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.
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 !
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.
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 🎉
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?
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
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.
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 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...
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.
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?
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)
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.
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.
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.
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......
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?
Why when I calculate Vout at a cutoff frequency I get 0.5 of Vin, not 0.707. for example lets say R = 1k, C = 1uF l, fc = 150. Xc = 1061. Vout = Vin*1061/R+1061, R is 1000. Vout = Vin*1000/2000 = 0.5 Vin
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
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 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.
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.
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.
Hello sir...i have one In this video your telled the one problem. In that problem fc=(1)/(2πRC) = (1)/(2*3.14*(10*10^3)*(0.1*10^-6))= i am getting value is 159.2356 hz interms of Khz = 0.159khz but your explained as 1.59khz how sir that.? Can you check and clarify.
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?
Dear friend, if you want the amplitude of the output, same as the input then a buffer amplifier will do and the filter is an active filter. If you use opamp based filter then amplitude can be more than unity....
I have a question. If we can achieve the same filter at different RC combinations, is it better to use smaller R or bigger R values and is it the same answer on high pass filter? Thanks
It actually depends, what is the impedance of the input stage (Source impedance). And at the same time, you need to take into account the operating frequency (for load). Because at the high frequencies, the impedance offered by the capacitor will be low. So, in that case, you need to select the capacitor value such that there is no loading effect due to the capacitor and accordingly select the R. And at the same time, R should be greater than the output impedance of the circuit which acts as an input to the filter. The best practice is to use a buffer circuit at the output of the filter to avoid the loading effect. And one can select the R according to the source impedance. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thanks. I actually use op amp before and after (with gain not just buffer). I use both low pass filter and high pass filter but not certain of the best values. I would appreciate if you can have a video or blog about this. Thanks
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 ?
Hi. May I know how to choose the optimum resistance and capacitance based on not only cut off frequency but also the time constant? (If lower cut off frequency, the time to reach steady state is longer right?)
Since the time is inversely proportional to frequency, so smaller the cut-off frequency, the longer will the settling time. Your actual output depends on your input signal shape and frequency.
Hello Thank for the video, one question I hope you can give me some guiding If there is one more resistor in the circuit parallel to the capacitor (to the right side of the Capacitor) and we are supposed to measure the output voltage across that resistor. how would that affect the filter's output and cut off freq and how should i think when calculating output voltage and cut off freq. Thank you in advance
Like I said, when you cascade two low pass filters, then the second low-pass filter will act as a load for the first filter. Let's say R1 and C1 are the values of resistor and capacitor for first filter and R2 and C2 are the values resistor and capacitor for the second filter. (and let's assume that R1=R2 and C1=C2). So, the input impedance of the second filter R2 +XC2 will be in parallel with XC1. And that will change the cut-off frequency of the filter. So, suppose if you are designing second order passive low-pass filter, then the good practice is to choose R2=10R1 and C2=C1\10 (So, that R1C1=R2C2 and the cutoff frequency of both filters will be same). And in such case when you connect two filters then loading effect will be negligible. (as, R2 +XC2 in new case is much higher than as it was previously, and when it is in parallel with XC1, it will not affect XC1) I hope it will clear your concept about loading effect.
The combination of resistor and capacitor acts as a voltage divider, if there is no resistor, output will be always zero, for all frequencies as current flow through capacitor is always zero.....
Thanks, I applied the example on octave but there is a weird thing Why outputvolt2 don't equal outputvolt1? code: --------------- r=1000; c=0.1*10^(-6); f=2000; inpt=10; Xc=1/(2*3.14*f*c) b=Xc/(r+Xc); a=Xc/(sqrt((r)^2+(Xc)^2)); a b fc=1/(2*3.14*r*c) outputvolt1=inpt*a outputvolt2=inpt*b outpt=inpt*(1/sqrt(2)) ---------------------------- result Xc = 796.18 a = 0.62287 b = 0.44326 fc = 1592.4 outputvolt1 = 6.2287 outputvolt2 = 4.4326 outpt = 7.0711
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!
Been searching all over yt for a good visually demonstration of what LPF and HPF does, and this was the best one. Thanks.
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
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!!!
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
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 !
never seen such kind of useful channel, keep making more videos
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.
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 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
Your way to teach is soooooo amazing that make learning easy 👌👌
Why all the good teaching videos are made by the teacher with the dominant Indian accent :D
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?
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
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.
Thank you very much. You are a genius. 👍👍🙏🙏🔝🔝👌👌
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.....
I just realized that free videos on youtube teach me better than my paid college.
Thank you so much for your valuable information
Surprisingly clear once you get used to in to's.
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
One of the best ways to explain... Thanks a lot
Awesome. Really enjoyed your straight to the point explanation
Thank you for the video, very nice explanation and went into depth about nth order filters and respective roll off.
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
sir i love the way u teach.thank u sir. I'm very happy
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....
Did someone notice how he says "this"?
nice video
Good explanation, keep uploading my dear friend 😊😊
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.
Really very good Narration, keep posting
Yes Mr. Shah, really nice....
Teaching is best 👍... Continue to new lectures
Yes Akansha, really good....
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?
Hi, how is the formula of the overall fc=1/2π(R1C1R2C2)^0.5 derived in the 2nd order low pass filter?
Very helpful video Sir. Thanku
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 😊😊
great video.Very clear explanation. Thankyou..
your videos are very informative. Plz add how to construct equivalent circuit models of various microwave filters, antennas etc.
your explanaion tips is toooo good
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.
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.
Thank you, this channel is brilliant.
i need this specs circuit preamp,
-2 (left and right) input 6 output
-working dc 12V
-all of outputs must be fullrange
?
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.
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..
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.
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......
Please explain me the LC FILTER with half wave & fullwave rectifier . I just love Ur explaination.
Nice explanations. Simple & to the point.
really good please make more vedios sir
Thank you so much for this i just enjoyed your lecture...
Thanks a lot.
Would you like to make a documentary on "the propagation constant" that has the term attenuation and phase shift.
please.....
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
Why when I calculate Vout at a cutoff frequency I get 0.5 of Vin, not 0.707. for example lets say R = 1k, C = 1uF l, fc = 150. Xc = 1061. Vout = Vin*1061/R+1061, R is 1000. Vout = Vin*1000/2000 = 0.5 Vin
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
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.
Thank you so much :) pls keep going
Thank you. I really appreciate your support.
Thank you for the 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
Good work. Please explain the Phasor diagrams also w.r.t. the phase differences.
Would you care to give an explanation of the 2order filter fc= 1/( 2PI sqrt(R1C1R2C2)) why is the formula like this?
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.
Excellent explanations. Two thumbs up.
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.
For first order we take 3 db fall.which db fall is taken for second order? Is it 6 db?
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?
Thanks..It was very helpful
Very nice explanation
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.
Hello. Which type of capacitors should we use? polarized (eletrolytical) or not-polarized?
Non polarized. Ceramic capacitors.
You must use ceramic (non polarised) capacitor......
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.
Its good
Can you please explain about digital electronics .
Hello sir...i have one In this video your telled the one problem. In that problem fc=(1)/(2πRC) = (1)/(2*3.14*(10*10^3)*(0.1*10^-6))= i am getting value is 159.2356 hz interms of Khz = 0.159khz but your explained as 1.59khz how sir that.? Can you check and clarify.
R is 1 kΩ, you have taken 10 kΩ. Please do calculation with 1kΩ, you will get it.
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
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.
can't we just add a unity buffer between the two low pass filters when using passive LPF, or would adding a buffer turn this into an active LPF?
Yes, it can be added. But yes when we use any active element then it's an active filter. But it won't provide any gain.
Dear friend, if you want the amplitude of the output, same as the input then a buffer amplifier will do and the filter is an active filter. If you use opamp based filter then amplitude can be more than unity....
I have a question. If we can achieve the same filter at different RC combinations, is it better to use smaller R or bigger R values and is it the same answer on high pass filter? Thanks
It actually depends, what is the impedance of the input stage (Source impedance). And at the same time, you need to take into account the operating frequency (for load). Because at the high frequencies, the impedance offered by the capacitor will be low. So, in that case, you need to select the capacitor value such that there is no loading effect due to the capacitor and accordingly select the R.
And at the same time, R should be greater than the output impedance of the circuit which acts as an input to the filter.
The best practice is to use a buffer circuit at the output of the filter to avoid the loading effect. And one can select the R according to the source impedance.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thanks. I actually use op amp before and after (with gain not just buffer). I use both low pass filter and high pass filter but not certain of the best values. I would appreciate if you can have a video or blog about this. Thanks
@@ALLABOUTELECTRONICS and I work around 5-30 Hz so I believe frequency is not a major factor here on capacitive impedance.
Yes, I think since you are using input and output buffer and operating at a low frequency, the resistor values won't impact much.
Dear Waleed, you may use input and output buffers to reduce or eliminate effect of loading.....
Super explanation thank you
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.....
sir your filter video is very exellant /good
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.
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.
What about the ESL and the self-inductance in REAL LIFE situations??
Hi. May I know how to choose the optimum resistance and capacitance based on not only cut off frequency but also the time constant? (If lower cut off frequency, the time to reach steady state is longer right?)
Since the time is inversely proportional to frequency, so smaller the cut-off frequency, the longer will the settling time. Your actual output depends on your input signal shape and frequency.
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)
Thanks...
What software is used in making the video?
Hello
Thank for the video, one question I hope you can give me some guiding
If there is one more resistor in the circuit parallel to the capacitor (to the right side of the Capacitor) and we are supposed to measure the output voltage across that resistor. how would that affect the filter's output and cut off freq and how should i think when calculating output voltage and cut off freq.
Thank you in advance
first, you need to find the tranfer function. From transfer function, you can easily find the cut-off frequency equation.
As you have mentioned that loading effect is difficult to overcome in second order filters. What is loading effect?
Like I said, when you cascade two low pass filters, then the second low-pass filter will act as a load for the first filter.
Let's say R1 and C1 are the values of resistor and capacitor for first filter and R2 and C2 are the values resistor and capacitor for the second filter. (and let's assume that R1=R2 and C1=C2). So, the input impedance of the second filter R2 +XC2 will be in parallel with XC1. And that will change the cut-off frequency of the filter.
So, suppose if you are designing second order passive low-pass filter, then the good practice is to choose R2=10R1 and C2=C1\10 (So, that R1C1=R2C2 and the cutoff frequency of both filters will be same). And in such case when you connect two filters then loading effect will be negligible. (as, R2 +XC2 in new case is much higher than as it was previously, and when it is in parallel with XC1, it will not affect XC1)
I hope it will clear your concept about loading effect.
yes, its cleared now..
Thanks
Dear Shafaq, you can refer my channel.....
for the low pass RC filter, can you say that at high frequencies the reactance of the capacitor is low causing a short in the circuit?
Yes, That's true.
Yes Thomas, exactly ....
great video, thank you so much
Sir ye network theory ke filters and analog filters same h kya?I am very confused please tell me?
Yes Sonu, both are same....
Sir, can you please tell what would happen if we remove the resistor from this circuit or in other words what is the need of resistor here?
Same doubt!!. Why do we need resistor
The combination of resistor and capacitor acts as a voltage divider, if there is no resistor, output will be always zero, for all frequencies as current flow through capacitor is always zero.....
Thanks, I applied the example on octave but there is a weird thing
Why outputvolt2 don't equal outputvolt1?
code:
---------------
r=1000;
c=0.1*10^(-6);
f=2000;
inpt=10;
Xc=1/(2*3.14*f*c)
b=Xc/(r+Xc);
a=Xc/(sqrt((r)^2+(Xc)^2));
a
b
fc=1/(2*3.14*r*c)
outputvolt1=inpt*a
outputvolt2=inpt*b
outpt=inpt*(1/sqrt(2))
----------------------------
result
Xc = 796.18
a = 0.62287
b = 0.44326
fc = 1592.4
outputvolt1 = 6.2287
outputvolt2 = 4.4326
outpt = 7.0711
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