1 rad/s and C = 1 F Then

the expressions for R^, R 2 , and R 3 can be given in terms of the desired

quality factor and passband gain We leave you to show (in Problem 15.45)

that for the prototype circuit, the expressions for Ri, R 2 , and R 3 are

Ri = Q/K,

Ri = Q/(2Q 2 - K), R3 = 2Q

Scaling is used to specify practical values for the circuit components This

design process is illustrated in Example 15.12

Designing a High-Q Bandpass Filter

Design a bandpass filter, using the circuit in Fig 15.26,

which has a center frequency of 3000 Hz, a quality

factor of 10, and a passband gain of 2 Use 0.01 /xF

capacitors in your design Compute the transfer

func-tion of your circuit, and sketch a Bode plot of its

mag-nitude response

Solution

Since Q = 10 and K = 2, the values for R u R2,

and R 3 in the prototype circuit are

Ri = 10/2 = 5,

R2 = 10/(200 - 2) = 10/198,

R3 = 2(10) = 20

The scaling factors are kf = 6OOO77- and

km = lOfyfc/ After scaling,

R ] = 26.5 kO,

R 2 = 268.0 a , R3 = 106.1 k a

The circuit is shown in Fig 15.27

Substituting the values of resistance and

capac-itance in Eq 15.56 gives the transfer function for

this circuit:

s 2 + 1885.05 + 355 X 106

It is easy to see that this transfer function meets the

specification of the bandpass filter defined in the

example A Bode plot of its magnitude response is

sketched in Fig 15.28

Figure 15.27 • The high-G bandpass filter designed in ExampLe 15.12

10

5

0

- 3

- 1 0

CQ

5 -15

- 2 0 -25

- 3 0 -35

- 4 0 /

/ /

t

/

V

j

•

t

1

1

1

Mill 1

,- 6 dB (gain 0

_

\

\

\

\

\

n

f 2 )

^

100 500 1000 5000 10,000 50,000100,000

/(Hz) Figure 15.28 • The Bode magnitude plot for the high-Q bandpass filter designed in Example 15.12

The parallel implementation of a bandreject filter that combines low-pass and high-low-pass filter components with a summing amplifier has the same low-<2 restriction as the cascaded bandpass filter The circuit in Fig 15.29 is

an active high-Q bandreject filter known as the twin-T notch filter because of

the two T-shaped parts of the circuit at the nodes labeled a and b

We begin the analysis of this circuit by summing the currents away from node a:

(Vn - V,)sC + (K - V0)sC + 2(V a -<rV 0 )

or

V a [2sCR + 2] - VjsCR + 2a] = sCRV; (15.60)

J_

sC

6

R

i

<vw-sC

b R

« vv^ '

1

2sC

f

o-K,

^ — • • — • K

• +

:(l-«r)J?

* <•

(TV„\(TR

Figure 15.29 • A high-Q active bandreject filter

Summing the currents away from node b yields

Vb-V, Vb - Va

R R + (Vb - <rV a )2sC = 0

or

Vb [2 + 2RCs) -V0[l + 2a RCs] = Vh (15.61) Summing the currents away from the noninverting input terminal of

the top op amp gives

V — K (V 0 -V a )sC+ - 2 - ^ - t = 0

or

From Eqs 15.60-15.62, we can use Cramer's rule to solve for V 0:

2(RCs + 1) 0 sCRV;

0 2(RCs + 1) V t -RCs - 1 0

K =

2(RCs + 1) 0 -(RCs + 2a)

0 2{RCs + 1) -(2trRCs + 1)

- A C * - 1 ffCj + 1

R2C2s2 + ARC{\ - a)s + 1'

Rearranging Eq 15.63, we can solve for the transfer function:

1

H(s) = f =

S? +

/?2C2

2 , 4(1 - o) 1

R2C2\

(15.63)

(15.64)

which is in the standard form for the transfer function of a bandreject filter:

S + 0)()

H(s) = -= -

s 2 + (3s + (4

Equating Eqs 15.64 and 15.65 gives

(15.65)

2,-2 '

R l C

(15.66)

P = 4(1 ~ cr)

In this circuit, we have three parameters (R, C, and a) and two design constraints (co a and /3) Thus one parameter is chosen arbitrarily; it is usu-ally the capacitor value because this value typicusu-ally provides the fewest commercially available options Once C is chosen,

R = 1 a> n C (15.68)

and

A(o (> AQ (15.69)

Example 15.13 illustrates the design of a high-Q active bandreject filter

Designing a High-Q Bandreject Filter

Design a high-Q active bandreject filter (based on

the circuit in Fig 15.29) with a center frequency of

5000 rad/s and a bandwidth of 1000 rad/s Use 1 /xF

capacitors in your design

Solution

In the bandreject prototype filter, co n = 1 rad/s,

R = 1 H, and C = 1 F As just discussed, once <o(,

and Q are given, C can be chosen arbitrarily, and R

and cr can be found from Eqs 15.68 and 15.69 From

the specifications, Q = 5 Using Eqs 15.68 and

15.69, we see that

R = 200 il,

a = 0.95

Therefore we need resistors with the values 200 ft

(/?), 100 ft (i?/2), 190 ft (0-/2),and 10 ft [(1 - a)R]

The final design is depicted in Fig 15.30, and the Bode magnitude plot is shown in Fig 15.31

Figure 15.30 A The high-Q active bandreject filter designed in Example 15.13

JLU

5

0

PQ

- 1 0

- 1 5

/

co (rad/s)

50,000 100,000

Figure 15.31 • The Bode magnitude plot for the high-Q active

bandreject filter designed in Example 15.13

/ A S S E S S M E N T PROBLEMS

Objective 4—Be able to use design equations to calculate component values for prototype narrowband, bandpass,

and bandreject filters

15.5 Design an active bandpass filter with Q = 8,

K = 5, and <o a = 1000 rad/s Use 1 /xF

capaci-tors, and specify the values of all resistors

Answer: R t = 1.6 kfl, R 2 = 65.04 a , R 3 = 16 kfl

NOTE: Also try Chapter Problem 15.60

15.6 Design an active unity-gain bandreject filter

with (o (} = 1000 rad/s and Q = 4 Use 2 jtF

capacitors in your design, and specify the values

of R and a

Answer: R - 500 Q, a = 0.9375

Practical Perspective

Bass Volume Control

We now look at an op amp circuit that can be used to control the

amplifica-tion of an audio signal in the bass range The audio range consists of signals

having frequencies from 20 Hz to 20 k H z The bass range includes

frequen-cies up to 300 Hz The volume control circuit and its frequency response are

shown in Fig 15.32 The particular response curve in the family of response

curves is selected by adjusting the potentiometer setting in Fig 15.32(a)

In studying the frequency response curves in Fig 15.32(b) note the

fol-lowing First, the gain in dB can be either positive or negative I f the gain

is positive a signal in the bass range is amplified or boosted I f the gain is

negative the signal is attenuated or cut Second, i t is possible to select a

Vs*

*—• * ,

Figure 15.32 A (a) Bass volume control circuit; (b) Bass volume control circuit frequency response

•V B

Figure 15.33 A The s-domain circuit for the bass

volume control Note that a determines the

potentiometer setting, so 0 £ a £ 1

response characteristic that yields unity gain (zero dB) for all frequencies in the bass range As we shall see, if the potentiometer is set at its midpoint, the circuit will have no effect on signals in the bass range Finally, as the frequency increases, all the characteristic responses approach zero dB or unity gain Hence the volume control circuit will have no effect on signals

in the upper end or treble range of the audio frequencies

The first step in analyzing the frequency response of the circuit in

Fig 15.32(a) is to calculate the transfer function V 0 /V s To facilitate this

calculation the s-domain equivalent circuit is given in Fig 15.33 The node voltages V a and V b have been labeled in the circuit to support node volt-age analysis The position of the potentiometer is determined by the

numer-ical value of a, as noted in Fig 15.33

To find the transfer function we write the three node voltage equations that describe the circuit and then solve the equations for the voltage ratio

V () /V s The node voltage equations are

(1 - a)R 2 aR2

+ K - K /? + ( K -V^sC^O;

+ (V b -V a )sC l + V b ~Vo

(1 - a)R 2 + aRo

0;

= 0

These three node-voltage equations can be solved to find V 0 as a function

of V s and hence the transfer function H(s):

' s

It follows directly that

-jRi + aR 2 + R&Cis) /?! + ( ! - a)R 2 + R&Cis'

_ - ( / ? ! + gR 2 + jcoR&Ci)

W [ # ! + ( 1 - a)R2 + jaRiRfit]'

Now let's verify that this transfer function will generate the family of fre-quency response curves depicted in Fig 15.32(b) First note that when

a = 0.5 the magnitude of H{j<o) is unity for all frequencies, i.e.,

|H0»)l \Ri + 0.5R2 | A, + 0.5R 2 + jo}RxR2Cx\ + jaR&Cil 1

When w = Owe have

\H(jO)\ = /¾ + aR 2

R t + (1 - a)R 2

Observe that \H(jO)\ at a = 1 is the reciprocal of \H(jO)\ at a = 0, that is

/¾ + Ri 1

\H{m\a=x =

* i \H(Ma=0'

With a little thought the reader can see that the reciprocal relationship

holds for all frequencies, not just <a = 0 For example a = 0.4 and a = 0.6

are symmetric with a = 0.5 and

#0' w )«=o.4 - -(/¾ + 0.4/¾) + fofr/frd

(/¾ + 0.6/¾) + j(oRiR 2Cx

while

Hence

tf(;w)a=0.6 - ( / ¾ + 0.6/¾) + j(oRiR 2Ci

(/¾ + 0.4/¾) + jo)RiR 2 C l

H(jo>) af= oA =

H(j<»)a-0.6

I t follows that depending on the value of a the volume control circuit can

either amplify or attenuate the incoming signal

The numerical values of R\,R 2 , and C] are based on two design

deci-sions The first design choice is the passband amplification or attenuation

in the bass range (as a>-*0) The second design choice is the frequency at

which this passband amplification or attenuation is changed by 3 d B The

component values which satisfy the design decisions are calculated with a

equal to either 1 or 0

As we have already observed, the maximum gain will be (/¾ + R 2 )/Ri

and the maximum attenuation will be R\/(R\ + /¾) I f we assume

(/¾ + R 2 )/R\ » 1 then the gain (or attenuation) will differ by 3 dB

from its maximum value when o> = 1 / / ¾ ^ This can be seen by noting that

R 2 C X « = i

1/¾ + R 2 + m 1/¾ + iR\\

/¾ + /¾

Ri + /1

and

H J R

2 Cy a=0

II + /11

1*1 + /*ll 1/¾ + R 2 + /Xi|

II + /11

1 //¾ + /¾

V 2 \ fll

Ri + R2

Ri

V2

+ /1

* i

*1 + /¾

/V07E; Assess your understanding of this Practical Perspective by trying

Chapter Problems 15.61 and 15.62

Summary

• Active filters consist of op amps, resistors, and capacitors

They can be configured as low-pass, high-pass, bandpass,

and bandreject filters They overcome many of the

disad-vantages associated with passive filters (See page 560.)

• A prototype low-pass filter has component values of

Ri = R2 = 1 H and C = 1 F, and it produces a unity

passband gain and a cutoff frequency of 1 rad/s The

prototype high-pass filter has the same component

val-ues and also produces a unity passband gain and a

cut-off frequency of 1 rad/s (See pages 561 and 562.)

• Magnitude scaling can be used to alter component

val-ues without changing the frequency response of a circuit

For a magnitude scale factor of k , the scaled (primed)

values of resistance, capacitance, and inductance are

R' = kmR, V = kmL, and C = C/km

(See page 564.)

• Frequency scaling can be used to shift the frequency

response of a circuit to another frequency region without

changing the overall shape of the frequency response

For a frequency scale factor of kf, the scaled (primed)

values of resistance, capacitance, and inductance are

R' = R, L' = L/kp and C = C/kf

(See page 564.)

• Components can be scaled in both magnitude and

fre-quency, with the scaled (primed) component values

given by

R' = k m R, L' = (k m /k f )L, and C = C/{k m k f )

(See page 564.)

• The design of active low-pass and high-pass filters can

begin with a prototype filter circuit Scaling can then be

applied to shift the frequency response to the desired

cutoff frequency, using component values that are

com-mercially available (See page 565.)

• An active broadband bandpass filter can be constructed

using a cascade of a low-pass filter with the bandpass

fil-ter's upper cutoff frequency, a high-pass filter with the

bandpass filter's lower cutoff frequency, and (optionally)

an inverting amplifier gain stage to achieve nonunity

gain in the passband Bandpass filters implemented in

this fashion must be broadband filters (wc.2 » <ac]), so

that the elements of the cascade can be specified

inde-pendently of one another (See page 568.)

• An active broadband bandreject filter can be

con-structed using a parallel combination of a low-pass filter

with the bandreject filter's lower cutoff frequency and a

high-pass filter with the bandreject filter's upper cutoff frequency The outputs are then fed into a summing amplifier, which can produce nonunity gain in the pass-band Bandreject filters implemented in this way must

be broadband filters {co c2 » o>c.i), so that the low-pass and high-pass filter circuits can be designed independ-ently of one another (See page 572.)

• Higher order active filters have multiple poles in their transfer functions, resulting in a sharper transition from the passband to the stopband and thus a more nearly ideal frequency response (See page 573.)

• The transfer function of an nth-order Butterworth low-pass filter with a cutoff frequency of 1 rad/s can be determined from the equation

H(s)H(s) =

=-K J K ' i + ( - i ) V »

by

• finding the roots of the denominator polynomial

• assigning the left-half plane roots to H(s)

• writing the denominator of H(s) as a product of

first-and second-order factors (See page 578-579.)

• The fundamental problem in the design of a Butterworth filter is to determine the order of the filter The filter specification usually defines the sharpness of

the transition band in terms of the quantities A , a p-,As> and <a r From these quantities, we calculate the smallest integer larger than the solution to either Eq 15.42 or

Eq 15.46 (See page 583.)

• A cascade of second-order low-pass op amp filters (Fig 15.21) with 1 JQ resistors and capacitor values cho-sen to produce each factor in the Butterworth poly-nomial will produce an even-order Butterworth low-pass filter Adding a prototype low-pass op amp filter will pro-duce an odd-order Butterworth low-pass filter (See page 581.)

• A cascade of second-order high-pass op amp filters (Fig 15.25) with 1 F capacitors and resistor values cho-sen to produce each factor in the Butterworth poly-nomial will produce an even-order Butterworth high-pass filter Adding a prototype high-pass op amp filter will produce an odd-order Butterworth high-pass filter (See page 585.)

• For both high- and low-pass Butterworth filters, fre-quency and magnitude scaling can be used to shift the cutoff frequency from 1 rad/s and to include realistic

component values in the design Cascading an inverting

amplifier will produce a nonunity passband gain (See

page 580.)

Butterworth low-pass and high-pass filters can be

cas-caded to produce Butterworth bandpass filters of any

order n Butterworth low-pass and high-pass filters can

be combined in parallel with a summing amplifier to

produce a Butterworth bandreject filter of any order n

(See page 585.)

If a high-(2, or narrowband, bandpass, or bandreject fil-ter is needed, the cascade or parallel combination will not work Instead, the circuits shown in Figs 15.26 and 15.29 are used with the appropriate design equations Typically, capacitor values are chosen from those com-mercially available, and the design equations are used

to specify the resistor values (See page 586.)

Problems

Section 15.1

15.1 Find the transfer function V 0/V; for the circuit

shown in Fig P15.1 if Zf is the equivalent

imped-ance of the feedback circuit, Z, is the equivalent

impedance of the input circuit, and the operational

amplifier is ideal

Figure PI5.1

15.2 a) Use the results of Problem 15.1 to find the

trans-fer function of the circuit shown in Fig PI5.2

b) What is the gain of the circuit a s w ^ 0?

c) What is the gain of the circuit as w —• oo?

d) Do your answers to (b) and (c) make sense in

terms of known circuit behavior?

Figure P15.2

15.3 Repeat Problem 15.2, using the circuit shown in

Fig P15.3

Figure P15.3

DESIGN PROBLEM

15.4 Design an op amp-based low-pass filter with a

cut-off frequency of 2500 Hz and a passband gain of 5 using a 10 nF capacitor

a) Draw your circuit, labeling the component val-ues and output voltage

b) If the value of the feedback resistor in the filter

is changed but the value of the resistor in the forward path is unchanged, what characteristic

of the filter is changed?

15.5 The input to the low-pass filter designed in

Problem 15.4 is 2 cos cot V

a) Suppose the power supplies are ±V CC What is the smallest value of V cc that will still cause the

op amp to operate in its linear region?

b) Find the output voltage when o> = a) c

c) Find the output voltage when w = 0.2(o c

d) Find the output voltage when oo = 5wt

15.6 a) Using the circuit in Fig 15.1, design a low-pass

filter with a passband gain of 10 dB and a cutoff frequency of 1 kHz Assume a 750 nF capacitor

is available

b) Draw the circuit diagram and label all components