3 ms·
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 w
by gamache 13y ago
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.
- fnordfnordfnord 13y agoDerp, thanks to both for the correction.
- lutusp 13y agoActually, if we look at two half-cycle waveforms, one sine and one square, like this: http://i.imgur.com/oE5NFZ9.png 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%2Cx%2C0%2Cpi%29 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.
- dpe82 13y agoIt also assumes a speaker's reactance is purely resistive. It's not.