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From this, (Square wave / Sine wave is more audible) I now understand that a Square soundwave will be louder than that of a Sine sound wave when played at the same frequency (for example 500 Hz). However, how about a sawtooth sound wave; which order from loudest to quietest is it placed?

I know that the Sine wave is only composed of the fundamental frequency and hence produces the lowest pitch - which is less sensitive to human ears. Square sound waves are composed of infinite odd harmonics, and sawtooth waves are composed of both odd and even harmonics. From experimenting, I have found that square sound waves are louder than sawtooth, which is louder than sine waves. What is the explanation behind this system of order?

Thankyou very much.

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    \$\begingroup\$ How are you matching the electrical levels of the three waveforms? If you're going by the peak values, then yes, the square wave will be the loudest, because it contains the most power (highest RMS voltage) of the three. But if you match the RMS values, then they should be similarly loud. However, there are psychoacoustic effects, too, in which a waveform that's rich in harmonics will seem louder than a pure or nearly-pure sine wave. \$\endgroup\$ – Dave Tweed Jul 13 '15 at 13:14
  • \$\begingroup\$ This may sound silly, but how do you calculate the RMS voltage / peak values for this? So, to put into more detail; I conducted an experiment where I put a speaker and a soundmeter at a distance away (say 10 cm) from each other, then played 1000 Hz Sine wave and measured the intensity of it. Similarly, I did this for 1000 Hz Square, and 1000 Hz Sawtooth. What I found was that the sine had an intensity of 94dB,square had 95dB and sawtooth had 89 dB. I want to investigate the effects of the waveforms on the intensity of sound.*Oh, and the amplitude was set at a constant. \$\endgroup\$ – user80922 Jul 15 '15 at 11:09
  • \$\begingroup\$ I was reading this: beausievers.com/synth/synthbasics/#sine And was confused as to why sawtooth would be less intense than square, as the amplitudes would add up to be much greater than that of square's amplitude. Or does the amplitude simply indicate the human perception of loudness - so the sawtooth is heard to be louder than the square, but in reality, the square has more energy than sawtooth - hence being recorded to be more intense in sound intensity level? Not quite sure if I make much sense, my apologies. \$\endgroup\$ – user80922 Jul 15 '15 at 11:47
  • \$\begingroup\$ What are you using to "play" the sounds? What kind of "soundmeter" are you using, and what are the settings on the meter? Some settings are for measuring the acoustic intensity, while others are intended to mimic the response of the human hearing system. \$\endgroup\$ – Dave Tweed Jul 15 '15 at 12:56
  • \$\begingroup\$ I am using a function generator program Audacity to generate the sounds. The soundmeter is just an app 'Decibel 10th' on my phone - update frequency set to 10 Hz, and calibration set to 0.2 dB. In a quiet room (during the experiment), the sound intensity was measured to be 52dB. \$\endgroup\$ – user80922 Jul 15 '15 at 13:10
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Loudness is not intensity or energy, but perception.

Energy in the 1 to 4 kHz band where the ear is most sensitive is perceived as louder than energy below 1 kHz.

High harmonics can be more attention-getting or even annoying than the actual SPL would suggest, though as a Scotsman I would make an exception for properly tuned bagpipes!

These factors combine to make a simple analysis unrealistic in analysing perceived loudness, as opposed to measured intensity.

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  • \$\begingroup\$ My apologies for making that rookie mistake of switching between 'loudness' and 'intensity'! I actually meant 'sound intensity' all throughout my question, sorry for the confusion and thanks for letting me know! \$\endgroup\$ – user80922 Jul 15 '15 at 11:11
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  1. sound intensity is a physical quantity i.e. the product of particle velocity and sound pressure
  2. loudness is a psychoacoustic quantity, i.e. human perception of sound pressure

Your question is ill-posed in the sense that you are confusing 1 and 2. In addition, sound intensity depends on the impedance of the sound field. There is no universally applicable way to determine what will be the intensity if you are just looking at a waveform in a computer. You would also need to specify RMS intensity or instantaneous intensity. I'd guess that RMS intensity will be proportional to RMS voltage at any point in space.

In addition, you cannot measure sound intensity using a smartphone (not a single app I know can do that).

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