
Recall the DTFT equations:

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

\begin{example}
1) Find the DTFT for the signal $x(n) = 1 $ $ \forall n$

\begin{align*}
x(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle } X(\omega)e^{i\omega n}d\omega  \\
1 &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle } X(\omega)e^{i\omega n}d\omega  \\
e^{i\omega 0} &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle } X(\omega)e^{i\omega n}d\omega  \\
e^{i\omega 0} &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle } A\delta(\omega)e^{i\omega n}d\omega  \\
e^{i\omega 0} &= \frac{1}{2\pi} Ae^{i\omega 0} \\
1 &= \frac{1}{2\pi} A \\
A &= 2\pi \\
\end{align*}

Hence, $X(\omega) = 2\pi\delta(\omega)$ for $\abs{\omega} \lt \pi$, or $X(\omega) = 2\pi\sum\limits_{k\in\Z}\delta(w-2\pi k)$. Lets take this a little further, utilizing the analysis equation:

\begin{align*}
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n)e^{-i\omega n} \\
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}e^{-i\omega n} \quad \mbox{let }x(n) = 1 \\
X(\omega) &= \sum \limits_{n=-\infty}^{\infty}e^{-i\omega n} = 2\pi\sum\limits_{k\in\Z}\delta(w-2\pi k) \\
\end{align*}

Hence we have the relation known as \emph{Poisson's Identity}:

$$
\sum \limits_{n\in\Z}e^{-i\omega n} = 2\pi\sum\limits_{k\in\Z}\delta(w-2\pi k) \\
$$

Notice that the left side is a continuous function where the right side is of discrete dirac deltas. Hence, the equality is not pointwise, but more that both representations treat other functions the same way when they come into contact. This relationship is not consistent with our intuition and therefore must be derived.

\end{example}

\begin{example}

2) Find the DTFT of the signal $x(n)=\sin(2\pi n/3)$.

\begin{align*}
x(n) &= \sin(2\pi n/3) \\
x(n) &= \frac{1}{2i}e^{i2\pi n/3} - \frac{1}{2i}e^{-i2\pi n/3} \\
\ftrans{x(n)} &= \ftrans{\frac{1}{2i}e^{i2\pi n/3}} - \ftrans{\frac{1}{2i}e^{-i2\pi n/3}} \\
\end{align*}

Now since the DTFT in linear, we can look at these individually. Consider an arbitrary complex exponential:

\begin{align*}f(n) &= \frac{1}{2\pi}\int_{\langle 2\pi\rangle }F(\omega)e^{i\omega n}d\omega \\
Re^{i\omega_0 n} &= \frac{1}{2\pi}\int_{\langle 2\pi\rangle }A\delta(\omega - \omega_0)e^{i\omega n}d\omega \\
Re^{i\omega_0 n} &= \frac{1}{2\pi}Ae^{i\omega_0 n} \\
R &= \frac{1}{2\pi}A \\
A &= 2\pi R \\
\end{align*}

This shows that $F(\omega) = 2\pi R \delta(\omega - \omega_0)$:

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

So in our case we can use the derived formula to find the coefficients:
\begin{align*}
\ftrans{x(n)} &= \ftrans{\frac{1}{2i}e^{i2\pi n/3}} - \ftrans{\frac{1}{2i}e^{-i2\pi n/3}} \\
X(\omega) &= \frac{\pi}{i}\delta(\omega - 2\pi n/3) - \frac{\pi}{i}\delta(\omega + 2\pi n/3) \\
\end{align*}

Remember that we have to restrict our domain of $\omega$ to $\abs{\omega} \lt \pi$ for the equation to hold.

\end{example}

DTFT of Complex Exponentials