What is the Fourier Transform of the Fourier Transform of the signal $x(t)$? 

$$\mbox{If } x(t) \ftp X(\omega), \mbox{ then } X(t) \ftp ? $$
\begin{align*}
x(t) &= \frac{1}{2\pi} \int_{\R}X(\omega)e^{i\omega t}d\omega \\
x(\omega) &= \frac{1}{2\pi} \int_{\R}X(t)e^{i\omega t}dt \quad \mbox{let } t=\omega \\
x(-\omega) &= \frac{1}{2\pi} \int_{\R}X(t)e^{-i\omega t}dt \quad \mbox{let } \omega=-\omega \\
2\pi x(-\omega) &= \int_{\R}X(t)e^{-i\omega t}dt  \\
2\pi x(-\omega) &= \ftrans{X(t)}  \\
\end{align*}

If you take the fourier transform of the fourier transform, you get the original signal back with a scaled factor.

\begin{nicebox}
$$x(t) \ftp X(\omega)$$
$$\Rightarrow$$
$$X(t) \ftp 2\pi x(-\omega)$$
\end{nicebox}

\begin{example}
What is the fourier transform of $x(t) = 1$?



We know that $\ftrans{\delta(t)} = 1$, so we can apply duality and find that $\ftrans{1} = 2\pi\delta(-\omega) = 2\pi\delta(\omega)$. In other words, a constant signal 1 corresponds to a dirac delta.  
\end{example}

\begin{example}
What is the fourier transform of $\cos(\omega_0 t)$?


First note that $\cos(\omega_0 t) = \frac{1}{2}e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t}$. Then recall a few properties of the CTFT:
\begin{align*}
\delta(t) &\ftp 1 \\
\delta(t-t_0) &\ftp e^{-i\omega t_0} \\
\end{align*}

Then by duality, we can say that
\begin{align*}
e^{-i\omega_0 t} &\ftp 2\pi\delta(\omega_0+\omega) &\quad \mbox{since } \delta(-\omega_0-\omega) = \delta(\omega_0+\omega) \\
e^{i\omega_0 t} &\ftp 2\pi\delta(\omega_0-\omega) \\
\end{align*}

Now putting it all together:

\begin{align*}
\cos(\omega_0 t) &= \frac{1}{2}e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t} \\
\ftrans{\cos(\omega_0 t)} &= \frac{1}{2}\ftrans{e^{i\omega_0 t}} + \frac{1}{2}\ftrans{e^{-i\omega_0 t}} \\
\ftrans{\cos(\omega_0 t)} &= \frac{1}{2}2\pi\delta(\omega-\omega_0) + \frac{1}{2}2\pi\delta(\omega+\omega_0) \\
\ftrans{\cos(\omega_0 t)} &= \pi\delta(\omega-\omega_0) + \pi\delta(\omega+\omega_0) \\
\end{align*}

\end{example}


CTFT Duality