Assume we have a time domain signal: s(i), i=1:N.

To apply a derivative filter to it, i.e. D*s, where * denotes the convolution, we can simply use the finite difference to approximate it as:

w(i) = ( s(i+1)-s(i) ) / dt;

where w is the signal after time derivative. My question is how to go back to s, i.e. apply an integral filter to w, i.e. I*w? Can I do it like this?

for i=2:N
    s(i) = s(i-1) + w(i)*dt;

Mathematically, your system could work, up to one point. A direct parallel can be made with continuous time derivatives.

In your case, to perfectly reconstruct your signal, you would need to also store s(1), because in the nonzero case, there will be an offset of s(1) between the original and reconstructed signal.

Solving ODEs in continuous time is exactly the same, you need an initial/boundary condition, otherwise you'll have an infinite amount of solutions that solve your ODE.

Another way to see the relation to continuous time: $\int \frac{dx(t)}{dt}dt = x(t)+C$, $C$ being the offset you have to determine by an initial condition.


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