Recall that a signal $x:\Z \to \C$ is periodic with fundamental period $p \in \N$ such that 
$$x(n+p) = x(n) \mbox{ } \forall n \in \Z$$
We can decompose a signal $x$ into a linear combination of complex exponentials, where $\omega_0$ is the fundamental frequency, given by 
$$ \omega_0 = \frac{2\pi}{p}$$ 
The signal $x$ can be written in terms of the spectral coefficients as weights on complex exponentials containing harmonics of the fundamental frequency $\omega_0$

$$ x(n) = \sum \limits_{k=\langle p \rangle}X_k\Psi_k(n) = \sum \limits_{k=\langle p \rangle}X_ke^{ik\omega_0 n} \quad \mbox{(synthesis equation)}  $$
Note that in reality, $x(n) \sim \frac{1}{p} \sum X_ke^{ik\omega_0 n}$, but the energy difference is the same, so in practice, we can use the equals sign. To learn more about this, research Dirichlet conditions and Gibb's phenomenon.




We can determine the coefficients with the analysis equation:

$$ X_k = \frac{1}{p} \sum \limits_{k=\langle p \rangle}x(n)e^{-ik\omega_0 n} \quad \mbox{(analysis equation)}  $$


\begin{example}
Given $x(n) = sin(2\pi/3)$ with fundamental period 3, why can't $e^{i\frac{\omega_0}{2}n}$ be a term in $x(n)$?


Assume there exists such a frequency and that it contibutes to the signal. 

\begin{align*}
e^{i\frac{\omega_0}{2}n} &= e^{i\frac{2\pi}{2p}n} \\
e^{i\frac{\omega_0}{2}n} &= e^{i\frac{2\pi}{6}n} \\
\Rightarrow p &= 6
\end{align*}

By assuming the frequency exists, we determined the period and arrived at a contradiction. We know that in general $p = 2\pi/\omega_0$ must be an integer, but this is not recommended as a proof. 
\end{example}

\begin{example}
Is $x(n) = \cos(n)$ a periodic signal? 


$\cos(n + 2\pi) = \cos(n)$ and $p = 2\pi \notin \Z$. Therefore this cosine function is \emph{quasi-periodic}, but not periodic.
\end{example}

\begin{example}
if $x(n) = \cos(\omega_0 n)$, what must $\omega_0$ be for this signal to be periodic? 

$$x(n+p) = \cos(\omega n + \omega p) $$

This implies that $\omega p = 2\pi k$ for $k \in \Z$. Then $\omega_0 = \frac{2\pi k}{p}$. Where $2k/p \in \Q$. 
\end{example}

Note that for discrete-time only the trigonometric functions must have a fundamental frequency that is a rational multiple of $\pi$. For continuous-time this doesn't matter.

\begin{example}
Given a signal that has a Kronecker delta every $p$ units, determine the harmonics and spectral coefficients. 

\begin{center}
\includegraphics{images/pulsetrains/discrete.ps}
\end{center}

We can determine the $k^{th}$ spectral coefficient with the analysis equation:

$$ X_k = \frac{1}{p} \sum \limits_{n=\langle p \rangle}x(n)e^{-ik\omega_0 n} = \frac{1}{p} \sum \limits_{n=\langle p \rangle}\delta(n-\ell p)e^{-ik\omega_0 n} = e^0 \cdot \frac{1}{p} = \frac{1}{p}$$

So in this case we find that $X_k = 1/p $ $ \forall k$. This means that all of the frequencies in the period, namely $o,\omega_0, 2\omega_0, \dots (p-1)\omega_0$, have an even contribution to the signal. 

\end{example}

Given a continuous-time signal $x(t)$ with period $p$, we still use can break the signal into a linear combination of complex exponentials, but there is a slight difference in the synthesis equation. For discrete-time, we sum over $p$ terms, whereas in continuous time, we sum over a countably infinite set of terms:

$$ x(t) = \sum \limits_{-\infty}^{\infty}X_ke^{ik\omega_0 t} $$

This also implies that we have a countably infinite set of frequencies $\{\dots, -2\omega_0,-\omega_0, 0, \omega_0, 2\omega_0,\dots\}$.



We can also write $x(t) = \sum X_k \Psi_k(t)$, and find the spectral coefficient in the same manner as in discrete-time using the inner product, thus 

$$ X_k = \frac{<\vec{x},\Psi_k>}{<\Psi_k,\Psi_k>} $$

Recall that the inner product is defined as

\begin{nicebox}
\begin{align*}
\langle f,g \rangle &= \sum \limits_{n=\langle p \rangle}f(n)g^*(n) &\quad \mbox{(Discrete-time)} \\
\langle f,g \rangle &= \int_{\langle p \rangle}f(t)g^*(t)dt \quad &\mbox{(Continuous-time)} 
\end{align*}
\end{nicebox}

\begin{claim}
For $k \neq \ell$, $\Psi_k(t) \perp \Psi_\ell(t)$
\end{claim}

\begin{proof}
\begin{align*}
\langle \Psi_k,\Psi_\ell \rangle &= \int_{\langle p \rangle} \Psi_k(t) \Psi_k^*(t) dt \\
\langle \Psi_k,\Psi_\ell \rangle &= \int_{\langle p \rangle} e^{i\omega_0 k t} e^{-i\omega \ell t} dt \\
\langle \Psi_k,\Psi_\ell \rangle &= \int_{\langle p \rangle} e^{i\omega_0 t (k - \ell)} dt \\
\langle \Psi_k,\Psi_\ell \rangle &= 
\begin{cases}
\int_{\langle p \rangle}dt & \mbox{ if } k = \ell\\
\int_{\langle p \rangle}e^{i\omega_0 t(k-\ell)} & \mbox{ if } k \neq \ell \\
\end{cases} \\
\langle \Psi_k,\Psi_\ell \rangle &=
\begin{cases}
p & \mbox{ if } k = \ell\\
0 & \mbox{ if } k \neq \ell \\
\end{cases}
\end{align*}
When $k \neq \ell$, the integral produces zero because we are integrating over a period. Consider when you integrate a cosine function from 0 to $2\pi$. You sum up as much positive value as you do negative, and it is cancelled out. 
\end{example}

Spectral Coefficients

\begin{enumerate}
\item Low-Pass Filter
$$ H(\omega) = \frac{1}{1-\alpha e^{i\omega}} \iff h(n) = \alpha^nu(n) $$
for $\alpha \in (0,1)$

\item High-Pass Filter
$$ H(\omega) = \frac{1}{1-\alpha e^{i\omega}} \iff h(n) = \alpha^nu(n) $$
for $\alpha \in (-1,0)$

\item All-Pass Filter
$$ H(\omega) = \frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}} \iff h(n) = \alpha^{n+1}u(n) - \alpha^{n-1}u(n-1) $$

\begin{claim}
$\abs{H(\omega)} = \abs{\frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}}} = 1$
\end{claim}
\begin{proof}
\begin{align*}
H(\omega) &= \frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}} \\
\abs{H(\omega)} &= \abs{\frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{1-\alpha e^{-i\omega}}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{e^{-\omega}}\abs{e^{i\omega}-\alpha}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{e^{i\omega}-\alpha}} \\
\abs{H(\omega)} &= 1 \quad \mbox{ since complex conjugate have equal magnitude} \\
\end{align*}
\end{proof}

\item Continuous-time High-Pass Filter / Notch Filter
$$ H(\omega) = \frac{i\omega}{i\omega + \alpha} \iff h(t) = 1 - \alpha e^{-\alpha t}u(t)$$
This filter becomes as notch-filter if $\alpha \approx 1$, and high-pass if $\alpha \approx 0$. 

\item Continuous-time Low-Pass Filter / Anti-notch Filter
$$ H(\omega) = \frac{1}{i\omega + \alpha} \iff h(t) = e^{-\alpha t}u(t) $$
This filter becomes an anti-notch filter if $\alpha \approx 0$, and low-pass if $\alpha \approx 1$. 

\item Notch Filter / Low Pass Filter
$$ H(\omega) = \frac{e^{i\omega}+1}{e^{i\omega}+\alpha} \iff h(n) = \alpha^nu(n)+\alpha^{n-1}u(n-1) $$
This filter is a notch if $\alpha \approx 1$, and is low-pass when $0 \lt \alpha \lt\lt 1$. 

\item Comb Filter
$$ H(\omega) = \frac{e^{i\omega N}}{e^{i\omega N}-\alpha} \iff h(n) = \sum \limits_{k=0}^\infty\alpha^k\delta(n-kN) $$
for $\abs{\alpha} \lt 1$
\end{enumerate}

We can generalize a systems effect on a complex exponential:

$$ e^{i\omega_0 n} \to \fbox{H} \to \abs{H(\omega_0)}e^{i\angle H(\omega_0)}e^{i\omega_0 n} =  \abs{H(\omega_0)}e^{i\lr{\omega_0 n + \angle H(\omega_0)}}  $$


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