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


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.