Fourier transform of $f[n]$ and $f[-n]$ Hi I am just wondering,
If I have a signal $f[n]\in \mathbb{C}^L$, i.e. $f$ is $L$-periodic, i can also define $h[n]=f[-n]$.
Is it true that the Fourier transform of $f$, say $\hat{F}$, and the Fourier tranform of $h$, say $\hat{H}$ is related by the equation below?
$$|\hat{F}[k]|=|\hat{H}[k]|$$
Any advice/suggestion is welcome! 
 A: This is not true in general.
Consider $f(x) = e^{ix}$ which is $2\pi$-periodic. Define $g(x) = f(-x)$ which is also $2\pi$-periodic.
Observe that
\begin{align*}
\widehat f(k)&= \frac{1}{2\pi}\int_0^{2\pi} f(x) e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{ix}e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{e(1 - k)x} \, dx \\
&= \begin{cases} 1 & \text{if } k = 1 \\ 0 & \text{otherwise} \end{cases}
\end{align*}
but then
\begin{align*}
\widehat g(k) &= \frac{1}{2\pi} \int_0^{2\pi} f(-x) e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{-ix}e^{-ikx} \, dx = \frac{1}{2\pi} \int_0^{2\pi} e^{-i(k + 1)x} \, dx \\
&= \begin{cases} 1 & \text{if } k = -1 \\ 0 & \text{otherwise} \end{cases}
\end{align*}
So in this case, we see $\vert \widehat f(k) \vert \neq \vert \widehat g(k) \vert$ for all $k$.
However, if $f$ is real, then the reality condition, $\widehat g(k) = \overline{\widehat f(k)}$ forces the result to hold.

Lemma: If $f$ is real and even, then $\widehat f$ is real and even.
Proof: Observe that
\begin{align*}
\widehat f(-k) &= \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-i(-k)x} \, dx = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ik(-x)} \, dx = -\frac{1}{2\pi} \int_0^{-2\pi} f(-u) e^{-iku} \, du \\
&= \frac{1}{2\pi} \int_0^{2\pi} f(u)e^{-iku} \, du = \widehat f(k)
\end{align*}
and so $\widehat f$ is even.
Observe that
\begin{align*}
\widehat f(k) &= \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ikx} \, dx \\
&= \frac{1}{2\pi} \int_0^{2\pi} f(x) \big[\cos(-kx) + i \sin(-kx) \big] \, dx\\
&= \frac{1}{2\pi} \int_0^{2\pi} f(x) \cos (kx) \, dx - \frac{i}{2\pi} \int_{-\pi}^\pi f(x) \sin(kx) \, dx\\
&= \frac{1}{2\pi} \int_0^{2\pi} f(x) \cos (kx) \, dx + 0
\end{align*}
The latter part is zero since $f(x) \sin (kx)$ is odd about the center, and we can switch from $0 \to 2\pi$ to $-\pi \to \pi$ because of translation invariance.
Lemma: If $f$ is real and odd, then $\widehat f$ is pure imaginary and odd.
Proof: Similar as above.
Theorem: Setting $g(x) = f(-x)$, we get $\widehat g = \overline{\widehat f}$.
Proof: Write $f$ as the sum of an even function and an odd function: $f = f_e + f_o$. Then $\widehat f = \widehat f_e + \widehat f_o$.
Observe that $g(x) = f(-x) = f_e(-x) + f_o(-x) = f_e(x) - f_o(x)$, hence $\widehat g = \widehat f_e - \widehat f_o = \overline{\widehat f}$.
