Why -3dB? (An alternate approach to explaining dBs) 

decades. I see what you did there

Hello, helloworld :)

First question: You are right in the first place (the "but..." comes later). The foremost intention of this tutorial was to explain when and why 10*log(x) and 20*log(x) is used and where the difference comes from. And this is simply because for one single measurement power is always proportional to the square of the voltage (or current). Or the other way around voltage/current is always proportional to the square-root of the power. Now let´s take a typical measurement situations: You have a source (like a frequency-generator) connected to a DUT ("Device under test"; e.g. a filter-block) and you measure the output-amplitude at the end of the DUT. Now you measure the output voltage with a constant load impedance (e.g. 50 Ohms, 600 Ohms or 1 Megohm typically) and on the other hand your source that feeds the DUT mandatorily has to have a constant output (source) impedance. That way it is justified to assume a constant impedance for your measurement so that the cancelling out of "R" is true. Give me an example of a measurement-setup where this is not the case and which complies with the specifications how this special measurement is to be done (there are certainly situations where you truly have to measure the RMS-power instead of the voltage to get the right dB-values) The factor SQRT(2) in this video comes from my memory only into play when I explain why the "-3dB"-point is so often chosen for the frequency-range of a device. -3dB is a relative drop of 1/2 = 0.5 in power and because voltage is porpotional to the SQRT of power a drop to 1/2 in power translates to a drop of SQRT(1/2)=1/SQRT(2) = 0.71 in voltage.

This error was already mentioned in one of the other comments.

@@KainkaLabs you keep saying that but I don't see any comment where you actually answered this. Is it 1/10th or not? Do we multiply by 10 or not? Why take all the time to make an educational video and then not explain a mistake in a comment that takes a few seconds to answer? I now have no confidence in the video cause I can't find an answer to this issue. All I find is "look at the other comments". I did. It didnt help.

@@kurchak hey sorry if the English isn't clear, If you still have the doubt about this Here's the approach, Just like 100cm turn to meters by replacing multiplier C(centi) as 10^-2 100cm=100*10^-2m. Same way, 10log(P1/P2)db Changes to bel as 10log(P1/P2)db=10log(P1/P2)*10^-1bel. The video just cleared some strong doubts to many just a small explaination of a thing in comments doesn't mean all the effort was gone. It would be better to be grateful for good content.

Still didn´t find the time to shoot that video-series :-( The last years were quite busy and I had nearly no time for further videos. But patience in the end will be rewarded.

@@KainkaLabs thank you, eagerly waiting for it

It depends only on if you look at the value in regard to "amplification" or "attenaution". Same value, alternate sign.

It´s an arbitrary though natural selection. (Optical) Bandpass-Filters are e.g. also specified at FWHM = Full Width at _half_ Maximum

they are both connected to the same circuit ad perhaps this makes the Resistance , the total resistance of the circuit, i'm just guessing here but that's the only explanation i have right now

If he doesn't take R2 = R1 , then it would be a different circuit....but we are comparing two voltages or current or power response for the same circuit.... But changing R2 and R1 will make different circuits/system....and hence you will compare two different systems's response at two different frequencies which might be of no use for any specific analysis....

Yes, I also wondered about the missing "l" in the unit bel and decibel

Look further down in the comments. I think I goofed this up.

@@KainkaLabs Your derivation is wrong. From 3:14 to 3:18, let me show you below. (1) According to defination from wikipedia : x = p1/p2, means power ratios; √x = √(p1/p2), means amplitude ratios; dB = 10log10 x = 10 log10 (p1/p2) (2) Your defination of the Bel = log 10 x = log10 (p1/p2) is wrong, because according to (1) above, you will know the Bel = 100 log10 x = 100 log10 (p1/p2)

@@KainkaLabs Thank you for your reply, the explanation in other parts is awesome

@KainkaLabs you keep saying that but I don't see any comment where you actually answered this. Is it 1/10th or not? Do we multiply by 10 or not? Why take all the time to make an educational video and then not explain a mistake in a comment that takes a few seconds to answer? I now have no confidence in the video cause I can't find an answer to this issue. All I find is "look at the other comments". I did. It didnt help.

You are right.

That´s easy first grade school physics

You are right. This was an error I made during the recording and which escaped my attention. Some people in the comments have noticed this error.

Of course I watch my videos: when editing them :-) When I write something outside the window I usually recognize this a few seconds later and pull the sheet up or down. Otherwise I insert a text-message in the video. Should be also the case here if you tell me at what time this has happened.

Read the other comments.

@@KainkaLabs you keep saying that but I don't see any comment where you actually answered this. Is it 1/10th or not? Do we multiply by 10 or not? Why take all the time to make an educational video and then not explain a mistake in a comment that takes a few seconds to answer? I now have no confidence in the video cause I can't find an answer to this issue. All I find is "look at the other comments". I did. It didnt help.

there are ten decilitres in 1 litre, similarly, there are ten dB in 1 Bel, hence the multiplication by ten.