On 11 May 2004 12:35:13 -0700,

http://www.velocityreviews.com/forums/(E-Mail Removed) (viswanath) wrote:

>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? If I implemented this in VHDL and

>used math_real library functions sin() and real variables, I would

>just be multiplying the two numbers at each time sample. This value

>would encompass the frequency of the sine wave. However I donot

>understand how I would be getting a difference in frequency and sum in

>frequency terms?

>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?
Firstly, find a table of trig identities:

http://www.google.com.au/search?q=trig+identities
Then find the one that looks like a product of sinusoids on one side

and a sum (or difference) of sinusoids on the other, e.g.

cos x - cos y = -2 sin( (x-y)/2 ) sin( (x + y)/2 )

Express this in terms of your original w1*t and w2*t :

cos((w1+w2)*t) - cos((w1-w2)*t) = -2 sin(w1*t) sin(w2*t)

The right hand side is the two sine waves you are multiplying

together, sample by sample.

The left hand side is the 'sum' of two sinusoids at the sum and

difference frequencies that produces identical values, sample by

sample.

If you apply a frequency selective filter, you can get rid of the

(w1+w2) term, and just leave the (w1-w2) term.

Note: this is not magic, just a trig identity. It may take a while to

get your head around it though.

(There is no need to worry about that Fourier guy yet.)

Please ask this sort of question in comp.dsp in future.

Regards,

Allan.