mixing sampled sine waves

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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.

Re: mixing sampled sine waves

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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.)

Re: mixing sampled sine waves
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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.


Tauno Voipio
tauno voipio @ iki fi

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