Hot answers tagged

5

Note that when you compute the DFT, you compute the dot product with a sine term and with a cosine term. In your third example, even though the dot product with the sine term is zero, the dot product with the cosine term isn't, so you do get spectral leakage, as you expected. The dot product with the sine term is zero because you have a whole number of ...


3

Dot returns zero, but I don't understand why. Why do we get zero here instead of a non-zero number Luck mostly. It's only zero because you chose the right phase difference. If you add any non-trivial phase to either one of the sine waves, you would get a non zero result. That's NOT the case for 5Hz and 6 Hz, where the dot product is zero regardless of the ...


2

If you want to learn about the DFT standalone you need to ditch the Hz. DFT exercise in the book Understanding digital signal processing 3 Ed Any two Sine or Cosine waves multiplied together are the same as the sum or the difference of two others. It's just a matter of getting the phases line up. $$ \cos(X)\cos(Y) = \frac{1}{2} \left( \cos(X+Y) + \cos(X-Y) \...


Only top voted, non community-wiki answers of a minimum length are eligible