Filters 10

8/8/2019 Filters 10

1/32

B. Eng. Modular Degree Analogue and Instrumentation UG3

ACTIVE FILTERS

1 Introduction

Filters are widely used in Electronics and Communications. Passive filters make use ofpassive devices only (capacitors, inductors and resistors) and are difficult to design. Activefilters use active devices (usually op amps) in addition to the passive components but withoutinductors. Design is comparatively straight forward.

P a s s b a n d P a s s b a n d

S t o p b a n d

S t o p b a n d T

r a

n

s m

i s s

i o

n

T r a

n

s m

i s s

i o

n

f r e q u e n c y f r e q u

Fig. 1.1 Fig. 1.2

Ideal transmission characteristics for low-pass and high-pass filters are shown in Figs 1.1& 1.2. The ideal characteristics for the band-pass filter are shown below in Fig. 1.3.

P a s s b a n d

S t o p b a n d S t o p b a n d

Tra

n

sm

iss

io

n

f r e q u

U p p e rL o w e r

Fig. 1.3

Real filters do not have such brick-wall like characteristics. A real low-pass filtercharacteristic is shown below in Fig. 1.4. We note the following important differencesbetween Fig. 1.1 and Fig. 1.4: transmission in the pass-band is not constant, there is atransition band between the pass-band and the stop-band, the transmission in the stopband is not zero but below some minimum value. The rippling in the pass-band is acharacteristic of the Chebyshev filters. It does not occur in all filters. Similarly, the ripplingin the stop-band occurs in some filters only.

Simple first order (or single-pole) active filters are shown in Fig. 1.5 (low-pass) and Fig. 1.6(high-pass) below. The circuits consist of a passive first order filter consisting of R and C,

followed by a non-inverting amplifier with gain of (1 + R2/R1).

Nick Holden BCU Ref.: lect2010\analogUG

8/8/2019 Filters 10

2/32

AI UG3 Active Filters Version 6.5 12/10/2010

A m a x

A m i n

P a s s b a n d S t o p b a n d

T r a n -s i t i o n -b a n d

T d B

0

0

sp

Fig. 1.4

Fig. 1.5 Fig. 1.6

Note that it is easy to remember which is which. The capacitor C in Fig. 1.6 does not pass

a signal at DC, so Fig. 1.6 must be the high pass filter. In both circuits the components Rand C fix the 3dB cut-off of the filter. The roll-off rate in these filters is only 20dB/decade(or 6dB/octave). If a faster roll-off rate is required then more complex filters must be used.

The transfer function for the low-pass filter is given by:

=

=

+ j

AjVin

VoutjTo

oVO

)(

where o = 1 / CR and AVO = (1 + R2/R1) is the non-inverter opamp gain.We note that when = 0 or if >o T approaches (1 + R2/R1).

=

=

+

oj

jAj

Vin

VoutjTVO

)(

Nick Holden BCU Ref.: lect2010\analogUG3 2

8/8/2019 Filters 10

3/32


Second order or two pole active filters are shown in Figs 1.7 and 1.8. They are often calledSallen-Key filters after their inventors.


8/8/2019 Filters 10

4/32


Both these have two capacitors. The roll-off rate is now 40dB/decade (12dB/octave) andthe 3dB cut-off angular frequency, o, is determined by the CR product. R1 and R2 may notnormally be chosen at random to fix the filter gain (1 + R2/R1) as this becomes animportant parameter in higher order filter design.

Fig. 1.7 Fig. 1.8

Again, Fig. 1.7 is the low-pass and Fig. 1.8 the high-pass using the simple reasoning used

before. The transfer function for the low-pass filter is:

( )

++==

Aj

AjVin

Vou tjT

oo

o

VO

VO

3)(

2

2

where o = 1 / CR and AVO = (1 + R2/R1).We note that when = 0 or if > o the transmission reduces to zero. The high-pass filter transferfunction is similar except that j and o are interchanged:

( )

++==

Aj

jAjVin

Vou tjT

oo VO

VO

3

)()(

2

2

The second order band pass filter is a mixture of the above two:

( )

++==

Aj

jAjVin

Vou tjT

oo

o

VO

VO

3

)()(2

This is often written as:

( )

++

==

jQ

j

jKj

Vin

VoutjT

oo 2

)()(

2

2


8/8/2019 Filters 10

5/32


Very often a roll-off rate much bigger than that produced by a second order filter (for >>o it is 40dB/decade for the low-pass) is required. Then higher order filters will have to beused. Generally, for low-pass and high-pass filter the roll-off rate is given by 20*NdB/decade well out in the stop band, where N is the filter order. These can be obtained bycascading first and second order filters. For example, a fifth order filter can be produced bycascading two second order filters and one first order filter. This is discussed under inmore detail under Butterworth and Chebyshev filters.


8/8/2019 Filters 10

6/32


2 Filter Transmission and Specification

If the transfer function T(j) = Vout(j)/Vin(j) is known, then the output can be calculatedfrom the input. In addition the function and/or attenuation can also be obtained:

( ) ( ) ( )jVinjTjVo =The gain function G()is defined by:( ) ( )jTG log1020 =Similarly, the attenuation function A() is given by:

( )jTA log 1020 =

To design a filter to a set of specifications we will have a set of transmissioncharacteristics. Fig. 1.4 shows realistic characteristics for a low-pass filter. Note that as nofilter has the ideal transmission of 0dB throughout the pass-band, so an upper bound Amax

(dB) is placed on the deviation. Amax

will vary depending on the application from a fractionof a dB to 3dB. Similarly, the transmission in the stop-band cannot be zero but it isrequired that the attenuation be at least Amin (dB). Values for Amin range from 20dB to about100dB. The transition band extends from the pass-band edge, p, to the stop-band edge,s. So, normally, the following will have to be specified:

(i) the pass-band edge, p;(ii) the maximum allowed deviation from 0dB in the pass-band, Amax;(iii) the stop-band edge, s; and(iv) the minimum stop-band attenuation, Amin.

The more closely the specifications approach the idea (brick-wall) response the higher

the order and the more complex the design.

3 Butterworth Filters

The above name is that of a mathematician who worked out the functions on which thisclass of filters is based. In the Butterworth low-pass filter the transmission reduces withfrequency; see Fig. 3.1. The magnitude function for an Nth-order low-pass Butterworthfilter with pass-band edge p is given by:

( ) ( )

p

NjT 221

1

+=

At = p this reduces to:

( )

21

1

+=

pjT

and the parameter determines Amax according to:

210max 120 log +=AConversely, given Amax, the value of may be found from:

11010

max = A Nick Holden BCU Ref.: lect2010\analogUG3 6

8/8/2019 Filters 10

7/32



8/8/2019 Filters 10

8/32


At the stop-band edge, s, the attenuation is given by:

( ) ( )

( )

+=

+=

ps

psA

N

N

s

22

10

22

10

110

1/120

log

log

We use this equation to find the filter order N which is the lowest INTEGER for which A( s) Amin.

0

T

1

p

211

+

Fig. 3.1

Fig. 3.2 shows the Butterworth response for first, second, fourth and eighth order filters for = 1. We note that for large N the response approaches the brick-wall response.

T

0p/

0 . 2

0 . 4

0 . 6

0 . 8

1 . 0

0 . 4 0 . 8 1 . 2 1 . 6 2 . 0

2

48

N = 1

Fig. 3.2

The poles of an Nth-order Butterworth filter may be determined from the construction inFig. 3.3.


8/8/2019 Filters 10

9/32


The poles lie on a circle of radius p(1/)1/N and are separated by /N but the first pole isat angle /2N from the j axis. The natural modes will all have frequency o = p(1/)1/N.Note that the poles occur in complex conjugate pairs except in the case of an odd-orderfilter where one of the poles will be real (corresponding to the first order filter).

j

N/

N

N

/

/ N/2

p

p

p

p

p

1

11

2

3

4

0

( )_

s - p l

Fig. 3.3

The transfer function will be of the type:

=pNspsps

NoKsT..........21

)(

where K is the dc gain of the filter. We will see that this can be implemented in terms offirst and second order filters.

EXAMPLE: Design a Butterworth low-pass filter that meets the following specifications: fp= 10kHz, Amax = 1dB, fs = 15kHz, Amin = 25dB and dc gain K = 10.

SOLUTION: Substituting Amax = 1dB into the equation on page 4 for , we get:

5088.012589.1110 101 = ==

The equation for A( s) on page 5 is then used to calculate the value of N. A( s) = 25 dB,so we have:

( )

+= ps

N22

1011025 log

But as 102.5 = 316.23 we have 2( s/ p)2N = 315.23. Substituting for , s and p wehave: (1.5)2N = 1217.68. So taking logs to base 10 we get: 2N*0.1761= 3.0855. Fromwhich N = 8.76. So we will have to use N = 9.

The Poles all have the same frequency given by:o = p(1/)1/N = 2 *10*103*(1/0.5088)1/9 = 6.773*104 rads/s.

Bearing in mind that /18 = 10and /9 = 20and that the poles are as shown in Fig. 3.3we have for the first pole:P1 = o(-cos80+ jsin80)= o(-0.1736 + j0.9848)


8/8/2019 Filters 10

10/32


And p9 will be the conjugate of p1, i.e.P9 = o(-0.1736 - j0.9848)

In the transfer function denominator, these two poles will be of the type:(s p1)(s p9) = (s + - j )(s + + j ) = (s + )2 + 2Note that and are BOTH positive, so that

(sp1)(sp9)= s2

+2 s+(2

+2

) = s2

+2*0.1736*s o+(0.17362

+0.98482

) o2

or, as (0.17362+0.98482) = 1:

(s p1)(s p9)= s2 + 0.3472s o + o2

The other conjugate pole-pairs may be shown to give:(s p2)(s p8)= s2 + s o + o2

(s p3)(s p7)= s2 + 1.5321s o + o2

(s p4)(s p6)= s2 + 1.8794s o + o2

and the real pole gives: (s p5) = (s + o).

So the complete transfer function will be:T(s) = 10 o9/[(s + o)(s2+0.3472s o+ o2)(s2+s o+ o2)*

*(s2+1.5321s o+ o2)(s2+1.8794s o+ o2)]

4 Design Implementation Using Sallen-Key Filters

Implementation is straight forward. The first term in the transfer function (T.F.)

denominator represents the first order filter with T.F. as on p.2, i.e.:

=

=

+ j

AjVin

VoutjTo

oVO

)(

For this one filter the gain AVO = (1 + R2/R1) may be chosen to suit the overall dc gain of theentire filter. So leaving the choice of R2 and R1 for the moment, we choose suitable valuesfor R and C from o = 1 / RC = 6.773*104 rads/s. Choosing a simple value for thecapacitor C = 1nF we get for R:

=

== koC

R 765.1410 4*773.6*10 9*1

11This value is suitable but C = 10nF and R = 1.4765k would also be O.K.

The above values will also be used for ALL the second order sections. It is important tonote that in the second order section the opamp gain A VO = (1 + R2/R1) is used to obtainthe multipliers of s o. Comparing the first second order term in the T.F.(s2+0.3472s o+ o2) with the T.F. of the filter:


8/8/2019 Filters 10

11/32


( )

++==

ssA

As

Vi n

Vo u tsT

oo

o

V O

V O

22

2

3)(

we find that for the first second order section we must have:

RRRAVO 2121333472.0 =+==

Hence: R2/R1 = 2 0.3472 = 1.6528. So we could use, for example, R1 = 10k and R2 =16.53k . We should note that the dc gain of this section will be AVO = (1 + R2/R1) = 2.653.

In exactly the same way, the values of the feedback resistors R2 and R1 can be calculatedfor the other THREE second-order sections. For the second one (i.e. U3) we have: R2/R1 =

2 1 = 1. So we could use, for example, R 1 = 10k

and R2 = 10k

. We should note thatthe dc gain of this section will be AVO = (1 + R2/R1) = 2.

The third one (U4) will be: R2/R1 = 2 1.5321 = 0.4679. So we could use, for example, R 1= 10k and R2 = 4.68k . We should note that the dc gain of this section will be AVO = (1 +R2/R1) = 1.468.

And finally for U5 R2/R1 = 2 1.8794 = 0.1206. So we could use, for example, R1 = 10kand R2 = 1.2k . We should note that the dc gain of this section will be A VO = (1 + R2/R1) =1.12.

So the gains of all four second order sections will be:1.12 * 1.468 * 2 * 2.653 = 8.724So for a dc gain of 10 the first order section will have to have a gain of 10/8.724 = 1.146 =(1 + R2/R1). Hence R2/R1 = 1.146 1 = 0.146. So we could again use: R1 = 10k and R2 =1.46k .

That completes the design for the ninth order Butterworth. It is shown below in Fig. 4.1.


8/8/2019 Filters 10

12/32


U 1

U 2

U 3

U 4U 5

Fig. 4.1.

Fig. 4.2

The frequency response of the filter is shown above in Fig. 4.2. We note that 19.998dB

corresponds to 9.9977 and that the specification of the filter is tighter than required. Thevalue for fp is very slightly larger and fs smaller than specified. This is a consequence ofmaking N = 9, rather than the theoretically required value of 8.76.


8/8/2019 Filters 10

13/32


5 Chebyshev Filters

Again named after a mathematician, these filters have the property that the transmissionripples in the pass-band see Fig. 5.1. The roll-off rate is much faster in the transitionband than for a Butterworth filter of the same order.

The magnitude of the transfer function for an Nth order low-pass Chebyshev filter withpass-band edge p is given by:

( )

forN

jT

p+

=

))/(cos(cos 221

1

1

and

( ) for

NjT

p+=

))/(cosh(cosh 221

1

1

At = p this reduces to:

( )

21

1

+=PjT

The parameter determines Amax, the pass-band ripple, according to:

210max

110

log+=

AConversely, given Amax, the value of may be found from:

11010

max = AAt the stop-band edge, s, the attenuation is given by:

( )

+= cosh 1[cosh 221log 1010 psNsAThis equation can be used with a calculator to determine the value of N to give a specifiedAmin by finding the lowest integer value of N for which A(s) Amin.

The transmission characteristics for a fourth and fifth order Chebyshev low-pass filter areshown below in Fig. 5.1.


8/8/2019 Filters 10

14/32


0 0pp

T

1

211

+

T

1

211

+

N = 4 N = 5

Fig. 5.1

The poles of the Chebyshev filter are given by:

.......,,3,2,11

sinh1

cosh2

12cos

1sinh

1sinh

2

12sin

1

1

NkforNN

k

NN

k

j

p

p

pk

=

+

=

+

The transfer function for the Chebyshev low-pass filter may be written as:

= pNspsps

NpK

sT N ..........212)( 1

where K is again the dc gain of the filter.

The design procedure is as for the Butterworth:(i) Find the value of required;(ii) Determine the value of N;(iii) Find the poles using the equation above;(iv) Determine the transfer function;(v) Implement design using Sallen-Key filters.

The Chebyshev provides a faster roll-off for a particular order than the Butterworth. Thedrawback is the ripple in the pass-band, which cannot be tolerated for some applications.

EXAMPLE: Design a Chebyshev low-pass filter that meets the specifications for theButterworth example, i.e.: fp = 10kHz, Amax = 1dB, fs = 15kHz, Amin = 25dB and dc gain K =10.

SOLUTION: Substituting Amax = 1dB into the equation on p10 for , we get:5088.012589.1110 101 = ==


8/8/2019 Filters 10

15/32


The equation for A( s) on p10 is then used to calculate the value of N. A( s) = 25 dB, sowe have:

(

+= cosh 1[cosh 221log 1010 sNsA

But as 102.5

= 316.23 we have 2

cosh2

[N cosh-1

( s/ p)] = 315.23. Substituting for , sand p, we find that for N=4.41. So we use N = 5, which gives A( s) = 29.9 dB. (N.B. aButterworth of order 9 was required for the same specification!)

The poles may be obtained from the formula as:

P1, P5 = p(-0.0895 j0.9901)P2, P4 = p(-0.2342 j0.6119)

P3 = p(-0.2895)

Using the same procedure as for the Butterworth filter, we get the following transferfunction:

T(s) = 10 p5/[8.1408(s + 0.2895 p)**(s2+0.4684s p+0.4293 p2)(s2+0.1789s p+0.9883 p2)]

where p = 2 *104 rads/s.


8/8/2019 Filters 10

16/32


Implementation in terms of Sallen-Key filters is left as an exercise. It should however benoted that neither the first order nor the second order denominator terms are in thestandard forms.

The first step is to reorganize them into the standard forms. For the first order one we will

have: (s +

o).Similarly, for the second order terms: [s2 + (3-Avo)s o + o2]. o and Avo will have to becalculated individually for each stage.

One of the consequences of this is that, unlike for the Butterworth, the 3dB cut-off o [=1/CR] is likely to be different for all the individual sections of the Chebyshev.

Fig. 5.2

N.B. Transmission at 10kHz is 1.007dB and at 15kHz it is 29.9dB. Again, as with theButterworth, the result corresponds to tighter specs than required. Note that the roll off rateof the Chebyshev fifth order is faster than that of the Butterworth ninth order!!


8/8/2019 Filters 10

17/32


6 Building Butterworth Filters from Butterworth Polynomials

6.1 Low Pass (with 0 = 1)

If a filter of a particular order is required the polynomials given below are the quickest wayof building them. They are given in terms of factors and in normalised form with p = 1.The table of polynomials is normalised to p = 1. This means that in the table, thequadratic terms (e.g. for N=2) should really read not (s2 + 1.4142s + 1) but (s2 +1.4142sp + p2). In addition, for the Butterworth filter with = 1, p = o. So the Transferfunction for a first order low-pass filter is:

=

=

+

sAs

V i n

Vo u tsT

o

o

V O)(1

where o = 1/CR. Sensible values of R should be used (it represents the input resistanceof the filter at high frequencies) and C calculated from o.

Similarly, the transfer function for a second order low-pass filter will be:

( )

++==

22

2

23

)(oo

o

sAs

As

Vi n

Vo u tsTV O

V O

where the (3 AVO) is the number given in the table below. So for the second orderButterworth low-pass filter (3 AVO) = 1.4142 and for the first second order section of theseventh order Butterworth filter (3 AVO) = 0.44505. We should note that in the case of thesecond order sections AVO or (1 + R2 / R1) is fixed because (3 AVO) is fixed. Whereas inthe case of the first order sections AVO may be chosen at will. We should however bear inmind that the 3dB cut-off for the first order section fh will be given by:

(1 + R2 / R1).fh = fT

where fT is the gain-bandwidth product for the opamp used. In the case of low-pass filtersfh should be much bigger than fo, the 3dB cut-off frequency of the filter.

Finally, it should be noted that the parameter = 1 in the Butterworth table below and o =p unlike on pages 4 7. So, in the case of the Butterworth filters (with = 1) the tablesprovide us only with the values of (3 AVO), where AVO=(1 + R2/R1).

Butterworth Polynomials (with 0 = 1)

1 (s + 1)2 (s2 + 1.4142s + 1)3 (s + 1)(s2 + s + 1)4 (s2 + 0.7654s + 1)(s2 + 1.848s + 1)5 (s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)6 (s2 + 0.5176s + 1) (s2 + 1.4142s + 1) (s2 + 1.9319s + 1)7 (s + 1) (s2 + 0.44505s + 1) (s2 + 1.2470s + 1) (s2 + 1.8019s + 1)8 (s2 + 0.3902s + 1) (s2 + 1.1111s + 1) (s2 + 1.6629s + 1) (s2 + 1.9616s + 1)9 (s + 1) (s2 + 0.3473s + 1) (s2 + s + 1) (s2 + 1.532s + 1) (s2 + 1.8794s + 1)


8/8/2019 Filters 10

18/32

8/8/2019 Filters 10

19/32


7 Building Chebyshev filters from Chebyshev Polynomials

7.1 Low Pass with specific values of

The procedure with Chebyshev filters is similar to that with the Butterworth. The table ofpolynomials is again normalised to 0 = 1. This means that in the table for the 0.1dB ripplefilter, the quadratic terms (e.g. for N=2) should really read not (s2 + 2.3724s + 3.314) but(s2 + 2.3724sp + 3.314p2). There is an additional complication in that the constant in thefirst order and the second order factors is not unity.

So the Transfer Function for a first order low-pass filter is now of the form:

=

=

=

++

s

A

as

Ks

Vin

VoutsTo

o

p

p

VO

1

1 1)(

where o = a1p = 1/CR and AVO = K1 / a1. Sensible values of R should be used (itrepresents the input resistance of the filter at high frequencies) and C calculated from o.

The situation is similar for the second order factors:

22

2

2

21

2

2

2

)3(

)(2

oovo

o

pp

p

sAs

A

asas

KsT

VO

++++

= =

where a2.p2 = o2 or p = o / a2 so that3 - AVO = a1 / a2.

EXAMPLE:

Suppose we want a five-pole 1dB ripple low-pass Chebyshev filter. The table gives thenormalised transfer function as: (s + 0.289)(s2 + 0.179s + 0.988)(s2 + 0.468s + 0.429).

Taking the first order section first: (s + 0.289 p). Here a1 = 0.289 and o = a1p. Supposethe pass-band edge fp is to be 10kHz, then p = 2 x x 10 x 103. Then we have:

o = a1p = 0.289 x 2 x x 10 x 103 = 1.816 x 104 = 1 / CR

Taking C = 0.1F, the R = 1 / [0.1 x 10 -6 x 1.816 x 104] = 550.7. [If this were deemed toosmall, then R could be increased by 10 and C decreased by 10].

Like with the Butterworth,the gain (1 + R2

/ R1

) can be chosen at will, e.g. to give aparticular overall gain. We should again bear in mind that the 3dB cut-off for the first ordersection fh will be given by:

(1 + R2 / R1) fh = fT


8/8/2019 Filters 10

20/32


where fT is the gain-bandwidth product for the opamp used. In the case of low-pass filtersfh should be much bigger than fo, the 3dB cut-off frequency of the filter. Here the overallgain is to be 10 and we leave the choice of R2 and R1 to the end.

Taking the first second order section with (s2 + 0.179 p s + 0.988 p2), we know that a1 =0.179 and a2 = 0.988 and p = 2 x x 10 x 103. So we have:

o = (a2) p = 0.994 x 2 x x 10 x 103 = 6.245 x 104 radians/s

As o = 1/CR and if we choose C = 0.1F then R = 1 / [C . o] = 160.13. This value of Ris rather low, so we increase R by 10 to 1.6k and decrease C by 10 to 0.01F = 10nF.Also,

3 - AVO = a1 / a2 = 0.179 / 0.988 = 0.179 / 0.994 = 0.18008Hence:

AVO = ( 1 + R2 / R1) = 3 0.18008 = 2.82and R2 / R1 = 1.82. So we could use, for example, R1 = 10k and R2 = 18.2k and thatwould complete the design of this particular section of the fifth order 1dB low-pass

Chebyshev.Finally, taking the second second order section with (s2 + 0.468p s + 0.429p2), we knowthat a1 = 0.468 and a2 = 0.429 and p = 2 x x 10 x 103. So we have:

o = a2 . p = 0.655 x 2 x x 10 x 103 = 4.115 x 104 radians/s

As o = 1/CR and if we choose C = 0.1F then R = 1 / [C . o] = 243. This value of R israther low, so we increase R by 10 to 2.43k and decrease C by 10 to 0.01F = 10nF.Also,

3 - AVO = a1 / a2 = 0.468 / 0.429 = 0.468 / 0.655 = 0.7145

Hence: AVO = ( 1 + R2 / R1) = 3 0.7145 = 2.2855

and R2 / R1 = 1.2855. So we could use, for example, R1 = 10k and R2 = 12.855k.

The design is now almost complete. If the gains of the first, second and third sections areK1, K2 and K3 then the overall gain K is given by:

K = K1 K2 K3 = 10

But the gain of the first second order section K2 is 2.82 and that of the second second

order section K3 is 2.2855. So, the gain of the first order section K1 for an overall gain of 10must be:

10 / [2.82 x 2.2855] = 1.552 = ( 1 + R2 / R1).

Hence: R2 / R1 = 0.552 and if R1 is again 10k then for the first order section R2 = 5.52k.

And that completes the design of the Chebyshev filter.

The first eight Chebyshev polynomials are given below for some values of ripple:

Chebyshev 0.1dB ripple ( = 0.1526)

1 (s + 6.552)2 (s2 + 2.3724s + 3.314)


8/8/2019 Filters 10

21/32


3 (s + 0.9694)(s2 + 0.9694s + 1.6898)4 (s2 + 0.5283s + 1.33)(s2 + 1.2755s + 0.6229)5 (s + 0.5389)(s2 + 0.3331s + 1.1949)(s2 + 0.872s + 0.6359)6 (s2 + 0.2294s + 1.1294) (s2 + 0.6267s + 0.6964) (s2 + 0.8561s + 0.2634)7 (s + 0.3768) (s2 + 0.1677s + 1.0924) (s2 + 0.4698s + 0.7532) (s2 + 0.6789s + 0.3302)8 (s2+0.128s+1.0695) (s2+0.3644s+0.7989) (s2+0.5454s+0.4162) (s2+0.6433s+0.1456)


1 (s + 2.863)2 (s2 + 1.425s + 1.516)3 (s + 0.626)(s2 + 0.626s + 1.142)4 (s2 + 0.351s + 1.064)(s2 + 0.845s + 0.356)5 (s + 0.362)(s2 + 0.224s + 1.036)(s2 + 0.586s + 0.477)6 (s2 + 0.1554s + 1.024) (s2 + 0.4142s + 0.5475) (s2 + 0.5796s + 0.157)7 (s + 0.2562) (s2 + 0.1014s + 1.015) (s2 + 0.3194s + 0.6657) (s2 + 0.4616s + 0.2539)8 (s2+0.0872s+1.012) (s2+0.2484s+0.7413) (s2+0.3718s+0.3872) (s2+0.4386s+0.08805)






8/8/2019 Filters 10

22/32




8.1 High pass Butterworth filters with 0 = 1

Construction is exactly as for the low-pass filters except that high-pass filter sections areused, i.e. Fig. 1.6 instead of Fig. 1.5 and Fig. 1.8 instead of Fig. 1.7. In other words,the frequency determining resistors are interchanged with the capacitors.


8/8/2019 Filters 10

23/32


8.2 Building high-pass Chebyshev filters with specific values of from ChebyshevPolynomials

Procedure is similar but more complicated than in the case of Butterworth Filters. Thefrequency determining R and C are again interchanged as for the Butterworth filterbutthe constants that determine o = 1/CR also need to be changed.

ForChebyshev filters the following transformation is required:s/p 1/(S/p),

where S is the Laplace variable for the high-pass filter and p is the pass band edgefrequency times 2 for the high-pass filter.

It is easy to show that this procedure converts the first order low-pass transfer function(T.F.) to a high-pass filter T.F. Let Ap / (s + p) be the T.F. of the filter, where is theconstant, the dc gain, given by s=0, is A/ and o = p. Dividing by p we get A / ( +s/p). Making the above substitution leads to:

ppp

p

p

S

SA

S

AS

S

SA

S

A

+=

+=

+

=

+

1

)/(

1/

/

/

1

At the end of the transformation we have a high-pass filter T.F. with a high frequency gainA/. The original constant has become 1/ so that o = (1/)p, so the transformationhas the following effect:

pp

p

S

SA

s

A

+

+

1

)/(

A similar but rather more complex procedure leads to a similar result for second ordersections, which shows that a low-pass T.F. A2p / [s2 + sp + 2p], i.e., one with a highfrequency gain of A/ and 2o = 2p transforms to:

22

2

22

2

1

)/(

pppp

p

SS

SA

ss

A

++

++

so that the high frequency gain is again A/ but 2o is now (1/)2p.

Note that the transformation in the first order term is equivalent to INVERTING theconstant . In the second order terms is again INVERTED and it also DIVIDES the sterm multiplier .

EXAMPLE: Design a third order high-pass 0.1dB Chebyshev filter with the pass bandedge frequency fp = 1kHz and dc gain K = 20.

SOLUTION: The T.F. denominator for the low-pass filter follows from p. 16 and is(s+0.9694p).(s2+0.9694sp+ 1.68982p).

Making the substitution s/p = 1/(S/p) is equivalent to putting s = 1/(S / 2

p) = 2

p/S.Substituting, we have:

[2p/S + 0.9694p].[ 4p/S2 + 0.9694(2p/S)p + 1.68982p ]


8/8/2019 Filters 10

24/32


We now multiply the first bracket by S and divide by p and for the second bracket wemultiply by S2 and divide by 2p. This leads to:

[p + 0.9694S].[ 2p + 0.9694pS + 1.6898S2 ]

We finally divide the first term by 0.9698 and the second by 1.6898. This leads to the highpass filter T.F. denominator:

[S + (1/0.9694)p].[ S2 + (0.9694/1.6898)pS + (1/1.6898)2p ]

We again note that the transformation is equivalent in the first order sections toINVERTING the constant. In the second order sections the constant which multiplies 2p isINVERTED and the same constant is used to divide the multiplier of the sp term.

So, for the second order section we have: o = (1/1.6898)p = 0.7693 . 2 .1000 =4.8335kr/s. For C = 0.1F we have R =1/(o.C) = 2.069k.

Also, (0.9694/1.6898)p = (3 AVO)o, or (3 AVO) = (0.9694/1.6898).(p/o) = 0.74574,so that AVO = 2.2543 and if R1 = 10k then R2 = 12.54k.

For the first order section we have: o=(1/0.9694)p=2.1000/0.9694 = 6.4815kr/s. For C= 0.1F we have R = 1/(o.C) = 1.543k.

An overall gain of 20 is required and the gain of the second order section is 2.2543 so thegain for the first order section will have to be 20/2.2543 = 8.872 = 1 + R 2/R1 so that ifR1=10k then R2 = 78.72k. This completes the design. The circuit is shown below in Fig.

8.1.

Fig. 8.1


8/8/2019 Filters 10

25/32


Fig. 8.2

The transmission of the filter is shown in Fig. 8.2. We note that the high-pass filter actuallyturns out to be a band-pass filter because of the finite bandwidth of the op amps. As thefirst order stage has the bigger gain it will be mainly responsible for determining thebandwidth. For a cut-off frequency of fh, a gain-bandwidth product of fT and a gain of 1 +R2/R1 we have:

(1 + R2/R1).fh = fT.

Here 1 + R2/R1 = 8.872 so that for fT = 1MHz, which is typical for the 741 op amp, we havefh = 113kHz, which is precisely the value measured above. [As the gain of the secondorder section is 2.2543, i.e. a quarter that of the first order section, the op amp cut-offfrequency for the second order section will be approximately four times higher, i.e. around450kHz.Hence, this op amp should be relatively unimportant in determining the overallupper 3dB cut-off frequency.]


8/8/2019 Filters 10

26/32


9 The Two Integratorloop Biquad

9.1 Theory

There are a number of second-order filters based on the opamp integrator transfer functionVout/Vin = (-o/s). They have the great advantage that all three types of filter output areavailable from the circuit.

Consider the T.F. for a high-pass filter:

22

2

oQoss

sK

Vin

Vhp

++=

Cross multiplying and dividing both sides by s2 we have:

VinKVhps

oVhp

s

o

QVhp =++

2

21

The second and third terms are obtainable by passing Vhp through integrators with timeconstants CR = 1/o.

Fig. 9.1

The problem of forming Vhp remains. If we rearrange the second equation by having Vhp onthe left and using the T.F. for the opamp integrator (-o/s), we have:

Vhso

so

Vhpso

QVinKVhp

+=1

This suggests, that Vhp may be obtained by the summing process below:

Fig. 9.2 Nick Holden BCU Ref.: lect2010\analogUG3 26

8/8/2019 Filters 10

27/32


So, combining Figs. 9.1 and 9.2, we get the diagram shown below in Fig. 9.3.

=

=

Fig. 9.3

Note that in the above circuit Vhp is the output of the summer, which gives the high-passT.F. Vhp / Vin. The signal at the output of the first integrator is (-o/s)Vhp which gives the T.F.below:

oQ

oss

soKVin

Vhp

so

Vin

Vhp

so =

++==

22

and the output of the second integrator is:

( )

Q

oss

oK

oQ

oss

sK

s

o

Vin

Vhpso

++=++=

2

2

22

2

2

22

9.2 Circuit Implementation of the KHN Biquad


8/8/2019 Filters 10

28/32


Fig. 9.4

Working from Fig. 9.4, we use the principle of superposition to obtain Vhp at the output ofU1 in terms of Vin, Vbp and Vlp. Note the substitution of (-o/s)Vhp for Vbp and(o/s)2Vhp forVlp.


8/8/2019 Filters 10

29/32


We get:

( )R1

RfVhp

s

o2

Vhps

o

R3R2

R2Vin

R3R2

R3

R1

Rf1Vhp

++

+

+=

Multiplying out:

s

R3R2

R2

R1

Rf1VinR3R2

R3

R1

Rf1Vhp +++

++=

Comparing the last term in the equation above with the last equation on p. 21 we musthave Rf/R1 = 1 or Rf = R1. Any arbitrary but sensible value will do here.

Comparing the second term on the right:

Vhps

o

R3R2

R22Vhps

o

Q

1 +

=

Therefore:

R2

R31

R2

R3R22Qor

R3R2

R22

Q

1 +=+

=

+

=

Hence:

12QR

R

2

3=

Either R2 or R3 is arbitrary. The other one must be chosen to give the correct Q.

Finally, equating the first terms:

VinKVinR3R2

R32 =+

or:( )

Q

12

2Q

24Q

12Q1

12Q2

112Q

1

2

R3

R21

2K ==

+=

+

=+

=

This is fixed and determined by the required Q.

Two integrator-loop biquads are very versatile and easy to design. Their performance is,however, quite badly affected by the finite bandwidth of the opamps used. There arespecial techniques for the compensation of these effects.

Low-pass, high-pass and band-pass functions are already available. Other types may beobtained by adding these three with appropriate weightings using an inverting adder. Aband-stop filter, for example, may be obtained by adding a low-pass and a high-pass filterfunction with equal weighting.


8/8/2019 Filters 10

30/32



8/8/2019 Filters 10

31/32


ACTIVE FILTER TUTORIAL EXAMPLES

1. Design a Butterworth low-pass filter with fP = 1kHz, fS = 3kHz, Amax = 1.5dB, Amin =20dB and dc gain K = 5.

[Ans.:

= 0.6423, N=2.5 (use N = 3), p1,p3 =

O[-0.5

j0.866], p2 = -

O. For freq.Selecting components C = 0.1 F, R = 1.373k . For second order stage R2/R1 = 1,e.g. R1 = 10k , R2 = 10k . For first order section R2/R1 = 1.5, e.g. R1 = 10k , R2= 15k .]

2. Design a Chebyshev low-pass filter with fP = 1kHz, fS = 2kHz, Amax = 1.5dB, Amin =35dB and dc gain K = 10.

[Ans.: = 0.6423, N=3.92 (use N = 4), p1,p4 = p[-0.11913 j0.9676], p2,p3 = p[-0.2876 j0.4008].

(s-p1)(s-p4) = s2 + 0.2383s P + 0.95044 P2 and (s-p2)(s-p3) = s2 + 0.5752s P +0.24335 P2.For first second order section (p1,p4) for freq. Selecting components C = 0.1 F, R =1.633k and R2/R1 = 1.7556, e.g. R1 = 10k , R2 = 17.556k .For second second order section (p2,p3) for freq. Selecting components C = 0.1 F,R = 3.226k and R2/R1 = 0.834, e.g. R1 = 10k , R2 = 8.34k .

For the gain control section R2/R1 = 0.979, e.g. R1 = 10k , R2 = 9.79k .]

3. Design a fifth order Butterworth low-pass filter from the Table of Butterworthpolynomials ( = 1) with fP = 3kHz and dc gain K = 25. How would the design bedifferent for a high-pass filter?

[Ans.: For freq. Selecting components C = 0.01 F, R = 5.30k . For the firstsecond order section R2/R1 = 1.382, e.g. R1 = 10k , R2 = 13.82k . For the secondsecond order section R2/R1 = 0.382, e.g. R1 = 10k , R2 = 3.82k . For first ordersection R2/R1 = 6.594, e.g. R1 = 1k , R2 = 6.594k .]

4. Design a 2dB ripple sixth order Chebyshev low-pass filter from the Table ofChebyshev polynomials with fP = 5kHz and dc gain K = 30.

[Ans.: For first second order section for freq. Selecting components C = 0.01 F, R= 3.239k and R2/R1 = 1.9045, e.g. R1 = 10k , R2 = 19.045k .

For second second order section for freq. Selecting components C = 0.01 F, R =4.36k and R2/R1 = 1.6484, e.g. R1 = 10k , R2 = 16.484k .For third second order section for freq. Selecting components C = 0.01 F, R =10.071k and R2/R1 = 0.89075, e.g. R1 = 10k , R2 = 8.9075k .

For the gain control section R2/R1 = 1.0627, e.g. R1 = 10k , R2 = 10.627k .]


8/8/2019 Filters 10

32/32


5. Given the transfer functions for the high-pass filter and the integrator, derive theschematic diagram (Fig. 8.4 in your notes) for the KHN biquad, showing how thehigh-pass, band-pass and low-pass transfer functions may be obtained from theinput.

6. Derive an expression for Vhp in terms of Vbp and Vlp for the circuit of Fig. 8.4.

Replace Vbp by (- o/s).Vhp and Vlp by ( o/s)2.Vhp and henceobtain suitable valuesfor the resistors and capacitors in Fig. 8.4 if a KHN biquad band-pass filter withcenter frequency fO = 1kHz and Q = 10 is required.

[Ans.: For the freq. selecting components C = 0.1 F, R = 1.592k . Rf = R1 =10k , R2 = 10k and R3 = 190k is a possible solution.]

7. Design a third order 1dB high-pass Chebyshev filter with the pass band edge at100Hz and a gain of 100.

[Ans.:For 2nd order: C = 0.1 F, R = 15.87k , R1 = 10k , R2 = 15.05k ; 1st order:C = 0.1 F, R = 7.862k , R1 = 10k , R2 = 389.2k ; is a possible solution.]

8. Design a fourth order 2dB high-pass Chebyshev filter with the pass band edge at

100Hz.

[Ans.:For one 2nd order: C = 1 F, R = 1.534k , R1 = 10k , R2 = 17.82k ; second

2nd order: C = 0.1 F, R = 7.492k , R1 = 10k , R2 = 9.24k ; is a possiblesolution.]

Filters 10

Documents

Transcript of Filters 10