The Continuous-Time Fourier Transform is defined 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}


The reason that we integrate over all real numbers for the synthesis equation in continuous-time is because $X(\omega)$ doesn't have the periodicity of the DTFT. Note that the $2\pi$ is a product of the units we are using for $\omega$, namely the fact that $\omega = \frac{2\pi}{f}$, where $f$ is in $Hz$. If we used $f=2\pi\omega$ instead of $\omega$, we could write the CTFT as

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

This notation is used more frequently in communication and circuit design, since $Hz$ is more intuitive than radians per second. If we look at the synthesis equation and rearrange some terms, we can see that we have a linear combination of complex exponentials whose fate is determined by $X(\omega)$, the part of the integral that distinguishes the contribution of each frequency:

$$ x(t) = \int_{-\infty}^{\infty}\lr{\frac{X(\omega)d\omega}{2\pi}}e^{i\omega t} $$

\begin{example}
Find the Fourier Transform of $x(t) = \delta(t)$.


To solve this, we can simply find that $\delta(t) \ftp 1$:
\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt \quad \mbox{(analysis)} \\
X(\omega) &= \int_{-\infty}^{\infty}\delta(t)e^{-i\omega t}dt \\
X(\omega) &= e^{0} \\
X(\omega) &= 1 \\
\end{align*}


\begin{nicebox}
$$ \delta(t) \ftp 1 $$
\end{nicebox}

\begin{example}
Find the Fourier Transform of $x(t) = \delta(t-T)$.



\begin{align*}
X(\omega) &= \int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt \quad \mbox{(analysis)} \\
X(\omega) &= \int_{-\infty}^{\infty}\delta(t-T)e^{-i\omega t}dt  \\
X(\omega) &= e^{-i\omega T}  \\
\end{align*}


\begin{nicebox}
$$\delta(t-T) \ftp e^{-i\omega T} $$
\end{nicebox}

\begin{example}
Find the Fourier Transform of $x(t) = 1$.

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We know that 0 is the frequency of $x(t) = 1$, we just don't know the magnitude. 
\begin{align*}
x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega \quad \mbox{(synthesis)} \\
1 &= \frac{1}{2\pi}\int_{-\infty}^{\infty}A\delta(\omega)e^{i\omega t}d\omega \\
1 &= \frac{A}{2\pi} \\
A &= 2\pi \\
\end{align*}

\begin{nicebox}
$$1 \ftp 2\pi\delta(\omega) $$
\end{nicebox}

\begin{example}
Find the Fourier Transform of $x(t) = e^{i\omega_0 t}$.


We that the only frequency in the signal is $\omega_0$, and again, we only need to find the amplitude of a shifted dirac delta.

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\begin{align*}
x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega \quad \mbox{(synthesis)} \\
e^{i\omega_0 t} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}A\delta(\omega-\omega_0)e^{i\omega t}d\omega \\
e^{i\omega_0 t} &= \frac{A}{2\pi}e^{i\omega_0 t} \\
A &= 2\pi \
\end{align*}

\begin{nicebox}
$$e^{i\omega_0 t} \ftp 2\pi\delta(\omega-\omega_0) $$
\end{nicebox}

\begin{example}
Find the impulse response of the ideal low pass filter with a pass band amplification of $\pi/b$:

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\begin{align*}
h(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}H(\omega)e^{i\omega t}d\omega  \\
h(t) &= \frac{1}{2\pi}\int_{-a}^{a}\frac{\pi}{b}e^{i\omega t}d\omega  \\
h(t) &= \frac{1}{2b} \lr{\left. \frac{e^{i\omega t}}{it}\right|_{-a}^{a}} \\
h(t) &= \frac{1}{2b} \frac{e^{i a t} - e^{-i a t}}{it} \\
h(t) &= \frac{\sin(at)}{tb} \\
\end{align*}

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What is $\int_{-\infty}^{\infty}\frac{\sin(at)}{tb}dt$? Upon first glance, this problem seems to require complex analysis, but we can simply note that $H(\omega) = \int_{-\infty}^{\infty}\frac{\sin(at)}{tb}e^{-i\omega t}dt$, so we can take $\int_{-\infty}^{\infty}\frac{\sin(at)}{tb}e^{0}dt = H(0) = \pi/b$. The Fourier Transform of the signal at 0 is the integral of the signal.


Continous-Time Fourier Transform