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Why does the derivative of an audio file act like a high-pass filter?

Yes both the time derivative $d/dt$ and the discrete time difference $x[n]-x[n-1]$ are indeed "high pass filters": consider the extreme case of the lowest frequency, which is DC or a ...
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10 votes

Why does the derivative of an audio file act like a high-pass filter?

$\frac{d}{dt}\sin(\omega t) = \omega\cos(\omega t)$ as a consequence of the chain rule. $\sin$ and $\cos$ are the same apart from the phase shift, so the derivative is a filter with a response that's ...
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6 votes

Why does the derivative of an audio file act like a high-pass filter?

@OverLordGoldDragon Your x(n)−x(n−1) expression is called a "first-difference discrete differentiator". Its frequency magnitude response is the blue dashed curve below:
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Confusion regarding frequency spectrum in MATLAB

The $fs$ you are getting from audioread is the sampling rate and has nothing to do with the actual frequency content of the signal you are analyzing other than the maximum frequency would be $fs/2$. ...
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3 votes

Why does the derivative of an audio file act like a high-pass filter?

You're doing $x(n) - x(n - 1)$, the finite difference / "discrete derivative", not to be confused with $\frac{d}{dt}x(t)$. The frequency response is where $H(\omega) = 1 - e^{-j 1\omega}$ ...
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1 vote

Why does the derivative of an audio file act like a high-pass filter?

In addition to the excellent existing answers we can also consider the continuous domain: If we have a signal as a fourier series (here in sine-cosine form) $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \...
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