Re: [AUDITORY] Question about level conversions

[Bob Masta <audio@xxxxxxxxxxx>]

Not on the topic you are asking about, but I notice you are using Gaussian 
noise.  Has anyone ever shown this to be detectable, compared to a uniform 
distribution?  There's a nearly 8 dB penalty for using it, in reduced ability 
to reach high levels without distortion.  See my "Comparing Noise 
Distributions" at <https://www.daqarta.com/dw_0agg.htm>

Best regards,

Bob Masta

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On 8 Mar 2023 at 12:38, Alejandro Osses wrote:

> 
> Dear list, hello!
> 
> I have a technical question about presentation levels and signal-to-noise 
> ratios (SNRs), as expressed in the current literature and in more classical
> studies (some of the papers I have recently investigated: Richards (1992, JASA -
> doi:10.1121/1.402831) and Ahumada et al (1975, JASA - doi:10.1121/1.380453)).
> 
> I have some thoughts, which I will share next, please feel free to comment on
> any of these aspects in terms of maths or concepts, or in terms of conceptual
> accuracy....
> 
> In classical papers (think also of the work by David Green and colleagues), SNRs
> are often expressed as energy with respect to N0, often reported as E/N0. In
> this expression E is typically the energy of a target signal (e.g., a tone)
> which is embedded in a noise, whose spectrum level is N0. I am interested to
> know the mathematical expressions needed to convert between noise levels to the
> appropriate target sound level. I will now give an example about how I am
> currently doing such a conversion.
> 
> The example: Let's say I have a Gaussian noise (duration=0.5 s) with 
> frequencies between 0 and 5 kHz, with a total RMS level of 70 dB SPL. On the
> other hand, my target sounds have a centre frequency of 500 Hz, a duration of
> 0.1 s, temporally centred in the noise. I want to know the level of the tone, if
> E_tone/N0 needs to be 11.8 dB (this is actually from Ahumada et al (1975)).
> 
> My calculations are as follows:
> *   Step 1: N0 = 33 dB/Hz (because N0 = lvl_noise - 10*log10(BW) = 70 - 
>     10*log10(5000) = 33 dB/Hz)
> *   Step 2: The energy of the tone E_tone can be calculated from the normalised 
>     level to a duration of 1 second: E_tone = lvl_tone - 10*log10(dur_tone) =
>     lvl_tone + 10 dB (using dur=0.1 s)
> *   Step 3: At the same time: E_tone needs to have a E_tone/N0 of 11.8 dB or
> E_tone 
>     = N0 + 11.8 dB
> *   From Step 3 and 4: N0+11.8 = lvl_tone + 10 -----> for the current values: 
>     lvl_tone = 33+11.8-10 = 34.8 dB
> So, my tone will have a level of 34.8 dB SPL for this example.
> 
> Am I correct with my calculation method or am I missing any important concept?
> 
> Thank you for the feedback,
> 
> Alejandro Osses
> 
>