I'm trying to understand the derivation of the zero order hold discretization method, and I have a couple of questions about some of the steps.
I think I understand the first part, this is just the solution to a system of linear equations using the matrix exponential and an integrating factor: $$ \begin{split} \dot{x}(t) &= A x(t) + B u(t) \\ \dot{x}(t) - A x(t) &= B u(t) \\ e^{-At} \dot{x}(t) - e^{-At} A x(t) &= e^{-At} B u(t) \\ \frac{d}{dt}\left(e^{-At} x(t)\right) &= e^{-At} B u(t) \\ e^{-At} x(t) &= \int_{-\infty}^{t} e^{-A\tau} B u(\tau) d\tau + c \\ x(t) &= e^{At}c + \int_{-\infty}^{t} e^{A(t-\tau)} B u(\tau) d\tau \end{split} $$
Next, the integration constant is determined using the initial condition $ x(0) $: $$ x(0) = e^{0}c + \int_{-\infty}^{0} e^{-A\tau} B u(\tau) d\tau $$ Now, I understand that for finite, unilateral signals, this simplifies to $ c = x(0) $, however, what happens if $ u(t) $ is a Dirac delta? Why would the integral be zero in that case?
Then the state is sampled at $ (k+1) T_s $: $$ \begin{split} x((k+1) T_s) &= e^{A (k+1) T_s} x(0) + \int_{-\infty}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} B u(\tau) d\tau \\ &= e^{A T_s} \left(e^{A k T_s} x(0) + \int_{-\infty}^{k T_s} e^{A(k T_s-\tau)} B u(\tau) d\tau \right) + \int_{k T_s}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} B u(\tau) d\tau \\ &= e^{A T_s} x(k T_s) + \int_{k T_s}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} B u(\tau) d\tau \\ \end{split} $$ Since the input is the result of a ZOH operation with the same sample time $ T_s $, the input is constant over a single time step: $ \forall t \in [k T_s, (k + 1) T_s): u(t) = u(k T_s) $, therefore, $ u(\tau) $ can be moved outside of the integral: $$ x((k+1)T_s) = e^{A T_s} x(k T_s) + \int_{k T_s}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} d\tau B u(k T_s) $$ $$ x[k+1] = e^{A T_s} x[k] + \int_{k T_s}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} d\tau B u[k] $$
The discrete system matrices then are: $$ \begin{split} A_d &= e^{A T_s} \\ B_d &= \int_{k T_s}^{(k+1) T_s} e^{A((k+1) T_s-\tau)} d\tau B \end{split} $$ My second question is if it's well-defined to solve this integral using substitution. The result seems to be correct, but it feels strange that the integration boundaries and the variable of integration suddenly become matrices themselves. Also, there seems to be something funny going on with the commutativity of the factor $ A^{-1} $ and the integral when substituting $ d\tau $.
$$ \begin{split} V &= A((k+1) T_s-\tau) \\ dV &= -A d\tau \Leftrightarrow d\tau = -A^{-1} dV \\ B_d &= -A^{-1} \int_{A T_s}^{0} e^V dV B \\ &= A^{-1} \left[e^V\right]_{0}^{A T_s} B \\ &= A^{-1} \left(e^{A T_s} - I\right) B \end{split} $$
If there are any other holes or mistakes in my reasoning, feel free to let me know.
