Good explanation, my only remarks> the phase curve's graph can be done much better--in the negative zone and Xc 's value is in Ohms, not just a number. Keep up the work !
it was the most helpful video i have ever watched. thank you so much!! A small favor, if u can provide some notes or the pages you worked on the video in the description of your video, it would be gold and so much appreciated.
The video was very informative and I understood many topics, however I got lost during the equations. If you could explain the equations in detail, that would be awesome. Thanks a lot for making this 🎉
नहीं....यदि आप इसे ठीक से हल करेंगे, तो आपको वही मिलेगा जो वीडियो में दिखाया गया है no....if you will properly solve it, some terms will get cancel out and you will get what is shown in the video
At 3:16 the output that u wrote was vout=vin*xc/(xc+R) ....But according to my knowledge since this is a A.C circuit, So vout=vin*xc/z...and Z=sqrt(xc^2+R^2)...then in denominator instead of xc+R we should have sqrt(x^2+R^2)...Let me know if I'm missing anything. P.O.V:- I'm a Aerospace Engineer trying to understand filters, for my work.
Hi ji, You are very good technically clarifying the circuit analysis. Can you please make one more video for the pi filter circuit design in a technical way.. i hope you wil be support to my request... Thanks in advance
At 9:10 you wrote 2(pi) x 10^3 x 0.1 x 10^-6. PLEASE BEAR WITH ME. If the resistance(R) is 1 kila ohm and the capacitance(C) is 0.1 µF. Why is it not written as 2(pi) x 1 x 0.1 and where did the third power, as well as the six power come from. Why is the sixth power negative? Also is omega a constant. If so, what is the constant value? If you don’t feel like answering all this, I understand can you copy the link to one of your other videos so I can learn where this stuff is coming from?
Very good explanation. Although it took me second to understand that your `7` is not an `F` and your Curls on your `V` was also hard to decipher sometimes.
At 3:13 I’m confused, I’m pretty new to this find of stuff so tell me if I’m wrong. But Vout is Xc•Vin/(Xc + R) which means that multiplying Vin by Xc, then dividing it by Xc, wouldn’t that just cancel out?
Dear Collin, it is the application of voltage division rule. Vin/(R+jXc) is the current through the circuit and product of this current and capacitive reactance will give output voltage......
You just need to know how to write the impedance in amplitude and phase form. You should also know what is reactance. That's it. Perhaps this video will help you. th-cam.com/video/HaFrY0qQ-NU/w-d-xo.html
From 5:35 to 5:57, May I ask how did you jump from the first equation to the next? Considering that the values will be different (As witnessed in Pythagoras Theorem). Really appreciate anyone's help on this! Have a great ahead.
R+ Xc is (R+1/(jwc)). As Xc= 1/jwc. If you consider the only magnitude, it will be Sqrt (R^2 + (1/w^2*C^2)). Or you can say it is sqrt(R^2 +Xc^2). I hope it will clear your doubt.
Vo = Xc * Vin /( Xc +R ) gives amplitude and phase information togather. |Xc / sqrt ( Xc^2 + R^2)| * |Vin| gives the amplitude of Vo. I hope, it will clear your doubt.
Here just the magnitude of the transfer function is considered. If you see the transfer function then it is Xc / Xc + R. Or 1/jwc / [ 1/jwc + R] = 1 / (1 + jwcR) So, if we just consider the magnitude then it is 1 / sqrt (1 + (wCR)^2). If you know about the complex number then you can easily correlate it. e.g if you have A + j B, then its magnitude is sqrt (A^2 + B^2) I hope, it will clear your doubt. If you still have any doubt then let me know here.
Sir,having a doubt.@11:09 in the graph of frequency response when there is the line parallel to frequency axis for low pass band,is that showing the amplitude of input and output are same(i mean is that showing unity gain?) cause if we think the low pass filter as RC circuit,then the amplitude of voltage across the capacitor won't be same as that of the input amplitude,right? Please help sir🙏
You mean in the passband right !! In that band, the gain almost remains constant. Of course as you rightly said, the reactance of the capacitor changes with frequency and so does the relationship between input and output. But in the passband,that change is very subtle. And when the scale is in dB, it hardly noticeable.
How can you obtain Vout when the input signal's frequency is not constant (like in an audio signal)? In other words, is there a way to calculate Vout(t) in terms of Vin(t), R, and C, for any arbitrary instant t?
Good explanation sir.. cut off frequency is 120hz, frequency band 20 to 120hz, its for a 200watts subwoofer, that is output from the amplifier sir.. 1) what could be the value of resistor and capacitor for either second order or third order? 2) How to find the resistor wattage? 3)substitute unit in the cut off frequency formula, ex: resistance in ohm or kilo ohm, capacitance value in farad or microfarad or nf etc,, If you help me solve this sir, it could be useful not only for me but also for everyone sir...
Could you please explain where the -3dB is coming from in the graph at 04:42? I get the cutoff frequency but where are we getting this number from? Thank you so much in advance :)
actually at fc the Vout is 0.707 of Vin so By using the Gain in dB formula which is 20 log (Vout/Vin) you get 20 log (0.707) which is nearly equal to -3 dB. That's how you get it.
Buddy ur work is great.I m big fan of urs buddy Really.I want to ask query, please explain what is the meaning of slope in frequency response? Kya iska matlab ye he k frequency badhega but gain 0 hota jayega? smjha nai pls explain bro.
It means how the fast the amplitude will decay with frequency. Steeper the slope, faster the amplitude will reduce as frequency increase. I hope now it will clear your doubt. :)
Thanx a lot bro... Now I get it.. I have one more query slope Banega hi? Kya esa nai ho skta LPF me fc cutoff frequency k Baad usse next wali frequency Ko reject krna shuru krde!? jaise ideal case m hota h wese!
Yes, Slope will remain. But you can make it steep by increasing the order of the filter. For the first order filter the slope is -20dB/dec, for second order it's -40 dB/ dec, and for the nth order filter is -20n dB/dec. So, by increasing the order of the filter, we can get close to the ideal filter response. But at the same time complexity of the circuit will also increase.
How is the phase lag zero for zero frequency (DC signal)? It will take a while to charge up the capacitor from 0V to DC voltage. Will it not account for phase lag?
For DC, frequency f=0 or w=0. So, phase tan-1 (wCR) will be zero. Of course, capacitor will take finite time to charge to specific value at DC. But it will come under transient time. Here we are taking about the phase lag in steady state condition. Therefore for DC, it will be zero. I hope it will clear your doubt.
The loading effect means, the second stage will act as a load for the first stage. And it will come in parallel with the first stage capacitor. Because of this load, the effective voltage which is appearing at the output of the first stage will reduce. That is the loading effect. To reduce the effect of loading, the value of load should be larger than the output impedance. In case of higher order passive filters, this effect is bound to happen. For more info, please check this video on input and output impedance. And in that video, I have explained it in detail. Here is the link: th-cam.com/video/7jw2_x8dyQ8/w-d-xo.htmlsi=cl7SXMXLBldL_Zmw
i have a question, if i have a time domain signal of a magnitude of 1V, and my low pass filter is of a magnitude of 0.4 , does passing the signal through a low pass filter will affect the time domain magnitude ?
t's basically how fast the capacitor charges.. There's a resistor there so.. it gives resistance. The charge curve is effected, but there's no load resistor to say how fast the capacitor can discharge. So... a specify threshold frequency is controlled or watched.. The resistor in front of the capacitor is kinda like a regulator valve for how much frequency at a specific voltage is allowed to enter or charge the capacitor. What medication can I take to remember all these equations.. lol.. seriously. I WILL learn.. If it takes me my whole life. I've learned a great deal, but more MUST be learned.
Hello friend, Vin/(R+jXc) is the current through the circuit. Output voltage is the product of current flow through the circuit and capacitive reactance. So Vo = jXc×Vin/(R+jXc)
5:12 Hey, like I don't know how to do math yet I know how to assemble, what would you say the best resistance and capacitance for an output of 3.86KHz??? 1/2πRC≈3,860 So far, the best I could get with this formula is ~3,881 with "R=1,000Ω" and "C=410uF" May you help, please? :')
Two gems of online electronics teaching. All about electronics and neso academy.
No doubt!👍
True
True, they are really diamonds!!
instablaster...
Yes
This has to be the best and most straight forward teaching of filters ever. Thanks a ton!
Been searching all over yt for a good visually demonstration of what LPF and HPF does, and this was the best one. Thanks.
Thanks man this videos made me pass my oral exam in electrical engineering
I finally understood the filters! Thanks a bunch and please keep making more videos. You are very helpful. Thanks again!
Thank you soo much sir..... M bahut time s ece k lie lecture dhud rhi thi ... .... thankyou again ❤️❤️❤️❤️❤️
whilst i am doing the studying, you have very much blesed me yaar, i would like to thank v much my friend
You are an amazing human being. Thank you for all of these videos!!!
never seen such kind of useful channel, keep making more videos
Your way to teach is soooooo amazing that make learning easy 👌👌
you saved me, my stupid course gave us a lab about filters and we have to turn in the report before we ever learn what they are even about
Welcome to the real world. Bitching doesn't help
Thank you very much. You are a genius. 👍👍🙏🙏🔝🔝👌👌
Thank you so much :) pls keep going
Thank you. I really appreciate your support.
Good explanation, my only remarks> the phase curve's graph can be done much better--in the negative zone and Xc 's value is in Ohms, not just a number. Keep up the work !
Good explanation, keep uploading my dear friend 😊😊
it was the most helpful video i have ever watched. thank you so much!!
A small favor, if u can provide some notes or the pages you worked on the video in the description of your video, it would be gold and so much appreciated.
The video was very informative and I understood many topics, however I got lost during the equations. If you could explain the equations in detail, that would be awesome. Thanks a lot for making this 🎉
You saved my butt with beautiful explanation. Thanks a ton !
Yes Mr. Paidi..... really nice explanation....
At 7:32 the equation for the angle should be 1/(wCR) as the tangent of the angle is opposite (Xc) over adjacent (R)
yes exactly
नहीं....यदि आप इसे ठीक से हल करेंगे, तो आपको वही मिलेगा जो वीडियो में दिखाया गया है
no....if you will properly solve it, some terms will get cancel out and you will get what is shown in the video
Thank you. Your explanation is direct and easy to understand.
Yes Raman, really nice.....
your way of teaching is very good.please add more and more videos
I will upload many more videos in the upcoming weeks.
Dear friend, you can refer my channel where you will find lot of videos.....
Awesome. Really enjoyed your straight to the point explanation
Teaching is best 👍... Continue to new lectures
Yes Akansha, really good....
At 3:16 the output that u wrote was vout=vin*xc/(xc+R) ....But according to my knowledge since this is a A.C circuit, So vout=vin*xc/z...and Z=sqrt(xc^2+R^2)...then in denominator instead of xc+R we should have sqrt(x^2+R^2)...Let me know if I'm missing anything.
P.O.V:- I'm a Aerospace Engineer trying to understand filters, for my work.
Really very good Narration, keep posting
Yes Mr. Shah, really nice....
Thank you for the video, very nice explanation and went into depth about nth order filters and respective roll off.
One of the best ways to explain... Thanks a lot
thank you so much.
good source to revise and mugg up, keep up the good work.
Dear aadil, dont mugg these topics, understand it, you will never forget it....
Hi ji,
You are very good technically clarifying the circuit analysis. Can you please make one more video for the pi filter circuit design in a technical way.. i hope you wil be support to my request... Thanks in advance
Thank you so much for your valuable information
Thank you, this channel is brilliant.
Why all the good teaching videos are made by the teacher with the dominant Indian accent :D
sir i love the way u teach.thank u sir. I'm very happy
At 9:10 you wrote 2(pi) x 10^3 x 0.1 x 10^-6. PLEASE BEAR WITH ME. If the resistance(R) is 1 kila ohm and the capacitance(C) is 0.1 µF. Why is it not written as 2(pi) x 1 x 0.1 and where did the third power, as well as the six power come from. Why is the sixth power negative? Also is omega a constant. If so, what is the constant value? If you don’t feel like answering all this, I understand can you copy the link to one of your other videos so I can learn where this stuff is coming from?
Nice explanations. Simple & to the point.
Excellent explanations. Two thumbs up.
Very good explanation. Although it took me second to understand that your `7` is not an `F` and your Curls on your `V` was also hard to decipher sometimes.
5:02 output will reduce by 20 db
Surprisingly clear once you get used to in to's.
Very helpful video Sir. Thanku
great video.Very clear explanation. Thankyou..
At 3:13 I’m confused, I’m pretty new to this find of stuff so tell me if I’m wrong. But Vout is Xc•Vin/(Xc + R) which means that multiplying Vin by Xc, then dividing it by Xc, wouldn’t that just cancel out?
Vin is not just divided by Xc, it is getting divided by (Xc + R). So, Xc will not get canceled out.
Dear Collin, it is the application of voltage division rule. Vin/(R+jXc) is the current through the circuit and product of this current and capacitive reactance will give output voltage......
your explanaion tips is toooo good
Did someone notice how he says "this"?
nice video
really good please make more vedios sir
Thank you so much for this i just enjoyed your lecture...
Thank you for the explanation!
i need this specs circuit preamp,
-2 (left and right) input 6 output
-working dc 12V
-all of outputs must be fullrange
?
I just realized that free videos on youtube teach me better than my paid college.
Excellent video! Subscribed.
Thanks..It was very helpful
on 7:12, aren't we taking the phase difference between voltage across resistor and capacitor? Which is clearly 90 degrees.
No, here we are taking the phase difference between the input and the output. The output is the signal across the capacitor.
NYC explaination❤
What concepts from network theory we need to cover to understand frequency response and vo/vin eqn ?
You just need to know how to write the impedance in amplitude and phase form. You should also know what is reactance. That's it.
Perhaps this video will help you.
th-cam.com/video/HaFrY0qQ-NU/w-d-xo.html
From 5:35 to 5:57,
May I ask how did you jump from the first equation to the next?
Considering that the values will be different (As witnessed in Pythagoras Theorem).
Really appreciate anyone's help on this!
Have a great ahead.
R+ Xc is (R+1/(jwc)). As Xc= 1/jwc. If you consider the only magnitude, it will be Sqrt (R^2 + (1/w^2*C^2)). Or you can say it is sqrt(R^2 +Xc^2).
I hope it will clear your doubt.
Hi, how is the formula of the overall fc=1/2π(R1C1R2C2)^0.5 derived in the 2nd order low pass filter?
great video, thank you so much
your videos are very informative. Plz add how to construct equivalent circuit models of various microwave filters, antennas etc.
Great video. I would like to know how a (Vout/Vin) = Xc/(xc + R) suddenly becomes Xc/(sqrt(Xc^2 + R^2). thanks in advance
Vo = Xc * Vin /( Xc +R ) gives amplitude and phase information togather. |Xc / sqrt ( Xc^2 + R^2)| * |Vin| gives the amplitude of Vo. I hope, it will clear your doubt.
Where does the square root of square root of R^2 + Xc^2 at @5:56 come from?
Here just the magnitude of the transfer function is considered. If you see the transfer function then it is Xc / Xc + R.
Or 1/jwc / [ 1/jwc + R] = 1 / (1 + jwcR)
So, if we just consider the magnitude then it is 1 / sqrt (1 + (wCR)^2).
If you know about the complex number then you can easily correlate it.
e.g if you have A + j B, then its magnitude is sqrt (A^2 + B^2)
I hope, it will clear your doubt. If you still have any doubt then let me know here.
At 4:15 , is it -20dB amplitude or power , please confirm. and when dB is mentioned how can I tell if it power or amplitude ?
It's amplitude. The power in dB is described as 10 log (P2/P1), while voltage in dB is described as 20 log ( V2/V1)
Very nice explanation
I didn't understand in 10:14 minutes
.. how Vout = |xc|/√(r²+Xc²) ?
plz help.
Look at the circuit diagram it's similar to voltage divider and therefore r2×Vin/(r2 + r1). Then replace r2 by Xc and r1 by R as given.
Sir,having a doubt.@11:09 in the graph of frequency response when there is the line parallel to frequency axis for low pass band,is that showing the amplitude of input and output are same(i mean is that showing unity gain?) cause if we think the low pass filter as RC circuit,then the amplitude of voltage across the capacitor won't be same as that of the input amplitude,right?
Please help sir🙏
You mean in the passband right !! In that band, the gain almost remains constant. Of course as you rightly said, the reactance of the capacitor changes with frequency and so does the relationship between input and output. But in the passband,that change is very subtle. And when the scale is in dB, it hardly noticeable.
How can you obtain Vout when the input signal's frequency is not constant (like in an audio signal)? In other words, is there a way to calculate Vout(t) in terms of Vin(t), R, and C, for any arbitrary instant t?
in 14:23 what do you mean by the term loaded
Super explanation thank you
at 5:03 the output reduces by a factor of 20 if frequency is increased by a factor of 10
It is in dB. If you convert the dB into normal amplitude then it will reduce by the factor of 10.
Good explanation sir.. cut off frequency is 120hz, frequency band 20 to 120hz, its for a 200watts subwoofer, that is output from the amplifier sir..
1) what could be the value of resistor and capacitor for either second order or third order?
2) How to find the resistor wattage?
3)substitute unit in the cut off frequency formula, ex: resistance in ohm or kilo ohm, capacitance value in farad or microfarad or nf etc,,
If you help me solve this sir, it could be useful not only for me but also for everyone sir...
We are all waiting for your valuable reply sir
Could you please explain where the -3dB is coming from in the graph at 04:42? I get the cutoff frequency but where are we getting this number from? Thank you so much in advance :)
actually at fc the Vout is 0.707 of Vin so By using the Gain in dB formula which is 20 log (Vout/Vin) you get 20 log (0.707) which is nearly equal to -3 dB. That's how you get it.
Please explain me the LC FILTER with half wave & fullwave rectifier . I just love Ur explaination.
how can we use LPF to turn PWM to analog signal?
Yes, you can. I think, in one of my old video on PWM I have already mentioned that. Please check that video for more info.
Fantastic! Thank you
Buddy ur work is great.I m big fan of urs buddy Really.I want to ask query, please explain what is the meaning of slope in frequency response? Kya iska matlab ye he k frequency badhega but gain 0 hota jayega? smjha nai pls explain bro.
It means how the fast the amplitude will decay with frequency. Steeper the slope, faster the amplitude will reduce as frequency increase.
I hope now it will clear your doubt. :)
Thanx a lot bro... Now I get it..
I have one more query slope Banega hi?
Kya esa nai ho skta LPF me fc cutoff frequency k Baad usse next wali frequency Ko reject krna shuru krde!? jaise ideal case m hota h wese!
Yes, Slope will remain. But you can make it steep by increasing the order of the filter.
For the first order filter the slope is -20dB/dec, for second order it's -40 dB/ dec, and for the nth order filter is -20n dB/dec.
So, by increasing the order of the filter, we can get close to the ideal filter response.
But at the same time complexity of the circuit will also increase.
Thanx yaraa...
Sara concept clear ho gya thanx bro.. khoob aage badho.. tarakki karo..
How is the phase lag zero for zero frequency (DC signal)? It will take a while to charge up the capacitor from 0V to DC voltage. Will it not account for phase lag?
For DC, frequency f=0 or w=0. So, phase tan-1 (wCR) will be zero. Of course, capacitor will take finite time to charge to specific value at DC. But it will come under transient time. Here we are taking about the phase lag in steady state condition. Therefore for DC, it will be zero.
I hope it will clear your doubt.
Good work. Please explain the Phasor diagrams also w.r.t. the phase differences.
Thank you, nice explanation (Y)
What is the loading effect at @14:20
The loading effect means, the second stage will act as a load for the first stage. And it will come in parallel with the first stage capacitor. Because of this load, the effective voltage which is appearing at the output of the first stage will reduce. That is the loading effect. To reduce the effect of loading, the value of load should be larger than the output impedance. In case of higher order passive filters, this effect is bound to happen. For more info, please check this video on input and output impedance. And in that video, I have explained it in detail.
Here is the link: th-cam.com/video/7jw2_x8dyQ8/w-d-xo.htmlsi=cl7SXMXLBldL_Zmw
Thank you sir
Would you care to give an explanation of the 2order filter fc= 1/( 2PI sqrt(R1C1R2C2)) why is the formula like this?
can you give the cut off frequency formula for nth order filter with R and C are different
Thanks...
What software is used in making the video?
For first order we take 3 db fall.which db fall is taken for second order? Is it 6 db?
in 6:52 you say that when you further simplify 1/2=((1/wc)^2)/(R^2+(1/wc)^2) to w=1/RC i dont understand how this makes any sense mathematically.
Its good
Can you please explain about digital electronics .
At 9:33 shouldn’t it be .159 KHz? Or am I missing something? Thanks in advance.
no hes right its 1.59khz. make sure you are using the right exponent for the capacitor it should be 10^-6.
(1/(2*pi*1000*0.1*10^-6))
@@sum1528 thanks for the help!
i have a question, if i have a time domain signal of a magnitude of 1V, and my low pass filter is of a magnitude of 0.4 , does passing the signal through a low pass filter will affect the time domain magnitude ?
If in the passband, the amplitude of the filter transfer function is 0.4, then even if you applied 1v signal, at the output you will get 0.4V signal.
thanks for clear concept
sir pls make a video on1) types of memory and its application2) flop flop 3)type of sensor
I have made a video on different types of memory.
t's basically how fast the capacitor charges.. There's a resistor there so.. it gives resistance.
The charge curve is effected, but there's no load resistor to say how fast the capacitor can discharge.
So... a specify threshold frequency is controlled or watched.. The resistor in front of the capacitor is kinda like a regulator valve for how much frequency at a specific voltage is allowed to enter or charge the capacitor.
What medication can I take to remember all these equations.. lol.. seriously.
I WILL learn.. If it takes me my whole life. I've learned a great deal, but more MUST be learned.
Thanks អរគុណ❤
excellent teaching thank u
Nice 😘❤️😍
Thank you sooo much I understood this concept very clearly
But one doubt how did we get
Vout=(xc/xc+R) ×Vin
This equation
Using the voltage divider rule.
@@ALLABOUTELECTRONICS thank you soo much
Hello friend, Vin/(R+jXc) is the current through the circuit. Output voltage is the product of current flow through the circuit and capacitive reactance. So Vo = jXc×Vin/(R+jXc)
@@circuitsanalytica4348 thank you 😊😊
Excellent
I have a question. Can I get the link to Your film to my students as a practical application of the filters?
Yes, you can.
Thanks a lot.
Would you like to make a documentary on "the propagation constant" that has the term attenuation and phase shift.
please.....
Thanks for shared!
5:12
Hey, like I don't know how to do math yet I know how to assemble, what would you say the best resistance and capacitance for an output of 3.86KHz???
1/2πRC≈3,860
So far, the best I could get with this formula is ~3,881 with "R=1,000Ω" and "C=410uF"
May you help, please? :')
You can use C = 10 nF and R around 4.1 KΩ. If required, you can use trim pot to adjust the value of resistor.
6:45 how Vout/Vin = 1/sqrt(2) ... how it came please let me know anyone..
using the definition of cutoff frequency or 3db frequency