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.

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  • $\begingroup$ what does y[n] and x[n] denotes in pre emphasis equation ? $\endgroup$ – Naveen Gabriel Jun 14 at 10:00
  • $\begingroup$ @NaveenGabriel: $x[n]$ is the input sequence, and $y[n]$ is the output sequence. $\endgroup$ – Matt L. Jun 14 at 10:25
  • $\begingroup$ Yes, I know that but what is the input sequence ? Is this equation is evaluated in frequency domain? In particular..what is "n" in "x" ? $\endgroup$ – Naveen Gabriel Jun 14 at 10:35
  • $\begingroup$ @NaveenGabriel: $x[n]$ and $y[n]$ are time domain sequences, $n$ is the time index. $\endgroup$ – Matt L. Jun 14 at 10:45
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    $\begingroup$ @NaveenGabriel: yes, that's basically it. $\endgroup$ – Matt L. Jun 14 at 10:57

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