We will also introduce a new representation of signals in the form of vectors. Consider the vector given by $\vec{x} = \left[ \begin{array}{r} 4 \\ 2 \\ \end{array} \right]$. This vector corresponds to a periodic signal $4\delta(n) + 2\delta(n-1)$. Notice that we can link a Finite Duration signal to a periodic signal by letting the entire duration be one period, and decompose a periodic signal conversely. 


Fourier analysis allows us to decompose signals into sums of orthogonal vectors. This signal $\vec{x}$ can also be written as the sum of the two vectors $\Psi_0 = \left[ \begin{array}{r} 1 \\ 0 \\ \end{array} \right]$, and $\Psi_1 = \left[ \begin{array}{r} 0 \\ 1 \\ \end{array} \right] $. Then we can see that $\vec{x} = 4\Psi_0 + 2\Psi_1$.


What if $\Psi_0 = \left[ \begin{array}{r} 1 \\ 1 \\ \end{array} \right]$ and $\Psi_1 = \left[ \begin{array}{r} 1 \\ -1 \\ \end{array} \right]$? \\

\begin{align*}
x(n) &= X_0\Psi(n) + X_1\Psi_1(n) \\
\left[ \begin{array}{r} x(0) \\ x(1) \\ \end{array} \right] &= X_0\left[ \begin{array}{r} \Psi_0(0) \\ \Psi_0(1) \\ \end{array} \right] + X_1 \left[ \begin{array}{r} \Psi_1(0) \\ \Psi_1(1) \\ \end{array} \right] \\ 
\left[ \begin{array}{r} 4 \\ 2 \\ \end{array} \right] &= 3\left[ \begin{array}{r} 1 \\ 1 \\ \end{array} \right] + 1 \left[ \begin{array}{r} 1 \\ -1 \\ \end{array} \right] \\ 
\end{align*}

In general, we could have used any orthogonal vectors, but the vectors $\{  \left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right],\left[ \begin{array}{c} 1 \\ -1 \\ \end{array} \right] \}$ are special.

\begin{align*}
\Psi_0 &= \left[ \begin{array}{c} \Psi_0(0) \\ \Psi_0(1) \\ \end{array} \right] =  \left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right] \\
\Psi_1 &= \left[ \begin{array}{c} \Psi_1(0) \\ \Psi_1(1) \\ \end{array} \right] =  \left[ \begin{array}{c} 1 \\ -1 \\ \end{array} \right] \\
\end{align*}

Where $\Psi_k(n) = e^{i k \omega_0 n}$. So when we use the basis $\{  \left[ \begin{array}{c} 1 \\ 1 \\ \end{array} \right],\left[ \begin{array}{c} 1 \\ -1 \\ \end{array} \right] \}$, we can decompose any DT signal of period 2 into linear combinations of $\Psi_0(n) = 1$ and $\Psi_1(n) = \lr{-1}^n$ in the Discrete Fourier Series expansion $X_0e^{i 0 \omega_0 n} + X_1e^{i\omega_0 n}$.


Let $\Psi = \{ \Psi_0, \Psi_1, \cdots , \Psi_{p-1} \}$ be an orthogonal basis. Then we can use the discrete fourier series exansion to decompose a signal $\vec{x}$ by projecting the signal onto the basis vectors, or writing the signal as a linear combination of the basis vectors:

$$\vec{x} = X_0\Psi_0 + X_1\Psi_1 + \cdots + X_{p-1}\Psi_{p-1}$$
In general the discrete fourier series expansion is
$$ \sum \limits_{k=0}^{p-1}X_ke^{ik\omega_0n} =  \sum \limits_{k=0}^{p-1}X_k\Psi_k  $$


We can determine the coefficients by the fact that the basis is orthogonal. We can take the dot product over the norm squared to find the cofficients. Given that $\vec{x} =  \left[ \begin{array}{c} x_0 \\ x_1 \\ \vdots \\ x_{p-1} \\ \end{array} \right]$, we can write this as a projection onto the orthogonal basis:

$$\vec{x} = \frac{\vec{x} \cdot \Psi_0}{\norm{\Psi_0}^2}\Psi_0 +  \frac{\vec{x} \cdot \Psi_1}{\norm{\Psi_1}^2}\Psi_1 + \cdots +  \frac{\vec{x} \cdot \Psi_{p-1}}{\norm{\Psi_{p-1}}^2}\Psi_{p-1} $$

The dot product is denoted as $a \cdot b$, which can also be written as 

$$a^Tb = a_0b_0 + a_1b_1 + \cdots + a_{p-1}b_{p-1} = \sum_{\ell=0}^{p-1}a_\ell b_\ell = a \cdot b$$

Here we can prove that the coefficients of the projection formula are equal to the dot product of the original signal and the respective basis vector or the norm of that vector squared:

\begin{proof}
We can write the signal as a linear combination of some orthogonal basis $\Psi = \{ \Psi_0, \Psi_1, \cdots , \Psi_{p-1} \}$.
\begin{align*}
\vec{x} &= X_0\Psi_0 + X_1\Psi_1 + \cdots + X_k\Psi_k  + \cdots + X_{p-1}\Psi_{p-1} \\
\vec{x}\cdot\Psi_k &= X_0\Psi_0\cdot\Psi_k + X_1\Psi_1\cdot\Psi_k + \cdots + X_k\Psi_k\cdot\Psi_k  + \cdots + X_{p-1}\Psi_{p-1}\cdot\Psi_k \\
\vec{x}\cdot\Psi_k &=  X_k\Psi_k\cdot\Psi_k \quad \mbox{since the dot product of any orthogonal vectors is zero} \\
\frac{\vec{x}\cdot\Psi_k}{\Psi_k\cdot\Psi_k} &= X_k \\
\frac{\vec{x}\cdot\Psi_k}{\norm{\Psi_k}^2} &= X_k \\
\end{align*}
\end{proof}

When writing a signal as a linear combination of an orthogonal basis, we are writing it as a sum of complex exponentials where $\Psi_k(n) = e^{ik\omega_0 n}$ some frequency component of the signal, $\omega_0 = 2\pi/p$ is the fundamental frequency and $p$ is the period of the signal.

If we have a period of 2, note that $\Psi_k(n+2) = e^{i\omega_0\lr{n+2}} = e^{i\omega_0 n}e^{i2\omega_0} = e^{i\omega_0 n}e^{i2\pi} = e^{i\omega_0 n} = \Psi_k(n)$. In general, for a period $p$, $\Psi_k(n+p) = \Psi_k(n)$.


How do we project complex-valued vectors onto each other? We cannot use the dot product for projections involving the dot product. For example, $\left[ \begin{array}{r} 1 \\ i \\ \end{array} \right] \cdot \left[ \begin{array}{r} 1 \\ i \\ \end{array} \right] = 0 \neq 2 = \norm{\left[ \begin{array}{r} 1 \\ i \\ \end{array} \right]}^2$. Say hello to the \emph{inner product}. Now we can define $\langle f,g\rangle = f^Tg^*$. This way we take the conjugate of the second term in the inner product. Now we can accomplish projections using the formula:

$$ X_k = \frac{\langle \vec{x},\Psi_k \rangle}{\langle\Psi_k,\Psi_k\rangle}  $$

Signals as Vectors