I am a computer engineer and I am not an expert in physics. I read about Sound Pressure Level from different references but I can't solve my problem. My question is: if we turn on the TV, is the sound pressure level of the room increased definitely? In other words, in the room we have different sounds; if the sounds in the room are louder than the sounds created by TV, does the sound pressure level increased or not? why?


3 Answers 3


In general, any added sound increases the total sound pressure level (SPL), even if this sound alone has a smaller SPL than that of the background sound of the room. This is right when the added sound is incoherent with the background sound.

However, the SPLs with and without the added sound could be considered as about equal when the increase of SPL can be too small to be perceived and/or to be measured; have a try with this online calculator here: if you add two sources of 60 dB SPL and 40 dB SPL, the total level is 60.043 dB SPL.

The sound envelop can quickly vary with time. As the SPL involves the calculation of an average over a duration (referred as time constant of integration or time-weighting in SPL meters), SPL may increase only at particular moments if this duration is in the range of the speech word duration.

In some particular cases of coherent sounds (active noise control or standing waves), the total SPL can even be reduced by adding a sound which phase is offset compared to that of the background sound.

If you are familiar with mathematical notations

At a given location, the change in SPL $\Delta L_{T,\Delta t}(t)$ between times $t$ and $t-\Delta t$ is proportional to:

$ \Delta L_{T,\Delta t} (t) \sim log_{10}( \frac{p_{rms,T}(t)}{p_{rms,T}(t-\Delta t)} )$

where $ p_{rms,T}(t) = \sqrt{ mean( p^2(t),T)}$ is the RMS of pressure $p(t)$ over duration $T$.

Then, adding an incoherent sound at $t$ (or between $t-T$ and $t$) increases the RMS pressure $p_{rms,T}(t)$ as compared to $p_{rms,T}(t - \Delta t)$, so $ \Delta L_{T,\Delta t}>0$ (i.e. the SPL increases)


Human hearing sensitivity is approximately logarithmic, leading to the use of decibel scale to measure sound level.

Decibels don't add up directly. For unrelated sounds, the formula is (see e.g. this source for details):

$$ SPL(A+B) = 10 \times \log_{10}\left(10^{SPL(A)/10} + 10^{SPL(B)/10}\right)$$

Some practical examples on how this works:

  • Normal 60 dB conversation is going on. Television is turned on at 60 dB volume level. Total sound level in room increases to 63 dB.
  • Vacuum cleaner is making 80 dB noise. Television is again turned on at 60 dB. Total sound level increases to 80.04 dB.

As you can see, quieter noises or even noises at the same sound level do not make a huge difference to the total sound level. Due to the logarithmic nature of hearing, loudest sound dominates.

One extra complication is human's ability to separate sounds by frequency. A high-pitched noise will be noticed even if there is a louder low-frequency sound in the background, and vice-versa, although masking may occur. Due to the human non-linear perception of loudness across frequency, A-weighting is used to better assess SPL when human hearing is of relevance.


Let me add to @Noil's answer a little bit of physical interpretation.

Every additional sound adds sound energy and the total sound energy increases.The resulting RMS sound pressure level is proportional to the square-root of the sound intensity which is nothing else than the sound energy per second arriving to your ears.

In how far humans notice any difference is another question that is also addressed by others.


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