We define the continuous-time fourier transform pair as:

\begin{nicebox}
\begin{align*}
x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega \quad &\mbox{(synthesis equation)} \\
X(\omega) &= \int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt \quad &\mbox{(analysis equation)}
\end{align*}
\end{nicebox}

Neither $x(t)$ or $X(\omega)$ has to be periodic.

\begin{example}

1) If $x(t) = \begin{cases} B & t<\abs{A} \\ 0 & \mbox{otherwise} \\ \end{cases}$, what is $\ftrans{x(t)}$?

\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t}dt \\
X(\omega) &= B\int_{-A}^{A} e^{-i\omega t}dt \\
X(\omega) &= B\left. \frac{e^{-i\omega t}}{-i\omega} \right|_{-A}^{A} \\
X(\omega) &= B\frac{e^{i\omega A} - e^{-i\omega A}}{i\omega} \\
X(\omega) &= \frac{2B\sin(\omega A)}{\omega} \\
\end{align*}

Note that $\lim_{\omega \to 0} \frac{2B\sin(\omega A)}{\omega} = \lim_{\omega \to 0} \frac{2AB\cos(\omega A)}{1} = 2AB$ by L'Hopital's Rule.

%\img{images/sinc/timedomain/new.ps}

Also note that $\lim_{A \to \infty} X(\omega) = \delta(\omega)$.

\end{example}


\begin{example}

2) If $x$ is $p$-periodic, then
\begin{align*}
x(t) &= \sum_{k\in\Z}X_ke^{ik\omega_0 t} \\
e^{i\omega_0 t} &\ftp 2\pi \delta(\omega - \omega_0)
\end{align*}

We can find $X(\omega)$:

\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-i\omega t}dt \\
X(\omega) &= \int_{-\infty}^{\infty} \lr{\sum_{k\in\Z}X_ke^{ik\omega_0 t}}  e^{-i\omega t}dt \\
X(\omega) &= \sum_{k\in\Z} X_k \int_{-\infty}^{\infty} e^{ik\omega_0 t}  e^{-i\omega t}dt \\
X(\omega) &= \sum_{k\in\Z} X_k \ftrans{ e^{ik\omega_0 t} } \\
X(\omega) &= \sum_{k\in\Z} X_k 2\pi\delta(\omega - k\omega_0) \\
\end{align*}

This tells us that we have an amount $X_k$ of frequency $k\omega_0$:
$$
X(\omega) = \sum_{k\in\Z} X_k 2\pi\delta(\omega - k\omega_0) 
$$

\end{example}

CTFT