Amazing video, thanks!!! I saw 20 videos of audio people talking about crossovers and hi-low pass filters, yours is the only one that really explains it well.
I was thinking about getting into some basic analog signal processing for radio signals as well as things like designing cross overs for custom speakers. This helped a lot. Thank you 🙏
Already covered questions related to it. Please check the second channel related to quiz. You will find there. In case, if you are not able to get it, then let me know here.
Thanks for the video. What did you mean when you mentioned the next stage will get loaded if you choose the value of R in Ohms? I just don't understand the concept!!
Hello Dean If value of R is very small, when a load is connected, effective value of input impedance of the next stage will be very small and effective value of output voltage falls or it leads to loading....
Loading effect is what occurs when you directly cascade the output port of one network with the input port of the next (without providing any buffer or active components in between). Basically, the input voltage at the second network is not what we would normally obtain as the output of the first network (here, filter) assuming infinite impedance. Here, the second filter would draw some current from the first filter and thereby, further reducing the gain of the first filter. So, in order to reduce this loading effect, the impedance of the second filter must be as high as possible.
Oh sir so in Rx phase shift oscillator we are using a 3 stage cascaded RC high pass filter in feedback path right So is it like they pass all frequency components of the noise above this cut off frequency of 1/2πRC but only this 1/2πRC frequency makes transfer function of feedback network Beta=1/29 and with A=29 we can satisfy Barkhausen criteria and they sustain And for all the ones above this frequency that has passed this feedback filter network do not satisfy Barkhausen criteria thus won't be sustained??? Sorry for such a long question
Dear Abhijith, in RC Coupled amplifier, for sustained oscillations barkhausens criteria must be satisfied. Total phase shift is 360 degree only for those frequency to which the oscillator is tuned to.
When phase information of Xc is not required and only amplitude of the reactance is required at that time I have used |Xc|= 1/wC. I hope now it will get clear to you.
sir , i was unable to understand the part where you assumed the value of Resistance as 10 Kohm.......what is the next level loaded you are reffering to?....plz do reply
when you connect the circuit next to the filter, the connected circuit will act as a load to the filter. Let's say the input impedance of the circuit which we want to connect to the filter is Zin, then the effective load or R-value for the high pass filter is R || Zin. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS but as far i know, when impedances are in parallel, net impedance gets reduced. so how does it increase the loading? please explain
Also at 11:39 of this video, you said, " So, now suppose, in your design if you want that at 1 KHz..." I'm not quite sure what we want at 1 KHz. "that" refers to what? Thanks!!
He meant if you don't want that magnitude of noise at 1KHz then higher order filters are needed to reduce the magnitude of that said 1KHz noise even further. By now you could have figured it out. This is for those who have the same doubt and opened your comment. Correct me if I am wrong.
When you connect the load to this filter (Let's say speaker), then the impedance of that load will be in parallel with resistor R. And in many cases, the (like in case of a speaker) the value of load is in ohms. So, that will change the effective value of the resistor R and hence the cut-off frequency also. Let's say in your filter design R is 1Kilo ohm and the load is 50 ohm then effective resistance R will be less than 50 ohms and your designed cut-off frequency will get changed. So, that is the loading effect on the filter. So, in your design value of R should be appropriate. If it's too small then it will draw too much current and if it is too high then value of capacitor will not small (and might be comparable to parasitic capacitance of the circuit) The best way to avoid loading effect is to isolate the load from the filter circuit, or in general, for any circuit it is true, and it can be easily done by using OP-AMP as a buffer. To know more about it you can check my video on the active low pass and high pass filter.
Hello, when a load is connected to the filter, effective value of R will change, leading to a shift in cutoff frequency, possibilty of loading. So to avoid that, it will be better if a buffer amplifier is used in between....
I have explained it in the Butterworth Filter video. Here is the link: ua-cam.com/video/lc6QT8VjqVc/v-deo.html If you still have any doubt after watching it, let me know here.
So would just the positive terminal coming off an amplifier go into this circuit and the resistor gets its own ground or does the positive and negative terminals from the amplifier go into this circuit?
when finding phase of transfer function how does tan (angle) = 1/(1-j/wrc) simplify to tan(angle) = 1/wrc I have looked into by trig books but can not find how this is solved. Thanks
If you have complex number, (A + jB) / (C + jD), then phase angle,= tan -1( B/A) - tan-1 (D/C) Comparing it with, 1/(1- j/wRC) Here in the numerator B=0, A= 1, C= 1 and D = -1/wRC If you put all these in the above expression then you will get it. I hope it will clear your doubt.
could you please inform me where to look for the formula (A + jB) / (C + jD)??? and also when you w=0 then angle =90!! but wouldn't that be tan-1(1/0*R*C)???? you cant divide by zero right???
Here high pass filter with a 10 kHz cut-off frequency is designed. So, f is already known. The value of R is selected as 10 kilo ohm. And with that value of R, the capacitor value has been found for 10 kHz cut-off frequency.
I can cut an existing rca cable in half, solder the resistor + capacitor into the two wires, put it all back together... And that will work fine, right?
In the video of input and output impedance, I have explained the loading effect. You may check that video. And still if you have any doubt then do let me know here. Here is the link: ua-cam.com/video/7jw2_x8dyQ8/v-deo.html
No, it is not Butterworth high pass filter. The Q should be equal to 0.707 for the Butterworth filter. Please check my video on Butterworth filter for more information about the Butterworth filter. At the latter part of the video, I have explained how to design the Butterworth high pass filter. Here is the link: ua-cam.com/video/lc6QT8VjqVc/v-deo.html
It comes from taking the magnitude of the values and the magnitude of a complex number |a+bi|=sqrt(a^2+b^2), all other values in the equation are real numbers therefor they don't change.
Amazing video, thanks!!! I saw 20 videos of audio people talking about crossovers and hi-low pass filters, yours is the only one that really explains it well.
you got me through my ee degree and here you are getting me through interviews
sir,
yr video on pASSIVE HIGH PASS FILTER AND AN EXAMPLE OF DESIGNING A CIRCUIT IS VERY GOOD AND CLEAR. GOOD EXPLANATION.
THANKS
S.VATSA, BANGALORE
bro you are actually a legend at teaching
Your intro sounds like your making makeup tutorials :D
well said Mr. president
I was thinking about getting into some basic analog signal processing for radio signals as well as things like designing cross overs for custom speakers. This helped a lot. Thank you 🙏
sir
yr video on BAND PASS FILTER AND YR EXPLANATION IS EXCELLENTLY BROUGHT OUT. IT HELPED ME TO UNDERSTAND BETTER.
THANKS
S.VATSA. BANGALORE
all about electronics is nothing but life saver
Yes Sledge, you are right.....
I'm glad to see this video,it solve my doubts about hpf thank u.........
Yes Ganesh, really nice video....
Superb video keep up the good work ✌
This was insanely helpful. Thanks!
Very nicely described... Thank you so much
nice explanation keep it up👍👍
Perfect explanation, thanks
Very well explained
Thanks.. Would be helpful if you could also solve some questions..
Already covered questions related to it. Please check the second channel related to quiz. You will find there. In case, if you are not able to get it, then let me know here.
Excellent
Excellent ❤
Sir why didn't we place the Buffer in Second Order Low Pass Filter as we have placed for the Second order High Pass Filter
Very well Narration
Thanks for the video. What did you mean when you mentioned the next stage will get loaded if you choose the value of R in Ohms? I just don't understand the concept!!
Hello Dean If value of R is very small, when a load is connected, effective value of input impedance of the next stage will be very small and effective value of output voltage falls or it leads to loading....
Loading effect is what occurs when you directly cascade the output port of one network with the input port of the next (without providing any buffer or active components in between). Basically, the input voltage at the second network is not what we would normally obtain as the output of the first network (here, filter) assuming infinite impedance. Here, the second filter would draw some current from the first filter and thereby, further reducing the gain of the first filter. So, in order to reduce this loading effect, the impedance of the second filter must be as high as possible.
please make videos on transfer function
Oh sir so in Rx phase shift oscillator we are using a 3 stage cascaded RC high pass filter in feedback path right
So is it like they pass all frequency components of the noise above this cut off frequency of 1/2πRC but only this 1/2πRC frequency makes transfer function of feedback network Beta=1/29 and with A=29 we can satisfy Barkhausen criteria and they sustain
And for all the ones above this frequency that has passed this feedback filter network do not satisfy Barkhausen criteria thus won't be sustained???
Sorry for such a long question
Dear Abhijith, in RC Coupled amplifier, for sustained oscillations barkhausens criteria must be satisfied. Total phase shift is 360 degree only for those frequency to which the oscillator is tuned to.
when you took the magnitude how come the denominator Xc and R became summation of square of it at 4:01 ?
Please explain it kindly
why you sometime use sometime Xc = 1/(jwC) sometimes Xc = 1/(wC) ? i am confused.
I have used Xc= 1 /wC, when I am talking about only amplitude of Xc, or I would say it is |Xc|.
ALL ABOUT ELECTRONICS I did not get..
When phase information of Xc is not required and only amplitude of the reactance is required at that time I have used |Xc|= 1/wC. I hope now it will get clear to you.
Hello Arya, the term "j" make it a complex number. Both are correct, first one is a vector (with j) and second one is scalar....
@@ALLABOUTELECTRONICS More information is needed here.
sir , i was unable to understand the part where you assumed the value of Resistance as 10 Kohm.......what is the next level loaded you are reffering to?....plz do reply
when you connect the circuit next to the filter, the connected circuit will act as a load to the filter. Let's say the input impedance of the circuit which we want to connect to the filter is Zin, then the effective load or R-value for the high pass filter is R || Zin.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS oo so net impedance will be the parallel combination of both ....now I get it
Thanku very much sir
@@ALLABOUTELECTRONICS but as far i know, when impedances are in parallel, net impedance gets reduced. so how does it increase the loading? please explain
Also at 11:39 of this video, you said, " So, now suppose, in your design if you want that at 1 KHz..." I'm not quite sure what we want at 1 KHz. "that" refers to what? Thanks!!
He meant if you don't want that magnitude of noise at 1KHz then higher order filters are needed to reduce the magnitude of that said 1KHz noise even further. By now you could have figured it out. This is for those who have the same doubt and opened your comment. Correct me if I am wrong.
What does it mean that the circuit might get loaded by the value of R at 8:34?
When you connect the load to this filter (Let's say speaker), then the impedance of that load will be in parallel with resistor R. And in many cases, the (like in case of a speaker) the value of load is in ohms.
So, that will change the effective value of the resistor R and hence the cut-off frequency also.
Let's say in your filter design R is 1Kilo ohm and the load is 50 ohm then effective resistance R will be less than 50 ohms and your designed cut-off frequency will get changed.
So, that is the loading effect on the filter.
So, in your design value of R should be appropriate. If it's too small then it will draw too much current and if it is too high then value of capacitor will not small (and might be comparable to parasitic capacitance of the circuit)
The best way to avoid loading effect is to isolate the load from the filter circuit, or in general, for any circuit it is true, and it can be easily done by using OP-AMP as a buffer.
To know more about it you can check my video on the active low pass and high pass filter.
That's a very good and clear explanation. I really appreciate it :')
Hello, when a load is connected to the filter, effective value of R will change, leading to a shift in cutoff frequency, possibilty of loading. So to avoid that, it will be better if a buffer amplifier is used in between....
What does it mean for a stage to get "loaded"?
I have explained it in the Butterworth Filter video.
Here is the link: ua-cam.com/video/lc6QT8VjqVc/v-deo.html
If you still have any doubt after watching it, let me know here.
what am i doing here? I am a taxi driver.
Did Uber drive you here?
saboor saboor xDdddd
Quarantine day: 8
No profession is barred from learning, silly goose. Now go wreak havoc with your new knowledge!
Unit of capacitance is Farad and not Faraday
Yes Jairam you are right....
Could you elaborate a bit about the j thing?
So would just the positive terminal coming off an amplifier go into this circuit and the resistor gets its own ground or does the positive and negative terminals from the amplifier go into this circuit?
when finding phase of transfer function how does tan (angle) = 1/(1-j/wrc) simplify to tan(angle) = 1/wrc I have looked into by trig books but can not find how this is solved. Thanks
If you have complex number, (A + jB) / (C + jD), then phase angle,= tan -1( B/A) - tan-1 (D/C)
Comparing it with, 1/(1- j/wRC)
Here in the numerator B=0, A= 1, C= 1 and D = -1/wRC
If you put all these in the above expression then you will get it.
I hope it will clear your doubt.
could you please inform me where to look for the formula (A + jB) / (C + jD)??? and also when you w=0 then angle =90!! but wouldn't that be tan-1(1/0*R*C)???? you cant divide by zero right???
At 2:50 why the cutoff frequency is 1/√2 of input voltage??
Because at cut-off frequency, the power reduced by 3 dB or the power becomes half. So in terms of voltage, it becomes 1/√2.
@@ALLABOUTELECTRONICS why the power reduce to 3dB
was expecting more mathematical problems
Iam not getting@ 9:23 How you select the frequency value for calculating the value of capacitor
Here high pass filter with a 10 kHz cut-off frequency is designed. So, f is already known. The value of R is selected as 10 kilo ohm. And with that value of R, the capacitor value has been found for 10 kHz cut-off frequency.
What about the phase change in higher order filters ?? 7:29
I love the accent. Is that Hindi?
Jefe Julio el Gringo Judeo yes it's hindi
Punjabi
I can cut an existing rca cable in half, solder the resistor + capacitor into the two wires, put it all back together... And that will work fine, right?
Hello Mr.Russell, It will work for frequencies greater than cutoff frequency....
How did R/(R+ 1/jwc) become 1/(1-j/wcR) at 6:03 ?
First divide both numerator and denominator by R and then put (-j) in place of (1/j) as they are same.
@@rupeshkarar You're a legend, thank you so much for explaining this. I have been stuck because of this for a week.
Didn't understand why you chose the resistance value as such.
Dear friend, you can refer my channel for more details.....
HPF leads LPF by 90 degrees.
for second order high pass give the phase response
Sir...what is meant by loading...which you mentioned in video...?
In the video of input and output impedance, I have explained the loading effect. You may check that video. And still if you have any doubt then do let me know here.
Here is the link:
ua-cam.com/video/7jw2_x8dyQ8/v-deo.html
what is the highest amount of freq. component the filter can pass?
Dear friend, highest frequency is infinite ideally....
nice information
Yeah Smruti really nice videos....
10:29
do you mean 20kΩ "rheostat"?
Sir what about LC filters...please make them if u can...and .thanks for these vedios....
Yes, in this analog filter series, I will also make a video on LC filters.
ALL ABOUT ELECTRONICS Thank you sir
ALL ABOUT ELECTRONICS why we shouldn't use 10 mega ohm.. Explain detail
Is this buttorworth high pass active filter...
No, it is not Butterworth high pass filter. The Q should be equal to 0.707 for the Butterworth filter.
Please check my video on Butterworth filter for more information about the Butterworth filter. At the latter part of the video, I have explained how to design the Butterworth high pass filter.
Here is the link:
ua-cam.com/video/lc6QT8VjqVc/v-deo.html
What here I was referring was for the higher order filter. (second order or more)
when we take magnitude we square the denominator but not numerator ..why?
It comes from taking the magnitude of the values and the magnitude of a complex number |a+bi|=sqrt(a^2+b^2), all other values in the equation are real numbers therefor they don't change.
I DONT UNDERSTAND ... HOW A LOW FREQUENCY IS CUT OFF?
Hello friend, can you be more specific, it didn't get you?
@@circuitsanalytica4348 yes i dont get it. A capasitors will just increase the volts and it will produce bass and treble at high db.
Oh my I found it
REASON FOR PHASE -- WHY MINUS TAN THETA ?
Bro didn't understand lot of things u did it by just explaining the derivation
Thank🙏💕🙏💕🙏💕🙏💕🙏💕🙏💕
what is the loading effect of which you are talking about in this video.
I have a separate video for that. Please check this video, you will get it.
ua-cam.com/video/7jw2_x8dyQ8/v-deo.html
Dear Siddharth, you can refer my channel to know about loading...
Vout=(R/sqrt(R^2+Xc^2))*Vin........ye Hona chahiye bhai....tune galat likha hai
if its 20db per decade, it should be 7.07/20= 0.535v isn it???
It's a logarithmic scale. Please check my video on Decibels. Your doubt will be get cleared.
ua-cam.com/video/ta1sUTiJNkY/v-deo.html
Dear Jeff, 20db per decade means change in gain is 20db in every decade. One decade, means multiples of 10s. ...
is a diode is called bufer, why?
A diode is not a buffer, an emitter follower is called as a buffer....
you pronounce that word, "a ten you ate" attenuate...
what is Omega(c)
omega is (2 x pi x frequency)
Great...
,👌
hol