Timeline for Decomposition of two equal and overlapping complex signals
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2021 at 19:34 | comment | added | Magne Lauritzen | I see. Well, I'm still not sure where to go from here, but at least now I know how to represent the signals in the frequency domain so that helps. I will work on it some more. | |
| Jun 15, 2021 at 22:10 | comment | added | Marcus Müller | the exponential term I moved out of the Fourier transform is just a constant factor, as it's independent of $t$, so that's "legal". | |
| Jun 15, 2021 at 21:02 | comment | added | Magne Lauritzen | Hi Marcus, thank you very much for taking the time to respond to my question and for helping me formatting it correctly. On the line where you take the Fourier transform of $R_{l,m}$, you seem to be saying that $\mathcal F\{x_0(t-\tau_m) e^{i\theta \tau_m}\}(f) = e^{i\theta \tau_m}\mathcal F\{x_0(t-\tau_m\}(f)$. But I don't think this is valid because solving the fourier transform of a signal shifted by $\tau$ involve doing the substitution $t' = t-\tau$, which also alters the $e$ term you move outside the fourier transform : $e^{i\theta \tau_m}$ becomes $e^{i\theta (t - t')}$ | |
| Jun 13, 2021 at 17:09 | history | edited | Marcus Müller | CC BY-SA 4.0 |
added 172 characters in body
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| Jun 13, 2021 at 17:01 | history | edited | Marcus Müller | CC BY-SA 4.0 |
added 172 characters in body
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| Jun 13, 2021 at 16:56 | history | answered | Marcus Müller | CC BY-SA 4.0 |