Fourier Transform: why time reversal in freq domain is the same as complex conjugate?

Question

have a function $x(-t)$, and I take the Fourier transform of this function:

$\mathcal{F} \Big[x(-t) \Big] = X(-\omega)$

Why does time reversal of the frequency $\omega$ equal the complex conjugate?

$X(-\omega) = X^{*}(\omega)$

Answer 1

So the exact equations will vary depending on whether your frequency and time domains are being viewed as discrete or continuous (e.g. DFT, CTFT), but all have the core property of some summation over $t$ of something similar to $x(t) * e^{i * t * w}$.

The complex conjugate comes from the fact that $e^{-w} = (e^w)^*$ By substituting w with -w and assuming the function x(t) is real, we can see that $\mathcal{X}^*(w)$ is roughly a sum of $x^*(t)(e^{i * t * w})^* = x(t) e^{-i * t * w}$. This summation/integration becomes then $\mathcal{X}(-w)$. Note that this assumes $x$ is real in time domain.

If you aren't sure why the complex conjugate is of the complex exponential is just a negative exponent, consider Euler's formula:

$ e^{i * t} = cos(t) + i*sin(t)$

$ (e^{i * t})^* = (cos(t) + i*sin(t))^* = (cos(t))^* + (i*sin(t))^*$

$cos(t)$ and $sin(t)$ are real, so:

$ (e^{i * t})^* = cos(t) + sin(t) * i^* = cos(t) - i * sin(t) $

Given that sine is odd ($sin(-t) = -sin(t)$) and cosine is even ($cos(-t) = cos(t)$):

$ (e^{i * t})^* = cos(t) - i * sin(t) = cos(-t) + i * sin(-t) = e^{-i * t}$

Answer 2

Your statement hols only if your function $f$ to which the Fourier transformation is applied is real:

$$(\mathcal{F}f)(t):=\int_{\mathbb{R}} e^{-2\pi i x t} f(x)\,dx$$

Then

$$ (\mathcal{F}f)(-t) =\int_{\mathbb{R}} e^{2\pi i x t} f(x)\,dx=\overline{\int_{\mathbb{R}} e^{-2\pi i x t} \overline{f}(x)\,dx}=\overline{(\mathcal{F}\overline{f})(t)}$$

If your function $f$ is real valued, then $\overline{f}=f$ and so $(\mathcal{F}f)(-t)=\overline{(\mathcal{F}f)(t)}$