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Continuous-Time Fourier Transform - Review

Module by: Ivan Selesnick

Xω=ℱxt=∫-∞∞xtⅇ-ⅈωtdt X ω ℱ x t t x t ω t (1)
  1. The FT is invertible.
    xt=ℱ-1Xω=12π∫-∞∞Xωⅇⅈωtdω x t ℱ X ω 1 2 ω X ω ω t (2)
  2. The FT naturally arises in many physical problems.
  3. The FT simplifies the analysis of LTI systems because it has a convolution theorem.
    xt*gt↔XωGω ↔ x t g t X ω G ω (3)
  4. A discrete version of the FT is available. The DFT for finite-length discrete-time signals; the DTFT for infinite-length discrete-time signals.
  5. The Fourier transform is based on the notion of frequency.

A disadvantage of the FT: If one reconstructs xt x t from only part of Xω X ω , x ~ t=∫SXωⅇⅈωtdt x ~ t t S X ω ω t the quality of x ~ t x ~ t depends on how fast |Xω| X ω decays.

∥ x ~ t-xt∥2=12π∥ X ~ ω-Xω∥=12π∫ω∉S|Xω|2dω x ~ t x t 2 1 2 X ~ ω X ω 1 2 ω ω S X ω 2 (4)
For signals that are concentrated in frequency, |Xω| X ω decays fast. However, for many natural signals (for example, a row from an image), |Xω| X ω decays slowly and x ~ t x ~ t is poor. For those signals, the FT does not provide an efficient representation. slow decay of  |Xω|↔Xω is not an efficient representation of  xt ↔ slow decay of   X ω X ω  is not an efficient representation of   x t

Advantages of efficient representation:

  1. compression
  2. fast algorithms
  3. estimation (denoising, enhancement, etc.)

The efficiency of the FT for representing a signal can be evaluated by plotting the reconstruction error verses how much of Xω X ω is used in the reconstruction. For implementation purposes, we use the DFT. Suppose xn x n , 0≤n≤N-1 0 n N 1 is a length-NN discrete signal. Let Xk X k , 0≤k≤N-1 0 k N 1 be the DFT of xn x n . If x l ~ n x l ~ n is the reconstruction of xn x n from the ll largest values of Xk X k , we can make a plot of el=∥ x l ~ n-xn∥ e l x l ~ n x n versus ll. The faster el e l decays, the more efficient the DFT is for representing xn x n .

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