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??
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1$\begingroup$ Look up the complex conjugation property of the Fourier Transform. This is a common analysis technique. $\endgroup$– RyanSep 3, 2021 at 23:00
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$\begingroup$ What happens when you change $j$ into $-j$? $\endgroup$– robert bristow-johnsonSep 4, 2021 at 0:11
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$\begingroup$ 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. $\endgroup$– Marcus MüllerSep 4, 2021 at 10:56
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1 Answer
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$\mathcal{F}\{x^*(t)\} = X^*(-f)$
Taken from the table of Fourier transform theorems on slide 6 of this document: https://www.comm.utoronto.ca/~dkundur/course_info/316/KundurFTProperties_handouts.pdf
Tables like these are useful in general. Please look it up yourself next time! :)
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$\begingroup$ 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. $\endgroup$ Sep 4, 2021 at 19:34
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2$\begingroup$ @robertbristow-johnson you made it wrong. shredEngineer had it perfectly right. Compare en.wikipedia.org/wiki/Fourier_transform#Conjugation $\endgroup$ Sep 4, 2021 at 19:38
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1$\begingroup$ @robertbristow-johnson I agree with Jazzmaniac $\endgroup$ Sep 4, 2021 at 21:11
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$\begingroup$ 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? $\endgroup$ Sep 5, 2021 at 6:41
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$\begingroup$ @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. $\endgroup$ Sep 5, 2021 at 15:39