Timeline for Does non-orthogonal multicarrier modulation use pulse shaping?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2022 at 21:18 | comment | added | Dan Boschen | In general there is no need to simulate an actual carrier for a linear system, you can have them all centered on 0 and use complex signals (that is typically the approach and significantly simplifies the processing). "Baseband I and Q" and then frequency translated to the subcarriers using the way I suggested with an exponential. | |
| Apr 14, 2022 at 21:16 | comment | added | Dan Boschen | I don't know if the center one was complex or not, but the other three were clearly real in the baseband equivalent signal which you get if you remove your carreir at 1000 and make it 0. Multiplying the baseband signal by cosine shifts your signal to +/-w (consider that cosine is made up of two the the $e^{j\omega t}$ terms I mentioned- look up Euler's Formula. The DFT is just showing you the carriers at $e^{j \omega t}$ and not cosines, hence the positive and negative frequency that you have relative to a carrier at 1000 Hz. | |
| Apr 14, 2022 at 21:09 | vote | accept | Jonah F | ||
| Apr 14, 2022 at 21:05 | comment | added | Jonah F | For this example, I have only modulated my signal onto a cosine carrier, as you have stated. With $e^j(\omega t)$ carrier modulation I have the PSD as imgur.com/a/vKh5TJS. Could you quickly comment on how you could tell my signal has 1 complex and 3 real carriers? I dont think I understand the difference - I thought multiplication by $\cos(\omega t)$ would simply shift my baseband signal up to $\omega$ in the frequency domain? | |
| Apr 14, 2022 at 20:02 | comment | added | Dan Boschen | A parenthetical comment on your 4 sub-carrier case that I didn't want to cloud in my answer-- this would not be the way to create 4 sub-carriers as you have 1 complex carrier and 3 real carriers (I assume, unless you destroyed the one in the center if this all was zero). To create 4 sub-carriers maintain a complex IQ waveform and frequency translate each to the subcarrier location using $e^{j\omega t}$ not $\cos(\omega t)$ as you have. | |
| Apr 14, 2022 at 19:56 | answer | added | Dan Boschen | timeline score: 1 | |
| Apr 14, 2022 at 19:42 | history | asked | Jonah F | CC BY-SA 4.0 |