1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $F[\dot{x}(t)] = j \omega X(\omega)$ 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
0
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ $\ne -\omega^2 X(\omega) \ast X(\omega)$ $\endgroup$ – 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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