# Discrete-time input matrix when one of the eigenvalues of the system matrix is zero

If we have a continuous time state-equation,

$$\dot{x}(t) = A x(t) + B u(t)$$

where $A \in \mathbb{R}^{n \times n}, x \in \mathbb{R}^{n\times1}, B \in \mathbb{R}^{n \times m}, u \in \mathbb{R}^{m \times 1}$, the analytical solution is obtained as

$$x(t) = e^{A t} x(0) + \int_0^t e^{A(t-\tau)}B u(\tau)\, d\tau$$

Converting this to discrete-time, with sampling period $T_s$, and assuming that all eigenvalues of $A$ are non-zero, real & distinct and $u$ remains constant within each sample interval, $k$ to $k+1$, we get the following difference equation,

$$x[k+1] = e^{A T_s} x[k] + \left[\int_0^{T_s} e^{A\tau}B \, d\tau\right] u[k]$$

which can be written as

$$x[k+1] = A_d x[k] + B_d u[k]$$

When the original matrix $A$ is invertible, the integral term corresponding to $B_d$ can be written as

$$B_d = A^{-1}(A_d - I)$$

The question is, What about the case when A is non-invertible? In particular, if

$$A=\begin{bmatrix}a_1 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ and $$B=\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix}$$

then one of the eigenvalues of $A$ is zero. In this case, how do I obtain the corresponding $B_d$?

$$\begin{array}{rl} \mathrm B_d &= \displaystyle\int_{0}^{T_s} \exp(\mathrm A t) \mathrm B \, \mathrm d t\\ &= \displaystyle\int_{0}^{T_s} \left(\sum_{k=0}^{\infty} \frac{t^k}{k!} \mathrm A^k \right)\mathrm B \, \mathrm d t\\ &= \displaystyle \sum_{k=0}^{\infty} \left( \int_{0}^{T_s} \frac{t^k}{k!} \, \mathrm d t \right) \mathrm A^k \mathrm B\\ &= \displaystyle \sum_{k=0}^{\infty} \frac{T_s^{k+1}}{(k+1)!} \mathrm A^k \mathrm B\\ &\approx T_s \left( \mathrm I + \frac{T_s}{2} \mathrm A \right) \mathrm B\end{array}$$

That equivalence is only valid if $A$ is invertible, hence it is a result but not a definition. This is typically done via the following relation

$$\exp\left(\begin{bmatrix}A & B\\0 &0\end{bmatrix}t\right) = \begin{bmatrix}e^{At} &\int e^{At}B\\0&I\end{bmatrix}=\begin{bmatrix}A_d & B_d\\0 &I\end{bmatrix}$$

where exp is the matrix exponential, and $t$ is the sampling period. $C, D$ matrices remain unchanged.

Shameless plug: I've recently finished these conversion tools in my python package. The code for other methods can be found https://github.com/ilayn/harold/blob/master/harold/_discrete_funcs.py#L139.

• Sorry. That answer is absolutely unclear. I just want to know how to compute $A_d$ and $B_d$ for my specific example. Can you expand on your answer ? May 24, 2018 at 9:28
• If you type expm([A B;zeros(m,n+m) in matlab you get B as the upper right block. May 24, 2018 at 9:29
• And how is the sample time $T_s$ influencing the discrete matrices? Simply taking the matrix exponentials will not produce their corresponding discrete-time equivalents May 24, 2018 at 9:31
• @Krishna Ah good catch I've forgot to add that. Give me a second May 24, 2018 at 9:31
• @Krishna I don't understand, the matrix exponential is defined for all matrices, nonsingularity is not required. Either manually obtain the matrix exponential and integrate or you can use the formulation in my answer. May 24, 2018 at 9:47