science-notes/source/signal_processing/dft.rst

Discrete Fourier Transform
----------------------------

Discrete Fourier Transform (DFT) is used to perform the Fourier transform of a discrete signal.
The DFT of a discrete complex/real signal :math:`\vec{x}=\{x_0,...,x_{N-1}\}` is:

.. math::
   X_k = \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi\frac{k}{N}n}

.. note::
   In this case, :math:`\frac{k}{N}` can be seen as the frequency of the wave used to determined
   the :math:`k^{th}` frequency :math:`X_k`.
   See `here <https://www.youtube.com/watch?v=mkGsMWi_j4Q&t=282s>`__ for more infos.

One important property of the DFT is its **periodicity**. The DFT has a period of :math:`N`.
Hence, all :math:`X_k` such that :math:`k > \frac{N}{2}` corresponds to negative frequencies:

.. math::
   X_{N-1} &= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi\frac{N-1}{N}n}

   &= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi (n - \frac{n}{N})}

   &= \sum_{n=0}^{N-1} x_n \cdot e^{-j2\pi n} \cdot e^{j2\pi\frac{n}{N}}

   &= \sum_{n=0}^{N-1} x_n \cdot e^{j2\pi\frac{n}{N}} \mathrm{~~as~~} \forall n: e^{-j2\pi n} = 1

   &=X_{-1}

This explanation is extracted from `here <https://dsp.stackexchange.com/questions/50505/why-is-the-second-half-of-the-fft-negative-frequencies>`__.
Note that, the frequency corresponding to :math:`\frac{N}{2}` is called `Nyquist_rate <https://en.wikipedia.org/wiki/Nyquist_rate>`__.
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.
This leads to a capturing effect of the negative frequencies.
See `here <https://www.youtube.com/watch?v=nOLuVNhBIg0>`__ for a visual explanation.
   
Inverse DFT
===========
   
The inverse DFT of a discrete complex/real signal :math:`\vec{x}={x_0,...,x_{N-1}}` is:

.. math::
   x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k \cdot e^{j2\pi\frac{k}{N}n}

.. note::
   If the signal is *real*, all the complex terms of the inverse DFT are suppose to cancel out.

In depth example
=================

Let's take a discrete signal :math:`\vec{x}=\{1,2,3,4\}`, sampled at a frequency :math:`f_s=10 Hz`.

.. literalinclude:: code/dft.R
   :language: R

Now let's do its inverse DFT:

.. literalinclude:: code/idft.R
   :language: R

Understanding how imaginary parts cancel out in during the inverse DFT is important.
The periodicity of the DFT plays a key role for that.
For our signal :math:`\vec{x}` we have:

.. math::
   \vec{X}=\{10+0j, -2+2j, -2+0j, -2-2j\}

See how :math:`X_1 = \overline{X_3}`. When computing :math:`x_n`, :math:`X_1` becomes:

.. math::
    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))

    &= {\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)}

In turn,  :math:`X_3` becomes:


.. math::
    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))

    &= {\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)}

Notice that:

.. math::
   \forall n:\, & cos(2\pi\frac{1}{4}n) = 0

   & cos(2\pi\frac{3}{4}n) = 0

   & sin(2\pi\frac{1}{4}n) = j

   & sin(2\pi\frac{3}{4}n) = -j

In particular, see how the symmetry of the DFT leads to opposite values.
Plugin in these new values whenever :math:`j` shows up gives the following:

.. math::
   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)

   &=  -2 \cdot cos(2\pi\frac{1}{4}n) + 2 -2 \cdot sin(2\pi\frac{1}{4}n)

   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)

   &= -2 \cdot cos(2\pi\frac{3}{4}n) - 2 + 2 \cdot sin(2\pi\frac{3}{4}n)

As these two terms get summed during the inverse DFT gives:

.. math::
   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)\\

   -2 \cdot cos(2\pi\frac{3}{4}n) \cancel{- 2} + 2 \cdot sin(2\pi\frac{3}{4}n)

An you are left with sum of real signals.