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

Question

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?

Answer 1

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.