I want to express a convolution of a known signal $b(t)=c(t)\ast h(t)$ as an ODE $b'(t)$ instead.
The kernel $h(t)$ takes a double exponential form $Au(t)(1-e^{-T_{on} t})e^{-T_{off} t}$.
After rearranging and doing Laplace transforms, I get a lone $s$ term, the inverse of which I understand would be $\delta'(t)+\delta(0)$. This ends up on the right hand side of $\beta'=b''$:
$$\beta' = \dots - b(0)(\delta'(t)+\delta(0))$$
Is my cause hopeless at this point?
Here is my work in detail:
EDIT:
DinosaurEgg solved it! Looks like I was close, but missing two pieces:
- I got $H(S)$ wrong, by forgetting the minus sign, I got $2s+\lambda+\kappa$ on top instead of $\lambda-\kappa$.
- It didn't occur to me to simply assume $b(0)=b'(0)=0$. This works just fine. In the screenshot below, you can see the ODE solution is a little bit off at the beginning when that assumption is violated (look at $b_\text{conv}/b_\text{ode}$), but quickly corrects itself.
The result is actually quite a bit simpler. To declutter the notation a bit I will set $\kappa=\tau_{\text{off}}^{-1}, \lambda=\tau_{\text{off}}^{-1}+\tau_{\text{on}}^{-1}$ and then one calculate the Laplace transform of the kernel $h(t)$, which is
$$H(s)=\frac{\lambda-\kappa}{(s+\kappa)(s+\lambda)}$$
which after clearing denominators, gives the following relation for $B(s)$
$$(s^2+(\kappa+\lambda)s+\kappa\lambda)B(s)=(\lambda-\kappa)C(s)$$
The reason why we wrote the relation this way is because Laplace transforms multiplied by integer powers of $s$ can be rewritten as Laplace transforms of derivatives of the function. In your case, noting that $b(0)=b'(0)=0$, we find that from standard properties of the Laplace transform that
$$\mathcal{L[b'(t)](s)}=s\mathcal{\mathcal{L}[b(t)](s)}~~,~~\mathcal{L[b''(t)](s)}=s^2\mathcal{\mathcal{L}[b(t)](s)}$$
which allows us to write the relation after undoing the Laplace transform as
$$b''(t)+(\kappa+\lambda)b'(t)+\kappa\lambda b(t)=(\lambda-\kappa)c(t)$$