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mathematics-physics-wiki/docs/en/physics/mathematical-physics/signal-analysis/discrete-fourier-transform.md

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The discrete Fourier transform

Theorem: sampling a signal with the impulse train makes the spectrum of the signal periodic.

??? note "Proof:"

Will be added later.

A bandlimited signal implies that its frequency components are zero outside the bandwidth frequency interval.

Theorem: if a signal has a bandwidth \omega_b \in \mathbb{R} then it can be completely determined from its samples at a sampling frequency \omega_s \in \mathbb{R} given by

\omega_s > 2 \omega_b.

??? note "Proof:"

Will be added later.

When the sampling frequency does not comply to this statement, the reconstruction of the spectrum will exhibit imperfections known as aliasing. The critical value of the sampling frequency is known as the Nyquist frequency.

The discrete time Fourier transform

Theorem: let f: \mathbb{R} \to \mathbb{C} be a signal with its sampled signal f_s(t) = f(t) \delta_{T_s}(t) for all t \in \mathbb{R} with sampling period T_s \in \mathbb{R}. Then the discrete time Fourier transform F: \mathbb{R} \to \mathbb{C} of f_s is given by

F(\Omega) = \sum_{m = -\infty}^\infty f[m] e^{-im\Omega},

for all \Omega \in \mathbb{R}. With \Omega = \omega T_s the dimensionless frequency and F_s(\omega) := F(\Omega).

??? note "Proof:"

Will be added later.

The discrete Fourier transform

Theorem: let f: \mathbb{R} \to \mathbb{C} be a signal and f_N: \mathbb{R} \to \mathbb{C} the truncated signal of f by N \in \mathbb{N} given by

f_N[m] = \begin{cases} f[m] &\text{ if } m \in {0, \dots, N - 1}, \ 0 &\text{ if } m \notin {0, \dots, N - 1}, \end{cases}

sampled by T_s \in \mathbb{R}. Its discrete Fourier transform F_N: \mathbb{R} \to \mathbb{C} is given by

F_N[k] = \sum_{m=0}^{N-1} f[m] \exp \bigg(-2\pi i \frac{km}{N} \bigg)

for all k \in \{0, \dots, N-1\}.

??? note "Proof:"

Will be added later.

We have that F_N[k] = F_N(k\Delta \omega) with \Delta \omega = \frac{2\pi}{N T_s} the angular frequency resolution.

Theorem: let F_N: \mathbb{R} \to \mathbb{C} be a spectrum of a signal truncated by N \in \mathbb{N} then its inverse discrete Fourier transform f_N: \mathbb{R} \to \mathbb{C} is given by

f[m] = \frac{1}{N} \sum_{k=0}^{N-1} F_N[k] \exp \bigg(2\pi i \frac{km}{N} \bigg)

for all m \in \{0, \dots, N - 1\}.

??? note "Proof:"

Will be added later.

Definition: therefore f_N and F_N with N \in \mathbb{N} form a discrete Fourier transform pair denoted by

f_N \overset{\mathcal{DF}}\longleftrightarrow F_N,

therefore we have

\begin{align*} &f_N[m] = \mathcal{DF}^{-1}[F_N[k]], \quad &\forall m \in {0, \dots, N - 1}, \ &F_N[k] = \mathcal{DF}[f[m]], \quad &\forall k \in {0, \dots, N - 1}. \end{align*}