Recall that we can write the frequency response of a system as

$$ H(\omega) = \sum \limits_{n\in \Z} h(n) e^{-i\omega n}  $$

Also, recall the $2\pi$-periodicity of the frequency response:

$$ H(\omega + 2\pi) = H(\omega) \forall \mbox{ } \omega  $$

The fourier series was written as $ x(t) = \sum \limits_{k=-\infty}^{\infty}X_ke^{ik\omega_0t} $, and $x(t+p) = x(t)$. We can combine these principles to come up with the DTFT. $x$ was $p$-periodic, $H$ is $2\pi$-periodic. 

\begin{align*}
x &\leftrightarrow e^{ik\omega t} \leftrightarrow X_n \\
H &\leftrightarrow e^{-i\Omega_0\omega n} \leftrightarrow h(n) \quad \mbox{where } \Omega_0 = 1 \\
\end{align*}

$X_n$ and $h(n)$ are the coefficients. $x$ is a periodic signal and so is $H$. $x(t)$ and $H(\omega)$ are functions over continuous variables.

\begin{align*}
x(t) & = \sum \limits_{n=-\infty}^{\infty}X_ne^{in\omega_0t} = \sum \limits_{n=-\infty}^{\infty} X_n\Psi_n\\
H(\omega) &=  \sum \limits_{n=-\infty}^{\infty}h(n)e^{-in\Omega_0\omega} = \sum \limits_{n=-\infty}^{\infty} H_n\Phi_n 
\end{align*}

Assume for now that $\langle \Phi_n, \Phi_m\rangle  = 0 $ when $n\neq m$. We want to find $h(n)$ in terms of $H(\omega)$. So we can project $H$ onto $\Phi_m$ to determine $h(m)$, or $H_m$:

\begin{align*}
\langle H, \Phi_m\rangle  &= \langle \sum_n\in\Z h(n)\Phi_n, \Phi_m\rangle  \\
\langle H, \Phi_m\rangle  &= \sum_n\in\Z h(n)\langle \Phi_n, \Phi_m\rangle  \\
\langle H, \Phi_m\rangle  &= h(m)\langle \Phi_m, \Phi_m\rangle  \\
h(m) &= \frac{\langle H, \Phi_m\rangle }{\langle \Phi_m, \Phi_m\rangle } \\
\end{align*}

We can always rely on the ratio of two inner products. $H(\omega)$ is a $2\pi$-periodic function over a continuous variable $\omega$, and $\Phi_m(\omega)$ is also a $2\pi$-periodic function over a continuous variable $\omega$.

\begin{align*}
\langle H, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }H(\omega)\Phi_m^*(\omega)d\omega \\
\langle H, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }H(\omega)e^{i\omega m}d\omega \\
\end{align*}

We still need to prove orthogonality and determine the value of $\langle \Phi_m, \Phi_m\rangle $:

\begin{align*}
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }\Phi_n(\omega)\Phi_m^*(\omega)d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }e^{-i\omega n}e^{i\omega m}d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle }e^{i\omega(m-n)}d\omega \\
\end{align*}


We have two cases, where the exponent is zero when $m=n$, and then when it is non-zero and $m\neq n$. The first is more simple. We are integrating 1 over a contiguous interval of length $2\pi$, hence the value is $2\pi$:
\begin{align*}
\langle \Phi_m, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle } dt = 2\pi
\end{align*}

This case requires a bit more work. Lets integrate.

\begin{align*}
\langle \Phi_n, \Phi_m\rangle  &= \int_{\langle 2\pi\rangle } e^{i(m-n)\omega} d\omega \\
\langle \Phi_n, \Phi_m\rangle  &= \left.\frac{e^{i(m-n)\omega}}{i(m-n)} \right|_{\langle 2\pi\rangle } \\
\langle \Phi_n, \Phi_m\rangle  &= \frac{e^{i(m-n)2\pi}- e^{0}}{i(m-n)} \\
\langle \Phi_n, \Phi_m\rangle  &= \frac{0}{i(m-n)} = 0\\
\end{align*}

Now we can come to a conclusion regarding the DTFT. 


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

$H(\omega)$ is $2\pi$-periodic, but $h(n)$ is not necessarily periodic. We can make a meaningful interpretation of the synthesis equation if we write the synthesis equation as a linear combination of complex exponentials:

$$ h(n) = \int_{\langle 2\pi\rangle }\lr{\frac{H(\omega)d\omega}{2\pi}}e^{i\omega m}  $$

A discrete-time function can be broken down into a linear combination of complex exponentials. How much of each complex exponential is present in the function is determined by the coefficient. What distinguished this coefficient is $H(\omega)$. So the fourier transform motivates the term \emph{filter} since it can break signals up into their constinuent frequencies and filter them in or out of the output signal.
