# Butterworth low pass filter design

### Butterworth low pass filter design.

### Analogue systems and instrumentations.

### ABSTRACT.

The purpose of this report is to explain both design and simulation procedures of a higher order low-pass filter comprising of cascaded Sallen-Key filters to obtain a filter circuit equivalent to a Butterworth low-pass filter of a certain order for which the design specifications are interpreted. The half power frequency and the actual attenuation at the edges of the pass band and the stop band were obtained by using Bode plotter in the Multisim simulation software.

### INTRODUCTION.

Low-pass filter design is an important procedure which has many applications in both electrical and electronic systems. The software used for the simulation procedure is version 8 of Multisim produced by National Instruments Incorporations. Notwithstanding the engineering principles and concepts, Multisim does numerical calculations and is capable of producing a Bode plot for a circuit which is very useful in filter design. Hence, it will grant the accuracy, enhance the flexibility and even increase the reliability of an actual circuit if implemented relying on its simulations. This report describes the design and simulation procedures of a Sallen-Key low-pass filter where the design specifications were interpreted for a Butterworth low-pass filter. According to the design specifications of the Butterworth low-pass filter, the calculations were done to obtain the order of the filter, half power frequency and attenuations at both pass band and stop band. This report further describes the computations and the calculations for the Q factor of the said Butterworth filter design. As the result for the order did round up to 10, and since Sallen-Key filters are of 2nd order, it is necessary to have 5 Sallen-Key filters connected in cascaded form in order to implement an equivalent filter circuit for a 10th order Butterworth filter. The half power frequency and the attenuation coefficients amax and amin were obtained by using Bode plotter from the said Multisim software.

### EXPERIMENTAL.

The design specifications for a Butterworth low-pass filter were interpreted. The values for amax, amin, ?p and ?s were specified. The values are as follows:

amax= 0.25 dB

amin= 15 dB

?p= 10000 rad/s

?s= 14000 rad/s

The simulation was done by using the version 8 of Multisim simulation software.

### RESULTS.

Since 1920, it has been a tradition for both amplifier and filter designers to consider gain and loss to be positive respectively. In order to segregate these, two symbols were used. These are interpreted as

a= -20logT dB for T=1

A=20logT dB for T>1

Hence the numerical value remains positive.

A pass band is defined to be one such that the attenuation is less than amax.

A stop band is defined to be one such that the attenuation is greater than amin.

(Valkenburg, 1982)

A brief analysis of second order systems is necessary in order to analyze the properties of a Butterworth filter. The characteristic equation of a second order system can be given as

s2+2??0s+?02=0; where ? is the damping ratio. By solving for an RLC circuit and equating the characteristic equation with the previous one leads to a relationship between the Q factor (quality factor) and the damping ratio. It can be observed as

?=12Q.

Hence this can be expressed in terms of Q,

s2+?0Qs+?02=0

(Scott, 1987)

The Q factor is an important parameter in designing filters. Simply the Q factor is the ratio of reactance to effective resistance either in an inductor or in a capacitor. (Green, 1955)

The Butterworth polynomial is described such that all poles lie on the unit circle of the s-plane.

Let a complex transfer function be multiplied by its conjugate function which will provide an even function. Hence, both the numerator and denominator are even functions. If the numerator is made constant in order to have no zeros, the function can be written as

Tnj?.2=A0B0+B2?2+B4?4+...+B2n(?2n)

When ?=0 and ?=8, all except B0 and B2n will be zero.

B2n=1/?0.2n

Therefore

|Tnj?|2=1/1+(??0)2n

When the frequency is normalized, i.e.?0=1

Tn?=121+?2n

Since amax, amin, ?p and ?s are given, it is possible to derive an expression for attenuation a.

a=10log101+?/?02n

An expression for ?0 can be derived as

?0=?10a/10-11/2n

This can be written for both pass band and stop band frequencies

?0=?s10amin/10-11/2n

?0=?p10amax/10-11/2n

By dividing the above equations, it is possible to obtain an expression for the order of the filter n.

The order of the filter can be found as

n=log1010amin10-110amax10-12log10?s/?p

By substituting the values given, it is possible to obtain the order of the filter.

n=log10101510-1100.2510-12log1014000/10000

n=9.28

The order can be rounded up to 10.

By substituting in one of the equations which gives the cut off frequency,

?0=?s10amin/10-11/2n

?0=140001015/10-11/20

?0=11798.47 rad/s

The attenuations at pass-band and stop-band can be calculated.

At pass band

a=10log101+?p/?02n

a(?p)=10log101+10000/11798.4720

a?p=0.15611 dB

At stop band

a(?s)=10log101+?s/?02n

a(?s)=10log101+14000/11798.4720

a?s=15 dB

The Butterworth pole locations must be calculated in order to obtain the values for Q factors in the 10th order filter.

The following function can be considered as the amplitude of the function generated by multiplying the complex transfer function with the complex conjugate of the said transfer function.

Tnj?2=11+(?/?0)2n

Normalizing the frequency such that ?0=1

Tnj?2=11+(?)2n

Tn(j?)2=Tnj? Tn(-j?)

J? can be written in terms of s, whilst ? can be written as s/j.

Hence,

TnsTn-s=11+sj2n=11+{s2n/-1n}=11+(-1)ns2n

let

BnsBn-s=1+(-1)ns2n

The poles can be found by the equation

1+(-1)ns2n=0

Bns gives the factors which produce roots for the left half of the s-plane, while Bn-s gives factors which produce roots for the right half of the s-plane.

The order of the Butterworth filter is 10 and hence the characteristic equation can be written as

1+-110s20=0

s20=-1

s=(-1)1/20

-1 can be written in complex polar form as

-1=cosp+jsinp

Then

s=(cosp+jsinp)1/20

By applying De Moivre's theorem

s=cos(p20)+jsin(p20)

Since the nth root of a number should have n values, the 20th root of -1 should have 20 values.

Hence it can be written as a general expression

s=cos2kp+p20+jsin2kp+p20

Where k=0,1,2,3,4,.........,19.

(Hall and Knight, 2006)

All 20 roots can be found by substituting values for k.

k=0

s=cosp20+jsinp20= 0.9877+j0.1564

k=1

s=cos3p20+jsin3p20= 0.8910+j0.4540

k=2

s=cos5p20+jsin5p20= 0.7071+j0.7071

k=3

s=cos7p20+jsin7p20= 0.4540+j0.8910

k=4

s=cos9p20+jsin9p20= 0.1564+j0.9877

k=5

s=cos11p20+jsin11p20= -0.1564+j0.9877

k=6

s=cos13p20+jsin13p20= -0.4540+j0.8910

k=7

s=cos15p20+jsin15p20= -0.7071+j0.7071

k=8

s=cos17p20+jsin17p20= -0.8910+j0.4540

k=9

s=cos19p20+jsin19p20= -0.9877+j0.1564

k=10

s=cos21p20+jsin21p20= -0.9877-j0.1564

k=11

s=cos23p20+jsin23p20= -0.8910-j0.4540

k=12

s=cos25p20+jsin25p20= -0.7071-j0.7071

k=13

s=cos27p20+jsin27p20= -0.4540-j0.8910

k=14

s=cos29p20+jsin29p20= -0.1564-j0.9877

k=15

s=cos31p20+jsin31p20= 0.1564-j0.9877

k=16

s=cos33p20+jsin33p20= 0.4540-j0.8910

k=17

s=cos35p20+jsin35p20= 0.7071-j0.7071

k=18

s=cos37p20+jsin37p20= 0.8910-j0.4540

k=19

s=cos39p20+jsin39p20= 0.9876-j0.1564

All 20 roots are not necessary in order to obtain the Q factor values albeit all of them lye on the unit circle of the s plane. The poles lying on the left part of the s-plane are only needed for the calculations to obtain a stable circuit. The poles lying on the left half of the s-plane are those of which the real part is negative. Hence the roots are when k=5,6,7,8,9,10,11,12,13 and 14.

The characteristic equation of a second order system is

s2+2??0s+?02=0

The parameter ? can be substituted by 12Q.

s2+?0Qs+?02=0

By frequency normalization ?0=1

s2+1Qs+1=0

It is clear that all 10 poles that lie on the left half of the s-plane (i.e. k=5 to k= 14) has its complex conjugate.

Suppose a root to be s=-a+jb and its conjugate s=-a-jb, then the factors can be written as

s+a-jbs+a+jb

This solves to

(s+a)2-(jb)2=s2+2as+a2+b2.

Butterworth filter is designed as all poles to be on a unit circle. Hence a2+b2=1. Thus making the factor equivalent to a second order characteristic equation. These two can be equated as:

s2+1Qs+1=s2+2as+1.

The coefficients of s can be equated and Q can be written in terms of the pole's real value 'a'.

Q=12a=12cos2kp+p20

k=0,1,2,...19.

The conjugate pairs, their respective real values and Q factors are as follows:

a Q (k=5,k=14) 0.1564 3.1969 (k=6,k=13) 0.4540 1.1013 (k=7,k=12) 0.7071 0.7071 (k=8,k=11) 0.8910 0.5612 (k=9,k=10) 0.9877 0.5062

This yields that it is possible to implement the Butterworth filter by cascading five 2nd order filters. This can be done by using Sallen-Key filters.

### Sallen Key low-pass filter

Let the output of the operational amplifier be vo, the input vi and voltage at the junction where R and mR joins be vx.

By Kirchoff's law,

vi-vxmR=vx-v0R=(vx-vo)/(1jwnc)

Bu virtual earth concept

v0=vxa+j?Rc

Solving the above equations,

?0=12mnRC

The Q factor is given by

Q=2mnm+1

From the above two equations 2mn can be removed while making m=1 and let assume R=10kO for design purposes.

RC=12Q?0

For n

Q=2mnm+1

Q=21n1+1

n=4Q2

Now it is possible to design five Sallen-Key filters in cascaded form in order to obtain an equivalent circuit to a 10th order Butterworth low pass filter for a cut-off frequency of 11798.47 rad/s.

Stage1:

Q1=0.51

C1=8.31nF

nC1=8.65nF

Stage2:

Q2=0.56

C2=7.57nF

nC2=9.49nF

Stage 3:

Q3=0.71

C3=5.97nF

nC3=12.04nF

Stage4:

Q4=1.1

C4=3.85nF

nC4=18.65nF

Stage5:

Q5=3.2

C5=1.32nF

nC5=54.25nF.

The circuit was designed using Multisim simulation software. The circuit was simulated in order to obtain the Bode-plot and then determine the cutoff frequency at -3dB point. The circuit was successfully simulated and the results were accurate.

A snapshot of the simulated circuit is shown below:

### The Bode plotter is shown below.

It is clear from the bode plot that at -2.962dB the cutoff frequency has a value of 1.870kHz which is equivalent to 11749.56rad/s. According to the calculations, at -3dB the cutoff frequency is 11798.47 rad/s. Hence, it can be concluded that the design was successful and accurate.

### CONCLUSION.

The report explains the design procedure for a 10th order low-pass filter employing Sallen-Key filters in order to implement an equivalent circuit for a 10th order Butterworth low-pass filter. The report further explains the procedures for both computing and calculating the Q factor by determining the poles of the transfer function. The cutoff frequency of the simulated design matches with the theoretical calculation of the same and therefore the design can be considered to be successful.

### REFERRENCES.

1. Valkenburg. Van. M. E. 1982, *Analog filter design, *New York: Oxford university press.

2. Scott. Donald. E. 1987,* An introduction to circuit analysis: a systems approach*, New York: McGraw-Hill Book Company.

3. Hall. H. S., Knight S. R. 2006, *Higher Algebra,* India: A.I.T.B.S. Publishers.

4. Green. E. I. 1955, The story of Q, *American Scientist,* 43, pp584-594.* *