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Generally speaking, the db scale is very useful for many practical situations, and this is overlooked in this critique.

As have been pointed out, it's just a power ratio on a logarithmic scale, but this has many benefits, the main one being that chaining gain/attenuation in a system is just a case of adding the db values together. 'We're loosing 4db in this cable, and the gain through this amp is 6db, so the output is 2db hotter than the input'. Talk to any sound engineer and you'll do this sort of thing successfully without necessarily understanding the science, so that's a massive win.

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remram 2 days ago | parent | next [-]

I didn't get that reading the article. The author acknowledge that ratios are useful, but that there are specific problems in how we use this unit and how we picked and express the reference scales.

> On the face of it, the idea makes sense.

Your specific example is a pure ratio so there's no problem with it (there is no reference). Apart from the fact that I have to guess whether you are measuring volts or watts through your cables, of course...

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BenjiWiebe a day ago | parent [-]

So what I learned today, via another comment and confirmed with a dictionary, is that decibels (unspecified reference) is NOT pure ratio - it's a _power_ ratio. So it will be Watts, not Volts.

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bitdivision a day ago | parent [-]

But that's why it's so confusing. Did they purposefully leave off the reference to indicate Watts, or did they mean volts (as I'd guess audio engineers typically do when talking about amplifiers).

	▲	BenjiWiebe a day ago \| parent [-]
		No, it isn't Watts if there's no reference. It's a ratio of powers. i.e. amplification or attenuation. To have an actual amount of power, you need to reference it to some particular amount of power, like dBm for power-ratio-relative-to-milliwatt.

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Aachen 2 days ago | parent | prev [-]

> this has many benefits, the main one being that chaining gain/attenuation in a system is just a case of adding the db values together. 'We're loosing 4db in this cable, and the gain through this amp is 6db, so the output is 2db hotter than the input'.

I'm not a sound engineer, so to check my understanding: would this not be the case for any other scale indicator?

If you have a cable that loses 4m and you're sending 6m into it, you'd not get 2m out?

(The m being milli here, as in millivolts or whatever unit would be useful here — leaving it unspecified to keep the comparison to dB as close as possible)

	▲	bitdivision a day ago \| parent \| next [-]
		No, because db is logarithmic. That's the key property that allows you to do addition instead of multiplication. Remember these are ratios here as well. So the original example in linear units would be: We're reducing the signal by 63% in the this cable, and the gain through this amp is 199.5%, so the output is... 0.63 * 1.995 = 1.26 126% which is 20log10(1.26) db or around 2db. -4 + 6 = 2 is a little easier to do.
	▲	BenjiWiebe a day ago \| parent \| prev [-]
		It's a ratio, not a fixed amount. The amount you lose depends on the amount you send.