well, the formula is as follow:

enter image description here

but i am pretty sure that this answer - zero is not correct, so could anyone help me out of this problem?

  • $\begingroup$ $F[\dot{x}(t)] = j \omega X(\omega)$ so $F[\dot{x}(t)^2] = (j \omega X(\omega)) \ast (j \omega X(\omega))$ $\endgroup$ – reuns Apr 10 '17 at 8:36
  • $\begingroup$ your answer is correct, and i can totally understand that, but i want to know the exact result of F[g(t)]. can you help me with that? $\endgroup$ – amos Apr 10 '17 at 10:43
  • $\begingroup$ Can you use what I wrote for obtaining the correct formula for $F[g(t)]$ ? $\endgroup$ – reuns Apr 10 '17 at 11:37
  • $\begingroup$ does that matter? but i still cannot find my mistake. $\endgroup$ – amos Apr 10 '17 at 13:58
  • $\begingroup$ i am sorry if that makes it not readable since i don't know about how to edit formula on the website, so i just upload the pictures. $\endgroup$ – amos Apr 10 '17 at 14:01

After discussion with user1952009, i have realized my mistake that i considered jω as a coefficient rather than a variant.

$F[x(t)\ddot x(t)]=-\omega^2X(\omega)*X(\omega)$

$F[\dot x(t)^2]=j\omega X(\omega)*j\omega X(\omega)\neq-\omega^2X(\omega)*X(\omega)$

  • $\begingroup$ $\ne -\omega^2 X(\omega) \ast X(\omega)$ $\endgroup$ – reuns Apr 12 '17 at 1:30
  • $\begingroup$ And click on 'edit' in the other questions for seeing how to type in latex $\endgroup$ – reuns Apr 12 '17 at 1:33
  • $\begingroup$ thanks again. i am in china, it seems that my network always break down when i upload pics.. $\endgroup$ – amos Apr 12 '17 at 1:40

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