2. Discrete Time Fourier Transform

Recall that if x(n) is absolutlely summable, that is, $\sum_{-\infty}^{+\infty} |x(n)| < \infty$ than its Discrete-Time Fourier Transform is given by:
\[
X(e^{j \omega})=\sum_{n=- \infty}^{+ \infty} x(n) \cdot e^{-j \cdot \omega \cdot n} \tag{3.1}
\]
This function transforms a discrete signal $x(n)$ into a complex-valued continuous function $X(e^{j \omega})$.

${\bf Matrix Implementation}$
If $x(n)$ is a finite duration sequence, we can use MATLAB to compute the Discrete Time Fourier Transform (DTFT) $X(e^{j\omega})$ numerically at any frequency $\omega$.

Let assume that the sequence x(n) has N samples between $n_1 \leq n \leq n_N$ we can define:
\[\omega_k =\frac{\pi}{M}k, \qquad k=0,1,....,M\]
which are (M+1) frequencies between $[0, \pi]$. Then (1) can be written as:

\[
X(e^{j\omega_k})=\sum_{r=1}^{N} x(n_r) \cdot e^{-j \cdot \omega_k \cdot n_r}  = \sum_{r=1}^{N} x(n_r) \cdot e^{-j \cdot \frac{\pi}{M} k \cdot n_r}, \qquad k=0,...,M \tag{3.2}
\]

Let's define the vector $\mathbf{X} = \begin{bmatrix} X(e^{j\omega_0}) \\ \vdots \\ X(e^{j\omega_M}) \end{bmatrix}$ where the k-th element $X(e^{j\omega_k})$ has the below expression:

\[X(e^{j\omega_k})= [ x(n_1) \cdot e^{-j \cdot \frac{\pi}{M}\cdot k \cdot n_1} + \ldots + x(n_N) \cdot  e^{-j \cdot \frac{\pi}{M}\cdot k \cdot n_N} ] \]

if we define the matrix $\mathbf{W}=\begin{bmatrix} W_0\\ \vdots \\W_M \end{bmatrix}$ where the k-th row $W_k$ has the below expression:

\[ W_k = \begin{bmatrix} e^{-j \cdot \frac{\pi}{M}\cdot k \cdot n_1}, \ldots, e^{-j \cdot \frac{\pi}{M}\cdot k \cdot n_N} \end{bmatrix} \]

and $\mathbf{x(n)} = \begin{bmatrix} x(n_1) \\ \vdots  \\ x(n_N) \end{bmatrix}$, we can write (3.2) as follow:

\[\mathbf{X} = \mathbf{W} \cdot \mathbf{x}\]

Let's consider the row vectors ${\bf k}$ and ${\bf n}$ that represent, respectively, the index of the frequency and for the time.
\[ \begin{matrix}
\mathbf{k}= \begin{bmatrix} 0, 1, \ldots, M\end{bmatrix}\\
\mathbf{n}= \begin{bmatrix} n_1, n_2, \ldots, n_N\end{bmatrix}
\end{matrix} \]
the product $\mathbf{k^T \cdot n}$ is equal to:
\[ \begin{bmatrix} 0 \\1 \\ \vdots \\M\end{bmatrix} \cdot \begin{bmatrix} n_1 & n_2 & \ldots & n_N\end{bmatrix} = \begin{bmatrix} 0 & 0 & \ldots & 0 \\ n_1 & n_2 & \ldots & n_N \\ \vdots & \vdots & \ddots & \vdots \\ M \cdot n_1 & M \cdot n_2 & \ldots & M \cdot n_N \end{bmatrix} \]
so we can rewrite $\mathbf{W}$

\[ \mathbf{W} = \begin{bmatrix} W_0 \\ W_1 \\ \vdots \\ W_M \end{bmatrix} = \begin{bmatrix} e^{-j \cdot \frac{\pi}{M}\cdot 0 \cdot n_1} \ldots e^{-j \cdot \frac{\pi}{M}\cdot 0 \cdot n_N} \\ \vdots \\ e^{-j \cdot \frac{\pi}{M}\cdot M \cdot n_1} \ldots e^{-j \cdot \frac{\pi}{M}\cdot M \cdot n_N} \end{bmatrix} =
 exp[-j \cdot \frac{\pi}{M} \cdot \mathbf{k^T \cdot n}] \]

finally the complete matrix form for DTFT is
\[\mathbf{X} = exp[-j \cdot \frac{\pi}{M} \cdot \mathbf{k^T \cdot n}] \cdot \mathbf{x} \tag{3.3} \]

Using of filter command in the convolution computation MATLAB

In the MATLAB exists the command filter that permits the resolution of difference equation numerically.
An important difference equation that express a link between input ad output in a LTI is
\[\sum_{k=0}^N a_ky(n-k) = \sum_{m=0}^M b_mx(n-m)\]
this equation can be solved using filter(b,a,x) where $b=[b_0 ... b_m],\; a=[a_0 ... a_k]$.
\[z(n) = conv(x,y) = x(n)*y(n) = \sum_{k=-\infty}^{+\infty} x(k)y(n-k)\]
Considering two sequence with finite length we can observe that:
\[filter(x,1,y)\Rightarrow 1\cdot z(n) =\sum_{m=0}^M x(m)y(n-m)\]
from below expression we can see that using filter function we can compute convolution of x and y but the result will have the dimension of x

1. Discrete Time Sinusoidal Signal

A discrete time sinusoid has this expression:
\[A\cos(\omega_0n + \theta),\qquad n \in]-\infty, +\infty[\tag{1.1}\]
where:
$A$ is the Amplitude
$\omega_0$ is the Frequency in radians per sample
$\theta$ is the Phase in radians.
We can express frequency with the variable $f_0$ defined by $\omega_0 = 2\pi f_0$
and the (1) becomes:
\[A\cos(2\pi f_0 n + \theta),\qquad n \in]-\infty, +\infty[\tag{1.2}\]
Now the frequency $f_0$ has dimensions of cycles per samples.

A discrete time sinusoid is periodic only if its frequency $f_0$ is a rational number.

A discrete time signal x(n) is periodic with period N (N>0) if and only if:
\[x(n+N) = x(n), \qquad \forall n\tag{1.3}\]
The smallest value of N is called fundamental period.For a sinusoid with frequency $f_0$ we should have:
\[\cos[2\pi f_0(N+n) + \theta] = \cos(2\pi f_0n+\theta)\]
This relation is true if and only if
\[2\pi f_0N=2k\pi\Rightarrow f_0=\frac{k}{N}\]
in other words the sinusoid is periodic if we can express the frequency as a rational.
The fundamental period is N (the value we obtain cancelling common factors so that k and N are relatively prime).
The absolute value of k is the number of cycles the phasor makes before going to the initial position.

EXAMPLE 1.a
Let's consider the discrete sequence $x(n) = \cos(0.3\pi n)$, we can see that $\omega_0=0,3\pi$ and so since $f_0=\frac{\omega_0}{2\pi}$ we obtain that $f_0=\frac{0.3\pi}{2\pi}\Rightarrow f_0=\frac{3}{20}$
The sequence x(n) is periodic and N=20

EXAMPLE 1.b
Let's consider the discrete sequence $x(n) = \cos(0.3n)$, we can see that $\omega_0=0,3$ and so since $f_0=\frac{\omega_0}{2\pi}$ we obtain that $f_0=\frac{0.3}{2\pi}\Rightarrow f_0=\frac{3}{20\pi}$
The sequence x(n) is not periodic

A discrete time sinusoid is periodic in frequence with period of $2\pi$
To prove this assertion, let us consider a sinusoid $\cos[\omega_0n+ \theta]$ and add $2\pi$ to $\omega_0$, we obtain:
\[\cos[(\omega_0+2\pi)n+ \theta]=\cos[\omega_0n+2\pi n+ \theta]=\cos[\omega_0n+ \theta] \]
in fact n is an integer number.
This result is more important because it permits us to study the sinusoid only in the frequency range of $2\pi$ (i.e. $-\pi\leq\omega\leq\pi$ or $-\frac{1}{2}\leq f \leq\frac{1}{2}$)