Fourier transform: Difference between revisions

From David's Wiki
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Let <math>F(s) = FFT(f(s))</math>.
Let <math>F(s) = FFT(f(s))</math>.


'''Linearity'''
'''Linearity'''<br>
<math>FFT(\lambda f + g) = \lambda FFT(f) + FFT(g)</math>
<math>FFT(\lambda f + g) = \lambda FFT(f) + FFT(g)</math>


'''Shift'''
'''Shift'''<br>
<math>FFT(f(x-a)) = e^{-i \omega_k a} F(\omega_k)</math>
<math>FFT(f(x-a)) = e^{-i \omega_k a} F(\omega_k)</math>
* Note <math>\omega_k = 2 \pi k/N </math>
* Note <math>\omega_k = 2 \pi k/N </math>


'''Similarity'''
'''Similarity'''<br>
<math>FFT(f(ax)) = |a|^{-1} F(s/a)</math>
<math>FFT(f(ax)) = |a|^{-1} F(s/a)</math>


'''Convolution Theorem'''
'''Convolution Theorem'''<br>
<math>f *g  \Leftrightarrow F(s) \times G(s)</math><br>
<math>f *g  \Leftrightarrow F(s) \times G(s)</math><br>
<math>f \times g \Leftrightarrow F(s) * G(s)</math>
<math>f \times g \Leftrightarrow F(s) * G(s)</math>


'''Parseval's Theorem'''
'''Parseval's Theorem'''<br>
<math>\int |f(x)|^2 dx = \int |F(s)|^2 ds</math>
<math>\int |f(x)|^2 dx = \int |F(s)|^2 ds</math>
See [[Wikipedia: Parseval's theorem]]<br>
See [[Wikipedia: Parseval's theorem]]<br>
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<math>\int f(x) g^*(x) dx = \int F(s) G^*(s) ds</math>
<math>\int f(x) g^*(x) dx = \int F(s) G^*(s) ds</math>


'''Autocorrelation'''
'''Autocorrelation'''<br>
<math>\int f(x') f^*(x' - x) dx' \Leftrightarrow |F(x)|^2</math>
<math>\int f(x') f^*(x' - x) dx' \Leftrightarrow |F(x)|^2</math>
This is a case of the convolution theorem.
This is a case of the convolution theorem.


'''IFFT'''
'''IFFT'''<br>
<math>IFFT(f) = (1/n) \bar{FFT}(\bar{f})</math>
<math>IFFT(f) = (1/n) \bar{FFT}(\bar{f})</math>


'''2D FFT'''
'''2D FFT'''<br>
<math>FFT2(f) = FFT_{x}(FFT_{y}(f)) = FFT_{y}(FFT_{x}(f))</math>
<math>FFT2(f) = FFT_{x}(FFT_{y}(f)) = FFT_{y}(FFT_{x}(f))</math>



Revision as of 15:04, 12 April 2023

The Fourier transform decomposes a signal (i.e. a time series) into multiple sine and cosine waves.


Background

Suppose we have signal\(\displaystyle f(x)\).
Then the Fourier transform \(\displaystyle \hat{f}(\xi)\)is defined as: \[\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-i2 \pi \xi x} dx\]

Recall that Euler's formula states: \(\displaystyle e^{ix} = cos(x) + i \sin(x)\). Hence, \(\displaystyle e^{-i2 \pi \xi x} = cos(-2 \pi \xi x) + i \sin(-2 \pi \xi x)\) In other words, the Fourier transform is the integral (i.e. alignment) of signal times some sine and cosine waves.

The inverse of the fourier transform is: \[f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{i2 \pi \xi x} d\xi\]

Discrete Fourier Transform

A naive DFT would compute the matrix of \(\displaystyle e^{-i2 \pi \xi x}\) and multiply it with the signal. This would take \(\displaystyle \mathcal{O}(n^2)\) time.
However, most languages have an FFT library which can compute the DFT in \(\displaystyle \mathcal{O}(n \log n)\) time.

In most languages, FFT is defined as: \[A_k = \sum_{m=0}^{n-1} f(x) \exp\ \left\{ -2 \pi i \frac{mk}{n} \right\}\] and IFFT is defined as: \[a_m = \frac{1}{n} \sum_{k=0}^{n-1} \hat{f}(\xi) \exp \left\{ 2 \pi i \frac{mk}{n} \right\}\]

That the main difference between the FFT and IFFT is the negative symbol in the exponent. You can implement IFFT as IFFT(x) = (1/len(x))*conj(FFT(conj(x)))[1].

Properties

Note \(\displaystyle \bar{x}\) refers to the complex conjugate Let \(\displaystyle F(s) = FFT(f(s))\).

Linearity
\(\displaystyle FFT(\lambda f + g) = \lambda FFT(f) + FFT(g)\)

Shift
\(\displaystyle FFT(f(x-a)) = e^{-i \omega_k a} F(\omega_k)\)

  • Note \(\displaystyle \omega_k = 2 \pi k/N \)

Similarity
\(\displaystyle FFT(f(ax)) = |a|^{-1} F(s/a)\)

Convolution Theorem
\(\displaystyle f *g \Leftrightarrow F(s) \times G(s)\)
\(\displaystyle f \times g \Leftrightarrow F(s) * G(s)\)

Parseval's Theorem
\(\displaystyle \int |f(x)|^2 dx = \int |F(s)|^2 ds\) See Wikipedia: Parseval's theorem

More generally, \(\displaystyle \int f(x) g^*(x) dx = \int F(s) G^*(s) ds\)

Autocorrelation
\(\displaystyle \int f(x') f^*(x' - x) dx' \Leftrightarrow |F(x)|^2\) This is a case of the convolution theorem.

IFFT
\(\displaystyle IFFT(f) = (1/n) \bar{FFT}(\bar{f})\)

2D FFT
\(\displaystyle FFT2(f) = FFT_{x}(FFT_{y}(f)) = FFT_{y}(FFT_{x}(f))\)

Short-time Fourier transform

The STFT applies FFT using a sliding window over the signal.
This produces a matrix of FFT values over time which allow you to see how the signal is changing.

References

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  1. Siembida, A. (2010, March 11). How to compute the IFFT using only the forward FFT. Adam Siembida Personal Webpage. Retrieved January 24, 2023, from https://adamsiembida.com/how-to-compute-the-ifft-using-only-the-forward-fft/