113 lines
4.3 KiB
ReStructuredText
113 lines
4.3 KiB
ReStructuredText
Discrete Fourier Transform
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----------------------------
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Discrete Fourier Transform (DFT) is used to perform the Fourier transform of a discrete signal.
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The DFT of a discrete complex/real signal :math:`\vec{x}=\{x_0,...,x_{N-1}\}` is:
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.. math::
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X_k = \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi\frac{k}{N}n}
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.. note::
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In this case, :math:`\frac{k}{N}` can be seen as the frequency of the wave used to determined
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the :math:`k^{th}` frequency :math:`X_k`.
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See `here <https://www.youtube.com/watch?v=mkGsMWi_j4Q&t=282s>`__ for more infos.
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One important property of the DFT is its **periodicity**. The DFT has a period of :math:`N`.
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Hence, all :math:`X_k` such that :math:`k > \frac{N}{2}` corresponds to negative frequencies:
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.. math::
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X_{N-1} &= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi\frac{N-1}{N}n}
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&= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi (n - \frac{n}{N})}
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&= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi n} \cdot e^{j2\pi\frac{n}{N}}
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&= \sum_{n=0}^{N-1} x_n \cdot e^{j2\pi\frac{n}{N}} \mathrm{~~as~~} \forall n: e^{-j2\pi n} = 1
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&=X_{-1}
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This explanation is extracted from `here <https://dsp.stackexchange.com/questions/50505/why-is-the-second-half-of-the-fft-negative-frequencies>`__.
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Note that, the frequency corresponding to :math:`\frac{N}{2}` is called `Nyquist_rate <https://en.wikipedia.org/wiki/Nyquist_rate>`__.
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As frequencies above :math:`\frac{N}{2}` cannot be captured by the signal due to the `Nyquist-Shannon <https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem>`__, a reasoning phenomenon starts to arise.
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This leads to a capturing effect of the negative frequencies.
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See `here <https://www.youtube.com/watch?v=nOLuVNhBIg0>`__ for a visual explanation.
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Inverse DFT
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===========
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The inverse DFT of a discrete complex/real signal :math:`\vec{x}={x_0,...,x_{N-1}}` is:
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.. math::
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x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k \cdot e^{j2\pi\frac{k}{N}n}
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.. note::
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If the signal is *real*, all the complex terms of the inverse DFT are suppose to cancel out.
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In depth example
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=================
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Let's take a discrete signal :math:`\vec{x}=\{1,2,3,4\}`, sampled at a frequency :math:`f_s=10 Hz`.
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.. literalinclude:: code/dft.R
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:language: R
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Now let's do its inverse DFT:
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.. literalinclude:: code/idft.R
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:language: R
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Understanding how imaginary parts cancel out in during the inverse DFT is important.
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The periodicity of the DFT plays a key role for that.
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For our signal :math:`\vec{x}` we have:
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.. math::
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\vec{X}=\{10+0j, -2+2j, -2+0j, -2-2j\}
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See how :math:`X_1 = \overline{X_3}`. When computing :math:`x_n`, :math:`X_1` becomes:
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.. math::
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X_1 \cdot e^{j2\pi\frac{1}{N}n} &= ({\color{blue}-2}+{\color{red}2j}) \cdot (cos(2\pi\frac{1}{4}n)+j\cdot sin(2\pi\frac{1}{4}n))
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&= {\color{blue}-2 \cdot cos(2\pi\frac{1}{4}n) -2j \cdot sin(2\pi\frac{1}{4}n)} + {\color{red}2j \cdot cos(2\pi\frac{1}{4}n) -2 \cdot sin(2\pi\frac{1}{4}n)}
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In turn, :math:`X_3` becomes:
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.. math::
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X_3 \cdot e^{j2\pi\frac{3}{N}n} &= ({\color{blue}-2}{\color{red}-2j}) \cdot (cos(2\pi\frac{3}{4}n)+j\cdot sin(2\pi\frac{3}{4}n))
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&= {\color{blue}-2 \cdot cos(2\pi\frac{3}{4}n) -2j \cdot sin(2\pi\frac{3}{4}n)} {\color{red}- 2j \cdot cos(2\pi\frac{3}{4}n) + 2 \cdot sin(2\pi\frac{3}{4}n)}
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Notice that:
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.. math::
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\forall n:\, & cos(2\pi\frac{1}{4}n) = 0
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& cos(2\pi\frac{3}{4}n) = 0
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& sin(2\pi\frac{1}{4}n) = j
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& sin(2\pi\frac{3}{4}n) = -j
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In particular, see how the symmetry of the DFT leads to opposite values.
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In this case, :math:`cos` waves are all 0.
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But opposite values would arise for other :math:`\vec{X}`.
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Plugin in these new values whenever :math:`j` shows up gives the following:
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.. math::
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X_1 \cdot e^{j2\pi\frac{1}{N}n} &= -2 \cdot cos(2\pi\frac{1}{4}n) + (-2j \cdot j) + (2j \cdot 0) -2 \cdot sin(2\pi\frac{1}{4}n)
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&= -2 \cdot cos(2\pi\frac{1}{4}n) + 2 -2 \cdot sin(2\pi\frac{1}{4}n)
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X_3 \cdot e^{j2\pi\frac{3}{N}n} &= -2 \cdot cos(2\pi\frac{3}{4}n) +(-2j \cdot -j) +(- 2j \cdot 0) + 2 \cdot sin(2\pi\frac{3}{4}n)
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&= -2 \cdot cos(2\pi\frac{3}{4}n) - 2 + 2 \cdot sin(2\pi\frac{3}{4}n)
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As these two terms get summed during the inverse DFT gives:
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.. math::
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X_1 \cdot e^{j2\pi\frac{1}{N}n} + X_3 \cdot e^{j2\pi\frac{3}{N}n} = -2 \cdot cos(2\pi\frac{1}{4}n) \cancel{+ 2} -2 \cdot sin(2\pi\frac{1}{4}n)\\
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-2 \cdot cos(2\pi\frac{3}{4}n) \cancel{- 2} + 2 \cdot sin(2\pi\frac{3}{4}n)
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An you are left with sum of real signals.
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