crazydspguy wrote:

> I have the following question regarding obtaining the DFT of a signal
> directly from its DCT representation:

> Say we are given the N-point DCT coefficients Xdct(k) of some real-valued
> signal x(n), but what we really want is the N-point DFT of x(n), Xdft(k).

There might be.  I would read the description in Numerical Recipes, 
which describes the differences, especially in boundary conditions,
between DFT and DCT.

> I need to obtain the power spectrum of the signal x(n) in the DFT domain,
> but I only have the DCT domain information.

For real source data and 2N input values, DFT gives you N sine 
(imaginary) terms and N (real) cosine terms.  DCT gives you 2N cosine 
terms.  I haven't gone through the math, but there might be some trig. 
identities between those terms.

Note, though, that there are some that don't exist in one or the other,
that is, they are fundamentally different.  The DCT boundary condition 
is that the derivative is zero at the ends, the DST has the function 
going to zero at the ends, where the DFT is a compromise between the 
two, with periodic boundary conditions.

You can go from a 2N point DCT to a 4N point DFT, where half the terms 
are zero, and consider that a DFT domain power spectrum.  The terms
that are zero are those that are not allowed by the DCT boundary 
conditions.  If they were supposed to be in the power spectrum, were 
lost by the DCT operation.

-- glen

Hello,

I have the following question regarding obtaining the DFT of a signal
directly from its DCT representation:

Say we are given the N-point DCT coefficients Xdct(k) of some real-valued
signal x(n), but what we really want is the N-point DFT of x(n), Xdft(k).


I need to obtain the power spectrum of the signal x(n) in the DFT domain,
but I only have the DCT domain information.

One method would be to take the inverse DCT and then perform the DFT, but
I am trying to avoid doing such a computation. Is there a direct method
for DCT to DFT transformation?

Any suggestions will be most appreciated!

Thanks,
crazydspguy