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
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.
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.
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 🎉
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
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 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?
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.
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.
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...
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......
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)
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. ruclips.net/video/HaFrY0qQ-NU/видео.html
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?
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.
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.
Class is very informative sir .I have question that a signal passes through a low pass filter with cut off frequency 10Khz then what is the minimum sampling frequency?
Good job explaining. Appreciate it. Can you kindly explain to me that is it possible to use one of these filters to chop down frequency coming out of a high frequency (HF) transformer @220v, to 50 Hz @220v +- 5%?
Filter selectively passes certain frequencies and removes the unwanted ones. In your case, if HF transformer does not contain the 50Hz frequency component then it certainly cannot be filtered out using filters. Filters cannot perform the frequency conversion, it can only select or reject the band of frequencies from the input. I hope you understood what I mean to say.
As the frequency increases, the impedance of c1 will increase. And for high cutoff frequencies, it will be in Mega ohm. And it can be considered as open circuit for such case. So R1 and R2 will be in series. Now because of that, the cutoff frequency of the second stage will get changed. It can be reduced by selecting R2 as atleast 10 times R1. Or that effect can be almost removed by using the active filter. I hope it will clear your doubt.
Great explanation 😊 just 1 question: on 12th min. you said with the increasing of order of the filter, amplitude at 2kHz is decreasing. But in the graph, it seems all amplitudes are the same. Could you explain?
Although it seems that, but at the given frequency if you see the amplitude on the vertical axis, it will be different. For example, at a 2kHz frequency, I have drawn a vertical line. Now, if you see the intersection of that line with the different line (Blue, Red, pink) and corresponding amplitude on the y-axis, it will be different. I hope it will clear your doubt.
In the RC exercise, you first found the reactance at cutoff frequency (796) and then used it to calculate the output voltage. But since input frequency is 2khz dont you have to recalculate reactance before finding output voltage?
At low frequencies (lesser than cut-off frequency), Xc will be high enough and can be assumed as open circuit for analysis. (And R1 and R2 will be in series). If R1 and R2 are comparable then R1 will change the cut-off frequency of second filter. By selecting the R2=10R1, the effect of R1 on second filter can be reduced. I hope it will clear your doubt.
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!
I finally understood the filters! Thanks a bunch and please keep making more videos. You are very helpful. Thanks again!
Thanks man this videos made me pass my oral exam in electrical engineering
Been searching all over yt for a good visually demonstration of what LPF and HPF does, and this was the best one. Thanks.
whilst i am doing the studying, you have very much blesed me yaar, i would like to thank v much my friend
Thank you soo much sir..... M bahut time s ece k lie lecture dhud rhi thi ... .... thankyou again ❤️❤️❤️❤️❤️
You are an amazing human being. Thank you for all of these videos!!!
never seen such kind of useful channel, keep making more videos
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
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 so much :) pls keep going
Thank you. I really appreciate your support.
Your way to teach is soooooo amazing that make learning easy 👌👌
You saved my butt with beautiful explanation. Thanks a ton !
Yes Mr. Paidi..... really nice explanation....
Thank you. Your explanation is direct and easy to understand.
Yes Raman, really nice.....
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 !
Awesome. Really enjoyed your straight to the point explanation
Why all the good teaching videos are made by the teacher with the dominant Indian accent :D
I just realized that free videos on youtube teach me better than my paid college.
Good explanation, keep uploading my dear friend 😊😊
Surprisingly clear once you get used to in to's.
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.
5:02 output will reduce by 20 db
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.
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.....
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 🎉
Thank you very much. You are a genius. 👍👍🙏🙏🔝🔝👌👌
Thank you for the video, very nice explanation and went into depth about nth order filters and respective roll off.
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
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.
sir i love the way u teach.thank u sir. I'm very happy
Teaching is best 👍... Continue to new lectures
Yes Akansha, really good....
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....
Really very good Narration, keep posting
Yes Mr. Shah, really nice....
One of the best ways to explain... Thanks a lot
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?
Did someone notice how he says "this"?
nice video
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.
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.
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.
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.
Nice explanations. Simple & to the point.
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.
Thank you so much for your valuable information
Thank you, this channel is brilliant.
Thank you so much for this i just enjoyed your lecture...
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.
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
Very helpful video Sir. Thanku
your explanaion tips is toooo good
really good please make more vedios sir
Please explain me the LC FILTER with half wave & fullwave rectifier . I just love Ur explaination.
your videos are very informative. Plz add how to construct equivalent circuit models of various microwave filters, antennas etc.
i need this specs circuit preamp,
-2 (left and right) input 6 output
-working dc 12V
-all of outputs must be fullrange
?
great video.Very clear explanation. Thankyou..
Excellent explanations. Two thumbs up.
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.
Good work. Please explain the Phasor diagrams also w.r.t. the phase differences.
Thanks a lot.
Would you like to make a documentary on "the propagation constant" that has the term attenuation and phase shift.
please.....
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......
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 😊😊
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.
Very nice explanation
why idian teacher speak english really hard to hear althought their content is so wonderful and helpful.
Coz we r Indian and not an American/Britisher
Its good
Can you please explain about digital electronics .
Thank you for the explanation!
Thanks..It was very helpful
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.
ruclips.net/video/HaFrY0qQ-NU/видео.html
there will be a root over in Xc + Rc on the Vout formula
do a series in Signals and systems
Hi, how is the formula of the overall fc=1/2π(R1C1R2C2)^0.5 derived in the 2nd order low pass filter?
Super explanation thank you
NYC explaination❤
nice video! would be easier to follow if you didn't erase the last part of your equations when you go into the next step.
Hai, you may refer my channel, explained in a different way.....
Excellent video! Subscribed.
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?
excellent teaching thank u
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)
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..
Thanks...
What software is used in making the video?
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.
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.
Class is very informative sir .I have question that a signal passes through a low pass filter with cut off frequency 10Khz then what is the minimum sampling frequency?
Do you mean to say, what should be the sampling frequency of the filtered signal?
@@ALLABOUTELECTRONICS will you plz tell me how can f=2000?
I didn’t understand.
@@moriomchowdhury9170 2khz input fq is f=2000hz
Great videos, do you provide notes?
Good job explaining. Appreciate it. Can you kindly explain to me that is it possible to use one of these filters to chop down frequency coming out of a high frequency (HF) transformer @220v, to 50 Hz @220v +- 5%?
Filter selectively passes certain frequencies and removes the unwanted ones. In your case, if HF transformer does not contain the 50Hz frequency component then it certainly cannot be filtered out using filters. Filters cannot perform the frequency conversion, it can only select or reject the band of frequencies from the input.
I hope you understood what I mean to say.
What about the ESL and the self-inductance in REAL LIFE situations??
please provide lecture on RL low pss, RL high pass filter....
great video, thank you so much
sir your filter video is very exellant /good
Careful from 14:04. please give some info for last portion.
As the frequency increases, the impedance of c1 will increase. And for high cutoff frequencies, it will be in Mega ohm. And it can be considered as open circuit for such case. So R1 and R2 will be in series. Now because of that, the cutoff frequency of the second stage will get changed. It can be reduced by selecting R2 as atleast 10 times R1. Or that effect can be almost removed by using the active filter.
I hope it will clear your doubt.
Would you care to give an explanation of the 2order filter fc= 1/( 2PI sqrt(R1C1R2C2)) why is the formula like this?
What program do you use to write things down (black screen with yellow letters)?
For first order we take 3 db fall.which db fall is taken for second order? Is it 6 db?
Thank you sir
Great explanation 😊 just 1 question: on 12th min. you said with the increasing of order of the filter, amplitude at 2kHz is decreasing. But in the graph, it seems all amplitudes are the same. Could you explain?
Although it seems that, but at the given frequency if you see the amplitude on the vertical axis, it will be different. For example, at a 2kHz frequency, I have drawn a vertical line. Now, if you see the intersection of that line with the different line (Blue, Red, pink) and corresponding amplitude on the y-axis, it will be different.
I hope it will clear your doubt.
Thank you :) Intersection of that vertical line with blue one cprresponds to 0. Should it be so?
In the RC exercise, you first found the reactance at cutoff frequency (796) and then used it to calculate the output voltage. But since input frequency is 2khz dont you have to recalculate reactance before finding output voltage?
The value of reactance is found at 2 kHz only. (10:05)
The cutoff frequency is 1.59 kHz.
Aah.. right. My mistake.
Can you also please explain LC low pass filter, and how to choose filters RC LR and LC by usage, when you have time?
Yes, whenever I get some time, I will make a video on it
Is there a derivation for the output phase? great vids though they help a lot!
thanks for the video and I have one doubt how loading effect will be reduced by making R2=10R1
At low frequencies (lesser than cut-off frequency), Xc will be high enough and can be assumed as open circuit for analysis. (And R1 and R2 will be in series). If R1 and R2 are comparable then R1 will change the cut-off frequency of second filter. By selecting the R2=10R1, the effect of R1 on second filter can be reduced.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS sorry I can't understand
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.