Computing a DCT using an FFT, Signal analysis 
Computing a DCT using an FFT, Signal analysis 
Dec 9 2005, 06:40
Post
#1


Group: Members Posts: 28 Joined: 15April 04 From: Australia Member No.: 13509 
I am trying to make sense of the output of the FFT when used to calculate a type2 DCT (DCTII). I managed to scrape together enough documentation that infomed me that, if I wanted to perform a DCTII using an FFT algorithm, then the data should be arranged to be real even symmetric, with every even indexed data element set to zero.
No problem there. In addition, the outputs are correct, but half the expected value (I can see the output needs scaling by 2 to compensate for the fact that every second sample of the input is zero). However, I am tyring to figure out how the output comes to be what it is? In particular, where the zero comes from (term 4 & term 12) [counting from 0]. Input Real : 0 x0 0 x1 0 x2 0 x3 0 x3 0 x2 0 x1 0 x0 Imag : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Output Real : ya yb yc yd 0 yd yc yb ya yb yc yd 0 yd yc yb Imag : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [grumble: looks good in fixed width font, but spaces are not turning out here] In my real application, I am using a 16 point DCT, which equates to a 64 point FFT, but the above is representative. Any hints? PS: I know there are better ways of doing DCTs, but in my implementation, I only have access to an FFT routine. Cheers,Owen. 


Dec 12 2005, 01:13
Post
#2


Group: Developer Posts: 1245 Joined: 16December 02 From: Australia Member No.: 4097 
I agree that there is often not enough documentation on how to do a simple DCT using an FFT. Though it is not the best, computationalwise, it is however interesting to see the relationship between the two, esp. why the DCT is used more often than the FFT in compression (the evensymmetry creates smoothness that results in less high frequency energy).
Anyway, I was interested in this problem a while ago as well and found this link to be the best I've found. Not sure if this is any help or not, but here is some MATLAB code I wrote during that time, which calculates the DCT using FFT. CODE clear all; close all; x=[1 2 3 4]; N=length(x); d = dct(x); x2=[4 3 2 1 1 2 3 4]; f=fft(x2); n = 0:2*N1; shift=exp(j*2*pi*(N0.5)*n/(2*N)); f2 = real(f./shift); d2 = f2(1:N)/sqrt(2*N); d2(1)=d2(1)/sqrt(2); disp(d); disp(d2); Notice how I don't do any zerosetting (no idea why you need to anyway). I just take my original signal, mirror it and then take the fft. Now the thing to note is that the Fourier transform of an even symmetry signal is real but we can't make true even symmetry signals on a computer, so we have to compensate for the phase shift caused by the timeshift of 3.5 samples, hence the division by the FFT of the variable 'shift'. Then I scale the coefficients appropriately (to make them orthonormal), and voila. Hope that helps in a little way EDIT: fixed a spelling mistake This post has been edited by QuantumKnot: Dec 12 2005, 04:55 


Dec 13 2005, 23:25
Post
#3


Group: Members Posts: 28 Joined: 15April 04 From: Australia Member No.: 13509 
Hi QuantumKnot,
Thanks for your reply. Indeed I had read the same page on using an FFT to do a DCT, and understood the reasoning behind having to mirror the coefficents before placing them into a FFT. I did a bit of experimenting, and created a "brute force" DCT using floating point maths on a PC. I had some code for a DCTII, and rewrote it to perform a DCTI. The results were quite different. I also examined the documentation on FFTW, in particular about the different types of DCT. They refer to them as 00, 01, 10, and 11. The 0 stands for no shift, the 1 stands for a halfsample shift. The first figure refers to the input, the last to the output. The FFTW routines REDFT00 = DCTI, REDFT10 = DCTII, REDFT01 = DCTIII, REDFT11 = DCTIV. Apparently, filling an FFT without padding but with mirroring (as you have done) is equivilant to a DCTI. Filling with padding and mirroring (as described in my first post) is supposed to be equivilant to a DCTII (which is what I need). However, this takes the FFT to 4n. Further documentation (wikipedia in fact) hints at this "For example, a typeII DCT is equivalent to a DFT of size 4n with realeven symmetry whose evenindexed elements are zero. One of the most common methods for computing this via an FFT (e.g. the method used in FFTPACK and FFTW)..." It goes on a bit further to say computations can be shorthened to O(nlogn) + O(n) if a preprocessing step of radix4 is used (O(n) steps) and the FFT is then reduced to size n. A decent saving. As I said before, my application is power impoverished hardware, where all I have is a 16bit integer FFT routine (which works surprisingly well I must say). Additionally BinDCT may be very fast, but it is also considerably less accurate. Again, thanks for your reply. A cursory look at your sample code has been enlightening, and I am about to print it and examine it in detail. Cheers and thanks, Owen. 


Dec 14 2005, 00:08
Post
#4


Group: Developer Posts: 1245 Joined: 16December 02 From: Australia Member No.: 4097 
QUOTE (Mustardman @ Dec 14 2005, 08:25 AM) Apparently, filling an FFT without padding but with mirroring (as you have done) is equivilant to a DCTI. That is weird since I've also compared the results of the above MATLAB code with FFTW's REDFT10 r2r mode and got the same results (with the appropriate scaling). If you check FFTW's user manual (p. 11), it says that a DCTII is symmetric and even around j = 0.5, which is exactly how I'm mirroring it above. That is, a sequence of abcd is mirrored as dcbaabcd, where if we align the right 'a' as 0, then the axis of symmetry (represented as a '') is at 0.5. The DCTI is symmetric around j = 0, so the sequence abcd is mirrored as bcdabcd, where the axis of symmetry is on the 'a'. This post has been edited by QuantumKnot: Dec 14 2005, 00:10 


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