Answers.Body: <p>The <span class="math-container">$10\log_{10}()$</span> refers to a measure for acoustic energy. This should be used when adding up different sources, because you may add acoustic energy and not acoustic amplitude. Energy is related to the square of the amplitude.
  <span class="math-container">$20\log_{10}()$</span> shoudl be used for amplitude.</p>
  
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Answers.Body: <p>All the answers made so far are correct.</p>
  <p>I only would like to add that the word deciBel has 'deci' in the word so that deciBel means 1 tenth of a Bel. As @fabi_hirono indicates deciBel was created for power ratios, so 10 log(Power/Power_ref) is always correct.</p>
  <p>For sound waves, the use 20 log(Pressure/Pressure_ref) is only correct in the far field. Btw, I would not call the far field assumption, or real impedance assumption, a 'minor' assumption, but a realistic one, if measurements are far away from point source.</p>
  
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Answers.Text: I thought the `20*log10(p/p_ref)` form was also valid for spherical waves? I know it is often used withouth making a plane wave assumption, at least in air. In Pierce - Acoustics (3rd Ed), Sec. 2.3.5, he says the Sound Intensity Level `10*log10(I/I_ref)` and Sound Pressure Level `20*log10(p/p_ref)` are approximately the same for plane AND spherical waves in air, at least. Did I miss something?
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Answers.Text: You are correct, that is why I had changed to 'far field' (plane-field was a mistake). In the nearfield Impedance get complex, so 20log10(.) does not work
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Answers.Body: <p>This is the formula to convert an acoustic pressure from Pa to dB:  <span class="math-container">$20\log_{10}(p_{RMS}/p_{ref})$</span> where:</p>
  <ul>
  <li><span class="math-container">$p_{ref}$</span> is a reference pressure corresponding to 0dB (e.g. 20µPa for the dB SPL scale)</li>
  <li><span class="math-container">$p_{RMS}$</span> is the root mean square of the pressure over a certain period of time</li>
  </ul>
  <p>This formula is equivalent to  <span class="math-container">$10\log_{10}(p_{RMS}^2/p_{ref}^2)$</span> which is sometimes used when talking about sound power or sound intensity because these quantities are related to pressures squared.</p>
  
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Answers.Text: The most commonly used reference pressure for media other than gases is 1µPa. See ASA [reference value for sound pressure](https://asastandards.org/terms/reference-value-for-sound-pressure-2/).
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Answers.UserDisplayName: Maxime BRU
Answers.Body: <p><strong>TL;DR:</strong></p>
  <p><strong>In practice, it shouldn't matter whether you are using <span class="math-container">$10\log_{10}($</span>power_ratio<span class="math-container">$)$</span> or <span class="math-container">$20\log_{10}($</span>pressure_ratio<span class="math-container">$)$</span> to report your results - they should give the same value in dB, as long as you are using them correctly:</strong></p>
  <ul>
  <li><a href="https://en.wikipedia.org/wiki/Decibel#Power_quantities" rel="noreferrer"><strong>when comparing powers/energies, use the <span class="math-container">$10\log_{10}$</span> formula</strong></a>;</li>
  <li><a href="https://en.wikipedia.org/wiki/Decibel#Root-power_(field)_quantities" rel="noreferrer"><strong>when comparing pressures/voltages, use the <span class="math-container">$20\log_{10}$</span> formula</strong></a>.</li>
  </ul>
  <hr />
  <p><strong>Long explanation:</strong></p>
  <p>Originally, decibels were used to describe a ratio of <em>powers</em> (in fact, <a href="https://en.wikipedia.org/wiki/Decibel#History" rel="noreferrer">the loss of electrical power in signals sent across telephone/telegraph wires</a>), and the definition of the decibel is the <span class="math-container">$10\log_{10}(W_{value}/W_{ref})$</span> formula, where <span class="math-container">$W$</span> denotes power.</p>
  <p>However, electrical power in Watts is relatively difficult to measure, so it's convenient to infer its value from other, easier-to-measure quantities, such as voltages or currents. These can be related using the expressions <span class="math-container">$W = V\cdot I = V^2/R = R\cdot I^2$</span> where <span class="math-container">$V$</span> is the voltage, <span class="math-container">$I$</span> is the current, and <span class="math-container">$R$</span> is the resistance/impedance. In this context, voltage and current can be called &quot;<a href="https://en.wikipedia.org/wiki/Power,_root-power,_and_field_quantities" rel="noreferrer">root-power quantities</a>&quot;, as their value depends on the square-root of the power.</p>
  <p>From that, making a few minor assumptions (linearity, constant resistance/impedance, and a few other minor details), <em>one can obtain the same value for the ratio of <em>powers</em> in decibels from the ratio of voltages squared</em> as <span class="math-container">$10\log_{10}(V_{value}^2 / V_{ref}^2)$</span>. Using logarithm rules, this can be simplified to  <span class="math-container">$20\log_{10}(V_{value}/ V_{ref})$</span>, which is a ratio of voltages (not squared). Now, <strong>one can obtain the same value in decibels either from the ratio of <em>powers</em> using  <span class="math-container">$10\log_{10}$</span> or from the ratio of <em>voltages</em> using <span class="math-container">$20\log_{10}$</span></strong>.</p>
  <p>In acoustics, the same principle applies: acoustic pressure <span class="math-container">$p$</span> in Pascals is the &quot;root-power&quot; quantity for acoustic power <span class="math-container">$W$</span> - i.e. acoustic power <span class="math-container">$W$</span> is proportional to acoustic pressure squared <span class="math-container">$p^2$</span>, although the actual expression depends on the propagation conditions (plane/spherical wave, air/water, etc). Therefore, one can obtain the same value in decibels from either <span class="math-container">$10\log_{10}(W_{value} / W_{ref})$</span>  or <span class="math-container">$20\log_{10}(p_{value} / p_{ref})$</span> .</p>
  
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Body: <p>Inspired by <a href="https://bioacoustics.stackexchange.com/q/337/131">this question</a> on dBs, I wanted to ask about a potentially dumb confusion I always have.</p>
  <p>When referring to an absolute measurement value, dBs always come with a reference. However, it is not always clear to me which one is being referred to. Sometimes, $10log_{10}$ is used, and sometimes $20log_{10}$ is used to report dB values. Papers typically don't explicitly state which dB they are reporting (I myself somehow too haven't done so in the past!).</p>
  <p>I often end up guessing whether 10 or 20 is being used as the 'prefix' for the dB calculation, based on what I know from previous studies. This guessing however fails of course when I'm reading papers from a new field.</p>
  <p>Are there any rules of thumb that can be used to decide if the authors mean $10log_{10}$ or $20log_{10}$? when reporting their measurement values?</p>
  
Tags: <analysis><measurement><db>
Title: $10\log_{10}(X)$ or $20\log_{10}(X)$ dB values - how to tell?
ThreadText: # $10\log_{10}(X)$ or $20\log_{10}(X)$ dB values - how to tell?
  
  (Asked by: username_0 on 2022-07-19)
  
  Inspired by [this question](https://bioacoustics.stackexchange.com/q/337/131) on dBs, I wanted to ask about a potentially dumb confusion I always have.
  
  When referring to an absolute measurement value, dBs always come with a reference. However, it is not always clear to me which one is being referred to. Sometimes, $10log\_{10}$ is used, and sometimes $20log\_{10}$ is used to report dB values. Papers typically don't explicitly state which dB they are reporting (I myself somehow too haven't done so in the past!).
  
  I often end up guessing whether 10 or 20 is being used as the 'prefix' for the dB calculation, based on what I know from previous studies. This guessing however fails of course when I'm reading papers from a new field.
  
  Are there any rules of thumb that can be used to decide if the authors mean $10log\_{10}$ or $20log\_{10}$? when reporting their measurement values?
  
  **Tags:** analysis, measurement, db
  
  
  ---
  
  ## 4 answers
  
  ### Answer by username_3 on 2022-07-19. Score: 19 | Accepted answer
  
  **TL;DR:**
  
  **In practice, it shouldn't matter whether you are using $10\log\_{10}($power\_ratio$)$ or $20\log\_{10}($pressure\_ratio$)$ to report your results - they should give the same value in dB, as long as you are using them correctly:**
  
  * [**when comparing powers/energies, use the $10\log\_{10}$ formula**](https://en.wikipedia.org/wiki/Decibel#Power_quantities);
  * [**when comparing pressures/voltages, use the $20\log\_{10}$ formula**](https://en.wikipedia.org/wiki/Decibel#Root-power_(field)_quantities).
  
  ---
  
  **Long explanation:**
  
  Originally, decibels were used to describe a ratio of *powers* (in fact, [the loss of electrical power in signals sent across telephone/telegraph wires](https://en.wikipedia.org/wiki/Decibel#History)), and the definition of the decibel is the $10\log\_{10}(W\_{value}/W\_{ref})$ formula, where $W$ denotes power.
  
  However, electrical power in Watts is relatively difficult to measure, so it's convenient to infer its value from other, easier-to-measure quantities, such as voltages or currents. These can be related using the expressions $W = V\cdot I = V^2/R = R\cdot I^2$ where $V$ is the voltage, $I$ is the current, and $R$ is the resistance/impedance. In this context, voltage and current can be called "[root-power quantities](https://en.wikipedia.org/wiki/Power,_root-power,_and_field_quantities)", as their value depends on the square-root of the power.
  
  From that, making a few minor assumptions (linearity, constant resistance/impedance, and a few other minor details), *one can obtain the same value for the ratio of *powers* in decibels from the ratio of voltages squared* as $10\log\_{10}(V\_{value}^2 / V\_{ref}^2)$. Using logarithm rules, this can be simplified to $20\log\_{10}(V\_{value}/ V\_{ref})$, which is a ratio of voltages (not squared). Now, **one can obtain the same value in decibels either from the ratio of *powers* using $10\log\_{10}$ or from the ratio of *voltages* using $20\log\_{10}$**.
  
  In acoustics, the same principle applies: acoustic pressure $p$ in Pascals is the "root-power" quantity for acoustic power $W$ - i.e. acoustic power $W$ is proportional to acoustic pressure squared $p^2$, although the actual expression depends on the propagation conditions (plane/spherical wave, air/water, etc). Therefore, one can obtain the same value in decibels from either $10\log\_{10}(W\_{value} / W\_{ref})$ or $20\log\_{10}(p\_{value} / p\_{ref})$ .
  
  
  ---
  
  ### Answer by username_1 on 2022-07-19. Score: 7
  
  The $10\log\_{10}()$ refers to a measure for acoustic energy. This should be used when adding up different sources, because you may add acoustic energy and not acoustic amplitude. Energy is related to the square of the amplitude.
  $20\log\_{10}()$ shoudl be used for amplitude.
  
  
  ---
  
  ### Answer by username_2 on 2022-07-19. Score: 7
  
  All the answers made so far are correct.
  
  I only would like to add that the word deciBel has 'deci' in the word so that deciBel means 1 tenth of a Bel. As @fabi\_hirono indicates deciBel was created for power ratios, so 10 log(Power/Power\_ref) is always correct.
  
  For sound waves, the use 20 log(Pressure/Pressure\_ref) is only correct in the far field. Btw, I would not call the far field assumption, or real impedance assumption, a 'minor' assumption, but a realistic one, if measurements are far away from point source.
  
  > **Comments:**
  > **username_3**: I thought the `20*log10(p/p_ref)` form was also valid for spherical waves? I know it is often used withouth making a plane wave assumption, at least in air. In Pierce - Acoustics (3rd Ed), Sec. 2.3.5, he says the Sound Intensity Level `10*log10(I/I_ref)` and Sound Pressure Level `20*log10(p/p_ref)` are approximately the same for plane AND spherical waves in air, at least. Did I miss something?  
  > **username_2**: You are correct, that is why I had changed to 'far field' (plane-field was a mistake). In the nearfield Impedance get complex, so 20log10(.) does not work  
  
  ---
  
  ### Answer by username_4 on 2022-07-19. Score: 5
  
  This is the formula to convert an acoustic pressure from Pa to dB: $20\log\_{10}(p\_{RMS}/p\_{ref})$ where:
  
  * $p\_{ref}$ is a reference pressure corresponding to 0dB (e.g. 20µPa for the dB SPL scale)
  * $p\_{RMS}$ is the root mean square of the pressure over a certain period of time
  
  This formula is equivalent to $10\log\_{10}(p\_{RMS}^2/p\_{ref}^2)$ which is sometimes used when talking about sound power or sound intensity because these quantities are related to pressures squared.
  
  > **Comments:**
  > **username_5**: The most commonly used reference pressure for media other than gases is 1µPa. See ASA [reference value for sound pressure](https://asastandards.org/terms/reference-value-for-sound-pressure-2/).  
  
  ---
  
  