Answers.Body: <p>Yes, you can assume dB SPL is based on RMS.</p>
  <p>As far as I understand, Sound Pressure Level (SPL) is always based on root-mean-square (RMS) measurements. This is obvious for signals that last longer than the analysis window.</p>
  <p>For shorter signals (transients, clicks, pulses, etc.) the term peak equivalent (peRMS) is used (e.g. <a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4906301/" rel="nofollow noreferrer">Laukli E, Burkard R. Calibration/Standardization of Short-Duration Stimuli. Semin Hear. 2015 Feb;36(1):3-10. doi: 10.1055/s-0034-1396923. PMID: 27516707; PMCID: PMC4906301.</a>), where the transient is replaced by a sinusoid, such that the 0-p or p-p pressures are equal to the sinusoidal peak levels. For symmetric transients, this corresponds to reducing the peak level  by 3 dB. For asymmetric transients the reduction will be a little bit more.</p>
  
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Answers.Body: <p>Yes, more often than not you can safely assume that all level quantities are calculated based on RMS values unless stated otherwise.</p>
  <p>It doesn’t make much sense to provide level measurements of recorded signals that are <strong>not</strong> calculated from the RMS value of the linear quantity. If this was not the case, the level value would correspond to a specific instance in time, which is definitely not very helpful most of the time. A possible exception could be the peak amplitude of transients but, as already mentioned, when a value corresponding to an instantaneous field quantity is used for the calculation of levels it will (and <strong>must</strong>) be stated.</p>
  <p>An example which could provide some insight is the <em>Sound Pressure Level</em> (re <span class="math-container">$\sim 20 \mu \textrm{Pa}$</span>) of a sinusoid. An artificial sinusoid of frequency <span class="math-container">$f = 1 \textrm{kHz}$</span> is created with a peak value <span class="math-container">$\sqrt{2} \approx 1.414$</span>. The <em><strong>instantaneous</strong></em> SPL is shown in the image below.</p>
  <p><a href="https://i.sstatic.net/f5iZtil6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f5iZtil6.png" alt="Instantaneous amplitude of sine wave" /></a></p>
  <p>Some things to note are that the signal peak is such that its RMS value will correspond to <span class="math-container">$1 \textrm{Pa}$</span> and the peak amplitude in <span class="math-container">$\textrm{dB}$</span> is <span class="math-container">$\sim 97 \textrm{dB}_{\textrm{SPL}}$</span>. Knowing that the sinewave has a crest factor of <span class="math-container">$3 \textrm{dB}$</span> this would give an SPL value based on its RMS equal to <span class="math-container">$97 \textrm{dB}_{\textrm{SPL}} - 3 \textrm{dB} = 94 \textrm{dB}_{\textrm{SPL}}$</span>, which is what you can find in many references to be the SPL value corresponding to a <span class="math-container">$1 \textrm{Pa}$</span> pressure perturbation.</p>
  <p>What is important to take from the image above is that the instantaneous SPL measurement does <strong>not</strong> provide any meaningful information. It has no relevance to the perception of sound and does not convey any useful information from an “energy” perspective, which renders it “not very useful” in the end.</p>
  <p>In an attempt to further support my case, let me remind you that RMS values are related to the energy/power of the signal, which, first of all, is of high importance and second, provides a very good comparison metric to use for all signals. Even power signals (those that, in a mathematical sense, have infinite energy but finite power) can be compared to other signals on the grounds of energy comparison, which is facilitated by the use of RMS values.</p>
  
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Body: <p>I'm kind of struggling with the many definitions of decibels. I understand that we can either measure the root-mean-square, or the 0-to-peak, or the peak-to-peak value. I also understand that dB SPL is the unit employed for sounds, but sometimes I find values in &quot;dB SPL RMS&quot;, or &quot;dB pe SPL&quot;, and sometimes only dB SPL.</p>
  <p>When I find dB SPL in the spec sheets of my equipment, can I safely assume this is dB SPL RMS?</p>
  
Tags: <decibels><rms>
Title: Is dB SPL always RMS?
ThreadText: # Is dB SPL always RMS?
  
  (Asked by: username_0 on 2024-05-28)
  
  I'm kind of struggling with the many definitions of decibels. I understand that we can either measure the root-mean-square, or the 0-to-peak, or the peak-to-peak value. I also understand that dB SPL is the unit employed for sounds, but sometimes I find values in "dB SPL RMS", or "dB pe SPL", and sometimes only dB SPL.
  
  When I find dB SPL in the spec sheets of my equipment, can I safely assume this is dB SPL RMS?
  
  **Tags:** decibels, rms
  
  
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  ## 2 answers
  
  ### Answer by username_2 on 2024-05-29. Score: 4 | Accepted answer
  
  Yes, more often than not you can safely assume that all level quantities are calculated based on RMS values unless stated otherwise.
  
  It doesn’t make much sense to provide level measurements of recorded signals that are **not** calculated from the RMS value of the linear quantity. If this was not the case, the level value would correspond to a specific instance in time, which is definitely not very helpful most of the time. A possible exception could be the peak amplitude of transients but, as already mentioned, when a value corresponding to an instantaneous field quantity is used for the calculation of levels it will (and **must**) be stated.
  
  An example which could provide some insight is the *Sound Pressure Level* (re $\sim 20 \mu \textrm{Pa}$) of a sinusoid. An artificial sinusoid of frequency $f = 1 \textrm{kHz}$ is created with a peak value $\sqrt{2} \approx 1.414$. The ***instantaneous*** SPL is shown in the image below.
  
  [![Instantaneous amplitude of sine wave](https://i.sstatic.net/f5iZtil6.png)](https://i.sstatic.net/f5iZtil6.png)
  
  Some things to note are that the signal peak is such that its RMS value will correspond to $1 \textrm{Pa}$ and the peak amplitude in $\textrm{dB}$ is $\sim 97 \textrm{dB}\_{\textrm{SPL}}$. Knowing that the sinewave has a crest factor of $3 \textrm{dB}$ this would give an SPL value based on its RMS equal to $97 \textrm{dB}\_{\textrm{SPL}} - 3 \textrm{dB} = 94 \textrm{dB}\_{\textrm{SPL}}$, which is what you can find in many references to be the SPL value corresponding to a $1 \textrm{Pa}$ pressure perturbation.
  
  What is important to take from the image above is that the instantaneous SPL measurement does **not** provide any meaningful information. It has no relevance to the perception of sound and does not convey any useful information from an “energy” perspective, which renders it “not very useful” in the end.
  
  In an attempt to further support my case, let me remind you that RMS values are related to the energy/power of the signal, which, first of all, is of high importance and second, provides a very good comparison metric to use for all signals. Even power signals (those that, in a mathematical sense, have infinite energy but finite power) can be compared to other signals on the grounds of energy comparison, which is facilitated by the use of RMS values.
  
  
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  ### Answer by username_1 on 2024-05-28. Score: 4
  
  Yes, you can assume dB SPL is based on RMS.
  
  As far as I understand, Sound Pressure Level (SPL) is always based on root-mean-square (RMS) measurements. This is obvious for signals that last longer than the analysis window.
  
  For shorter signals (transients, clicks, pulses, etc.) the term peak equivalent (peRMS) is used (e.g. [Laukli E, Burkard R. Calibration/Standardization of Short-Duration Stimuli. Semin Hear. 2015 Feb;36(1):3-10. doi: 10.1055/s-0034-1396923. PMID: 27516707; PMCID: PMC4906301.](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4906301/)), where the transient is replaced by a sinusoid, such that the 0-p or p-p pressures are equal to the sinusoidal peak levels. For symmetric transients, this corresponds to reducing the peak level by 3 dB. For asymmetric transients the reduction will be a little bit more.
  
  
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