Pre-emphasis: What is connection between parameter a and 6db/oct? I know that the standard pre-emphasis function is $y[n] = x[n]-\alpha\cdot x[n-1]$ and $\alpha$ is often set as $0.97$.

However, when I was learning speech processing using Praat, the pre-emphasis factor in it is $6\ \rm dB/oct$ which means the signal emphasise $6\ \rm dB$ per octave increase. That's what confused me.

I've read lots of questions and answers, most of them are explaining how to calculate the value of $\alpha$ or the $\mathcal Z$-transformation. But What I'm confused is the connection between $\alpha$ and $6\ \rm dB/oct$.

I found this equation in the notes of Praat:


  The pre-emphasis factor $\alpha$ is computed as:
  $$
\alpha = \exp(-2\pi F \Delta t)
$$
  where the $F$ above which the spectral slope will increase by $6\ \rm dB/oct$, $\Delta t$ is the sampling period of the sound


According to above notes, it seems that the $\alpha$ only influences the "beginning frequency" of pre-emphasis but not "how much it changes".

Hope I explained myself clearly enough.
You're right that the value $\alpha$ will not influence the slope of the filter (approx. $6$ dB/octave). That slope is determined by the filter order, which is $1$ in this case (1 delay element in the filter implementation). The value $\alpha$ only influences the filter's cut-off frequency.