Closed form solution for Double Integrators dynamics
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I have a relatively specific problem. Consider a linear system, $dot{x}(t) = Ax(t)+Bu(t)$, where $A in mathbb{R}^{n times n}$ and $B in mathbb{R}^{n times 1}$ are the constant matrix and constant vector that defines the system.
The closed-form for a linear system with a constant control input $u(t) := u_k, forall tin [0, T]$ has the following form:
$x(t) = e^{At}x(0) + int_{0}^{T} e^{A(t-tau)} dtau Bu$. If $A$ is non-singular (invertible), then $int_{0}^{T} e^{A(t-tau)} dtau $ can be rewrite to $A^{-1}(e^{AT}-I)$, where $I$ is an identity matrix with appropriate dimension. At this point, I can rewrite $x(t) = e^{At}x(0) + A^{-1}(e^{AT}-I) Bu$, which does not have the integral.
But a lot of systems in real-life have a singular matrix $A$ for its dynamics ($A$ is not invertible ), such as a double integrators with $A = [0,1;0,0]$. In this case, are there any alternatives for me to write its closed-form solution that does not contain integral?
linear-algebra control-theory
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
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add a comment |
$begingroup$
I have a relatively specific problem. Consider a linear system, $dot{x}(t) = Ax(t)+Bu(t)$, where $A in mathbb{R}^{n times n}$ and $B in mathbb{R}^{n times 1}$ are the constant matrix and constant vector that defines the system.
The closed-form for a linear system with a constant control input $u(t) := u_k, forall tin [0, T]$ has the following form:
$x(t) = e^{At}x(0) + int_{0}^{T} e^{A(t-tau)} dtau Bu$. If $A$ is non-singular (invertible), then $int_{0}^{T} e^{A(t-tau)} dtau $ can be rewrite to $A^{-1}(e^{AT}-I)$, where $I$ is an identity matrix with appropriate dimension. At this point, I can rewrite $x(t) = e^{At}x(0) + A^{-1}(e^{AT}-I) Bu$, which does not have the integral.
But a lot of systems in real-life have a singular matrix $A$ for its dynamics ($A$ is not invertible ), such as a double integrators with $A = [0,1;0,0]$. In this case, are there any alternatives for me to write its closed-form solution that does not contain integral?
linear-algebra control-theory
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
$endgroup$
add a comment |
$begingroup$
I have a relatively specific problem. Consider a linear system, $dot{x}(t) = Ax(t)+Bu(t)$, where $A in mathbb{R}^{n times n}$ and $B in mathbb{R}^{n times 1}$ are the constant matrix and constant vector that defines the system.
The closed-form for a linear system with a constant control input $u(t) := u_k, forall tin [0, T]$ has the following form:
$x(t) = e^{At}x(0) + int_{0}^{T} e^{A(t-tau)} dtau Bu$. If $A$ is non-singular (invertible), then $int_{0}^{T} e^{A(t-tau)} dtau $ can be rewrite to $A^{-1}(e^{AT}-I)$, where $I$ is an identity matrix with appropriate dimension. At this point, I can rewrite $x(t) = e^{At}x(0) + A^{-1}(e^{AT}-I) Bu$, which does not have the integral.
But a lot of systems in real-life have a singular matrix $A$ for its dynamics ($A$ is not invertible ), such as a double integrators with $A = [0,1;0,0]$. In this case, are there any alternatives for me to write its closed-form solution that does not contain integral?
linear-algebra control-theory
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
$endgroup$
I have a relatively specific problem. Consider a linear system, $dot{x}(t) = Ax(t)+Bu(t)$, where $A in mathbb{R}^{n times n}$ and $B in mathbb{R}^{n times 1}$ are the constant matrix and constant vector that defines the system.
The closed-form for a linear system with a constant control input $u(t) := u_k, forall tin [0, T]$ has the following form:
$x(t) = e^{At}x(0) + int_{0}^{T} e^{A(t-tau)} dtau Bu$. If $A$ is non-singular (invertible), then $int_{0}^{T} e^{A(t-tau)} dtau $ can be rewrite to $A^{-1}(e^{AT}-I)$, where $I$ is an identity matrix with appropriate dimension. At this point, I can rewrite $x(t) = e^{At}x(0) + A^{-1}(e^{AT}-I) Bu$, which does not have the integral.
But a lot of systems in real-life have a singular matrix $A$ for its dynamics ($A$ is not invertible ), such as a double integrators with $A = [0,1;0,0]$. In this case, are there any alternatives for me to write its closed-form solution that does not contain integral?
linear-algebra control-theory
linear-algebra control-theory
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
asked Dec 3 '18 at 17:57
axpilllaxpilll
32
32
add a comment |
add a comment |
1 Answer
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You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$
e^{begin{bmatrix}A & B \ 0 & 0end{bmatrix}t} =
begin{bmatrix}e^{A,t} & int_0^t e^{A(t-tau)} dtau,B \ 0 & Iend{bmatrix}.
$$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$
e^M = I + M + tfrac{1}{2}M^2 + tfrac{1}{3!}M^3 + cdots.
$$
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$
e^{begin{bmatrix}A & B \ 0 & 0end{bmatrix}t} =
begin{bmatrix}e^{A,t} & int_0^t e^{A(t-tau)} dtau,B \ 0 & Iend{bmatrix}.
$$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$
e^M = I + M + tfrac{1}{2}M^2 + tfrac{1}{3!}M^3 + cdots.
$$
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
add a comment |
$begingroup$
You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$
e^{begin{bmatrix}A & B \ 0 & 0end{bmatrix}t} =
begin{bmatrix}e^{A,t} & int_0^t e^{A(t-tau)} dtau,B \ 0 & Iend{bmatrix}.
$$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$
e^M = I + M + tfrac{1}{2}M^2 + tfrac{1}{3!}M^3 + cdots.
$$
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
add a comment |
$begingroup$
You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$
e^{begin{bmatrix}A & B \ 0 & 0end{bmatrix}t} =
begin{bmatrix}e^{A,t} & int_0^t e^{A(t-tau)} dtau,B \ 0 & Iend{bmatrix}.
$$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$
e^M = I + M + tfrac{1}{2}M^2 + tfrac{1}{3!}M^3 + cdots.
$$
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$
e^{begin{bmatrix}A & B \ 0 & 0end{bmatrix}t} =
begin{bmatrix}e^{A,t} & int_0^t e^{A(t-tau)} dtau,B \ 0 & Iend{bmatrix}.
$$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$
e^M = I + M + tfrac{1}{2}M^2 + tfrac{1}{3!}M^3 + cdots.
$$
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 3 '18 at 23:41
Kwin van der VeenKwin van der Veen
5,3552826
5,3552826
add a comment |
add a comment |
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