In continuous-time, we have a \emph{continuous-time fourier series} (CTFS) expansion of a signal denoted as

$$ x(t) = \sum \limits_{k=-\infty}^{\infty}X_k e^{i k \omega_0 t} = \sum \limits_{k=-\infty}^{\infty} X_k \Psi_k(t) $$

where $x$ is periodic with fundamental period $p$, and where the fundamental frequency $\omega_0 = 2\pi/p$ and $x(t+p) = x(t)$ $\forall t$.

When we pass a complex exponential through a system $H$, we find the following relationship:

$$ e^{i\omega n} \to \fbox{H} \to H(\omega)e^{i\omega n} $$

where the frequency response of the system $H(\omega)$ is defined as

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

Well, it turns out that $H(\omega)$ is the \emph{Discrete-Time Fourier Transform} (DTFT) of $h(n)$, also known as the analysis equation. Note that $\Psi_k(\omega) = e^{-k\Omega_0\omega}$, but we leave out $\Omega_0$ because it is equal to $2\pi/p=1$.


We know that $H(\omega)$ is periodic, and thus $H(\omega + 2\pi k) = H(\omega)$. Consider the following

\begin{align*}
H(\omega) &=  \sum \limits_{n=-\infty}^{\infty} H_k \Psi(\omega) \\
\end{align*} 
So we are looking for the spectral coefficients $h(n) = H_n$ $ \forall n$. We can solve for this, but first we must
\begin{enumerate}
\item Define the appropriate inner product
\item Prove mutual orthogonality of $\Psi_k$ and $\Psi_\ell$.
\end{enumerate}

The continuous-time fourier series lives in the universe of $p$-periodic functions of a continuous variable $t$.

\begin{nicebox}
\begin{align*}
\langle f,g\rangle &= \int_{\langle p\rangle}f(t)g^*(t)dt
\end{align*}
\end{nicebox}

The discrete-time fourier transform lives in the universe of $2\pi$-periodic functions of a continuous variable $\omega$.

\begin{nicebox}
\begin{align*}
\langle F,G\rangle &= \int_{\langle 2\pi \rangle}F(\omega)G^*(\omega)d\omega
\end{align*}
\end{nicebox}

\begin{claim}
$\Psi_k(\omega) \perp \Psi_\ell(\omega)$ if $k \neq \ell$
\end{claim}
\begin{proof}

\begin{align*}
\langle \Psi_k,\Psi_\ell\rangle  &= \int_{\langle 2\pi\rangle }\Psi_k(\omega)\Psi_{\ell}^*(\omega)d\omega \\
\langle \Psi_k,\Psi_\ell\rangle  &= \int_{\langle 2\pi\rangle }e^{-ik\omega}e^{i\ell\omega}d\omega \\
\langle \Psi_k,\Psi_\ell\rangle  &= \int_{\langle 2\pi\rangle }e^{-i\lr{k-\ell}\omega}d\omega \\
\langle \Psi_k,\Psi_\ell\rangle  &= \int_{\langle 2\pi\rangle }\cos\lr{(k-\ell)\omega}d\omega -i\int_{\langle 2\pi\rangle }\sin\lr{(k-\ell)\omega}d\omega\\
\langle \Psi_k,\Psi_\ell\rangle  &= 0 -i0\\
\end{align*}
 
\end{proof}

The last statement is true because anytime we integrate over a full period $2\pi$, $\forall m \in \Z\setminus \{0\}$, 

$$  \int_{\langle 2\pi\rangle}\cos\lr{m\omega}d\omega = \int_{\langle 2\pi\rangle}\sin\lr{m\omega}d\omega = 0 $$

The reason that this works for all $m  \in \Z\setminus \{0\}$ is because we will integrate over $\abs{m}$ cycles, which is still always zero. Now that we have a defined inner product and can assume orthogonality, we can solve for the impulse response, otherwise known as spectral coefficients:

\begin{align*}
H &= \sum \limits_{k=-\infty}^{\infty}h(k)\Psi_k \\
\langle H,\Psi_n\rangle &= \langle \sum \limits_{k=-\infty}^{\infty}h(k)\Psi_k, \Psi_n\rangle \\
\langle H,\Psi_n\rangle &= h(n)\langle \Psi_n, \Psi_n\rangle \\
\frac{\langle H,\Psi_n\rangle}{\langle \Psi_n, \Psi_n\rangle}&= h(n) \\
\end{align*}
The spectral coefficient $h(n)$ (a.k.a. impulse response) is the ratio of inner products: the inner product of $H$ and $\Psi_n$, over the norm of $\Psi_n$ squared, which we can show that $\norm{\Psi_n}^2$ is $2\pi$:

\begin{align*}
\norm{\Psi_n}^2 &= \langle \Psi_n,\Psi_n \rangle \\
\norm{\Psi_n}^2 &= \int_{\langle 2\pi\rangle}e^{-i\omega n}e^{i\omega n}d\omega \\
\norm{\Psi_n}^2 &=  \int_{\langle 2\pi\rangle}d\omega \\
\norm{\Psi_n}^2 &= 2\pi
\end{align*}

Now we can solve for the impulse response:

\begin{align*}
h(n) &= \frac{\langle H,\Psi_n\rangle}{\langle \Psi_n, \Psi_n\rangle} \\
h(n) &= \frac{\langle H,\Psi_n\rangle}{\norm{\Psi_n}^2} \\
h(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle} H(\omega)\Psi_n(\omega)d\omega \\
\end{align*}

$h(n)$ demonstrates the inverse transform relationship. Here we have the DTFT transform pair:

\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}

If we rearrange the synthesis equation slightly, we can analyze its meaning.

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

We can write a signal as a linear combination of an uncountable infinite set of complex exponentials. Furthermore, an uncountable infinite set of frequencies can contribute to $h(n)$. Now we have the possibility of a continuum of frequencies that can potentially contribute to $h(n)$, not just integer multiples. An infinity of harmonics contribute to the overall signal.



What distinguishes each frequency within $h(n)$ is $H(\omega)$. The coefficients of the linear combination are $\frac{H(\omega)d\omega}{2\pi}$, but $d\omega/2\pi$ is common among all frequencies, so $H(\omega)$ is what determines how much of frequency $\omega$ is present.

\begin{example}
$x(n) = \delta(n)$, find $X(\omega)$. 



We can use the analysis equation:

$$ X(\omega) = \sum \limits_{-\infty}^{\infty}\delta(n)e^{-i\omega n} = 1$$

Thus, this is a fourier transform pair:

$$ \delta(n) \ftp 1 $$
\end{example}

\begin{example}
Consider an ideal low-pass filter that kills all frequencies except frequencies in $[-A,A]$. What is the impulse response $h(n)$?

\begin{center}
\includegraphics{images/idealpass/idealpass.ps}
\end{center}

Use the discrete-time fourier transform.
\begin{align*}
h(n) &= \frac{1}{2\pi} \int_{\langle 2\pi \rangle}H(\omega)e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \int_{-A}^{A}e^{i\omega n} d\omega \\
h(n) &= \frac{1}{2\pi} \lr{\left. \frac{e^{i\omega n}}{in}\right|_{-A}^{A}} \\
h(n) &= \frac{1}{2\pi} \lr{\frac{e^{iAn}-e^{-iAn}}{in}} \\
h(n) &= \frac{1}{2\pi} \frac{2i\sin (An)}{in} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Note that we could have also written this as
\begin{align*}
h(n) &= \frac{1}{2\pi}\left[ \int_{-A}^{A}\cos(\omega n) d\omega + i  \int_{-A}^{A}\sin(\omega n) d\omega\right] \\
h(n) &= \frac{1}{2\pi}\int_{-A}^{A}\cos(\omega n) d\omega  \\
h(n) &= \left. \frac{1}{2\pi} \sin(\omega n) \right|_{-A}^{A} \\
h(n) &= \frac{1}{\pi} \frac{\sin (An)}{n} \\
\end{align*}

Since sine is an odd function, then integrating from $-A$ to $A$ gives us zero, so we are left with integrating the cosine only. What happens to the impulse response $h(n)$ as $A$ approaches $\pi$?

$$ \lim_{A\to\pi} \frac{\sin (An)}{\pi n} =  \frac{\sin (\pi n)}{\pi n}  $$

Now for all $n \neq 0$, $h(n) = 0$, because we are feeding a sine function multiples of $\pi$, and $h(0)=1$ by l'Hopital's rule. This means that as $A$ approaches $\pi$, the impulse response becomes the identity element of convolution: 

$$  \lim_{A\to\pi} \frac{\sin (An)}{\pi n} = \delta(n) $$

\end{example}

\begin{example}
Find the frequency response for $x(n) = \delta(n-1)$. 




$$ X(\omega) = \sum \limits_{n=-\infty}^{\infty}\delta(n-1)e^{-i\omega n} = e^{-i\omega} $$

Now we know the following fourier transform pairs:
\begin{align*}
\delta(n) &\ftp 1 \\
\delta(n-1) &\ftp e^{-i\omega} \\
\delta(n-N) &\ftp e^{-i\omega N} \\
\end{align*}
\end{example}

\begin{example}
Find the impulse response $g(n)$ of the frequency response $G(\omega)$.

$$ G(\omega) = \begin{cases}
e^{-i\omega/2} & \mbox{if }\abs{\omega}<\pi \\
 \mbox{periodically replicates outside} \\
\end{cases}
  $$


\begin{align*}
g(n) &= \frac{1}{2\pi} \int_{\langle 2\pi\rangle}e^{-i\omega/2}e^{i\omega n}d\omega \\
g(n) &= \frac{1}{2\pi} \int_{-\pi}^{\pi}e^{-i\lr{\omega/2 - \omega n}}d\omega \\
g(n) &= \frac{1}{2\pi} \left. \lr{\frac{e^{-i\lr{\omega/2 - \omega n}}}{-i\lr{1/2 -n}}}\right|_{-\pi}^{\pi} \\
g(n) &= \frac{1}{2\pi} \frac{2i\sin(\pi n-\pi/2)}{i(1/2-n)} \\
g(n) &= \frac{\sin(\pi n-\pi/2)}{\pi(1/2-n)} \\
\end{align*}
This is an example of a \emph{Half-Sample Delay System}
}

\begin{example}
$$ F(\omega) = \begin{cases}
e^{-i\omega \alpha} & \mbox{if }\abs{\omega}\lt \pi \\
 \mbox{periodically replicates outside} \\
\end{cases}
  $$

From the previous example, we know that 

$$ f(n) = \frac{\sin(\pi(n-\alpha))}{\pi(n-\alpha)} $$

Show that $f(n)$ reduces to $\delta(n-\alpha)$ whenever $\alpha \in \Z$.

\end{example}
