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.

37 comments

  • Mayur Shah

    Very nice

    Reply
  • Umar Kaleem

    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..

    Reply
  • Umar Kaleem

    plz see at 13:27

    Reply
  • PRABHAKAR DAS

    where are the remainig videos of filters?

    Reply
  • Clive Stephenson

    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

    Reply
  • arsman safdar

    Bro feedback per capistor cu lagty hain

    Reply
  • Isaac Osahon

    great explanation

    Reply
  • Jupiter Tanha

    thank you very much. it helped me alot

    Reply
  • red cat

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

    Reply
  • Varun Nagpal

    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)

    Reply
  • Sree Sanjanaa Bose.S

    Can u please provide the note for the derivation

    Reply
  • Omer Ahmed

    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

    Reply
  • maria maahi

    Please provide note for derivation

    Reply
  • lrylrylry lry

    Thanks.

    Reply
  • Adib Othman

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

    Reply
  • ALL ABOUT ELECTRONICS

    The link for the derivation of the transfer function for the sallen key filter topology.
    https://drive.google.com/open?id=1igXSBw6Rmb_HtWzGtKW9q41lELzbY-V7

    Reply
  • mr. amp 007

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

    Reply
  • sreevani i

    sir, please provide note for derivation

    Reply
  • Joey Mac

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

    Reply
  • L. JENIPHAR JENI

    please explain the digital electronics and microprrocessor & microcontroller and DSP

    Reply
  • Joey Mac

    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

    Reply
  • ABHIJIT PANIGRAHY

    I need the derivation

    Reply
  • Ronak Agarwal

    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.

    Reply
  • Arsal 's

    Is transfer function just another name of gain ?

    Reply
  • Shazad Ali

    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!

    Reply
  • Saiprakash Putrevu

    Nice explanation sir…

    Reply
  • Ban Al Mandalawi

    thanks a lot

    Reply
  • papa smurf

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

    Reply
  • melvin infant

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

    Reply
  • LousyPainter

    Awesome cool video! Big Thumbs up.

    Reply
  • sahil das

    Can we get more mathematical examples regarding this?

    Reply
  • siddhant kumar sinha

    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

    Reply
  • Amit Ghosh

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

    Reply
  • enas kusuma

    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.

    Reply
  • Dhiraj Kumar Sahu

    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?

    Reply
  • lenin chakravarthy

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

    Reply
  • john moor

    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.

    Reply

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