Frequencies, Amplitudes, Log Scales, dB and dBm - ECE Topics #7

In this episode, we cover a wide range of topics including how and why logarithmic scales are used in frequency response graphs, and related measures of dB and dBm. We use all of these to deal with orders of magnitudes found in the physical world. While processes and the scale of values in the world is often exponential in nature, we use logs to deal with it for reasons explained. Examples in the audio and radio/microwave fields are presented together with some history and theory.

[@sp5xyz]

Brilliant, thanks!

[@MegawattKS]

You're welcome! Glad you liked it.

[@dominicestebanrice7460]

Remarkably useful content; concise yet comprehensive.
Question re; the reference piano music at the end: which clip should my ears perceive as "half as loud", the -10dB level or the -20dB level? I didn't grasp the whole 10db/20dB thing for physics/SPL vs audio "power" (and my old ears are surely screwed from decades of listening to AC/DC & Motorhead anyway!). I suspect the -20dB level is supposed to be half as loud as the reference level, right?

[@MegawattKS]

Thanks. That's good to hear (pun intended ;-) On the question: The half-as-loud should be the -10 dB one. But its very subjective, by definition. Interestingly when I was playing it back on my PC, the -10 dB section sounded a bit less than half as loud to me as the reference section - maybe 1/3'ish. And at -30, it seemed almost gone. But when I played it on my better speakers, it seemed about half as loud at -10 dB - agreeing with what the literature suggests. I suspect it depends on how loud stuff is when listening, maybe ?

[@MegawattKS]

I was a little worried about the 10 vs 20 dB thing when making the video but opted for leaving out the somewhat dry math derivation :-) It's confusing for sure. The 10 vs 20 multiplier in the formulas doesn't change the dB value. Its just how you get to dB from amplitude or power. If you already have a power ratio - then you use 10 log. But if it's voltage, its 20 log. That's because with a voltage ratio of V2/V1, the associated power ratio is (V2^/R)/(V1^2/R) = (V2/V1)^2, and 10 log [(V2/V1)^2] = 20 log (V2/V1) by the rules of logarithms...

[@user-ss1pc6mh8k]

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