## Butterworth Filter : Design of Low Pass and High Pass Filters

Hey, friends welcome to the YouTube channel

all about electronics. So, in the last video, we had seen that different

types of filter approximations which are quite commonly used in the filter design.So, in

this video, we will see the Butterworth filter approximation in a detail and we will also

see how to design the higher order low pass as well as the high pass filter using this

Butterworth filter approximation. So, now in the last video, we had seen that

this Butterworth filter has a very flat passband and the roll-off of this filter is at the

rate of 20n dB per decade. So, suppose if you are designing 4th order

filter then the roll-off of that filter will be at the rate of 80 dB per decade or if you

are designing 8th order filters then that filter will be having a roll-off at the rate

of 160 dB per decade. So, first of all, let us see how to design

the second order Butterworth low pass filter and based on that we will see how to design

the higher order Butterworth filters. So, if you see the transfer function of any

second order low pass filter then it can be given by this expression, where here this

k represents the gain of that particular filter and this omega n represents the cutoff frequency

of that particular filter. And if you write this expression in terms

of the quality factor Q and it can be written in this way. So, first of all, let us understand how to

get this transfer function for the second order low pass filter and then after we will

see the criteria for the Butterworth filter design. So, in the earlier videos of RC low pass filter,

we had seen that just by cascading the 2 first order low pass filters we can get the second

order low pass filter. So, first of all, let us derive the transfer

function for this second order filter. So, for the first order filter, we know that

the output by input can be given by Xc divided by Xc+R. Now here Xc is nothing but one divided

by j into Omega C. And suppose if we represent is j omega in s-plane that is j Omega is equal

to S, then we can write this Xc as one divided by SC. So, we can write this expression as one divided

by SC, divided by one over SC plus R. Or simply we can write it as 1 divided by 1 plus RCS. So, now if we have 2 such filters which are

cascaded then we can write that Vout by Vin as one divided by one plus R1C1 into S, into

one divided by one plus R2C2 into S. And if you rearrange this expression then we can

write it as, one divided by R1C1 into R2C2, divided by S square, plus R1C1 plus R2C2,

divided by R1R2 into C1C2 into S, plus one divided by R1R2 into C1C2. so now if you compare

this expression with the generalized expression of second order filter then in need you can

see that the transfer function of the second order filter is in this form. Where here this Omega not is equal to one

divided by under root R1R2 into C1C2.So, in this way you can see that the transfer function

of any second order low pass filter will be off this form. So now, whenever we are cascading the two

first order low pass filter to design second order filter then the cutoff frequency of

the second order filter will not be same as the first order filter. Let us say in this design R1 is equal to R2

is equal to R and C1 is equal to C2 is equal to C, then the cutoff frequency FC will be

equal to one divided by 2 pi into RC. So now, for this filter, the cutoff frequency

of the second order filter should be same as the first order filter but rather if you

see the cutoff frequency of the second order filter, it will be somewhat get shifted. So the cutoff frequency of this second order

filter, let’s say Omega 2C will be equal to omega into under root 2 to the power 1

by 2, – 1. Now, suppose if you’re cascading the n such

first order low pass filters, then the cutoff frequency of that nth order filter will be

equal to omega C into under root 2 to the power 1 by n, -1. So as you can see the cutoff frequency of

the nth order filter will get shifted by this amount. So, suppose if you are designing the second

order filter of 1 kilohertz then the cutoff frequency will not be one kilohertz but it

will somewhat different. While in case of the Butterworth filter design,

if you see the cutoff frequency will remain the same whether it is a first order filter

or any higher order filter. So in this way, as you can see, just by cascading

this first order low pass filters we cannot design this Butterworth filter. Apart from that in the design when all the

values are equal that is R1 is equal to R2 is equal to R, and C1 is equal to C2 is equal

to C, then this transfer function will become, k into Omega c square, divided by S square

plus 2 divided by RC into S, Plus Omega c square. So, here this 1 over Rc is nothing but Omega

C, so for this design if you see the value of Q at the most will be equal to 0.5. So, now this value of Q basically represent

the amount of peaking around the corner frequency or cutoff frequency fC. So if the value of Q increases you will see

the more amount of peaking around the cutoff frequency and as the value of Q reduces you

will see the higher amount of attenuation at the cutoff frequency.So at that time the

filter design is known as the Butterworth filter design because whenever Q is equal

to 0.707 then at cutoff frequency you will see the amplitude will be equal to 1 by √2

times the maximum value. So, simply just by cascading the 2 first order

RC low pass filter we cannot design second order butterworth filter. Because to design this butterworth filter,

we required the higher value of Q and that is only possible when we have a some sort

of positive feedback from the output side, and that is only possible when we have a some

sort of active component in our circuit. So one search filter topology which is quite

commonly used for the butterworth filter design is the sallen key filter topology. So, as you can see this topology, in this

topology we have a positive feedback from the output side or let me redraw this circuit. So, as you can see here, here we have a positive

feedback from the output to this capacitor C1 so this is known as the unity gain sallen

key low pass filter design. So, suppose we want gain then we can connect

the feedback register between the inverting end and the output side. And if you see the transfer function of this

sallen key filter topology then it can be given by this expression. So just by applying the Nodal analysis at

these nodes we can easily achieve this transfer function. So try to get this transfer function by yourself

and if you have any difficulty then do let me know in the comments section. I will provide the note for this derivation. So, now suppose if you compare this transfer

function, with a generalize form of the second order filter that is k into omega c square

divided by S square plus Omega c into S divided by Q plus Omega c square then we will have

a Omega c that is equal to one divided by under root R1R2 into C1 into C2. And the value of Q will be equal to under

root R1R2 into C1 into C2 divided by R1C1 plus R2C2 plus R1C1 into 1 minus k. So, in this way as you can see by choosing

the value of this gain, and R1R2 and C1C2, we can decide the quality factor Q of that

particular filter or if want to simplify the design then let us take the case of when R1

is equal to R2 is equal to R and C1 is equal to C2 is equal to C, so in such case the transfer

function can be given by the expression k into Omega C square divided by S square plus

S into Omega C into 3 – K , plus Omega c square. Where omega C is nothing but 1 over RC. So, now as you can see here for this design

the quality factor Q can be given by one divided by 3 – K. So when R1 is equal to R2 is equal to R and

C1 is equal to C2 is equal to C, then the quality factor Q only depends upon the value

of the gain. So, just by adjusting the gain of this filter

we can get the different values of the Q. But the main limitation of this design is

that you cannot have a gain more than 3 because as the gain approaches the value of 3 then

the quality factor Q will approach the value of infinity and then the system will become

unstable. So for this design, the value of k will be

always less than 3. Now we had seen that for the butterworth filter

design the value of Q should be equal to 0.707. So we can say that 0.707 should be equal to

1 dividede by 3 – k or we can say that 1.414 that is equal to 3-k.or if we simplify it

then we will get k as 1.586. Now we know that the gain of this filter k

will be equal to 1 plus R4 by R3 and the value of k should be equal to 1.586. So for the butterworth filter design the ratio

of R4 by R3 is equal to 0.586. So let us say if you have R3 that is equal

to 10 kilo ohm then value of R4 should be equal to 5.86 kilo ohm for the butterworth

filter design. So, in this way by just changing the value

of gain we can design the butterworth filter design using this sallen key filter topology

and in this way just by cascading the multiple such second order filters, we can design the

higher order filters. So, to design this higher order filters we

generally used to use this polynomials. So, let us understand how we have arrived

to this polynomials. So, in this transfer function let us say we

are designing the low pass filter for the frequency Omega is equal to 1 radian per second. So, this term 1 over R1C1 into R2C2 will be

become 1. So we can write this expression as k divided

by s square plus. And we know that the quality factor Q for

the butterworth filter design is 0.707. So we can write the expression as k divided

by s square plus 1.414 s plus 1. So, that is the polynomial for the second

order butterworth filter design for the frequency of 1 radian per second. So, in this way, if we want to design the

higher order filters then just by using this filter polynomials we can design the higher

order filters. So let us try to design the third order butterworth

low pass filter, which is having a transfer function of S plus1 into, S square plus s

plus 1. So for that we need to cascade the second

order filter with the first order filter or in terms of the filter design if we say, we

need to cascade the second order Sallen key butterworth low pass filter with the first

order RC low pass filter. And in this design we will use this polynomial

that is s + 1 into s S^2+ s + 1. Let us say, in this design we want to design

the cutoff frequency of this filter as 1 kilohertz. So, instead of having this S is equal to one

now in the design we will have a cutoff frequency of 1 kilohertz. So, we know that fc is equal to 1 divided

by 2 pi into RC. So let us assume the value of C that is equal

to 0.1 microfarads and the value of R will be equal to 1 divided by 2 pi into fc. So, if you put this values then we will get

R as 10 ^ 4 divides by 2pi, and that is equal to 1.59 Kilo ohm. So in this way just by choosing the value

of R1, R2, and R5 as 1.59 kilo-ohm and value of C1 C2 and C5 as 0.1 microfarads, we have

ensured that this filter will have a cutoff frequency of 1 kilohertz. So, apart from that we also need to ensure

that the quality factor Q of this filter will be equal to 1. Now to achieve this value of Q is equal to

one, in this filter design the value of one over 3 – K should be equal to one, or to say

the value of k will be equal to two. Or in another way, we can say R4 divided by

R3 plus one should be equal to 2, or we can say that R4 is equal to R3.So let us assume

that the value of R4 and R3 is 10-kilo-ohm. So just by using these values, we can ensure

that we will have a third order low pass Butterworth filter which is having a cutoff frequency

of 1 kilohertz. So, in this way just by using this polynomial

we can design the any order Butterworth low pass filter with our given particular frequency. So, if you understood so far how to design

this Butterworth low pass filters then you will easily understand that how to design

the Butterworth high pass filters.So, in this Butterworth high pass filter design, we just

need to interchange the position of this capacitor and resistor. So as you can see here we have interchanged

the position of this capacitor and resistor. So this same Sallen key topology will act

as a second order high pass filter, and if you see the transfer function of this high

pass filter then it will be given by this expression and in this expression whenever

you put the value of R1 is equal to R2 is equal to R and C1 is equal to C2 is equal

to C than the denominator of this transfer function will be identical to that of the

low pass filter. So, now the transfer function Vout by Vin

will be equal to K into S square divided by S square plus s into omega C into 3 – K plus

Omega c square. So this is how the transfer function of the

high pass filter can be given. So, now in this design, the value of Q should

be equal to 0.707 for the Butterworth filter design. And for that design again you will find that

the ratio of R4 by R3 should be equal to 0.586, so the procedure which we have forward for

the low pass filter design the same procedure can be followed for this high pass filter

design. So, just by interchanging the value of this

capacitor and the resistor, we can design this Butterworth high pass filters, and in

this design, we can use the same polynomials which we had used for the low pass filter

design. So, suppose if you want to design the third

order filter then just my cascading the second order filter with the first order filter we

can design the third order filter. Likewise, just by cascading the two second-order

filters, we can design the fourth order Butterworth high pass filter. And for this design, we can use the same polynomials

which we had used for the low pass filter design. So as an exercise try to design the fourth

order Butterworth high pass filter which is having a cutoff frequency fc of 10 kilohertz

and do let me know your design values in the comment section below. So, I hope in this video you understood how

to design this Butterworth low pass as well as the high pass filter using this sallen

key filter topology. Similarly, in the upcoming videos, we will

see how to design the Bessel as well as the Chebyshev filters using this Sallen key

filter topology.

7:43 am

Very nice

7:45 pm

very nice video thanks a lot..

but one thing i couldn't understand i.e. in (s^2+s+1) Wc=1. so how fc=1khz…it should be (1/2*3.14*wc) and if we take fc=1khz then (wc=2*3.14*fc). so the Q will not be = 1.. plz clarify it as soon as possible..

7:47 pm

plz see at 13:27

6:10 pm

where are the remainig videos of filters?

9:12 am

hello , I am glad to have found your channel , I am looking at building some DIY speakers for myself and want have a 4 way setup, I have little to no electronics background but read the following website

http://education.lenardaudio.com/en/06_x-over.html which explains the problem with the crossover but gives no solution to designing 4 way active systems.

I have 2 questions

1 . is an op amp a hi fi quality component or does quality sound in active crossover have to be done with transistors on second order and 3rd order filter

2 have you got circuit diagrams for a complete 4 way active crossover with formulas i would need to calculate the components according to my desired frequecy points?

much appreciated, i can send my email if desied

regards Clive

4:33 pm

Bro feedback per capistor cu lagty hain

4:42 am

great explanation

11:20 am

thank you very much. it helped me alot

4:53 pm

Do you have a video on schmit trigger circuit. I am unable to find in your channel.

6:06 pm

Remark at at 6:07: For 2nd order

If Q = 1/sqrt(2) then filter type is butteworth (no ripples in passband) and

if Q > 1/sqrt(2) then filter type is chebyshev (ripples in passband)

1:24 am

Can u please provide the note for the derivation

8:30 pm

expression for transfer function of sallen key is not correct. It should be

K/(s2(R1R2C1C2)+s(R1C1+R2C1+R1C2(1-K))+1)

Check:

http://www.ti.com.cn/cn/lit/an/sloa024b/sloa024b.pdf

2:00 pm

Please provide note for derivation

12:53 pm

Thanks.

7:08 am

what if cut off freq. 50Khz ?? how many value of R1, R2, R5, R3, R4 ??

5:04 pm

The link for the derivation of the transfer function for the sallen key filter topology.

https://drive.google.com/open?id=1igXSBw6Rmb_HtWzGtKW9q41lELzbY-V7

9:09 am

I am gonna put this on my breadboard…. Wanna see how well it works

2:43 pm

sir, please provide note for derivation

2:16 am

really good explaination. Now I know why I cant get the gain I was trying to get out of my filter. Thanks Boss!

4:38 am

please explain the digital electronics and microprrocessor & microcontroller and DSP

1:36 pm

There's a lot of math and I watched the video a couple of times to understand. This can all be simplified to a) using sallen-key topology b) Using Fc = 1/(2x PI x R x C) choose Fc , C and solver for R c) Keeping the gain between 1 and 2.5

2:12 pm

I need the derivation

5:42 pm

Just by cascading two low pass filter we get the same cutt off frequency as first order . And for butterworth also ,cuttoff frequency should be same for all orders of filters. Then what is the point to design butterworth filter? If we design then why we can't design using two low pass filters? As u said in video the cutt-off frequency for low pass filter is differ from order to order…How? Overall very difficult to analyze the need of butterworth filter.

7:33 am

Is transfer function just another name of gain ?

1:55 pm

Dude… like thank you so much I was lost until I heard that beautiful voice shower me with knowledge.

Thank you so much, this helps me so much in my ee2320 class!

10:21 am

Nice explanation sir…

12:39 pm

thanks a lot

10:12 am

Sir how do we design the filter for frequency other than one

5:30 pm

How did you arrived the polynomial for various orders of filter?

11:51 pm

Awesome cool video! Big Thumbs up.

2:35 am

Can we get more mathematical examples regarding this?

6:25 am

For 4th order what value of quality factor should i choose to solve. Like looking at the chart there are two values 0.7654 and 1.8478. Plzz help

8:45 am

I think i am half done by seeing this video, all i required now is to watch it again.

1:47 am

2:57 not that easy to formulate the cascaded RC network, unless you put buffer in the first stage. That because, the impedance of the first stage will change if you connect the second stage.

6:36 pm

Sir, please clear my doubt…you said that Butterworth filter has a higher value of Q over passive filters, then what is the advantage of having a Higher Q value?

3:03 am

More informative videos. Is there a single book for non-electronic student to learn about filter design for radio design?

11:29 am

good stuff but too much on transfer functions and not enough on the actual filters and their operation in a practical application. would be great if you did a video on how to build practical circuits and when to use the various types for the best results. like your vids but a little less mathematics would not be a bad thing.