Computing the discrete-time difference equation when the system matrix is non-invertible
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If we have a continuous time equation,
$$ dot{x}(t) = A x(t) + B u(t)$$
where $A in mathbb{R}^{n times n}, x in mathbb{R}^{ntimes1}, B in mathbb{R}^{n times m}, u in mathbb{R}^{m times 1}$, the analytical solution is obtained as
$$x(t) = e^{A t} x(0) + int_0^t e^{A(t-tau)}B u(tau), dtau$$
Converting this to discrete-time, with sample-time $T_s$, and assuming that all eigen-values of $A$ are real & distinct and $u$ remains constant within each sample interval, $k$ to $k+1$, we get the following difference equation,
$$x[k+1] = e^{A T_s} x[k] + left[int_0^t e^{Atau}B , dtauright] u[k]$$
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 my $A=begin{bmatrix}a_1 & 0 & 0 \ 0 & a_2 & 0 \ 0 & 0 & 0 end{bmatrix}$ and $B=begin{bmatrix}b_1 \ b_2 \ b_3 end{bmatrix}$. How do I obtain the corresponding $B_d$?
differential-equations dynamical-systems matrix-equations
asked May 23 at 19:51
Krishna
1438
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up vote
2
down vote
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If we have a continuous time equation,
$$ dot{x}(t) = A x(t) + B u(t)$$
where $A in mathbb{R}^{n times n}, x in mathbb{R}^{ntimes1}, B in mathbb{R}^{n times m}, u in mathbb{R}^{m times 1}$, the analytical solution is obtained as
$$x(t) = e^{A t} x(0) + int_0^t e^{A(t-tau)}B u(tau), dtau$$
Converting this to discrete-time, with sample-time $T_s$, and assuming that all eigen-values of $A$ are real & distinct and $u$ remains constant within each sample interval, $k$ to $k+1$, we get the following difference equation,
$$x[k+1] = e^{A T_s} x[k] + left[int_0^t e^{Atau}B , dtauright] u[k]$$
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 my $A=begin{bmatrix}a_1 & 0 & 0 \ 0 & a_2 & 0 \ 0 & 0 & 0 end{bmatrix}$ and $B=begin{bmatrix}b_1 \ b_2 \ b_3 end{bmatrix}$. How do I obtain the corresponding $B_d$?
differential-equations dynamical-systems matrix-equations
asked May 23 at 19:51
Krishna
1438
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If we have a continuous time equation,
$$ dot{x}(t) = A x(t) + B u(t)$$
where $A in mathbb{R}^{n times n}, x in mathbb{R}^{ntimes1}, B in mathbb{R}^{n times m}, u in mathbb{R}^{m times 1}$, the analytical solution is obtained as
$$x(t) = e^{A t} x(0) + int_0^t e^{A(t-tau)}B u(tau), dtau$$
Converting this to discrete-time, with sample-time $T_s$, and assuming that all eigen-values of $A$ are real & distinct and $u$ remains constant within each sample interval, $k$ to $k+1$, we get the following difference equation,
$$x[k+1] = e^{A T_s} x[k] + left[int_0^t e^{Atau}B , dtauright] u[k]$$
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 my $A=begin{bmatrix}a_1 & 0 & 0 \ 0 & a_2 & 0 \ 0 & 0 & 0 end{bmatrix}$ and $B=begin{bmatrix}b_1 \ b_2 \ b_3 end{bmatrix}$. How do I obtain the corresponding $B_d$?
differential-equations dynamical-systems matrix-equations
asked May 23 at 19:51
Krishna
1438
If we have a continuous time equation,
$$ dot{x}(t) = A x(t) + B u(t)$$
where $A in mathbb{R}^{n times n}, x in mathbb{R}^{ntimes1}, B in mathbb{R}^{n times m}, u in mathbb{R}^{m times 1}$, the analytical solution is obtained as
$$x(t) = e^{A t} x(0) + int_0^t e^{A(t-tau)}B u(tau), dtau$$
Converting this to discrete-time, with sample-time $T_s$, and assuming that all eigen-values of $A$ are real & distinct and $u$ remains constant within each sample interval, $k$ to $k+1$, we get the following difference equation,
$$x[k+1] = e^{A T_s} x[k] + left[int_0^t e^{Atau}B , dtauright] u[k]$$
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 my $A=begin{bmatrix}a_1 & 0 & 0 \ 0 & a_2 & 0 \ 0 & 0 & 0 end{bmatrix}$ and $B=begin{bmatrix}b_1 \ b_2 \ b_3 end{bmatrix}$. How do I obtain the corresponding $B_d$?
differential-equations dynamical-systems matrix-equations
differential-equations dynamical-systems matrix-equations
asked May 23 at 19:51
Krishna
1438
asked May 23 at 19:51
Krishna
1438
asked May 23 at 19:51
Krishna
1438
asked May 23 at 19:51
Krishna
1438
asked May 23 at 19:51
Krishna
1438
1438
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