I'm trying to understand the following formula in this paper:

enter image description here

Sadly, the paper does not define most of the symbols it is using, so my guesses so far are:

  • $T_{tr}$ refers to the sampling interval
    • i.e. $\frac{1}{44100} s$
  • $\phi$ refers to phase
    • specifically: $\phi(k_{tr}, (n-1)h)$ refers to the angle of the complex vector representing frequency $k_{tr}$ for the FFT from time point $(n-1)h$ to $nh$
  • so basically this is saying: out of all FFT frequency bins, only include those frequencies $k_{tr}$ where above equation holds true

I tried googling around to see if there's a standard definition of "transient frequency bins" (like there is for FFT frequency bins) this is referencing, but no success. I also don't understand

  • Why would one compare radians $\phi$ with a unit of time $T_{tr}$?
    • Is this maybe a stand-in for 0 to account for sampling errors?
  • Why isn't $T_{tr}$ just called $T$ as it seems to be a property of the signal and unrelated to the transient bins?
    • (If this is not the sampling interval, then how would one obtain $T_{tr}$?)

How does this formula define $K_{tr}$?


1 Answer 1


without reading the paper,

Eqn 4 has the form of a central difference form of the second derivative. It is a numerical second derivative compared to a threshold. It differs from the standard form in that it centers on $n-1$.


Whatever $\phi$ is, it is the positive change compared to a threshold.

  • $\begingroup$ well the second derivative of phase is the rate of change of frequency. the sweep rate. but if these are phases of a frequency component at adjacent sample times, this sweep rate (and second derivative) is radians per sample per sample.and cannot be compared dimensionally to $\frac{1}{44100}$ second. $\endgroup$ Commented Jul 31, 2018 at 5:00

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