Why -3dB? (An alternate approach to explaining dBs)
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- čas přidán 11. 01. 2018
- Why is there specifically a -3db point chosen for frequency response?
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Why is the dB-definition sometimes 10*log(x) and sometimes 20*log(x)?
If you ever wondered about some of the mysteries about the decibel, this tutorial is for You
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I’ve been tossing these terms around for decades. Now I actually understand what I once only knew.
Thank you.
decades. I see what you did there
Thank you so much for this detailed explanation!
There are bad teachers, there are worst teachers, good teachers, great teachers and there's you.
Your contribution to my knowledge is greatly appreciated, many thanks!
Hope the world would be always with teachers like you.
Tears of joy.! Thankyou kind sir. I wish i had my signal & system teacher like you during my uni time.
Great video for any experimentalists and designers in electronics, optics, acoustics or else. Instructor clarifies the confusion around the terms dB, dBm, dBV....etc. you encounter on datasheets and elsewhere in your career. He has put this valuable summary clearly and understandably together that most textbooks only throw out here and there to confuse readers. Thank you for posting such useful videos !
I wish I had you as my teacher at high school … thank you for your explanation … it’s fantastic
You are a legend. I am currently studying EE and I just got started with electronic filters, so it is really helpful.
This excellent video cleared up a lot if confusion with terms I used but was unclear about. Put much more succinctly than most courses ever do.
You are now officially a SUPERHERO- MR FANTASTIC- Thank you so much for an amazingly straight forward way of explaining dB-
Thank you for explaining all in a very clear and understandable way sir.
one of the most practical 3dB instruction videos...👍👍 thanks a lot...😉 a good one to watch again.
THANK YOU!! I've never seen these concepts explained so well!
Thank you for demystifying the -3dB that have been haunted me for quite some time.
Thank you for this detailed - and practical - explanation! I've been using decibels in the professional audio world for decades now, but there have been certain applications that always escaped my grasp. Now I understand why, cheers!
Great explanation as always. Thanks!
Very well explained better than any book or boring lecture at uni years ago.
Even with that hard topic, very well explained that I could understand it with only watching once, highly appreciate your effort.
Very well explained Sir! Thanks very much for teaching us all.
Very helpful tutorial. Thanks for sharing.
Das ist lustig. Ich habe deine Internetseite über die letzten Jahre (oder sogar Jahrzehnte?) immer wieder besucht (bzw. hat sie mir Google vorgeschlagen, wenn ich nach Informationen gesucht habe ;) ) und jetzt habe ich dich hier zufällig auf CZcams gefunden. Zu einem Thema das ich prinzipiell schon verstanden habe, aber nach diesem Video vollkommen klar (und in meinem Hirn endlich geordnet ;) ) ist. Danke sehr!
Very good and detailed explanation of db. Thank you sir.
Thanks a lot for this video,been confused about this for so long.
This is Gold,,,Where have you been so long,, God bless who ever you are
that's really helpful!! Thank you sensei!
excelent description thank you very much.
Thank you so much. I've gone from memorising to actually understanding.
Thanks for the video!
Nicely taught. Thankyou.
Thank you So much very great explanation!!!
A great explanation sir
thank you kainka labs
Thank you, I feel complete now.
Hello, helloworld :)
very nice talk
Good job man 👍
00:06 Understanding the concept of decibels and why -3dB is used.
03:35 Understanding power reference and comparison in electronics.
07:12 Decibels are based on logarithm of the ratio of voltages squared
10:36 dBs are used to define frequency response limits
13:43 Understanding the concept of -3dB as a ratio to the reference point.
17:20 Minus 3dB represents a power decrease by a factor of 1/2.
20:01 Understanding the significance of -3dB in audio applications
23:12 Understanding the significance of dBs in measurement.
26:19 Decibels represent a ratio of power or voltage levels.
30:01 Filters with -20dB per decade have a linear fall-off.
33:30 Understanding dB values and conversions
36:40 dB is used for loudness measurements and has special frequency filters.
Crafted by Merlin AI.
Thanks for the explanation - Very helpful! db = deci - Bel
thank you so much sir 🙏
Thank you very much.
i have been looking for that a very long ago
you are awesome
Quite explanatory and informative. Thank you, sir
You are number one!
Nope, I'm not copying Vegeta! (Okay, I AM! But, c'mon! It's awesome!)
In North America we use E as in Electromotive Force in Ohm's Law calculations. P = E·I. V is used for most everything else.
For the definition of power P=V^2/R, how do you know the R in the numerator and denominator are the same value and thereby can be canceled out?
Also, how come when going from power ratio to voltage or current ratio, you always take factor of sqrt(2)? HOw does it working out mathematically?
Thanks in advance!
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.
Thanks
12:21 start -3dB explanation
Very nice video , great job explaining all that. most of the times in books we see this expression 10 log (P2/P1) dB and not 10 log (P1/P2) (which is what you are using here . to my understanding both are correct. So in what case do we use each expression? obviously the number in db will be the same but the sign will different .
It depends only on if you look at the value in regard to "amplification" or "attenaution". Same value, alternate sign.
At 39:13 you mentioned about lectures on impedances, I tried in the playlist but couldn't pinpoint it, can you please share the link of that video
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
If a deci-Bel is one TENTH of a Bel; Shouldn't we DIVIDE (Log in base 10, etc...) by ten? Rather than multiplying it?
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.
thanks
Sir,I understood why we are choosing-3dB value to get half power.But why we need to go for half power frequency?
It´s an arbitrary though natural selection. (Optical) Bandpass-Filters are e.g. also specified at FWHM = Full Width at _half_ Maximum
Maybe not that important but the name Bell is with two l's (Alexander Graham Bell) so its a bit odd that the decibel is correctly written as decibel.
Yes, I also wondered about the missing "l" in the unit bel and decibel
So, do we always admit R2 = R1 when using the deci-Bel?
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....
I follow this. But as soon as I go away me be like deciwhat all over again
Demonstrate that the doubling of the transmission frequency or the distance between the transmitting and receiving antenna attenuates the received power by 6 dB. !!!!!!! we need a solution if possible from every one who sees it
That´s easy first grade school physics
👍
I can't understand timeline 3:14, the derivation is wrong, because if you refer to the relationship between g and kg, 1kg=1000g, here the 1decibel=0.1bel, another ex. 1 decilitre= 0.1litre, so the relationship will be 1dB=0.1bel=0.1*log(10)(P1/P2)
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.
Really good teaching, please use V, lol
20log(thanks)😘
Why did he, a German speaker, pronounce Euler's constant as "yu-ler's" @6:02? Is it not pronounced "oi-lers"? I just want to pronounce it correctly. Am I wrong?
You are right.
Strange. deci-meter is meter divided by 10, but deci-bel is bel multiplied by 10...
🙏🙏🌹🌹🌹🌹
isn't deci 1/10? a deci liter is 1/10 of a liter. Why is a deci-Bel 10 times a Bel instead of 1/10 a Bel?
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.
Hey Mr., do you ever watch your own videos? I bet you don't. Because on this particular video you seem to be writing something important under the graph and of course you write so far down the canvass thus no one can see it, and clearly you haven't noticed it, otherwise you would have corrected it. Am I wrong?
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.
Please, can someone comment on my evidently misguided thought process:
To me, if a Bel = log (P/Px), and I want to DEFINE WHAT A DECIBEL IS--(RECOGNIZING THAT "DECI" = 1/10TH)--then 1/10th of a Bel = 1/10 x log (P/Px). ??? This is NOT 10 x log(P/Px) as was stated in very 1st part of video!? HELP! I hope someone is viewing this video and can comment. THANK YOU IN ADVANCE!!!
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.