Answers.Body: <p>great question.</p>
  <blockquote>
  <ol>
  <li>Is it appropriate to include spreading compensation in dB units?</li>
  </ol>
  </blockquote>
  <p>Yes, if I've understood you correctly. Keep in mind that dB are &quot;unitless&quot;, and are only meaningful with respect to a reference point. For instance, 90 dB SPL re 20 uPa means that the sound pressure was 90 dB above a sound pressure of 20 uPa. It can also be useful to remember that adding or subtracting dB is equivalent to multiplying (or dividing) in non-logarithmic units. Thinking through the units without dB transformation will ensure your units line up correctly.</p>
  <p>In the case of your spreading compensation, you would like to check for deviation relative to the expected level under a scenario of cylindrical attenuation. You can compare observed and expected levels, and consider their difference to be some additional term that is not explained by cylindrical spreading: <code>Observed_dB = Original_dB + Geometric_attenuation_dB + Unexplained_dB</code> or equivalently <code>Observed_pressure = Original_pressure * Geometric_attenuation_factor * Unexplained_factor</code> where <code>Geometric_attenuation_dB= 20*log10(Geometric_attenuation_factor)</code></p>
  <blockquote>
  <p>I am used to presenting pressure as dB re 1m, but is this still the correct reference level with the spreading compensation included?</p>
  </blockquote>
  <p>You can choose the reference point arbitrarily, and <code>r_0=1 meter</code> is a standard choice. However, keep in mind that your spreading compensation formula has that reference distance embedded within it, even though you haven't written it out explicitly above: <code>10*log10(r/r_0)</code> , which is only <code>10*log10(r)</code> if r_0=1. (Note that <code>10*log10(r/r_0)</code> is equivalent to <code>20*log10(sqrt(r/r_0))</code>, which is similar to what you wrote above, but if r_0 does not equal 1, your formula above would become incorrect.)</p>
  <p>This <a href="https://dosits.org/science/advanced-topics/cylindrical-vs-spherical-spreading/" rel="nofollow noreferrer">page</a> has a good overview of attenuation due to cylindrical or spherical spreading.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2024-01-26T21:37:57.443Z
Answers.LastActivityDate: 2024-01-26T21:37:57.443Z
Answers.UserDisplayName: Sam Lapp
Answers.Body: <p>Sound propagation is adressing two effects</p>
  <ul>
  <li>spreading</li>
  <li>absorption</li>
  </ul>
  <p>If the propagation model can run without absorption, then you can compensate for assumed spreading law. In order to see the leaking effect you must make sure that absorption coefficient is set to zero.</p>
  <p>Compensating for spreading you get dB mismatch between two models which increases with range. You present this in form of dB/m or dB/km. The later allows you a direct comparison with the absorptions that typically is dB/km.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-07-24T17:23:22.913Z
Answers.LastActivityDate: 2023-07-24T17:23:22.913Z
Answers.UserDisplayName: WMXZ
Body: <p>I would like to present propagation modelling results in dB with spreading effects removed. The spreading is largely cylindrical, so I am plotting <span class="math-container">$20 \log_{10}\Big(p(r, z) \sqrt{r}\Big)$</span> instead of <span class="math-container">$20 \log_{10}p(r,z)$</span>.</p>
  <p>My two questions are:</p>
  <ol>
  <li><p>Is it appropriate to include spreading compensation in dB units?</p>
  </li>
  <li><p>I am used to presenting pressure as dB re 1m, but is this still the correct reference level with the spreading compensation included?</p>
  </li>
  </ol>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-07-20T04:36:07.023Z
Comments.Text: What do you wanted to show, attenuation only or also deviation from speading law (you say "mostly" cylindical)?
Comments.UserDisplayName: WMXZ
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-07-21T16:52:12.090Z
Comments.Text: I am interested in acoustic propagation in a leaky duct, so I want to show if energy is in the duct or not. I guess that can be thought of as deviation from a cylindrical spreading assumption.
Comments.UserDisplayName: nedlrichards
Tags: <propagation>
Title: dB units of spreading compensated pressure
ThreadText: # dB units of spreading compensated pressure
  
  (Asked by: username_0 on 2023-07-19)
  
  I would like to present propagation modelling results in dB with spreading effects removed. The spreading is largely cylindrical, so I am plotting $20 \log\_{10}\Big(p(r, z) \sqrt{r}\Big)$ instead of $20 \log\_{10}p(r,z)$.
  
  My two questions are:
  
  1. Is it appropriate to include spreading compensation in dB units?
  2. I am used to presenting pressure as dB re 1m, but is this still the correct reference level with the spreading compensation included?
  
  **Tags:** propagation
  
  > **Comments:**
  > **username_1**: What do you wanted to show, attenuation only or also deviation from speading law (you say "mostly" cylindical)?  
  > **username_0**: I am interested in acoustic propagation in a leaky duct, so I want to show if energy is in the duct or not. I guess that can be thought of as deviation from a cylindrical spreading assumption.  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_2 on 2024-01-26. Score: 1 | Accepted answer
  
  great question.
  
  > 1. Is it appropriate to include spreading compensation in dB units?
  
  Yes, if I've understood you correctly. Keep in mind that dB are "unitless", and are only meaningful with respect to a reference point. For instance, 90 dB SPL re 20 uPa means that the sound pressure was 90 dB above a sound pressure of 20 uPa. It can also be useful to remember that adding or subtracting dB is equivalent to multiplying (or dividing) in non-logarithmic units. Thinking through the units without dB transformation will ensure your units line up correctly.
  
  In the case of your spreading compensation, you would like to check for deviation relative to the expected level under a scenario of cylindrical attenuation. You can compare observed and expected levels, and consider their difference to be some additional term that is not explained by cylindrical spreading: `Observed_dB = Original_dB + Geometric_attenuation_dB + Unexplained_dB` or equivalently `Observed_pressure = Original_pressure * Geometric_attenuation_factor * Unexplained_factor` where `Geometric_attenuation_dB= 20*log10(Geometric_attenuation_factor)`
  
  > I am used to presenting pressure as dB re 1m, but is this still the correct reference level with the spreading compensation included?
  
  You can choose the reference point arbitrarily, and `r_0=1 meter` is a standard choice. However, keep in mind that your spreading compensation formula has that reference distance embedded within it, even though you haven't written it out explicitly above: `10*log10(r/r_0)` , which is only `10*log10(r)` if r\_0=1. (Note that `10*log10(r/r_0)` is equivalent to `20*log10(sqrt(r/r_0))`, which is similar to what you wrote above, but if r\_0 does not equal 1, your formula above would become incorrect.)
  
  This [page](https://dosits.org/science/advanced-topics/cylindrical-vs-spherical-spreading/) has a good overview of attenuation due to cylindrical or spherical spreading.
  
  
  ---
  
  ### Answer by username_1 on 2023-07-24. Score: 0
  
  Sound propagation is adressing two effects
  
  * spreading
  * absorption
  
  If the propagation model can run without absorption, then you can compensate for assumed spreading law. In order to see the leaking effect you must make sure that absorption coefficient is set to zero.
  
  Compensating for spreading you get dB mismatch between two models which increases with range. You present this in form of dB/m or dB/km. The later allows you a direct comparison with the absorptions that typically is dB/km.
  
  
  ---
  
  