Is there mathematical relationship between the FFT and DCT transforms

I need to model the relationship between the DCT and DFT transforms (If it exists). I mean real signal $$x \to y = \textrm{DCT}(x) \to z = \textrm{DFT}(y)$$, so I need to get the relationship between the $$x$$ and $$z$$ if possible.

For more details:

Assume I have a real signal $$x[m],\ m=1,2,….N$$, the $$N$$-point IDCT transform of the signal $$x[m]$$ is $$X[n]$$ which can be written as follows (let's ignore the coefficients for simplicity):

$$X[n] = \sum_{m=1}^N x[m] \cos\left( \frac{(2m+1)n\pi}{2N} \right)$$

I need to get the relationship between the FFT of $$X[n]\$$ and the signal $$x[m]$$, so taking the $$2N-$$point DFT for the signal $$X[n]\$$ (I upsampled the signal to have the DFT of each point resulted from the IDCT), that will give:

$$Y[v] = \sum_{n=1}^{2N} X[n] e^{-\frac{j2\pi mn}{2N}} = \sum_{n=1}^{2N} \left[ \sum_{m=1}^N x[m] \cos\left( \frac{(2m+1)n\pi}{2N} \right) \right] e^{-\frac{j2\pi mv}{2N}}$$

with some mathematical operations, we can get:

$$Y[v] = \sum_{n=1}^{2N} \left( \left[ \sum_{m=1}^N x[m] \cos\left( \frac{(2m+1)n\pi}{2N} \right) \right] \cos\left(\frac{2\pi mv}{2N}\right) - j\left[\sum_{m=1}^N x[m] \cos\left( \frac{(2m+1)n\pi}{2N} \right) \right] \sin\left(\frac{2\pi mv}{2N}\right) \right)$$

I am trying to get the relationship between the $$Y[v]$$ and $$x[m]$$ from the relationship above if it exists. How can we get it?

• Try to derive it in terms of matrices
– I.M.
May 23 at 0:18
• I have transformed your equations into latex. Please check if there is an error. I noticed that in your two last equations, your first sum must use $n$ index. May 23 at 8:22
• @AlexTP, Yes, they are right now. Thank you for correcting it. May 23 at 8:24
• Hint for your question: DCT is a DFT of signals with some symmetry characteristics. For the general case, the matrix forms will help. May 23 at 8:27
• @AlexTP I am trying to formulate it in terms of matrices, but I also couldn't get a logic expression for that. May 23 at 8:29

You have some confusion in the indices, maybe the correct expression would be

$$Y[k] = \sum_{n=1}^N \left[ \sum_{m=1}^N x[m] \cos\left( \frac{(2m+1)n\pi}{2N} \right) \right] e^{-\frac{j2\pi k n}{N}}$$

The DCT is the real part of the odd-indices coefficients of double length DFT

Expressing DCT in terms of complex harmonics

$$z[n] = \sum_{m=1}^N x[m] \left( \exp\left( j \frac{(2m+1)n\pi}{2N} \right) + \exp\left( -j \frac{(2m+1)n\pi}{2N} \right) \right)$$

If you take $$X(n)$$ being the fourier transform of $$x$$, in the reals instead of the integers, you could write $$z[n] = X(m+1/2)+X(-m-1/2)$$, $$X(n)$$ has Hermitian symmetry and the result agrees with the fact that $$z[n]$$ is real.

I think that the most efficient way to express the relation between $$Y[m]$$ and $$x[m]$$, is that $$Y = DFT(DCT(x))$$, unless you want to upsample your signal.

The denominator $$2N$$ in the exponential, the summation is from $$1$$ up to $$N$$, and the numerators may be even, so cannot simplify the exponent, the way to go is to compute a $$2N$$-samples, DFT.

If you create an intermediate signal with odd samples $$v_{2i+1} = x_i + x_{N - i}$$ and even samples $$v_{2i}=0$$, and compute $$V = DFT^{-1}(v)$$, then the first $$N$$ elements of $$V$$ equals to $$z$$, and the $$DFT(V) = DFT(DFT^{-1}(v)) = v$$

• What's about if we upsampled the signal, it means we take DFT with $2N$ points, right? in that case, can we get a relationship between $z[n]$ and $x[m]$? May 23 at 14:01
• You are right, I must up-sample the signal, and what I wanted to put in the question is to get the DFT with $2N-$point DFT. But, even in that case, I am not sure if I can get the relationship between $z[n]$ and $x[m]$. Additionally, $x[m]$ is always a real signal May 23 at 14:09
• I tried also to add that notice into the question. . May 23 at 14:21
• $z_{i}=V_{i}$ for $i \le N$ and since $v$ is real $V$ has Hermitian symmetry, i.e. $V_i = conj(V_{2N-i})$
– Bob
May 23 at 19:46