# taking FFT from DCT

There is a speech signal, $y[n]$, in the time-domain.

To take it to the frequency-domain, DCT has been got from $y[n]$ (w = dct(y) in MATLAB).

After it, I applied LPC on this signal ($w[k]$) in the frequency-domain to find the coefficients. Tthe following process was used:

yfft = fft(w);
yidft = ifft(yfft.*conj(yfft));
a = levinson(yidft,p);


My question is: what domain does the signal resulting from applying FFT to a DCT (i.e. $w$) belong to? Also, taking the FFT from a signal leads to the frequency-domain, right?

• I don't think the discrete Fourier transform of the discrete cosine transform has a specific name. Jan 9 '18 at 9:24
• The most similar thing I can think of is MFC. However, there are some operations in the middle that you didn't perform (the filter bank and the logarithm), so I don't think this could be what you are looking for. Maybe this is some kind of non-logarithmic based cepstrum in a non-mel-frequency scale...? Jan 9 '18 at 15:13
• you can't apply FFT to DFT data, coz both fft and dft lead to frequency domain and should applied to a signal represented in other basis .
– Issa
Jul 10 '18 at 10:55

The DCT of a sequence $y[n]$ is given by: $$W[k] = \sum_{n=0}^{N-1} y[n] \cos\left[\frac{\pi}{N} \left(n + \frac{1}{2} \right)k \right] \ , \ \ k=0,..,N-1$$

The reason why the DCT domain is in Hz $\left[\frac{1}{s} \right]$ is that the cosine's argument has to be dimensionless. In order to achieve this, $k$ has the necessary unit to cancel out the unit of $n$. In this case, given that $n$ is in the time domain, $k$ will be in the frequency one.

If we call $X[l]$ the result of taking the FFT of $W[k]$ we get: $$X[l] = \sum_{k=0}^{N-1} W[k] e^{\frac{-2\pi j}{N} k l } \ , \ \ l=0,..,N-1$$ Analogously, the unit of $l$ will be the one that makes the argument $\frac{-2\pi j}{N} k l$ dimensionless. Follows from this that the sequence $X[l]$ is again in the time domain.

You can even find a closed expression for the sequence $X[l]$ by recalling the relationship between the DCT and the FFT, and using the property: $$x_n =\mathrm{DFT}\{\mathrm{DFT}\{x_{N-n}\}\} /N$$