# Understanding the two different forms of the Cepstrum

I'm learning about Cepstrum and I see that there are two different forms to calculate it:

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

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

In Wikipedia, it is mentioned that these two are the same, with differing only in a scaling factor. How could FT and inverse FT be the same? And I'm not asking for proof. Rather an intuition to understand how is this possible. Especially, since it says that "The frequency is a measure of time". I mean if I consider the first formula, I understand how Cepstrum is in time but if it is a spectrum of the spectrum (the second formula), then I don't see how it could be in the time domain!

• Feb 23 at 23:22
• @lennon310, "quefrency" is actually correct (reference).
– Ash
Feb 24 at 0:38

An interesting way to think about Fourier transforms and their inverses is as rotations in time-frequency space. The Fractional Fourier transform (FrFT) is a generalization of the Fourier transform where you can transform a signal into a partial-time/partial-frequency representation. Take the example sourced from the Wikipedia page:

When the parameter $$\alpha=\pi/2$$, the FrFT simplifies to the Fourier transform (bottom right picture), demonstrating the transformation from a purely time-domain to a purely frequency-domain. If $$\alpha=-\pi/2$$, this is equivalent to the inverse Fourier transform. Taking either of these operations twice results in a rotation of $$\alpha=\pi$$ which would correspond to a time-reversed time-domain signal. The reason they say "quefrency is a measure of time" is because the resulting rotation of the Fourier transforms (and inverses) of both equations end up with $$\alpha=0$$ or $$\alpha=\pi$$.

From the images above, its clear that the Fourier transform and its inverse are simply 180 degree rotations of each other and would simply have mirrored symmetry about the origin. The outer squared absolute value of both formulas would find the magnitude of these spectra (cepstra?) to be equivalent for real-valued inputs, which $$\log(|F(f(t))|^2)$$ always is real. The difference in scale factor only exists if the Fourier transform is not a unitary transform.

How could FT and inverse FT be the same?

Because they are almost identical. In fact you can implement an inverse FT as with a forward FT just by conjugating, i.e.

$$\mathscr{F}^{-1}(f(x)) = \mathscr{F}(f^{*}(x))$$

The forward FT expresses a frequency domain signal as a weighted sum time domain complex exponentials. The inverse FT expresses a time domain signal as a weighted sum of frequency domain exponentials. The math behind each one is essentially the same.

Since the log magnitude is real, the conjugation doesn't do anything so both the forward or inverse transform will yield the exact same result.

I mean if I consider the first formula, I understand how Cepstrum is in time but if it is a spectrum of the spectrum (the second formula), then I don't see how it could be in the time domain!

The Cepstrum is neither in time nor in frequency: it has it's own domain typically called "quefrency" and that's the most useful way to look at it.