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mixing sampled sine waves

Hi, I have a question regarding mixing discrete sine waves. If you have two sine waves sin(w1*t) and sin(w2*t) and they are sampled at the same rate. If you are mixing them in a receiver operation, we are supposed to get at the output of the mixer the sum and difference of frequencies. But it is just the values that we are multiplying isn't it, at the sampled time instants? How do we end up getting a difference frequencies and sum frequencies which have to be low pass filtered? I have read from trigonometry and analog communications but somehow I am missing some essence here. Could you please let me know how the above is possible? I would greatly appreciate a response. Thanks, Viswanath

Reply to
viswanath
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That depends on what you mean by a mixer. A usual mixer in a receiver is not an additive thing. It is a non-linear circuit that essentially multiplies one signal by the other. To see how you get the sum and difference frequencies, just look at the trig identity for

sin(a + b) and sin(a - b)

and see how they relate to sin(a) * sin(b)

-Robert Scott Ypsilanti, Michigan (Reply through this forum, not by direct e-mail to me, as automatic reply address is fake.)

Reply to
Robert Scott

The analog receiver double-balanced mixing is just a signed multiplication of the both sample streams. To get meaningful sum and difference frequencies, all frequencies (f1, f2, f1 + f2, f1 - f2) must be below the Nyquist limit (half of the sample rate).

Apply the sine sum formula to sin(w1*t)*sin(w2*t) to see the desired frequency components.

HTH

Tauno Voipio tauno voipio @ iki fi

Reply to
Tauno Voipio

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