is there any alternative method for taking second order derivative in signals?

Question

Eg:

1.Z=1+i*5 (in quadratic form),

2.s(t)=Zexp(iw*t), (i'm multiplying two signals)

now, i want to take second order derivative, yes i know the usual way like this, Equation

but is there any other alternative form to find 2nd order derivative?,

and by doing that i want to extract my original signal ("z", which's given above ) does any one have idea about this ?

I tried this way it's worked but i want reduce it more complex.

After double derivative i will calculate instantaneous angular frequency like this W=(mean((y'')^2)/mean((s(t))^2)) and i will get result as I=2*mean[(s(t)*cos(w*t))] and I=-2*mean[(s(t)*sin(w*t))] this both I and Q are our original signal Z=1+i*5

Answer 1

Assuming you have a time series of values $s_k$, there are two ways of approaching this.

1. Analytical method: If you know exactly that your signal is of the form $$ s(t) = Ze^{i \omega t} = Z \cos{\omega t} + iZ \sin{\omega t} $$ Then you can just calculate the derivative: $$ \frac{d^2 s}{dt^2} = -Z\omega^2 e^{i\omega t} = -\omega^2 s(t) $$ Or, in simple terms, you just have to multiply the original signal by $-\omega^2$. Then the second derivative $y$ in every point is $$ y_k = -\omega^2 s_k $$
2. Approximate method: If you don't know that your signal is exactly of the mentioned form, you have to indirectly measure the second time derivative, by calculating it at each point. The most straightforward way is the finite difference method, and for live calculation you have to use the backwards difference. If we set the second derivative to be $y$ and use the first-order accurate backwards difference $$ y_k \approx \frac{1}{\Delta t^2}\left(s_k - 2s_{k-1} + s_{k-2} \right) $$ This method uses only the "history" of the signal ($s_{k-1}$ and $s_{k-2}$). It's rather simple but gives an approximate answer. In this equation, $\Delta t$ is the time step used.