- Thread Starter
- #621
Don't quite understand this part. If a human is listening to two tones, one at 50Hz and another 500Hz at 80 phones, and perceives these two tones to be equal in loudness, doesn't this mean that the 50Hz tone is 20dB higher than the 500Hz to achieve that equal perception? That's what I thought the FM-curve is saying (for 80 phons).Not quite - bassoon will play as intended, with no requirements to be 20 dB louder than the violin.
If my understanding above is incorrect, then why would we have to keep the volume below the highest bass SPL the speaker can produce? This made sense to me based on the point I'm referring to above, but if that point is wrong, then I don't understand this part either.Yes.
For the bolded part (short story): Yes-ish, sort of.
Big yes.
No!
Use FM curves only for determining how much bass boost you need at lower SPL than "original, normal" SPL. Recording already have captured the correct balance between bass and mid/highs.
If "normal" SPL captured in recording was 80 phons, when you reproduce it at 80 phons you don't need any bass boost!
If you want to reproduce this recording at 100 dB, you likely will want to reduce bass slightly (nobody is doing that!).
The text in red I don't understand. Let's say normal recording is captured at 80 phons where the 50Hz and 500Hz tones are perceived equally. To me, this means the 50Hz waves must have a higher amplitude and thus be at +20db (100dB) to be perceived equally to 500Hz. Now if we try to digitally reproduce it at 100 phons instead of 80 phons, doesn't this mean that the 500Hz is at 100dB and in order to perceive the same loudnss at 100 phons, the 50Hz must be at 110dB+ at this volume?