# Fourier Transform of the conjugate of a complex function [closed]

Given that $$x(t)$$ Fourier Transforms to $$X(f)$$. What happens when you Fourier Transform $$x^*(t)$$, and your $$x(t)$$ and $$x^*(t)$$ are both complex functions??

• Look up the complex conjugation property of the Fourier Transform. This is a common analysis technique.
– Ryan
Sep 3, 2021 at 23:00
• What happens when you change $j$ into $-j$? Sep 4, 2021 at 0:11
• This question appears to be homework. Complete answers to homework are off-topic, but specific questions about homework are acceptable if they include enough detail. Please edit the question to include more background about what you don't understand. Sep 4, 2021 at 10:56

$$\mathcal{F}\{x^*(t)\} = X^*(-f)$$
• the $-f$ was not correct. it's either negate $f$ or conjugate $X(f)$, but not both. all you're doing when you conjugate $x(t)$ is swapping $j$ and $-j$ everywhere. Sep 4, 2021 at 19:34
• i agree with Jazz, too. i had it wrong. although $j$ and $-j$ are swapped in $x^*(t)$, they are not in the $e^{-j2 \pi f t}$ factor in the Fourier Transform integral. but if $t$ is swapped with $-t$ in $x^*(t)$ and not in $e^{-j2 \pi f t}$, does that finagle it back again? Sep 5, 2021 at 6:41
• @robertbristow-johnson Please don't grab your left ear with your left hand by wrapping your left arm around your head first. There is no need to reverse time, why not write $$\exp(-j2\pi ft) = \exp(-(-j)2\pi (-f)t)$$ which shows that $f$ gets negated and the final answer is the complex conjugate of what we had previously: $\mathcal F\{x^*(t)\} = X^*(-f)$ exactly as shredEngineer wrote. Sep 5, 2021 at 15:39