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Is there a limit on how loud an audio file can be? There probably is a limit to stop people from breaking their ears, but I don't know what it is.

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    Unfortunately for people's ears, there's no way a file on the computer can prevent someone from mechanically turning their amplifier up to 11... – grawity May 11 '18 at 15:11
  • No. And for that matter what is considered digitally “loud” is really dependent on playback devices. If you have a basic system speaker outputting the audio, the loudest sound might just sound the same as any other sound. Perhaps it would cause the speaker to crackle. But pump the same audio file through a real stereo or sound system and then you could probably blow your ears out. And for that matter many personal audio devices have built in sound limiters can can—sometimes—be activated/deactivated that protects a person’s ear from audio that is too loud. – JakeGould May 11 '18 at 15:18
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    @JakeGould yes there is a limit. See my answer. – LPChip May 11 '18 at 16:46
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Given that we talk about a .wav file, yes there is a limit.

When you look at a waveform in an editor, you will see that the editor has a screen in which the waveform is drawn, and you can see the top and bottom of the curve for most waveforms.

If you can see the curve entirely in the window, then the sound will not distort. Normalizing a waveform will stretch it up so its as big as it can get while still fitting in this window.

This produces a good quality sound, and is fairly loud. It is probably going to be around the same loudness as other sounds, as other people do the same. You have to keep in mind that waveforms have different dynamics and that does affect the volume too.

If you would play a normal sine waveform that fits the window perfectly, you get a certain volume. If you now pretend that the window is bigger, and you increase the volume of the sine waveform, the sound actually does get louder. You will see that at the left and right side of the top curve, the waveform is round, but at the middle top, instead of forming a nice half circle, its now cut off by the window you're viewing it in. The waveform itself has changed shape. This causes the sound to slightly distort, but in return, the volume is higher.

If you continue to raise the volume, that sine will get higher beyond the window, which deforms the waveform even more.

This is what it would look like: enter image description here Source of the image

As you increase the waveform, the width at the top and bottom increase, which makes the sound louder, but at some point, you will turn the sine waveform into a square waveform, which is its limit. At that point, the waveform will not deform any further and you have reached the possible maximum amplification of the audio. Because this is a dynamic process, given that a waveform has different shapes (for example, you cannot amplify a square waveform as much as a sine waveform, but in nature, a square waveform is already far louder than a sine waveform) there is easy answer to say how much you can actually increase the volume. There are mathematical formulas available though, but that goes outside of the scope of SuperUser.

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    That image is really showing a DC offset rather than pure level gain. That would not only distort where it's clipping, but would also be holding the speakers at a non-zero 'resting point' all the time, rapidly building heat. (It would also, I totally agree, sound bloody 'orrible ;) – Tetsujin May 11 '18 at 18:56
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    This answer is wrong. I could record someone whispering, and another person shouting. I could easily adjust recording levels so that your method of looking at the waveform would be useless at determining which recording was the whispering versus the shouting. – sawdust May 11 '18 at 22:58
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    "I'm an audio engineer" -- You only seem to be familiar with playback, and completely fail to mention the concept of recording level. – sawdust May 12 '18 at 20:03
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    @Damon oh, you definitely clip in a digital audio file. Open a Wave file, normalize it first, so it fits best in the audio editor and is as loud as it can be without clipping. Now apply an amplify effect to the waveform so it clips and Save the wave file, open it back, and now lower the volume. Tada, the clip is still there. – LPChip May 13 '18 at 9:58
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    This limit applies to recording and reproduction (and is correct if you are talking about either of those processes), but has nothing to do with what is stored in a .wav file (which is what the question asks about). The maximum value in a .wav file can be associated with any loudness, or amplitude during processing. The answer is technically correct for what it states, but doesn't really answer what was asked in the question. – fixer1234 May 14 '18 at 20:12
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What is the loudest sound that can be stored in a .wav file?

As loud as you want (assuming that your microphone and electronics do not distort), since proper recording technique dictates that the input level should be adjusted to prevent input overload/clipping.
IOW the recording level should be adjusted to match the loudness of the source being recorded.

Every recording medium has a limited dynamic range, i.e. the difference between the loudest and quietest sounds that can be saved. Typical recordings of music and speech need to capture low-level sounds for realistic reproduction, so the recording level is adjusted appropriately.
For a given recording level there is a maximum sound level that can be recorded.
However if you need to record loud sounds, the recording level can be increased so that louder sound levels can be recorded, but with a corresponding loss of low-level sounds due to the limited dynamic range.

If recordings levels could not be adjusted to match the source, then audio recordings of very soft sounds or very loud sounds (e.g. Saturn V rocket launches) would not be possible. Fortunately recordings levels can be adjusted, so there are recordings of very loud events like rocket launches.

Is there a limit on how loud an audio file can be?

Technically no, because the recording does not maintain the the absolute sound levels of the source.
Instead the recording, whether analog or digital, captures the relative sound levels of the source, e.g. whether one instrument is playing louder than other, or someone shouting over someone whispering.
The "loudness" of an audio clip is really determined on playback, and with the amount of power and speaker efficiency you have for sound reproduction.

The sound levels in recordings are all relative to a maximum input level typically designated 0dB (i.e. zero deciBels) aka the recording level.
The maximum acoustical loudness that could be recorded would be dictated by the microphones and electronics employed.
The recording engineer would/should apply the necessary attenuation (or gain) to the input signal so that the maximum signal during the recording will register as 0dB.

If the 0dB level is set higher than necessary, then the full dynamic range offered by the sampling word size is not utilized, and softer sounds will lose resolution.
If the 0dB level is set lower than required, then once loud sounds exhaust whatever headroom that is available, the recorded signal will clip.

So there is no "limit" to the loudest sound that can be recorded simply because the "maximum" sound is expected to be adjusted to the 0dB level.
To forestall any unexpected input louder than the 0dB level, modern recording techniques use compressors and limiters, usually a detriment to audio quality.

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    This is correct as far as recording sound, but needs to be supplemented to fully answer the question: what is the loudest sound that can be stored in a .wav (regardless of how it got there). For that, you need to look at the other extreme, the minimum sound that can be discriminated from background noise. The number of bits define the possible signal to noise ratio. Each bit gives you approximately 6 db. So 16 bits per channel gives you the potential for the loudest sound to be 96 db louder than the softest. (cont'd) – fixer1234 May 12 '18 at 3:56
  • The .wav format is limited to a 4 GB filesize, and a bit depth of 16 is good enough for most purposes. But you could use a higher bit depth. 24 bits would give you a maximum of 144 db louder than the minimum (this exceeds the capabilities of recording equipment so it would have to be created synthetically). But that's just the signal you feed to your reproduction equipment, which must be capable of reproducing the levels. Further reading: en.wikipedia.org/wiki/WAV, en.wikipedia.org/wiki/Audio_bit_depth. – fixer1234 May 12 '18 at 3:56
  • The technically no is incorrect. It doesn't matter how you look at it, there will be a point in a wavefile where you simply can't go louder, which means, technically yes, there is a limit. See my answer. Yes, I did use the wrong image there, as I couldn't find another, better suited image. I'm an audio engineer and I've done extensive testing to see what the limits are. Once you blow up sound as far as you technically can, there will be a point where making the sound louder simply does not change the sound at all anymore. – LPChip May 12 '18 at 10:54
  • True, it will be extremely loud if you play it at normal volume on your speakers. But if you lower the volume of your speakers to a low level so you can listen to the volume, you will find out, that, although very distorted, possibly beyond recognision, depending on what you started with, there will be a point where the sound does not become louder. In fact, the closer you get to that maximum, the more you need to add loudness to notice any change. – LPChip May 12 '18 at 10:56
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    @sawdust, I was trying to keep it consistent with the question. Yeah, loudness really relates to the recording and reproduction, which is why I saw your answer as correct. "Loudness" in the question is the wrong term because a .wav file doesn't encode or store loudness, it stores dynamic range. Re: Saturn launch, you can record a 150dBA sound, store it in a 16 bit file, and maybe reproduce it at 150dBA, but the 16 bit file can't hold the 150dBA dynamic range. I'll just post another answer to try to more accurately cover it. – fixer1234 May 12 '18 at 19:40
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What is the loudest sound that can be stored in a .wav file?

There is no "loudest" sound that can be stored in a .wav file because .wav files don't store "loudness". I'll elaborate below.

Is there a limit on how loud an audio file can be?

Not a loudness limit, but there is a limit to dynamic range, which relates to loudness.

Loudness

Loudness is what you hear, and it can be measured as an absolute value based on sound pressure. For context, the sound pressure scale starts at 0 dB as the threshold of normal human hearing; loudness of 120-130 dB causes pain.

Only the original sound and the reproduced output have loudness. The loudness of the reproduced output is limited only by the capability of the equipment; you can make it as loud as the equipment can handle, regardless of how loud the original sound was or how it was stored.

Amplitude

During the recording and reproduction processes, the sound is represented by an electrical waveform. The amplitude of the waveform represents the loudness. Whatever equipment is processing the signal has some maximum amplitude it can handle.

If the waveform has more amplitude than the equipment can process, the excess portion can't be processed, so it is essentially chopped off at the limit. That's the clipping @LPChip describes. So the representation of loudness during these processes is constrained to an equipment-specific upper limit.

During the recording process, the amplitude limit is used to ensure that the loudest source sound is accurately recorded (not clipped). Amplitudes are referenced to the upper limit of the equipment, which is labeled zero, and everything below that is a negative relative value. That's the 0 dB @sawdust refers to. (Note that this is unrelated to the 0 dB hearing threshold mentioned earlier.) So by design, whatever was the maximum loudness of the original sound corresponds with the maximum of what is represented in a specific .wav file, but that maximum can be any loudness.

During reproduction, the amplitude limit of the equipment determines the maximum loudness it can output; i.e., the loudest that you can output the "loudest" sound represented in the .wav file; the output loudness limit is a function of the reproduction equipment rather than what is stored in the .wav file.

Dynamic Range

To ensure that the loudest sounds are captured and reproduced accurately, and the quietest sounds aren't lost in the background noise, sound recording and reproduction need to address both extremes. The relationship between the quietest sound that will be reproduced and the loudest is referred to as dynamic range.

The .wav File

The .wav file encodes the dynamic range; i.e., relative loudness rather than actual loudness. It is encoded as a numeric value using the available bit depth (more bits can store a larger number); the number of bits defines the dynamic range that can be stored. Each bit provides 6 dB of dynamic range. The typical depth of 16 bits allows a dynamic range of 96 dB, which is sufficient to cover most sound recording requirements.

You could use more bits and store a bigger dynamic range, but a given bit depth defines the limit. That is, how much louder the loudest sound can be compared to the quietest sound, but not how loud the loudest sound will be reproduced.

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In wave files, the "loudness" of a sound is the function of amplitude of the waveform, which is made up of samples. The bitrate of the file determines the maximum/minimum values for each sample.

8-bit audio: -128 to 127

16-bit audio: -32760 to 32760

32-bit audio: -1.0f to 1.0f (floating point).

The "loudest" sounds would be any wave that oscillates between the maximum and minimum values for a sample at your given bitrate. (E.G. -1.0f and 1.0f for 32 bit audio)

Further reading: https://blogs.msdn.microsoft.com/dawate/2009/06/23/intro-to-audio-programming-part-2-demystifying-the-wav-format/

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