0
$\begingroup$

Is there a way to compute the inverse discrete cosine transform (type-2) by leveraging either a DCT, FFT, or IFFT algorithm? I have seen ways to compute the DCT using FFTs, and I've seen ways to compute IFFT using FFT. I can't quite find a simple example with description.

$\endgroup$
1
$\begingroup$

Have a look at Fast DCT Algorithm (PDF Version).

It has both DCT and Inverse DCT using DFT (FFT).
They show how to do a DCT and Inverse without the reflection trick.

The standard (Less efficient then above) way doing so is:

numElements = 10;

vX = randn(1, numElements);

disp(vX); %<! Diusplay the input signal

vDctRef = dct(vX);

% Forward DCR using FFT
vXX     = [fliplr(vX), vX]; %<! Mirroring
vXDft   = fft(vXX);

vGrid = [0:((2 * numElements) - 1)];

vShiftGrid = exp(-1j * 2 * pi * (numElements - 0.5) * vGrid / (2 * numElements));

vXDft2 = real(vXDft ./ vShiftGrid);

vDct    = vXDft2(1:numElements) / sqrt(2 * numElements);
vDct(1) = vDct(1) / sqrt(2);

disp(vDctRef)
disp(vDct)

% Inverse DCT Using FFT
vDct(1) = vDct(1) * sqrt(2);
vDct    = vDct * sqrt(2 * numElements);

vXDft = [vDct, 0, -fliplr(vDct(2:numElements))] .* vShiftGrid;
vXX = ifft(vXDft);

vX = real(fliplr(vXX(1:numElements)));
disp(vX);

A reference code for the IDCT by the more efficient method (As in reference):

numElements = 10;

% Signal (DCT Coefficients)
vXDct = randn(1, numElements);

% Reference Inverse IDCT
vXRef = idct(vXDct);

% Inverse IDCT Using FFT
vGrid = [0:(numElements - 1)];

vShiftGrid      = exp((1j * pi * vGrid) / (2 * numElements));
vShiftGrid      = vShiftGrid * sqrt(2 * numElements);
vShiftGrid(1)   = vShiftGrid(1) / sqrt(2);


vTmp = vShiftGrid .* vXDct ;
vTmp = real(ifft(vTmp));

vX = zeros(1, numElements);

for ii = 0:((numElements / 2) - 1)
    vX((2 * ii) + 1) = vTmp(ii + 1) ;
    vX((2 * ii) + 2) = vTmp(numElements - ii) ;
end

disp(vXRef);
disp(vX);

Another reference is given by Lecture Notes for the Course MAT-INF2360 - Fourier Theory, Wavelet Analysis and Non Linear Optimization.
Have a look at 4.2 Efficient implementations of the DCT at page 115.

Enjoy...

Remark
When it is written DCT above it refers to DCT Type II.

$\endgroup$
  • $\begingroup$ Works great! thank you. I got hung up initially because I was using the DCT equations from wikipedia, which does not have the Kronecker delta and sqrt(2/N) scale, as opposed to the MATLAB implementation, which does. Once I realized this, I could get matching results from yours. $\endgroup$ – user2913869 Aug 18 '18 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.