Suppose we have a periodic signal with fundamental period 2.

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We can then write the signal $x$ as a linear combination of two signals $\Psi_0(n)$, and $\Psi_1(n)$:

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Therefore, we can write

$$x(n) = 2\Psi_0(n) + 3\Psi_1(n)$$

Here is another way to decompose the signal:

\begin{align*}
\left[ \begin{array}{r} 2 \\ 3 \\ \end{array} \right]  &= 2  \left[ \begin{array}{r} 1 \\ 0 \\ \end{array} \right] + 3  \left[ \begin{array}{r} 0 \\ 1 \\ \end{array} \right] \\
x &= x(0)\Psi_0 + x(1)\Psi_1
\end{align*}

Fourier analysis is  a way of decomposing signals in terms of some special signals that tell us information:

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What we want is to write $x$ as a linear combination of these signals:

$$ x = X_0\Psi_0 + X_1\Psi_1 $$

So to find $X_0$, we can project $\vec{x}$ onto $\Psi_0$:
 
\begin{align*}
\langle\vec{x}, \Psi_0\rangle &= \langle X_0\Psi_0 + X_1\Psi_1, \Psi_0\rangle \\
\langle\vec{x}, \Psi_0\rangle &= X_0\langle\Psi_0,\Psi_0\rangle +X_1\langle\Psi_1,\Psi_0\rangle \\
\langle\vec{x}, \Psi_0\rangle &= X_0\langle\Psi_0,\Psi_0\rangle + 0\quad \mbox{(because of orthogonality)}\\
X_0 &= \frac{\langle\vec{x}, \Psi_0\rangle }{\langle\Psi_0,\Psi_0\rangle}\\
\end{align*}

Fourier analysis says if we have a signal with period $p$, then we can represent the signal with complex exponentials that are harmonically related by a fundamental frequency $\omega_0 = 2\pi/p$:

\begin{nicebox}
$$ x(n) = \sum \limits_{k=0}^{p-1}X_ke^{ik\omega_0 n}  $$
\end{nicebox}

We can also write this relationship as 

\begin{align*}
x(n) &= \sum \limits_{k=0}^{p-1}X_k\Psi_k(n)  \quad \mbox{where } e^{ik\omega_0 n} = \Psi_k(n) \\
\end{align*}

Or even more simplified:

$$ x = \sum \limits_{k=\langle p\rangle}X_k\Psi_k  $$

% Examples

Let $p=2$, then $\omega_0 = 2\pi/2 = \pi$. 
\begin{align*}
\Psi_0(n) &= e^{i0\omega_0n} = 1 \\
\Psi_1(n) &= e^{i\omega_0n} = e^{i\pi n} = \lr{-1}^n \\
\end{align*}

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How can we determine $\Psi_0$ and $\Psi_1$? Since the fundamental period is two, we need only one contiguous interval of 2. Lets use 0 and 1:

\begin{align*}
\Psi_0(0) = 1 \\
\Psi_0(1) = 1 \\
\Psi_1(0) = \lr{-1}^0 = 1 \\
\Psi_1(1) = \lr{-1}^1 = -1 \\
\end{align*}

Hence the basis vectors are as shown:

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What frequencies can be present in a signal $\vec{x}$? The harmonics are given by a signal with period $p$ are $\{0,\omega_0,2\omega_0,\dots,(p-1)\omega_0\}$. Notice that $p\omega_0$ is missing. This is because $e^{ip\omega_0n}=e^{i2\pi n} = e^{i0n}$. Keep in mind that $\omega_0 = 2\pi/p$. In fact, the complex exponential basis of vectors is periodic with respect to $n$ and their indices $k$. 

\begin{align*}
\Psi_k(n+p) &= e^{ik\omega_0(n+p)}=e^{ik\omega_0n}e^{ik\omega_0p}=e^{ik\omega_0n} = \Psi_k(n) \\
\Psi_{k+p}(n) &= e^{i(k+p)\omega_0n} = e^{ik\omega_0n}e^{ip\omega_0n} = \Psi_k(n) 
\end{align*}

So we know that $\Psi_{-1}(n) = e^{-i\omega_0n} = \Psi_{p-1}(n) = e^{i(p-1)\omega_0n }$. So we can see that in general, we can write out a discrete fourier series expansion over any continuous set of $p$ integers. The notation used is
$$ x(n) = \sum \limits_{k=\langle p \rangle} X_ke^{ik\omega_0n}$$


Decomposition of Signals