Subject:[AUDITORY] Question about level conversionsFrom:Alejandro Osses <ale.a.osses@xxxxxxxx>Date:Wed, 8 Mar 2023 12:38:44 +0100--0000000000001acdfd05f661f94d Content-Type: text/plain; charset="UTF-8" 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 --0000000000001acdfd05f661f94d Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable <div dir=3D"ltr"><div>Dear list, hello!</div><div><br></div><div>I have a t= echnical question about presentation levels and signal-to-noise ratios (SNR= s), as expressed in the current literature and in more classical studies (s= ome 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)). <= br></div><div><br></div><div>I have some thoughts, which I will share next,= please feel free to comment on any of these aspects in terms of maths or c= oncepts, or in terms of conceptual accuracy....<br></div><div><br></div><di= v>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. <b>I a= m interested to know the mathematical expressions needed to convert between= noise levels to the appropriate target sound level</b>. I will now give an= example about how I am currently doing such a conversion.</div><div><br></= div><div>The example: Let's say I have a Gaussian noise (duration=3D0.5= s) with frequencies between 0 and 5 kHz, with a total RMS level of 70 dB S= PL. 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 le= vel of the tone, if E_tone/N0 needs to be 11.8 dB (this is actually from Ah= umada et al (1975)).</div><div><br></div><div>My calculations are as follow= s: <br></div><div><ul><li><b>Step 1</b>: N0 =3D 33 dB/Hz (because N0 =3D lv= l_noise - 10*log10(BW) =3D 70 - 10*log10(5000) =3D 33 dB/Hz)</li><li><b>Ste= p 2</b>: The energy of the tone E_tone can be calculated from the normalise= d level to a duration of 1 second: E_tone =3D lvl_tone - 10*log10(dur_tone)= =3D lvl_tone + 10 dB (using dur=3D0.1 s)</li><li><b>Step 3</b>: At the sam= e time: E_tone needs to have a E_tone/N0 of 11.8 dB or E_tone =3D N0 + 11.8= dB</li><li><b>From Step 3 and 4</b>: N0+11.8 =3D lvl_tone + 10 -----> f= or the current values: lvl_tone =3D 33+11.8-10 =3D 34.8 dB <br></li></ul><d= iv>So, my tone will have a level of 34.8 dB SPL for this example.</div><div= ><br></div><div>Am I correct with my calculation method or am I missing any= important concept?<br></div></div><div><br></div><div>Thank you for the fe= edback, <br></div><div><br></div><div>Alejandro Osses<br></div><div><br></d= iv><div><br></div></div> --0000000000001acdfd05f661f94d--

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