# 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
Commented Sep 3, 2021 at 23:00
• What happens when you change $j$ into $-j$? Commented 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. Commented 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. Commented 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? Commented 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. Commented Sep 5, 2021 at 15:39