# Standardizing amplitude (volume) of an audio signal at a certain frequency before & after taking the derivative of that signal?

I would like to take the derivative of an audio signal, which as I understand creates a 6 dB/oct upward sloping filter as per this thread: What phase rotation occurs when you take the derivative of an audio signal?

Simple Fourier Transform propery for differentiation is $$\mathscr{F}\Big\{\frac{\partial}{\partial t}x(t)\Big\}=j\omega \cdot\mathscr{F}\Big\{x(t)\Big\}$$ Differentiation in time corresponds to multiplication with $$j\omega$$ in frequency. Hence the +6dB/octave slope. The phase shift is a constant 90 degrees for all frequencies.

However, if you want the amplitude to be matched before/after this process at a given frequency, eg. 80 Hz say, how can one do so?

ie. How does one calculate the gain multiplier value to equalize the amplitude at a given frequency from before/after such differentiation?

• Well, you know that the amplitude of the derivative is $\omega$ times larger than the value of your signal… All you have to do then is scale by $\frac{1}{\omega}$. This is a quite well known issue in the Array Processing field where Differential Arrays suffer from high noise amplification at the low frequencies due to this very problem (the fact that you have to boost significantly to equalise the amplitude across the spectrum). Commented Sep 5 at 10:10

Let $$x(t)$$ be a sine wave of amplitude $$A$$ and frequency $$f_0$$

$$x(t) = A_x\cos(2\pi f_0 t)$$

The derivative is

$$y(t) = \frac{\partial}{\partial t}x(t) = 2\pi \cdot f_0 \cdot A_x\sin(2\pi f_0 t) = A_y\sin(2\pi f_0 t)$$

If we want a scaled version of the derivative to have the same amplitude, i.e. $$g \cdot A_y = A_x$$ we simply get

$$g = \frac{A_x}{A_y} = \frac{1}{2\pi f_0 }$$

Note that physically this is an ill posed problem. The signal and it's derivative have different units. If the signal is in Volts the derivative will be in Volts per second. Comparing amplitudes in different units is non-sensical so they CANNOT have the same amplitude.

• I might just point out that, physically, a device that acts as a differentiator will have an RC time constant in the mix that will scale the differentiated signal and bring it back dimensionally to have the same units as the input signal and then the amplitudes can be compared. If done digitally, the sample rate will get in there. Commented Sep 5 at 15:08