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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 signalf_s(t) = f(t) \delta_{T_s}(t)
for allt \in \mathbb{R}
with sampling periodT_s \in \mathbb{R}
. Then the discrete time Fourier transformF: \mathbb{R} \to \mathbb{C}
off_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 andF_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 andf_N: \mathbb{R} \to \mathbb{C}
the truncated signal off
byN \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 transformF_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 byN \in \mathbb{N}
then its inverse discrete Fourier transformf_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
andF_N
withN \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*}