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Can mixing/adding two band-limited signals create any frequencies above Nyquist?

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  • $\begingroup$ Mixing and adding are not the same. Adding is a Linear operation that will not alter the frequency content; mixing will (at least in the typicall interpretation of the word "mixing"). $\endgroup$
    – rrogers
    Mar 29, 2018 at 10:54
  • $\begingroup$ @rrogers Can you elaborate on the difference between adding/mixing $\endgroup$
    – user17127
    Mar 29, 2018 at 11:05
  • $\begingroup$ Better yet, so I don't confuse you (sometimes my mental shorthand miscommunicates); why don't you state what you mean by the two terms? Then I can talk about the different uses. $\endgroup$
    – rrogers
    Mar 29, 2018 at 11:48
  • $\begingroup$ @rrogers Personally I mean the same. Mixing 2 signals means to add them. $\endgroup$
    – user17127
    Mar 29, 2018 at 11:50
  • $\begingroup$ Then you are in the audio type of Engineering I guess. In the RF domain, it implies a multiplication of signals. $\endgroup$
    – rrogers
    Mar 29, 2018 at 12:14

1 Answer 1

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Ideally no, adding 2 signals should result as a simple sum.

provided you don’t clip

DSP systems have finite dynamic range. A double float has a very large dynamic range so it is not usually a consideration.

Fixed point arithmetic has uniform quantization but the finite dynamic range requires attention.

Floating point is not a uniform quantization, so arithmetic also introduce rounding, which are discontinuous on small scales.

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    $\begingroup$ This question might be tricker than initially obvious. A band limited signal is not necessarily symmetrically band limited around f=0. So if two band limited signals that have been sampled and could be reconstructed individually do not share the same base band, the addition of the samples makes reconstruction impossible and leads to aliases in both reconstructed bands. $\endgroup$
    – Jazzmaniac
    Mar 21, 2018 at 13:11
  • $\begingroup$ What happens if an harmonic in the band-limited signal at Nyquist frequency is added to a 90 degrees out-of-phase replica of it? $\endgroup$
    – user17127
    Mar 26, 2018 at 14:31
  • $\begingroup$ I suggest you post a new question. More people will be interested and you are more likely to get a quality answer $\endgroup$
    – user28715
    Mar 26, 2018 at 14:36

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