Zusammenhang zwischen DFT und DCT/MDCT

[pumpkin (Gast)]

Hallo,

hat jemand einen guten Link/Buchtip wo man den Betreff gut 
nachvollziehen kann? Google spuckt nichts gutes aus. Habs mir schon 
selber plausibel gemacht und möchte sicher gehen ob ich auf dem 
richtigen Weg bin.

pumpkin

[pumpkin (Gast)]

"Oppenheim, Allen V.; Schafer, Roland W.; Buck, John R.: Zeitdiskrete 
Signalverarbeitung; 2. Auflage; München 2004"

Damit kann man einigermaßen arbeiten. Weitere Quellenangaben im Text.

[T. H. (pumpkin)]

Schon länger her, aber ich finde die folgende Erklärung zum Thema DFT 
<-> DCT gut. Quelle:

  http://wapedia.mobi/en/Sine_and_cosine_transforms

Eine gute Erklärung zum Thema DFT <-> DCT gibts auch hier:

  http://www.fftw.org/fftw3_doc/Real-even_002fodd-DFTs-_0028cosine_002fsine-transforms_0029.html#Real-even_002fodd-DFTs-_0028cosine_002fsine-transforms_0029


Gute Literatur zum Thema MDCT gibts hier:

  "On the Relationship Between MDCT, SDFT and DFT", Wang et al.
  http://www.comp.nus.edu.sg/~wangye/papers/On%20the%20Relationship%20Between%20MDCT,%20SDFT%20and%20DFT.pdf


Umrechnung zwischen MDCT und DFT:

  "Improved Coding Techniques Using Estimated Spectral Magnitude and
   Phase Derived from MDCT Coefficients", Cheng et al.

  http://www.wipo.int/pctdb/images4/PCT-PAGES/2005/322005/05073960/05073960.pdf


  "Packet Loss Concealment for Audio Streaming Based on the GAPES
   Algorithm", Ofir & Malah

  http://sipl.technion.ac.il/new/Research/Publications/Graduates/Hadas_Ofir/AES_Convention/Article/AES_118_Article_327_Ofir_and_Malah_final.pdf


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In mathematics, the Fourier cosine transform is a special case of the 
continuous Fourier transform, arising naturally when attempting to 
transform an even function. Consider the general Fourier transform:

$$F(\omega) = \mathcal{F}(f)(t) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t) e^{-i\omega t}\,dt.$$

We may expand the integrand by means of Euler's formula:

$$F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)(\cos\,{\omega t} - i\,\sin{ \,\omega t})\,dt,$$

or, written as the sum of two integrals:

$$F(\omega)=\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\cos\,{\omega t} \,dt - \frac{i}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(t)\sin\,{\omega t}\,dt.$$

Now notice that if we assume f(t) is an even function, the product 
f(t)cosωt is also even whilst the product f(t)sinωt is an odd function. 
Since we are integrating over an interval symmetric about the origin 
(i.e. -∞ to +∞), the second integral must vanish to zero, and the first 
may be simplified to give:

$$F(\omega)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty f(t)\cos\,{\omega t} \,dt,$$

which is the Fourier cosine transform for even f(t). It is clear that 
the transformed function F(ω) is also an even function, and a similar 
analysis of the general inverse Fourier transform yields a second cosine 
transform, namely:

$$f(t)= \sqrt{\frac{2}{\pi}} \int\limits_{0}^\infty F(\omega)\cos\,{\omega t} \,d\omega.$$

Note that the numerical factors in the transforms are defined uniquely 
only by their product, as discussed for general continuous Fourier 
transforms.