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I'm trying to figure this out. Assume lower case is in $(x,y)$ and upper case is in $(fx,fy)$.

I understand 2 forward FFT's lead to a sign reversal.

FFT$(t(x,y)) = T(fx,fy)$, FFT$(T(fx,fy)) = t(-x,-y)$

I also know that there are sign reversals in FFT or IFFT's of conjugates.

FFT$(t*(x,y))$ = $T*(-fx,-fy)$

What happens when I take a forward FFT of a conjugate in the fourier plane?

FFT($T*(fx,fy))$ = ?

My intuition is that it will be $t*(x,y)$ but I'm not sure how the math works. Any help would be appreciated, thank you!

pmdaly
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As long as you are applying the Fourier transform to the complex conjugate of some function (whatever it is), you'll get the reversed-conjugate as output: $$\widehat{\bar g}(\xi) = \overline{\widehat g(-\xi)} $$ Indeed, $$\int e^{-ix\xi}\bar g(x)\,dx = \overline{\int e^{-ix(-\xi)} g(x)\,dx} = \overline{\widehat g(-\xi)}$$ It does not matter what the function $g$ represents; it could itself be the Fourier transform of something.