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)
Hi, I found a mistake in your video, at 2:51 you can not simply calculate the transfer function by multiplying the functions from the two RC lowpass filters, because the first transfer function changes when you connect a load. You can't just use a normal voltage divider there. You have to calculate the transfer function again using Kirchhoff which adds an extra term, however, it does't change much of the final result. When setting R1 = R2 = R and C1 = C2 = C the new transfer function simply has a 3 instead of a 2 as a factor at 5:29 .
for cascading first circuit to the second, there should a coupling circuit having infinite input resistance and zero output resistance. because of the loading things got different. you are right!
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!
I'm in a 400 level biomechanics class and have to apply a low pass butterworth filter to an EMG signal on a spread sheet for data smoothing. I'm still way lost, this stuff is going over my head.
2:46 bro the second low pass portion will draw some current from the first one, you cannot simply multiply the transfer function of individual filters In such case. Proper nodal analysis will give you the correct transfer function.
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
Choosing R1=R2 and C1=C2 is a more simplified way of choosing the components, but it limits your choice of Q. By choosing 4 separate values of those components, you can actually adjust Q. You don't need to adjust gain with the additional 2 resistors. His video has a lot in it, but I don't think he explained that.
Can you explain why exactly the terms in the denominator of transfer function are what they are? Like how exactly do we know that 1/R1R2C1C2 in the equation was wc^2, the term with s was equal Wc/Q and so on? Are these found from observation? Do we make the assumptions first and then find out from the equation that the gain from transfer function for w = wc becomes equal Q?
4th order filter will be the cascade connection of two 2nd order filters. Hence, there will be two polynomials for each stage. One second order filter will have Q= 0.7654 while the second stage will have Q= 1.8478. So, or the first stage, you can decide the values of R and C by considering Q= 0.7654 and similarly for the second stage you can decide the value by considering Q= 1.8478. I hope it will clear your doubt.
Well if you take any textbook, you won't get much solved examples. There will be practice problems at the end of the chapter. But you can purchase the book for gate,ies or any other competitive exam which covers many solved examples. There you will get many solved examples. Also you can check the second chanel ALL ABOUT ELECTRONICS- QUIZ, which is dedicatedly created for the solved examples on various topics.
If i wanted to add a gain to the 1st order part (the 2nd stage) of the 3rd-order filter, then i would use Q=1 to find R6 and R7 (like R3 and R4 of the 1st stage), is it correct? Since the 3rd order polynomial is (s2+s+1)(s+1) --> we see from the first part, s2+s+1 that Q=1 for the 1st stage, and we see from the second part, s+1 that Q=1 for the 2nd stage. Please let me know if i'm mistaken.
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.
Very good Is it not possible to cascade 4 operationals with gain 1 and without the R3 and R4 to adjust the gain, connecting the output directly to the negative input ?
if I use the S domain to calculate 3dB frequency of first order low pass filter I will have different result form using 1/(2piRC), can you explain me why???Like the example on the first low pass filter video. Using S domain method I have 659Hz and using 1/(2piRC) i got 1,59kHz.
Since the polynomial for the 4th order butterworth filter is (s^2 + 0.7654s + 1)(s^2 + 1.8478s + 1), how would you determine the Q value for the system? Would there be two separate Q values or could you somehow simplify for a single Q value?
Hi, if I am required to calculate the number of order for Butterworth high pass filter, is it normal for me to get a negative value? How do I actually calculate it?
Thank you for excellent explaination sir..... A small suggestion form my side ... It is good if subtitles r placed little bit low we couldn't see equations clearly
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.
I think you haven't watched the video of Butterworth filter properly. When you cascade the two RC filters of same cut-off frequency, then the resultant cut-off frequency won't be the same. Also for Butterworth filter Q= 0.707. Which you will not get just by cascading the two RC filter stages.
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?
For Butterworth filter, Q is 0.707. While for the normal passive filter it is at the most 0.5. Like I said, in the normal passive filter, if you simply cascade the two stages, then the cut-off frequency of the cascaded filter will less than the first-order filter. But due to the higher Q in Butterworth filter (due to positive feedback), even for the higher-order filters, the cut-off frequency remains the same. That's the main advantage.
The higher Q is accomplished by the transfer function. Positive feedback in the Sallen-Key topology is one way to achieve that, but it isn't the only way.
I don't think you can cascade two passive filters like you showed. It will cause loading of the first filter. You will need a buffer between the filters.
at first you told that for butterworth filter, we need small q-factor. Then you said we need positive feedback to get large q-factor. So isn't it contrary?
Suppose the cut-off frequency (Wc= 50000 rad /sec) then the value of R1, R2, and R5 will change. C1, C2, and C5, as well as R4 and R3, will remain same.
But If Wc chagnes, to mantain the polinominal function the same, Q has to change, and that makes some ripple appear in the pass band, I just simulated that... How can I choose a value for Wc and keep Q=0.707 (the value to have no ripples or peaks at Fc)
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..
Yes, that is correct. Instead of fc, it should have been wc= 1000 rad/sec, because the polynomial equations have angular frequency. Thanks for pointing it out. 👍 So, if we take wc= 1000 rad/sec, then design values will get changed now. C1=C2=C5= 0.1 uF and R1=R2=R3=R4=R5= 10 Kilo Ohm.
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.
expression for transfer function of sallen key is not correct. It should be K/(s2(R1R2C1C2)+s(R1C1+R2C1+R1C2(1-K))+1) Check: www.ti.com.cn/cn/lit/an/sloa024b/sloa024b.pdf
Very fast paced video .Infact all your videos are fast paced and difficult to understand. It seems like you have memorized the whole lecture and youre just orally communicating .There is no concept and basic explanation whatsoever.
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 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
Well, I had gone through the site you mentioned. It appears that it requires four different filters with different cut-off frequency. Well, I do not have any circuit diagram, but it can be designed using Op-amp. As Op-amp has very high bandwidth. So, for audio applications, it can be used. But only thing with Op-amp is that all Op-amps are not able to drive the speakers. As they have limited output current. So, it depends on your design, what output you require from the speaker. (e.g 1W. 2W etc) well in such case you can use driver IC's at the output of the Op-amp to drive the speaker. Now, about the design, it seems you will requie one low pass filter, two bandpass filters and one high pass filter. You can go for the second order filters, which will give you 12db/octave of roll-off rate.
Just try not to rush.... The fact that you have that typical "Indie english accent" makes it not easy to understand it... So speak more slowly (10 times)... For the rest it's good explained.
The link for the derivation of the transfer function for the sallen key filter topology.
drive.google.com/open?id=1igXSBw6Rmb_HtWzGtKW9q41lELzbY-V7
If you have other filter notes, kindly share them.
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)
Hi, I found a mistake in your video, at 2:51 you can not simply calculate the transfer function by multiplying the functions from the two RC lowpass filters, because the first transfer function changes when you connect a load. You can't just use a normal voltage divider there. You have to calculate the transfer function again using Kirchhoff which adds an extra term, however, it does't change much of the final result. When setting R1 = R2 = R and C1 = C2 = C the new transfer function simply has a 3 instead of a 2 as a factor at 5:29 .
Yes bro, the transfer function in this video was wrong
for cascading first circuit to the second, there should a coupling circuit having infinite input resistance and zero output resistance. because of the loading things got different. you are right!
What is the value of w(omega)?
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!
I'm in a 400 level biomechanics class and have to apply a low pass butterworth filter to an EMG signal on a spread sheet for data smoothing. I'm still way lost, this stuff is going over my head.
Thank you. The process of deriving formulas and drawing conclusions for higher-order filter design is very interesting
really good explaination. Now I know why I cant get the gain I was trying to get out of my filter. Thanks Boss!
If you want some gain, only active filter can do it....
2:46 bro the second low pass portion will draw some current from the first one, you cannot simply multiply the transfer function of individual filters In such case. Proper nodal analysis will give you the correct transfer function.
Yes bro. In this video the transfer function of second order low pass filter which was determined is fully wrong
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
Choosing R1=R2 and C1=C2 is a more simplified way of choosing the components, but it limits your choice of Q. By choosing 4 separate values of those components, you can actually adjust Q. You don't need to adjust gain with the additional 2 resistors. His video has a lot in it, but I don't think he explained that.
What is the value of w(omega)?
4:20
Do you mean cut off frequency gets shifted because of loading effect ? And from where the w_nc is derived ?
What is the value of w(omega)?
Can you explain why exactly the terms in the denominator of transfer function are what they are? Like how exactly do we know that 1/R1R2C1C2 in the equation was wc^2, the term with s was equal Wc/Q and so on?
Are these found from observation? Do we make the assumptions first and then find out from the equation that the gain from transfer function for w = wc becomes equal Q?
At 7:30 the second term in the denominator should be S[R1C1+R2C2+R1C1(1-k)/R1R2C1C2]
There is a minor typo in case anyone was confused
That part is correct. I guess the correction will be at 8:05 It'll be R1C2 in the denominator
What is the value of w(omega)?
Sir can u please explain how the frequency shifts by cascading? i.e. Wnc=Wc*√(2^(1/n)-1)
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
4th order filter will be the cascade connection of two 2nd order filters.
Hence, there will be two polynomials for each stage.
One second order filter will have Q= 0.7654 while the second stage will have Q=
1.8478. So, or the first stage, you can decide the values of R and C by considering Q= 0.7654 and similarly for the second stage you
can decide the value by considering Q= 1.8478.
I hope it will
clear your doubt.
Yes, thank you very much. My doubt is fully cleared.
@@ALLABOUTELECTRONICS so the coefficient of s from the table is Q? isn't it 1/Q?
13:22 why Q must be equals to 1? isnt the value multiplying s = wc/Q so wc/Q should be equal to 1?
wc is 1 for that equation so Q will also be 1.
What is the value of w(omega)?
What is the w(omega) value? Is it 1?
SIR -PRAY SUGGEST BOOK WITH MANY SOLVED EXAMPLES -THANK U SIR FOR EXCELENT LECTURE
Do you want me to suggest the book for a particular topic or in general for the electronics.
@@ALLABOUTELECTRONICS SIR SOLVED PROBLEMS OP AMP SOLVED EXAMPLES AMPLIFIERS +SOLVED EXAMPLES SEMI CONDUCTORS- THANK U SIR
Well if you take any textbook, you won't get much solved examples. There will be practice problems at the end of the chapter.
But you can purchase the book for gate,ies or any other competitive exam which covers many solved examples.
There you will get many solved examples.
Also you can check the second chanel ALL ABOUT ELECTRONICS- QUIZ, which is dedicatedly created for the solved examples on various topics.
@@ALLABOUTELECTRONICS thank u very much sir -amarjit -advocate
delhi high court
If i wanted to add a gain to the 1st order part (the 2nd stage) of the 3rd-order filter, then i would use Q=1 to find R6 and R7 (like R3 and R4 of the 1st stage), is it correct?
Since the 3rd order polynomial is (s2+s+1)(s+1) --> we see from the first part, s2+s+1 that Q=1 for the 1st stage, and we see from the second part, s+1 that Q=1 for the 2nd stage.
Please let me know if i'm mistaken.
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.
Exactly bro...
Can someone please tell me what the omega constant is. Or sum
please explain the digital electronics and microprrocessor & microcontroller and DSP
Very good
Is it not possible to cascade 4 operationals with gain 1 and without the R3 and R4 to adjust the gain, connecting the output directly to the negative input ?
if I use the S domain to calculate 3dB frequency of first order low pass filter I will have different result form using 1/(2piRC), can you explain me why???Like the example on the first low pass filter video. Using S domain method I have 659Hz and using 1/(2piRC) i got 1,59kHz.
I think i am half done by seeing this video, all i required now is to watch it again.
How can I construct 2nd order high Butterworth filter of cut off frequently 20 khz?
Is this digital signal processing? Because I never had electronic components in the subject
This is analog Butterworth Filter. In DSP, what you have studied is Digital filter.
This type of filters are used for DSP.....
Sir can I have video on vaccum photodiode.photo multipliers.microchannels...if there any plz share the link...
Vaccum photodiode.... ?
Since the polynomial for the 4th order butterworth filter is (s^2 + 0.7654s + 1)(s^2 + 1.8478s + 1), how would you determine the Q value for the system? Would there be two separate Q values or could you somehow simplify for a single Q value?
The Q-factor is only defined for second order filters.
What is the value of w(omega)?
Have you posted full playlist on filter design ?
There is seperate playlist on analog filters.
th-cam.com/play/PLwjK_iyK4LLCQkfK92vdh3gAXoaOXXQDu.html
I want to design filter with 05.hz to 15khz cuttoff.. pls let me know your suggestions
Design a low pass filter of 15K and high pass filter of 0.5 Hz.....
thank you for posting these videos! Good knowledge refresher!
Hi, if I am required to calculate the number of order for Butterworth high pass filter, is it normal for me to get a negative value? How do I actually calculate it?
How did you arrived the polynomial for various orders of filter?
You can find it from the datasheet/application note of the filter. I used it from the one of the application note of the Analog Devices.
@@ALLABOUTELECTRONICS so it cannot be derived uh...
Of course it can be derived. See the Wikipedia article, for example.
Thank you for excellent explaination sir..... A small suggestion form my side ... It is good if subtitles r placed little bit low we couldn't see equations clearly
You can turn it on or off the subtitles manually in the settings. Moreover on the desktop, you can drag it anywhere on the screen.
What is the value of w(omega)?
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.
I think you haven't watched the video of Butterworth filter properly. When you cascade the two RC filters of same cut-off frequency, then the resultant cut-off frequency won't be the same.
Also for Butterworth filter Q= 0.707. Which you will not get just by cascading the two RC filter stages.
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?
For Butterworth filter, Q is 0.707. While for the normal passive filter it is at the most 0.5. Like I said, in the normal passive filter, if you simply cascade the two stages, then the cut-off frequency of the cascaded filter will less than the first-order filter. But due to the higher Q in Butterworth filter (due to positive feedback), even for the higher-order filters, the cut-off frequency remains the same.
That's the main advantage.
@@ALLABOUTELECTRONICS sir does having higher Q help us in achieving brick wall response?
The higher Q is accomplished by the transfer function. Positive feedback in the Sallen-Key topology is one way to achieve that, but it isn't the only way.
You told that 3-k will be valid for k less than 3 what will be the value of Q for k>3 as 1/(k-3) =Q
As K approaches 3, Q tends to infinity and the system becomes unstable. K has to be less than 3 for a stable system and finite Q.
I don't think you can cascade two passive filters like you showed. It will cause loading of the first filter.
You will need a buffer between the filters.
That is what, I have said at 4:22. The cut-off frequency won't remain same. But it will get shifted.
at first you told that for butterworth filter, we need small q-factor. Then you said we need positive feedback to get large q-factor. So isn't it contrary?
How can I design butterworth band pass filter?
Design butterworth LP and butterworth HP filter and cascade it....
This video Helped me a lot. Thanks mate! Keep up with the great work.
What was the middle thing ...?
I am gonna put this on my breadboard.... Wanna see how well it works
what if cut off freq. 50Khz ?? how many value of R1, R2, R5, R3, R4 ??
Suppose the cut-off frequency (Wc= 50000 rad /sec) then the value of R1, R2, and R5 will change. C1, C2, and C5, as well as R4 and R3, will remain same.
ALL ABOUT ELECTRONICS thanksss soo much
But If Wc chagnes, to mantain the polinominal function the same, Q has to change, and that makes some ripple appear in the pass band, I just simulated that... How can I choose a value for Wc and keep Q=0.707 (the value to have no ripples or peaks at Fc)
I would recommend you to use this tool from analog devices for the filter design. It will be very helpful to you.
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..
Yes, that is correct. Instead of fc, it should have been wc= 1000 rad/sec, because the polynomial equations have angular frequency. Thanks for pointing it out. 👍
So, if we take wc= 1000 rad/sec, then design values will get changed now. C1=C2=C5= 0.1 uF and R1=R2=R3=R4=R5= 10 Kilo Ohm.
Can we get more mathematical examples regarding this?
I will post a Quiz related to it in the community Post soon.
Do check that section for more quiz.
Okk sir thank you...
Your statement that the Butterworth filter cannot be implemented passively is incorrect. Look up the Cauer topology.
Do you have a video on schmit trigger circuit. I am unable to find in your channel.
No, it is yet to be made. It will be covered in the ongoing lecture series on the op-amp.
Hello Rakesh, you can find it in my channel....
what does "S" represent?
jw
@@Pablo-lc5dq Hi , jw stand for ?
@@caydengoh2799 j multiplied by omega. j is the imaginary unit and omega is 2*pi*frequency.
S= jw where j is the imaginary part and w is the angular frequency...
We need design for linkwitz riley filter
Can u please provide the note for the derivation
Yes, I will update it very soon in the description.
where are the remainig videos of filters?
Next video will be on filter design. Chebyshev and Bessel Filter Design.
Nice video, nicely explained....
Awesome cool video! Big Thumbs up.
Nice explanation sir...
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.
Exactly bro....
expression for transfer function of sallen key is not correct. It should be
K/(s2(R1R2C1C2)+s(R1C1+R2C1+R1C2(1-K))+1)
Check:
www.ti.com.cn/cn/lit/an/sloa024b/sloa024b.pdf
The equation in the video is correct. The position of C1 and C2 is different in the video and in the pdf. That's why I think you got confused.
Thank you sir, too usefull vedio
Yes bro, really nice video....
wao it deserves a love react .
thank you very much. it helped me alot
Thankyou somuch 👍🏻
Please provide note for derivation
please wait for a couple of days. I will update it in the description.
I need the derivation
Please check the link in the description or the pinned comment for the link of the PDF file.
Good explaination. How about 6th Order low pass?
It is similar to second order, cascade of multiple stages....
Very fast paced video .Infact all your videos are fast paced and difficult to understand. It seems like you have memorized the whole lecture and youre just orally communicating .There is no concept and basic explanation whatsoever.
I agree with you. Im floating and videos are supposed to make it easier
Reduce the playback speed. I watched at 0.75x and it was well paced. Enjoy!
U can use his presentation as notes
I second that. One ore two more examples and a more thorough derivation of the formulas would have made understanding a lot easier.
What is the value of w(omega)?
w=1 rad/sec for all cases
and what will happen if w changes
why we assumed it's 0.14µF
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
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
Well, I had gone through the site you mentioned.
It appears that it requires four different filters with different cut-off frequency.
Well, I do not have any circuit diagram, but it can be designed using Op-amp. As Op-amp has very high bandwidth. So, for audio applications, it can be used. But only thing with Op-amp is that all Op-amps are not able to drive the speakers. As they have limited output current. So, it depends on your design, what output you require from the speaker. (e.g 1W. 2W etc)
well in such case you can use driver IC's at the output of the Op-amp to drive the speaker.
Now, about the design, it seems you will requie one low pass filter, two bandpass filters and one high pass filter.
You can go for the second order filters, which will give you 12db/octave of roll-off rate.
Good explanation !
What is the value of w(omega)?
sir, please provide note for derivation
I have already provided the note. Please check the description or the pinned comment.
You will even find it on the community tab of the channel.
plz see at 13:27
Can you upload a video M drive filter
Just try not to rush.... The fact that you have that typical "Indie english accent" makes it not easy to understand it... So speak more slowly (10 times)...
For the rest it's good explained.
Who is here after gate 2021 electrical engineering syllabus ?
great explanation
I am proof positive that not all Indians are good at math
Thank you sir
stop the background music please
thanks a lot
Very nice
Thanks.
thank you sir
What is the value of w(omega)?