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$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$

Can someone explain how the arithmetics works here? (My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$")

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    Images: modulation - https://dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_174940.jpg demodulation - https://dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_175218.jpg2012-08-23

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As has been pointed out in comments, for this equation to hold for all $x$, we need to have $d_1=d_2=0$. The reference to being removed by filtration doesn't refer to an arithmetic operation inherent in this equation, but to the application of a low-pass filter indicated in the second image linked to in a comment.

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    Alright, that makes sense. Thank you!2012-08-23