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0 pointsgamache12y ago0 comments

Assuming the parent is talking about equal maximum amplitude of the sine and square waves, the square wave RMS voltage will be sqrt(2), or 1.4, times the sine wave RMS voltage. And since power is proportional to voltage squared, the square wave will carry sqrt(2)^2, or 2, times the sine wave power.

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lutusp12y ago

Actually, if we look at two half-cycle waveforms, one sine and one square, like this:

http://i.imgur.com/oE5NFZ9.png

(Just the left 1/2 of the graph for this example)

The integral of the sine waveform with a peak value of 1, on the interval 0 x < pi, is :

http://www.wolframalpha.com/input/?i=integrate%28sin%28x%29%...

= 2

The integral of the square wave on the same interval is = pi

So the ratio increase in average voltage at the speaker (comparing the sine to the square) is pi/2 = 1.57. The increase in speaker power is (pi/2)^2 = 2.46.

dpe8212y ago

It also assumes a speaker's reactance is purely resistive. It's not.

fnordfnordfnord12y ago

Derp, thanks to both for the correction.

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