Timeline for Finding out modulation index and DC offset
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 11, 2017 at 21:44 | history | suggested | Gilles | CC BY-SA 3.0 |
improved formatting + additional tag
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| Dec 11, 2017 at 21:24 | review | Suggested edits | |||
| S Dec 11, 2017 at 21:44 | |||||
| Feb 3, 2013 at 10:58 | history | tweeted | twitter.com/#!/StackSignals/status/298022747248926721 | ||
| Feb 3, 2013 at 10:27 | vote | accept | Samuel | ||
| Feb 3, 2013 at 9:59 | answer | added | Deve | timeline score: 6 | |
| Feb 3, 2013 at 9:30 | comment | added | Samuel | The solution is actually write Ac/2 =4. and cm mean c times m, c stand for dc offset and m stand for modulation index. and the amplitude of the test signal is 6V. | |
| Feb 3, 2013 at 9:19 | comment | added | Deve | From $Ac/2 = 4$ follows $c = 8/A$, not $c = 4$. Are you sure that $m = cm/(2c)$? And what do you mean by $cm$? Is this the amplitude $c_\mathrm{m}$ of the AC part of the test signal? | |
| Feb 3, 2013 at 9:06 | comment | added | Samuel | this is the provided solution. | |
| Feb 3, 2013 at 8:45 | comment | added | Deve | Is the solution only "like this" or is it definitely the correct solution? | |
| Feb 3, 2013 at 6:57 | comment | added | Samuel | and This question is made by my teacher, so no reference. | |
| Feb 3, 2013 at 6:57 | history | asked | Samuel | CC BY-SA 3.0 |