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Consider you have 50 Hz both square wave as well as sine wave put on to a same speaker at different time. Which signal is more audible and what is the reason behind that ?

We know that by taking the Fourier series of the square wave we get multiple periodic sinusoidal wave, each having the frequency as the multiple of the fundamental frequency. So which will be more audible, whether a sine wave having single frequency or the square wave?

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    \$\begingroup\$ What are their respective levels? Same amplitude, same (rectified) average, or same RMS value, for instance? \$\endgroup\$ – stevenvh Oct 6 '12 at 18:30
  • \$\begingroup\$ It is not relevant point here. I already got the answer. Thank you.. \$\endgroup\$ – arun Oct 16 '12 at 20:02
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The square wave will be more audible. 50 Hz is fairly low and most speakers will not reproduce that frequency very well. Since a sine wave will have only 50 Hz there may not be that much audio reaching the human ear, and even then the human ear will not respond very efficiently to it. A square wave, on the other hand, will have lots of harmonics that the speaker will reproduce very well and the harmonic frequencies will be spread right through the ideal frequency range for human hearing (300Hz to 3 KHz).

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    \$\begingroup\$ +1 for great info...tried your suggestion with 14KHz and I really feel,ideal human hearing range is (300hz-3KHz).... \$\endgroup\$ – perilbrain Oct 5 '12 at 19:09
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    \$\begingroup\$ Thank you.I knew that it was square, but the reason behind it was not clear to me. :D \$\endgroup\$ – arun Oct 5 '12 at 19:13
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    \$\begingroup\$ In fact, most times when people "hear a 50 (or 60) Hz hum", they are really hearing some combination of the harmonics. \$\endgroup\$ – gbarry Oct 5 '12 at 19:45
  • \$\begingroup\$ It's worth pointing out that while this answer is 100% correct, the question is a bit ambiguous: are the sine and square waves equal peak voltage or RMS? \$\endgroup\$ – Bjorn Roche Oct 6 '12 at 1:37
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    \$\begingroup\$ I also find sources of sine waves much harder to locate than for other sounds. \$\endgroup\$ – starblue Oct 6 '12 at 7:26
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In the specific example you gave, the square wave is much louder; mostly for physiological reasons.

1) Into the same load, a square wave will deliver twice the power as a sine wave of the same peak voltage. This is same as saying the square wave has an RMS voltage equal to its peak value; whereas a sinewave has an RMS value of 0.707 (actually sqrt(2)/2) times the peak value.

2) More importantly, the sensitivity of the human ear is a strong function of frequency. The telephone operates from 300Hz - 3KHz (give or take). That is where the intelligible content is and it is also where the ear has a good sensitivity. The human ear is very insensitive to 50Hz. The same power level at 500Hz would be MUCH loader. Though the energy of the square wave in the harmonics (odd harmonics only; 3, 5, 7, 9..) has no more power than the fundamental, it will be perceived as much louder because of the ear's greater sensitivity to the higher frequencies.

The international standard ISO-226 identifies the accepted equal-loudness contour for the human ear as a function of frequency and absolute sound pressure level. Equal-loudness contours are often referred to as "Fletcher-Munson"' curves, after the earliest researchers, but those studies have been superseded and incorporated into newer standards.

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"Which signal is more audible" is more ambiguous to be answered but sinusoidal waves are melodious in nature while those in square form will feel like drilling your tymphanic membrane.Looking from that point, square waves will dominate.

However if you are interested with such experiments then I will suggest you an open source alternative "Audacity". Install this software and navigate to Generate menustrip->Tone and choose the right waveform of your choice while entering desired amplitude and frequency.

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  • \$\begingroup\$ The reason was between 300 Hz to 3 KHz is only audible to human ear as explained by @David Kessner \$\endgroup\$ – arun Oct 5 '12 at 19:17
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    \$\begingroup\$ David had already gave your answer, while I was posting so I better thought to put some extra information than continuing the same .... :) \$\endgroup\$ – perilbrain Oct 5 '12 at 19:19
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Driving a resistive load with a square wave at a particular voltage will deliver twice as much power as driving a resistive load with a sine wave of the same peak voltage. A speaker, however, does not represent a resistive load; the current required to change a speaker's position quickly is much greater than what's required to change its position slowly. To displace air with a perfect square wave would require an infinite amount of power(*). In practice, trying to drive a speaker with a perfect square wave will result in air motion which is not a perfect square wave. While the actual observed behavior will vary depending upon the type of speaker one is trying to drive, driving a speaker with a square wave of a certain peak voltage will take a lot more power than driving it with a sine wave of the same voltage and frequency, and in general the closer the speaker is able to get to a perfect square wave, the more power will be consumed in the attempt.

(*) If air were a lossless medium, it would be possible to recover all the energy one put into the air that echos back before it is removed by something else; the power one one would have to put in would thus equal the power that other things took out. In practice, though, the closer to a square wave one tries to get, the greater the losses. Instantaneously moving air would require moving it with infinite force against a finite distance; a speaker would be unable to recapture much of that energy. Indeed, because of the speaker's electrical resistance stopping the speaker quickly would likely require pushing more energy into the speaker, rather than passively recovering energy from it.

If one's objective is to achieve the maximum perceived with a given available voltage, using a square wave or sawtooth wave would be a good way to do that. If the objective is to receive the maximum perceived loudness for a given amount of power consumed, other waveforms may be better (which waveforms are best depends upon the designs of the speaker and the driver circuit).

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