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!
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.