Why is taking the inverse Fourier transform the same as taking the Fourier transform when computing the cepstrum?

Question

Usually, the cepstrum of a signal is introduced as the result of taking the logarithm of its power spectrum, then applying the inverse Fourier transform:

$$C_p = |F^{-1}\{\log(|F\{f(t)\}|^2)\}|^2$$

It is also sometimes called a "spectrum of a spectrum" since it can also be defined as:

$$C_p = |F\{\log(|F\{f(t)\}|^2)\}|^2$$

I'm really having trouble understanding how the two are equivalent (up to a scaling factor), and this is I get the impression quite important to understanding why in other processing steps (e.g. mel frequency cepstrum) a forward transform is taken at the end instead of an inverse transform...

Answer 1

The only difference between these two formulas is $\mathcal{F}$ and $\mathcal{F}^{-1}$, so let's take a look at the definitions of DFT and IDFT (we'll see this property in discrete domain)

$$ X[k] = \sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn} $$

$$ x[n] = \frac{1}{N} \sum_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn} $$

In our common sense $x[n]$ is a time sequence and $X[k]$ is a frequency sequence, however you can also treate DFT and IDFT as two transforms that can be applied to any sequence, in this case, $\log(|\mathcal{F}\{x[n]\}|^2)$, let's denote it as $f[m]$. The DFT and IDFT of $f[m]$ are respectively

$$ \mathcal{F}\{f[m]\} = \sum_{m=0}^{N-1}f[m]e^{-j\frac{2\pi}{N}km} $$

$$ \mathcal{F}^{-1}\{f[m]\} = \frac{1}{N} \sum_{m=0}^{N-1}f[m]e^{j\frac{2\pi}{N}mn} $$

There are two main differences, a scaling factor $1/N$ and a conjugation. Since $f[m]$ is a real sequence, we can derive that $$ \mathcal{F}\{f[m]\} = N \big[\mathcal{F}^{-1}\{f[m]\}\big]^* $$

Finally take the norm of these two sequences and we get two equivalent sequences with a scaling factor.