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.
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)
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.
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!
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.
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.
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?
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
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.
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.
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 ?
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?
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 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.
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.
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?
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.
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.
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.
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)
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.
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.
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.