Electric Circuits, 9th Edition P61 docx

Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency by starting with a fourth-order cascade consisting of prototype low-pass filters and then scaling t

Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency by starting with a fourth-order cascade consisting of prototype

low-pass filters and then scaling the components by kt = wc/0.435 to

place the cutoff frequency at any value of w c desired

Note that we can build a higher order low-pass filter with a nonunity gain by adding an inverting amplifier circuit to the cascade Example 15.7 illustrates the design of a fourth-order low-pass filter with nonunity gain

Example 15.7 Designing a Fourth-Order Low-Pass Op Amp Filter

Design a fourth-order low-pass filter with a cutoff

frequency of 500 Hz and a passband gain of 10 Use

1 fxF capacitors Sketch the Bode magnitude plot

for this filter

Finally, add an inverting amplifier stage with a gain

of Rf/Ri = 10 As usual, we can arbitrarily select one

of the two resistor values Because we are already using 138.46 11 resistors, let /¾ = 138.46 ft; then,

Solution

We begin our design with a cascade of four

proto-type low-pass filters We have already used

Eq 15.23 to calculate the cutoff frequency for the

resulting fourth-order low-pass filter as 0.435 rad/s

A frequency scale factor of kf = 7222.39 will scale

the component values to give a 500 Hz cutoff

fre-quency A magnitude scale factor of k, n = 138.46

permits the use of 1 (JLF capacitors The scaled

com-ponent values are thus

R = 138.46 0 ; C = l/xF

R f = 10/¾ 1384.6 tt

The circuit for this cascaded the fourth-order low-pass filter is shown in Fig 15.18 It has the transfer function

H(s) = - 1 0 7222.39

s + 7222.39

The Bode magnitude plot for this transfer func-tion is sketched in Fig 15.19

Figure 15.18 A The cascade circuit for the fourth-order low-pass filter designed in Example 15.7

30

20

PQ

—

10

-20

-30

\

\

\

10 50 100 500 1000

/(Hz)

500010,000

Figure 15.19 A The Bode magnitude plot for the fourth-order low-pass

fitter designed in Example 15.7

By cascading identical low-pass filters, we can increase the asymptotic

slope in the transition and control the location of the cutoff frequency, but

our approach has a serious shortcoming: The gain of the filter is not

con-stant between zero and the cutoff frequency oo c Remember that in an

ideal low-pass filter, the passband magnitude is 1 for all frequencies below

the cutoff frequency But in Fig 15.16, we see that the magnitude is less

than 1 (0 dB) for frequencies much less than the cutoff frequency

This nonideal passband behavior is best understood by looking at the

magnitude of the transfer function for a unity-gain low-pass nth-order

cas-cade Because

H(s)

the magnitude is given by

1//0)1 =

(* + "«,)"'

co cn Va? + (x)l r

\/(co/a> cn ) 2 + 1

(15.24)

As we can see from Eq 15.24, when (o <5C (o a „ the denominator is

approximately 1, and the magnitude of the transfer function is also nearly 1

But as a) —* (o cn , the denominator becomes larger than 1, so the magnitude

becomes smaller than 1 Because the cascade of low-pass filters results in

this nonideal behavior in the passband, other approaches are taken in the

design of higher order filters One such approach is examined next

A unity-gain Butterworth low-pass filter has a transfer function whose

magnitude is given by

\H(ja>)\ = , =====, (15.25)

V l + (co/coc ) 2 "

where n is an integer that denotes the order of the filter.1

When studying Eq 15.25, note the following:

1 The cutoff frequency is o) c rad/s for all values of n

2 If n is large enough, the denominator is always close to unity when

a) < co c

3 In the expression for | H(ja>)\, the exponent of co/co c is always even

This last observation is important, because an even exponent is required

for a physically realizable circuit (see Problem 15.26)

Given an equation for the magnitude of the transfer function, how do

we find H(s)? The derivation for H(s) is greatly simplified by using a

proto-type filter.Therefore, we set o) c equal to 1 rad/s in Eq 15.25 As before, we

will use scaling to transform the prototype filter to a filter that meets the

given filtering specifications

To find H(s), first note that if N is a complex quantity, then

\N\ 2 = NN*, where N* is the conjugate of N It follows that

But because s = jco, we can write

Now observe that s 2 = — a? Thus,

\H(j<o)\ 2 =

1 +

1 +

1 +

a> 2 "

1

(<o 2 y

1

is2)"

1

1 + ( - l ) V '

or

H(s)H(-s) = , , , - ^ (15.28)

1 + (-l)'V

The procedure for finding H(s) for a given value of n is as follows:

1 Find the roots of the polynomial

1 + ( - l ) \ v2* = 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

Example 15.8 illustrates this process

1 This filter was developed by the British engineer S Butterworth and reported in Wireless

Engineering 7 (1930): 536-541

15.4 Higher Order Op Amp Filters 579

Example 15.8 Calculating Butterworth Transfer Functions

Find the Butterworth transfer functions for n = 2

and n = 3

Solution

For n = 2, we find the roots of the polynomial

1 + ( - l ) V = 0

Rearranging terms, we find

s4 = - 1 = 1/180°

Therefore, the four roots are

*i = 1 / 4 T = 1/V5 + j/V2,

s 2 = 1/135° = - 1 / V 2 + j/V2,

s 3 = 1/225° = - 1 / V 2 + - / / V 2 ,

s 4 = 1/315° = 1/V2 + - / / V 2

Roots s 2 and y3 are in the left-half plane Thus,

H{$) = (s + 1/V2 - j/V2)(s + 1/V5 + ; / V 5 )

_ 1

" (52 + V5s + 1)'

For n = 3, we find the roots of the polynomial

l + ( - i ) V = o

Rearranging terms,

s 6 = 1/ry = 1/360°

Therefore, the six roots are

s t = l / o : = l,

.v2 = 1/60° = 1/2 + / V 5 / 2 ,

53 = 1/120° = - 1 / 2 + / V 3 / 2 ,

s 4 = 1/180° = - 1 + / 0 ,

.v5 = 1/240° = - 1 / 2 + - / V 3 / 2 ,

.vA = 1/300° = 1/2 + - ; ' V 3 / 2 Roots S3, \4, and s$ are in the left-half plane Thus,

H ^ ~ (s + 1)(5 + 1/2 - j\/3/2)(s + 1/2 + / V 3 / 2 )

1

" Cv+l)(.v2 + v+ 1)'

We note in passing that the roots of the Butterworth polynomial are always equally spaced

around the unit circle in the s plane To assist in the

design of Butterworth filters, Table 15.1 lists the

Butterworth polynomials up to n = 8

TABLE 15.1 Normalized (so that co c = 1 rad/s) Butterworth Polynomials up to the Eighth Order

/i wth-Order Butterworth Polynomial

1 (s + 1)

2 ( r + V2.v + 1)

7 (5 + 1)(52 + 0.4455 + 1)(52 + 1.2475 + 1)(52 + 1.8025 + 1)

8 (52 + 0.3905 + 1)(52 + 1.111s + l)(s2 + 1.6663s + l)(s2 + 1.962s + 1)

Butterworth Filter Circuits

Now that we know how to specify the transfer function for a Butterworth

filter circuit (either by calculating the poles of the transfer function directly

or by using Table 15.1), we turn to the problem of designing a circuit with

Figure 15.21 A A circuit that provides the second-order

transfer function for the Butterworth filter cascade

such a transfer function Notice the form of the Butterworth polynomials

in Table 15.1 They are the product of first- and second-order factors;

therefore, we can construct a circuit whose transfer function has a Butterworth polynomial in its denominator by cascading op amp circuits, each of which provides one of the needed factors A block diagram of such

a cascade is shown in Fig 15.20, using a fifth-order Butterworth poly-nomial as an example

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

so all odd-order Butterworth filter circuits must have a subcircuit that

provides the transfer function H(s) = 1/(5 + 1) This is the transfer

function of the prototype low-pass op amp filter from Fig 15.1 So what remains is to find a circuit that provides a transfer function of the form

H(s) = l/(s 2 +bis + 1)

Such a circuit is shown in Fig 15.21 The analysis of this circuit begins

by writing the s-domain nodal equations at the noninverting terminal of

the op amp and at the node labeled V a :

y - y V - V

+ (V a - V 0 )sC x + ^ ^ = 0,

R

V 0 sC 2 +

R

Vn - V*

(15.29)

(15.30)

Simplifying Eqs 15.29 and 15.30 yields

(2 + RC h s)V a - ( 1 + RC lS )V 0 = V, , (15.31)

Using Cramer's rule with Eqs 15.31 and 15.32, we solve for V a :

2+RC^s V,

- 1 0

V;

i?2C,C252 + 2RC 2 s + 1 (15.33)

Then, rearrange Eq 15.33 to write the transfer function for the circuit in Fig 15.21:

*<'> £

1

R 2 CiC 2

s L + s +

(15.34)

1

s 2 + 0 6 1 8 5 + 1

1

5 2 + 1.6185+ 1

Figure 15.20 A A cascade of first- and second-order circuits with the indicated transfer

functions yielding a fifth-order low-pass Butterworth filter with o) = 1 rad/s

15.4 Higher Order Op Amp Filters 5 8 1

Finally, set R = 1 fi in Eq 15.34; then

r + —s + C\ C\C

2

Note that Eq 15.35 has the form required for the second-order circuit

in the Butterworth cascade In other words, to get a transfer function of

the form

s2 + bh<> + 1

we use the circuit in Fig 15.21 and choose capacitor values so that

Z?i = —- and 1 =

(15.36)

We have thus outlined the procedure for designing an /ith-order

Butterworth low-pass filter circuit with a cutoff frequency of wt = 1 rad/s

and a gain of 1 in the passband We can use frequency scaling to calculate

revised capacitor values that yield any other cutoff frequency, and we can

use magnitude scaling to provide more realistic or practical component

values in our design We can cascade an inverting amplifier circuit to

pro-vide a gain other than 1 in the passband

Example 15.9 illustrates this design process

Example 15.9 Designing a Fourth-Order Low-Pass Butterworth Filter

Design a fourth-order Butterworth low-pass filter

with a cutoff frequency of 500 Hz and a passband

gain of 10 Use as many 1 k£l resistors as possible

Compare the Bode magnitude plot for this

Butterworth filter with that of the identical cascade

filter in Example 15.7

Solution

From Table 15.1, we find that the fourth-order

Butterworth polynomial is

( r + 0.7655 + 1)(52 + 1.848^ + 1)

We will thus need a cascade of two second-order

fil-ters to yield the fourth-order transfer function plus

an inverting amplifier circuit for the passband gain

of 10 The circuit is shown in Fig 15.22

Let the first stage of the cascade implement

the transfer function for the polynomial

(s 2 + 0.765s + 1) From Eq 15.36,

Cj = 2.61 F,

C2 = 0.38 F

Let the second stage of the cascade implement the transfer function for the polynomial

(s 2 + 1.8485 + 1) From Eq 15.36,

C3 = 1.08 F,

C4 = 0.924 F

The preceding values for C^, C2, C3, and C4 yield a fourth-order Butterworth filter with a cutoff frequency of 1 rad/s A frequency scale factor of

kf = 3141.6 will move the cutoff frequency to

500 Hz A magnitude scale factor of k m = 1000 will permit the use of 1 kft resistors in place of 1 ft resistors The resulting scaled component values are

R = 1 kft,

C, = 831 nF,

C2 = 121 nF,

C3 = 344 nF,

C, = 294 nF

Finally, we need to specify the resistor values in the

inverting amplifier stage to yield a passband gain

of 10 Let 7?, = 1 kil; then

R MR, = 10 k O

Figure 15.23 compares the magnitude

responses of the fourth-order identical cascade

fil-ter from Example 15.7 and the Butfil-terworth filfil-ter

we just designed Note that both filters provide a passband gain of 10 (20 dB) and a cutoff frequency

of 500 Hz, but the Butterworth filter is closer to an ideal low-pass filter due to its flatter passband and steeper rolloff at the cutoff frequency Thus, the Butterworth design is preferred over the identical cascade design

Figure 15.22 • A fourth-order Butterworth filter with non-unity gain

30

20

10

P2

Bu tei

!

worth

\ \

\ \

;de :as<

\

\

ntical :ade

\

\

-10

-20

-30

10 50 100 500 1000 500010,000

/(Hz)

Figure 15.23 • A comparison of the magnitude responses for a

fourth-order low-pass filter using the identical cascade and Butterworth designs

The Order of a Butterworth Filter

It should be apparent at this point that the higher the order of the Butterworth filter, the closer the magnitude characteristic comes to that of

an ideal low-pass filter In other words, as n increases, the magnitude stays

close to unity in the passband, the transition band narrows, and the magni-tude stays close to zero in the stopband At the same time, as the order increases, the number of circuit components increases It follows then that

15.4 Higher Order Op Amp Filters 583

a fundamental problem in the design of a filter is to determine the

small-est value of n that will meet the filtering specifications

In the design of a low-pass filter, the filtering specifications are usually

given in terms of the abruptness of the transition region, as shown in

Fig 15.24 Once A p , co p , A x , and &>s are specified, the order of the

Butterworth filter can be determined

For the Butterworth filter,

20 log,,, 1

V l + co p "

= -101og1 0(l + 4 " ) ,

A s = 20 logio"

(15.37)

V l + cuj"

It follows from the definition of the logarithm that

Now we solve for <a np and to" and then form the ratio (toj(o p )" We gel

o)sV Vl0"°-M« - 1

1

(T,

where the symbols a s and a p have been introduced for convenience

From Eq 15.41 we can write

n logu,(ws/o;p) = logH)(«-s/crp),

or

n =

JoginWa>) logm(a>5/&)„) (15.42)

We can simplify Eq 15.42 if o) p is the cutoff frequency, because then A p

equals - 2 0 log]()V2, and <r p = 1 Hence

login o\

One further simplification is possible We are using a Butterworth

fil-ter to achieve a steep transition region Therefore, the filfil-tering

specifica-tion will make 10~aM* » l.Thus

Therefore, a good approximation for the calculation of n is

-0.05/4,

Note that w5/w/} = f s /f p - so we can work with either radians per second or

hertz to calculate n

\H(ja>)\ dB

Stop band

Figure 15.24 • Defining the transition region for a

low-pass filter

The order of the filter must be an integer; hence, in using either

Eq 15.42 or Eq 15.46, we must select the nearest integer value greater than the result given by the equation The following examples illustrate the usefulness of Eqs 15.42 and 15.46

a) Determine the order of a Butterworth filter that

has a cutoff frequency of 1000 Hz and a gain of

no more than - 5 0 dB at 6000 Hz

b) What is the actual gain in dB at 6000 Hz?

Solution

a) Because the cutoff frequency is given, we know

a v = 1 We also note from the specification that

10 0 1 ( 50) is much greater than 1 Hence, we can use Eq 15.46 with confidence:

(-0.05)(-50) log10(6000/1000) = 3.21

Therefore, we need a fourth-order Butterworth filter

b) We can use Eq 15.25 to calculate the actual gain

at 6000 Hz The gain in decibels will be

K = 20 log ID 1

V l + 6* -62.25 dB

a) 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 less than

the passband magnitude at 5000 Hz

b) Determine the cutoff frequency of the filter

(in hertz)

c) What is the actual gain of the filter (in decibels)

at 5000 Hz?

Solution

a) Because the cutoff frequency is not given, we use

Eq 15.42 to determine the order of the filter:

<rp = Vl0-a,<-10> - 1 = 3,

= \/1(ro.i(-60) - i ps loot),

ojtop = fs/fp = 5000/500 = 10,

log10(1000/3)

* " log10(10) " 2 > 5 2'

Therefore we need a third-order Butterworth filter to meet the specifications

b) Knowing that the gain at 500 Hz is — 10 dB, we can determine the cutoff frequency From Eq 15.25 we can write

-101og10[l + (co/co c f) = - 1 0 ,

where u> = 10007T rad/s Therefore

1 + (w/o>c)6 = 10,

and

0)

= 2178.26 rad/s

It follows that

f c = 346.68 Hz

c) The actual gain of the filter at 5000 Hz is

K = - 1 0 log10[l + (5000/346.68)6]

= -69.54 dB

15.4 Higher Order Op Amp Filters 585

Butterworth High-Pass, Bandpass, and Bandreject Filters

An nth-order Butterworth high-pass filter has a transfer function with the

nth-order Butterworth polynomial in the denominator, just like the nth-order

Butterworth low-pass filter But in the high-pass filter, the numerator of the

transfer function is s n, whereas in the low-pass filter, the numerator is 1

Again, we use a cascade approach in designing the Butterworth high-pass

filter The first-order factor is achieved by including a prototype high-pass

filter (Fig 15.4, with R { = R 2 = 1 XI, and C = 1 F) in the cascade

To produce the second-order factors in the Butterworth polynomial,

we need a circuit with a transfer function of the form

H(s) = s 2

s 2 + bis + I

Such a circuit is shown in Fig 15.25

This circuit has the transfer function

H(s) = - f = - — — (15.47)

s 2 + ^—:s •+ R 2 C R X R 2 C 2

Setting C = 1 F yields

.v2

H(s) = — (15.48)

s 2 + —s +

Thus, we can realize any second-order factor in a Butterworth polynomial

of the form (s 2 + b x s + 1) by including in the cascade the second-order

circuit in Fig 15.25 with resistor values that satisfy Eq 15.49:

At this point, we pause to make a couple of observations relative to

Figs 15.21 and 15.25 and their prototype transfer functions

l/(s 2 + b x s + 1) and s 2 /(s 2 + b^s + 1) These observations are

impor-tant because they are true in general First, the high-pass circuit in Fig

15.25 was obtained from the low-pass circuit in Fig 15.21 by interchanging

resistors and capacitors Second, the prototype transfer function of a

high-pass filter can be obtained from that of a low-high-pass filter by replacing s in

the low-pass expression with \/s (see Problem 15.48)

We can use frequency and magnitude scaling to design a Butterworth

high-pass filter with practical component values and a cutoff frequency

other than 1 rad/s Adding an inverting amplifier to the cascade will

accommodate designs with nonunity passband gains The problems at the

end of the chapter include several Butterworth high-pass filter designs

Now that we can design both nth-order low-pass and high-pass

Butterworth filters with arbitrary cutoff frequencies and passband gains,

we can combine these filters in cascade (as we did in Section 15.3) to

pro-duce nth-order Butterworth bandpass filters We can combine these filters

in parallel with a summing amplifier (again, as we did in Section 15.3) to

produce nth-order Butterworth bandreject filters This chapter's problems

also include Butterworth bandpass and bandreject filter designs

Figure 15.25 A A second-order Butterworth high-pass filter circuit