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Can a sinusoid with unknown frequency be constructed from other sinusoids with known frequencies? Are there any theorems for this problem?

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can I represent it with other sinusoids of known frequencies.

Generally speaking: no. Sine waves are orthogonal. That's the whole idea behind the Fourier Transform.

You can change the frequency of a sine wave only with non-linear operations, which is probably not what you mean.

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  • $\begingroup$ Aliased signals kind of fit this category, which definitely are the result of a non-linear operation. For a discretized measurement, a sinusoid of unknown frequency, $f$, is represented by $f_a=|2 m f_s - f|$ for some integer $m$ such that $|f_a| < f_s$ where $f_s$ is the sample rate of the measurement (ref). $\endgroup$
    – Ash
    Oct 31, 2022 at 15:00
  • $\begingroup$ Sampling is actually a linear process. It's time-varying, which is how it can move frequency components around, but it obeys superposition so it is linear. So is DSB modulation (i.e., multiplying by a sine wave). $\endgroup$
    – TimWescott
    Oct 31, 2022 at 15:13
  • $\begingroup$ Thanks @TimWescott. I found a previous discussion on this topic which shares the same mix up. I agree that sampling is a linear process, but can we say the same for the mapping $f\rightarrow f_a$? $\endgroup$
    – Ash
    Oct 31, 2022 at 15:39
  • $\begingroup$ @Ash No, because strictly speaking the sampling you mean happens in time, not in frequency, and the signals you samples have to by no means be well-behaved enough to allow for a definition of "frequency" that makes sense, and sampling would still be linear. But, yes, for continuous monoperiodic signals, the mapping $\text{sampling in time}: f_{\text{continuous time signal}} \mapsto f_{\text{time-sampled signal}}$ is proportional. You will just notice that it's not a mapping from the real numbers into the real numbers, $\endgroup$ Oct 31, 2022 at 16:30
  • $\begingroup$ but into a compact interval of frequency, with addition rules that make it a ring. (Linearity cannot work on frequency alone. that already doesn't work in the time-continuous domain: $e^{j2\pi f_0 t}$ has the same frequency as $e^{j2\pi f_0 t}$, but you tell me how you end up with a signal with twice the frequency if you add these?) $\endgroup$ Oct 31, 2022 at 16:34

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