# Conversion between DCT to DFT frequency domains

It is well known that the DFT can be interpreted as a filter bank. The same can be done for the DCT, for instance Fig. 3.29, pag. 129 in  shows a plot of the frequency response of a DCT filter bank. I am wondering how this plot (and similar plots for other transforms like DST, etc.) was obtained. For the DFT, we have the prototype filter $H_0(z)=\sum_{n=0}^{N-1}z^{-n}$. The frequency response of the $k$-th DFT bin is $H_k(z)=H_0(z\, e^{-j\frac{2\pi}{N}k})$. Using the transformation $z=e^{j\omega}$ we can readily find the frequency response for all bins (see the plot below). As far as I know there is no z-transform equivalent for the DCT, so it's not clear to me how that plot was obtained. Any ideas?

 Malvar, H.S., Signal processing with lapped transforms, 1992, Artech House

• Not sure if I understand your confusion, but as far as I can see, it is simply the prototype filter and all its cosine-modulated versions. The prototype is rectangular in the time domain, i.e. a sinc function in the frequency domain. – Matt L. May 10 '13 at 9:32

## 2 Answers

Both are linear transforms. Both can be expressed as a matrix. These matrices are invertible. So you can have the product of one matrix with the inverse of the other. In practice, you could not actually calculate inverse matrices, so you'd have to stick with solvers. However, such products are a little op-art. Coursey GNU octave. A DCT bin filter's response depends on both the frequency and phase of the input. Thus a frequency response alone in informationally incomplete. Adjacent DCT bins overlap in frequency, but with differing phase responses at the overlap frequencies. Thus it may take 2 DCT bins to convey or respond to both the frequency and phase of any input sinusoid.

The plot may be for only one phase of input per bin. If the phase is fixed for any given frequency, then there can be a response that strictly depends on frequency.

• I agree that the plot is incomplete as it only shows the magnitude and not the phase of the transfer function. My question however is different. – Arrigo May 10 '13 at 0:41