What are Decibels and why are 3dB and dBm important?

[@mnada72]

You explained and clarified the concept of dB swiftly and left me wondering 🤔 why it was confusing in the first place ! Thank you

[@iain_explains]

Glad it was helpful!

[@luckyboy-ih5hd]

This channel means a lot to me! Most of the knowledge is simply not explained in the university where I am studying. Thank you. I hope you continue to do what you are doing and will do.

[@iain_explains]

You're very welcome! I'm glad the videos are helping.

[@sivasankars7365]

Most of the simple things I have never known before. What a neat and simple explanation. Really great. Thank you.

[@iain_explains]

Glad it was helpful!

[@dcchillin4687]

I have a "teach yourself" teacher and have been struggling with filters and amplifiers, this video really help explain why we use db and what he meant when he keeps talking about the 3db point

[@iain_explains]

I'm glad it helped.

[@Matchat11]

Hello, I really doubted of miself during this video until reading the description that notice the typo, your videos are so good , I think it should be nice to reupload this one without this error because it keeps this notions of levels in dB still very confusing for the novice while the purpose is to clarify. Thank you for all your content anyway:)

[@iain_explains]

Yes, it's annoying, I know. It's just a minor "typo" in scaling. It would be good if KZbin allowed me to edit the video to fix it, but once something is published, it's published.

[@Simplisanjana]

world need teachers like you sir. Thank you so much sir.

[@iain_explains]

Thanks for your nice comment. I'm glad you like the videos.

[@BlackBlast]

02:00 How can multiplying the value of bels by ten be equal to a 10th of a bel (a 'deci'bel so to speak)? If you multiply a bel by 10 isn't it 10 times as much? Like a giga or a terabel?

[@BlackBlast]

Haha, nevermind i just found the reason behind this. Greetings from elementary school 😂

[@iain_explains]

Glad you worked it out. And I'm happy the video prompted your question. I'm sure you're not the only person to wonder about that. It's one of the reasons I like making these videos.

[@nimrodgazit]

And what’s the answer? Multiplying the log by ten will result with 10bells

[@nimrodgazit]

It’s also doesn’t make sense if you talk about meters. 1deci meter doesn’t equal to 1metets

[@nimrodgazit]

@@iain_explains hi, can you please

[@Hadkek]

Very well explained! Clean and simple and get right to the point.

[@iain_explains]

Thanks. Glad you liked it.

[@bhaswardutta8493]

This really helped during studying Analog Filters. Thanks a lot Sir!

[@iain_explains]

That's great to hear. I'm glad you found the video useful.

[@amoltejal]

I wish you were my maths teacher in my higher secondary school....thanks for your efforts and enthusiasm to teach.....

[@iain_explains]

I'm so glad you like the videos.

[@elsbbbb1]

A very clear explanation, thank you.

[@iain_explains]

Glad it was helpful!

[@moontasirrafique5278]

you cleared a very important point (why 3db isnt halfway down the line). Silly me, thats easy and shoud've understood beforehand. Thank you sir!

[@iain_explains]

Glad it helped!

[@PCOE0112]

Great video exactly what I was looking for!

[@iain_explains]

Great to hear!

[@kristhompson8112]

As a person who has work in pro sound, finally the penny drops as to why we use decibels not "bels" to mathematically explain the values we are playing with, scratching my head what was taught in class around dB- SPL ( sound pressure levels) something to do with the lowest level the human ear can perceive. K regards and 73's

[@iain_explains]

Yes, it's all a matter of scale.

[@suhailbhat1996]

Why do multiply log by 20 sometimes that is also called dB. And for 3dB bandwidth it is also like the value of frequency at which magnitude response falls down to 1/squareroot(2) of its of peak value. In terms of 20log that gives the same result as you obtained in 10log

[@iain_explains]

Decibels are 10log_10 of a ratio of powers. For electrical signals, P=V^2/R. So if, instead of powers, you have a ratio of two voltages (across the same resistor), then the power ratio is (V_2/V_1)^2, and the square comes out the front in a log function, as a multiple - hence the 20, instead of the 10. In other words, 10log_10(P_2/P_1)=20log_10(V_2/V_1).

[@suhailbhat1996]

@@iain_explains thank you for clearing it. Your videos are super helpful.

[@hamidalimohammadi7255]

Thank you Iain, very well explained. I am just confused about using 10log vs 20log. I mean as you explained, the 20log is used for Voltage ratios, but in this example (Vout/Vin) you have used 10log not 20log. Maybe it was just a typo or I misunderstood?

[@iain_explains]

Ahhh. Very sorry. You're right, it's a typo. I should have used power everywhere, not voltage. ... or I should have replaced the 10's with 20's. That's annoying. Sorry.

[@chenwu7667]

Can you comment, when people consider 20*log10(V2/V1) (dB)?

[@iain_explains]

Please take a look at the description below the video. It explains that I made a typo in the video. I think it will answer your question.

[@tpmbe]

Good info but the way the information provided here can be confusing as you only speak about amplitudes, you don't introduce the 20 Log V1/V2 when it is relevant to comparing the power of the signal, and this confusion seems to come up quite a lot. Most engineering formulae use 20 Log V1/V2 and 10 Log P1/P2 when related to power in dB. I was also confused as you began to introduce dBm which is actually the ref of power related to 1mW instead of using the amplitude reference of 1 dbmV. Thanks.

[@iain_explains]

Yes, sorry about that. If you read the notes in the description below the video, I've explained it (someone pointed this out before). Pity I can't redo the video to update it - KZbin doesn't allow it.

[@tpmbe]

@@iain_explains Excellent Thank You, Great Videos

[@JyotiradityaSarkar-rk6xe]

Amazing. Awed at such a simple yet effective explanation. But it would be better if the concept of half power at 3dB cut-off could be incorporated. Nonetheless, this is worth admiring.

[@iain_explains]

Sorry, I'm not sure what you mean. I do talk about that, from the 4:26min mark onwards. Perhaps you're referring to the fact that I say "voltage" everywhere, when I should have been saying "power". As I note in the description below the video, I had a total brain fade the day I made that video. Sorry about that ... unfortunately KZbin doesn't allow me to re-upload a revised/corrected version.

[@iain_explains]

I've fixed it on the Summary Sheet, here: drive.google.com/file/d/1qtvfyv05Ju4HvLKRcfHdfdZfQUrMaPHh/view

[@monimonish302]

Sir can you explain the topic about middlebrook criterion. For choosing the inductor and capacitors to design smps

[@tuongnguyen9391]

Can you do a video explain why people prefer semilogy to plot the BER ?

[@iain_explains]

Thanks for the suggestion. I'll put it on my "to do" list. In short, it's because it's easier to see the BER region of interest.

[@cringeydramaaccount8390]

I'm ashamed to say all that I got from this was that "bel" was named after Alexander Graham Bell...y'all math wizards are crazy! (appreciate the video)

[@Simplisanjana]

it was very helpful sir....but why do we need to half the power sir?

[@iain_explains]

We don't need to halve the power. It's just a reference level that can be used. For example, in filter design, we want the frequency response to roll off between the "in band" frequency range and the "out of band" frequency range. The "half power" frequency gives us an indication of the slope of the roll-off.

[@paulpkae]

V is not a power. When working with "volts" it should be 20 x log (V/Vref).

[@iain_explains]

Yes, thanks. It's explained in the notes below the video.

[@Karnataka_Bengaluru]

Hi, why did you multiply by 1/2 ?

[@iain_explains]

Because at that point in the video I was talking about an example where the signal power has dropped by a half. I was showing that that factor corresponds to a 3dB drop.

[@renegaed]

incredibly well explained. thank you

[@iain_explains]

Glad it was helpful!

[@einekleineente1]

Also helpfull to know is that one can quickly convert from dBm to dB by simply substracting -30. E.g. 5dBm = -25 dB

[@iain_explains]

Yes, that's often helpful to know.

[@medolmshaoshasafetyman]

So is it safe to say that every 3db increase, the gain is doubled? But can you explain gain more?

[@iain_explains]

Yes, that's right. This video might help with respect to "gain": "What are Antenna Gain, EIRP, and Friis Equation?" kzbin.info/www/bejne/o3nSY2CcartrjJY

[@LIM_Khuan]

good explanation, thanks

[@iain_explains]

Glad you liked it

[@Aradhya_this_side]

very well explained!!

[@iain_explains]

Glad it was helpful!

[@abhayraj695]

Thanks for the video...Helped me alot

[@iain_explains]

Glad to hear that

[@paulwood3460]

10 log P1/P2 = Power Gain db, whereas -> 20 log V1/V2 = Voltage Gain db - typo in video.

[@iain_explains]

Yes, sorry about that. There's a note about it in the description below the video.

[@nekelly123]

I don't think it is a typo. The 10/20 applies for converting between power/voltage gains. A 6dB power gain is a 3dB voltage/current gain.

[@TekCroach]

When you deal with voltage or current ratios, you must multiply it with 20, not with 10. It is 10-fold when the ratio is a power ratio. Why is it 20 in case of voltage or current? It is like this. Power ratio in terms of Watt: db = 10log(OutputPower/InputPower) Power ratio in terms of Voltage: db = 10log{(OutputVoltage^2/OutputResistance)/(InputVoltage^2/OutputResistance)} Assuming input resistance = output resistance db = 10log(OutputVoltage^2/InputVoltage^2) = 10log(OutputVoltage/InputVoltage)^2 = 2x10log(OutputVoltage/InputVoltage) =20log(OutputVoltage/InputVoltage) In the same way, power ratio in terms of Current is 20log(OuputCurrent/InputCurrent) because P=(I^2)(R).

[@iain_explains]

Yes, you’re right, sorry, there’s a “typo” in my video. I previously added a note of explanation in the description under the video, explaining what you’ve said here.

[@davadoff]

@@iain_explains’why is half sometimes -6dB instead of -3dB?’ That’s what brought me here. If you do an updated video, put 3dB and 6dB in the title. I read the explanation in the description but I didn’t intuitively understand it. Half the voltage into a fixed resistor is a quarter of the power: is that simply why half the voltage is -6dB? (because -6dB power equals a quarter the power) So, a 1v RMS sine wave into an amplifier with voltage gain of 2 = 2v RMS output and +6dB gain? Into a speaker it would be 4 times the power and sound 4 times louder, compared to the 1v signal?

[@larry_0487]

Oh my goodness. U saved by college life!!!

[@iain_explains]

I'm so glad you've found it helpful.

[@skystalker2]

omg Bel was a unit

[@TheManosV]

You are God! Finally understood it!

[@iain_explains]

I'm glad the video was helpful. Have you noticed my webpage, where there are lots more videos on related topics? iaincollings.com

[@HopeJustice2024]

Thank you very much

[@iain_explains]

You are welcome

[@under_normal_conditions]

thanks!

[@regorRegor]

Thank you. It's the log math. I get it now.

[@regorRegor]

And roar, the graph is logarithmic and that's why the -3dB point is not half way on the scale. Roar, I never saw that. Thank you. I never picked up on that. Roar again.

[@iain_explains]

I'm so glad the video has been helpful.

[@k.chriscaldwell4141]

Superb! Thanks.

[@iain_explains]

Glad you liked it! You might like to check out some of the other videos on the channel as well. I've recently made a webpage with a full categorised list: iaincollings.com

[@naushabashaik6268]

Superb!!

[@jaffarahmadi7690]

The best

[@no-de3lg]

Thanks for your content professor i need to build my foundation in order to understand fourier transformation i have strong desire to learn things and u offer me this i wish i have money so I could donate to u open donations page so when i have credit card I would donate to u

[@iain_explains]

I'm glad you like the videos on the channel, and have found them useful.

[@evaristomusonda6612]

Quit interesting video and was just interested to see how you would explain it in a simple way. But the video is misleading because it deals with voltage ratios instead of power ratios. And I feel this negates the usefulness of this video to some. I recommend you redo this video and post the corrected version! (I know you have put up an explanation to this... :-)!)

[@iain_explains]

Yes, I agree. However KZbin does not allow replacement videos to be uploaded over the top of previous videos (I'm sure there are good reasons for this). I ended up deciding that the benefit people get from the reach that this video has achieved will outweigh the benefit of removing it and uploading another video that wouldn't have anywhere near the same viewer reach. After all, it's only a factor of 2, and very easy to understand after reading the comment below the video.

[@evaristomusonda7760]

@@iain_explains Understood. You are a great teacher Iain, I wish to recommend your website to my students.

[@ahemednor5201]

🙏🌹

[@iain_explains]

👍