$F_1(\omega)$ is the Fourier Transform of $f_1(t)$. $F_2(\omega)$ is the Fourier TRansform of $f_2(t)$. Can I obtain the Fourier Transform ($F_3(\omega)$) of

$$ f_3(t) = \frac{f_1(t)}{f_2(t)} $$

directly from $F_1(\omega)$ and $F_2(\omega)$? I mean, is there anything similar to the equivalence between Multiplication in the Time Domain to Convolution in the Frequency Domain but for the Division operation?

  • $\begingroup$ What you're probably looking for is called deconvolution; there's no single allways-working method for doing that, but plenty of nice approaches and examples :) $\endgroup$ Jan 12, 2018 at 13:03
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    $\begingroup$ If $1/f_2(t)$ exists, then you can convolve its FT with $F_1(\omega)$ to get $F_3(\omega)$. $\endgroup$ Jan 12, 2018 at 15:42

1 Answer 1


For division there is no equivalent to the duality between multiplication and convolution. Note that from the existence of the Fourier transforms of $f_1(t)$ and $f_2(t)$, you cannot conclude anything about the existence of the Fourier transform of $f_3(t)=f_1(t)/f_2(t)$. So $f_3(t)$ might not even have a Fourier transform, and if it exists, it cannot be expressed in any simple way in terms of $F_1(\omega)$ and $F_2(\omega)$.

  • $\begingroup$ i guess that, given $F_1(\omega)$ and $F_2(\omega)$, we can guess at $F_3(\omega)$ such that $$ F_1(\omega) = F_2(\omega) \circledast F_3(\omega) $$ it's a deconvolution problem, but i think it would be easier to transform back to the time domain and divide. $\endgroup$ Jan 12, 2018 at 23:37
  • $\begingroup$ @robertbristow-johnson: Sure it's deconvolution, but there's no way to "directly obtain" (as asked in the question) $F_3(\omega)$ from $F_1(\omega)$ and $F_2(\omega)$. In fact, $F_3(\omega)$ might not even exist. $\endgroup$
    – Matt L.
    Jan 13, 2018 at 7:50
  • $\begingroup$ you're right, Matt. $F_3(\omega)$ likely will not exist if there are any $t$ such that $f_2(t)=0$. $\endgroup$ Jan 13, 2018 at 8:14
  • $\begingroup$ Thank you all. You have confirmed what I was affraid. In fact, F3 exists, that's not the problem in this case, but I have not been alble to find a direct way to calculate it in order to avoid 'numerical noise' related with the conversion to time domain in order to do the division. $\endgroup$
    – Teresa M.
    Jan 15, 2018 at 15:39
  • $\begingroup$ @TeresaM.: Yes, unfortunately there's no simple way around this problem. If the answer was helpful you can upvote/accept it (by clicking on the check mark to its left), thanks! $\endgroup$
    – Matt L.
    Jan 15, 2018 at 15:50

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