I'm trying to make a process noise matrix in continuous time. But i can't seem to find a clear definition of what "unit" the matrix should contain in continuous time.
From our control book we have $V_{2d}=V_{2c}/T$, where $V_{2}$ is our measurement noise matrix. From the web i found the conversion for discrete time standard deviation to continuous time: $V_{1c}=V_{1d}/Ts$, where $V_1$ is process noise. Should this contain the variance, noise power or standard deviation? However isn't this different from the conversion of $V_2$, why is this?
The Band-Limited White Noise block from simulink/matlab takes in a noise power, $N_p = \sigma^2\cdot Ts$. So the question is... why specifically noise power? Normally power is the energy over time, is this also the case for noise power, if yes, isn't noise energy the same as variance?
If you have some sources for the information about it, i would happily look through it. It is very confusing to work with it without clear definitions.
The reverse, so from continuous time to discrete time, can be achieved using (with zero order hold for $u(t)$)
\begin{align} A_d &= e^{A_c\,T}, \\ B_d &= \int_0^T e^{A_c\,\tau} d\tau\,B_c, \\ C_d &= C_c, \\ D_d &= D_c, \\ W_d &= \int_0^T e^{A_c\,\tau}\,W_c\,e^{A_c^\top\,\tau} d\tau, \\ V_d &= V_c\,T, \end{align}
where the subscripts $c$ and $d$ stand for the continuous and discrete time state space models respectively and $T$ the discretization time step size. The continuous time model would be
\begin{align} \dot{x}(t) &= A_c\,x(t) + B_c\,u(t) + w(t), \quad w(t) \sim \mathcal{N}(0,W_c), \\ y(t) &= C_c\,x(t) + D_c\,u(t) + v(t), \quad \ \ v(t) \sim \mathcal{N}(0,V_c), \end{align}
such that the discrete time model becomes
\begin{align} x_{k+1} &= A_d\,x_k + B_d\,u_k + w_k, \quad w_k \sim \mathcal{N}(0,W_d), \\ y_k &= C_d\,x_k + D_d\,u_k + v_k, \quad \ \ v_k \sim \mathcal{N}(0,V_d). \end{align}
When all matrices are constant during the discretization time step the expression for $W_d$ can be simplified to the following impliciet equation
$$ A_c\,W_d + W_d\,A_c^\top = A_d\,W_c\,A_d^\top - W_c. $$
So if $A_d$, $A_c$ and $W_d$ are known the above equation is just a discrete Lyapunov equation in $W_c$.