[Physics] Where does the energy of noise cancellation go

acousticsenergy

This is related to this question on Aviation.SE about noise cancelling headsets.

We know that soundwaves can be superimposed on each other. Therefore, by taking an existing sound wave and producing a second wave which is exactly its inverse, the two will sum up to zero. This is how active noise cancelling headset works – by "listening" to the exterior sound with a microphone, then reverting it in its speaker.

The problem is, where does the energy go? The noise sound is an energy source. Generating an inverted sound on the drums of the speaker is also an energy source (from the electricity which powers the speaker). So the total energy output cannot be zero. But we hear silence! Where does that energy go?


EDIT: I'm more interested in a theoretical answer, assuming a perfect world with no energy loss in converters / out-of-sync waves. Although an answer giving real world converter efficiencies is also useful.

Best Answer

The speaker mimics a black body radiator/absorber. If the speaker were 100% efficient and the impedance between the diaphragm and air perfectly matched, then no power need be sent to the speaker and all the sound power would return to the circuit driving the speakers, where its up to the design of the circuit what to do with it (for example, some class D drivers would return the power to the supply, while typically linear circuits would waste the power in a resistor). However, not only are speakers intrinsically inefficient, the diaphragm is horribly matched to the air. So to make the speaker behave as if its a perfect absorber, power must be sent to the speaker, where most is wasted as heat in the coils and magnet. The vast majority of original air power that was to be cancelled is also wasted as heat in the speaker, but a tiny fraction makes its way back to the driving circuit, and then its up to the circuit. But because the efficiency is far below 50%, far more power is spent than regained.

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