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

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...

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.