# Fourier transform of function division in time domain

$F_1(\omega)$ 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?

• 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 :) – Marcus Müller Jan 12 '18 at 13:03
• If $1/f_2(t)$ exists, then you can convolve its FT with $F_1(\omega)$ to get $F_3(\omega)$. – AnonSubmitter85 Jan 12 '18 at 15:42

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 $F_2(\omega)$.
• 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. – robert bristow-johnson Jan 12 '18 at 23:37
• @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. – Matt L. Jan 13 '18 at 7:50
• you're right, Matt. $F_3(\omega)$ likely will not exist if there are any $t$ such that $f_2(t)=0$. – robert bristow-johnson Jan 13 '18 at 8:14