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Forceaudio


 
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Author Topic: The Clipping effect test  (Read 5665 times)
todd.brust
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« Reply #15 on: February 25, 2009, 08:17:03 pm »

I had the same question too but i read an article a while ago that helped out..  DC indicates direct current with either a positive or negative voltage.  With a square wave, there is a sine wave with harmonics and the average of these combined wave makes a square wave.  So a square wave really contains a bunch of different sine waves.

I am not sure if I understand the last part of the question correctly, the wording got me.

If I do understand correctly, my answer would be that the coil would act like a resistor if there was straight DC going through it.  With any type of AC, the resistance of the coil changes with frequency.

Someone else may be more qualified to go more in depth with this one though.
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« Reply #15 on: February 25, 2009, 08:17:03 pm »

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95Honda
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« Reply #16 on: February 27, 2009, 10:09:50 pm »

The amp won't put out any more power, anything negligable at least.  Even if the coil does act like a resistor for the mS/uS the driver attemps to be at the peak of the sqaure wave (there is ringing, dampening, back emf and all other kinds of things that probably won't let the driver act like a true resistor for that brief period of time) it still would be a resistor with a value a little lower than the overall inpedance of the driver.

I am not certain, but I am also pretty sure that a driver won't behave linearly enough above say 20-30Hz (guessing here) to really produce a square wave, let alone behave linearly enough to produce all those harmonics perfectly.  Drivers still have dampening, the combination of the motor and suspention still shape the way they could attempt to produce a square wave.  

The bottom line, I think, is, square or sine, the driver still moves just as far from +/-, sweeps the same amount of air and cools the coice coil very similary.  And coils themselves (no added forced cooling) could care less what kind of signal is applied....
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