Automatic control of linear system with non independent inputs
$begingroup$
I'm currently trying to (re)learn automatic control 2 years after having finished college, and I'm having a hard time. I'm trying to control a simplistic lunar lander in a 2d space.
The only way to control it is to change its angle and/or thrust power, so I have the following state-space representation:
$$
x=begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4
end{bmatrix}
$$
$$
begin{gather}
dot{x}
=
begin{bmatrix}
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{bmatrix}
x;+
begin{bmatrix}
0 & 0 \
0 & 0 \
1 & 0 \
0 & 1
end{bmatrix}
u
end{gather}
$$
Where:
- x$_1$ is the horizontal position
- x$_2$ is the vertical position
- x$_3$ is the horizontal velocity
- x$_4$ is the vertical velocity
And u is the command, accelerations on respectively the horizontal and vertical coordinates.
My problem is the acceleration is unidirectional, and its horizontal and vertical components are correlated:
$$
begin{gather}
u
=
begin{bmatrix}
atimescos{theta} \
atimessin{theta} - g
end{bmatrix}
end{gather}
$$
$a$ is the acceleration provided by the thruster and $theta$ is the angle of the lander. $g$ is the gravity constant.
Since the system is linear, I could calculate the feedback control with the gain K, but I realize this K would consider the two input commands are independent.
I've been stuck on this for a few days, I've tried including $theta$ and $a$ in $x$ so $u$ could simply be an angular velocity and an acceleration derivative, which would be independent, but then the system becomes nonlinear and I have no idea how to solve it.
Maybe my approach is wrong, but I'm honestly lost here, could anyone explain to me what I did wrong or how I could take this input restriction into account in the feedback loop?
ADDENDUM 1: I'm considering for the sake of simplicity that I can freely and instantly change $theta$ and $a$.
control-theory
asked Dec 11 '18 at 15:49
AlderiAlderi
134
$endgroup$
add a comment |
$begingroup$
I'm currently trying to (re)learn automatic control 2 years after having finished college, and I'm having a hard time. I'm trying to control a simplistic lunar lander in a 2d space.
The only way to control it is to change its angle and/or thrust power, so I have the following state-space representation:
$$
x=begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4
end{bmatrix}
$$
$$
begin{gather}
dot{x}
=
begin{bmatrix}
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{bmatrix}
x;+
begin{bmatrix}
0 & 0 \
0 & 0 \
1 & 0 \
0 & 1
end{bmatrix}
u
end{gather}
$$
Where:
- x$_1$ is the horizontal position
- x$_2$ is the vertical position
- x$_3$ is the horizontal velocity
- x$_4$ is the vertical velocity
And u is the command, accelerations on respectively the horizontal and vertical coordinates.
My problem is the acceleration is unidirectional, and its horizontal and vertical components are correlated:
$$
begin{gather}
u
=
begin{bmatrix}
atimescos{theta} \
atimessin{theta} - g
end{bmatrix}
end{gather}
$$
$a$ is the acceleration provided by the thruster and $theta$ is the angle of the lander. $g$ is the gravity constant.
Since the system is linear, I could calculate the feedback control with the gain K, but I realize this K would consider the two input commands are independent.
I've been stuck on this for a few days, I've tried including $theta$ and $a$ in $x$ so $u$ could simply be an angular velocity and an acceleration derivative, which would be independent, but then the system becomes nonlinear and I have no idea how to solve it.
Maybe my approach is wrong, but I'm honestly lost here, could anyone explain to me what I did wrong or how I could take this input restriction into account in the feedback loop?
ADDENDUM 1: I'm considering for the sake of simplicity that I can freely and instantly change $theta$ and $a$.
control-theory
asked Dec 11 '18 at 15:49
AlderiAlderi
134
$endgroup$
$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29
add a comment |
$begingroup$
I'm currently trying to (re)learn automatic control 2 years after having finished college, and I'm having a hard time. I'm trying to control a simplistic lunar lander in a 2d space.
The only way to control it is to change its angle and/or thrust power, so I have the following state-space representation:
$$
x=begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4
end{bmatrix}
$$
$$
begin{gather}
dot{x}
=
begin{bmatrix}
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{bmatrix}
x;+
begin{bmatrix}
0 & 0 \
0 & 0 \
1 & 0 \
0 & 1
end{bmatrix}
u
end{gather}
$$
Where:
- x$_1$ is the horizontal position
- x$_2$ is the vertical position
- x$_3$ is the horizontal velocity
- x$_4$ is the vertical velocity
And u is the command, accelerations on respectively the horizontal and vertical coordinates.
My problem is the acceleration is unidirectional, and its horizontal and vertical components are correlated:
$$
begin{gather}
u
=
begin{bmatrix}
atimescos{theta} \
atimessin{theta} - g
end{bmatrix}
end{gather}
$$
$a$ is the acceleration provided by the thruster and $theta$ is the angle of the lander. $g$ is the gravity constant.
Since the system is linear, I could calculate the feedback control with the gain K, but I realize this K would consider the two input commands are independent.
I've been stuck on this for a few days, I've tried including $theta$ and $a$ in $x$ so $u$ could simply be an angular velocity and an acceleration derivative, which would be independent, but then the system becomes nonlinear and I have no idea how to solve it.
Maybe my approach is wrong, but I'm honestly lost here, could anyone explain to me what I did wrong or how I could take this input restriction into account in the feedback loop?
ADDENDUM 1: I'm considering for the sake of simplicity that I can freely and instantly change $theta$ and $a$.
control-theory
asked Dec 11 '18 at 15:49
AlderiAlderi
134
$endgroup$
I'm currently trying to (re)learn automatic control 2 years after having finished college, and I'm having a hard time. I'm trying to control a simplistic lunar lander in a 2d space.
The only way to control it is to change its angle and/or thrust power, so I have the following state-space representation:
$$
x=begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4
end{bmatrix}
$$
$$
begin{gather}
dot{x}
=
begin{bmatrix}
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{bmatrix}
x;+
begin{bmatrix}
0 & 0 \
0 & 0 \
1 & 0 \
0 & 1
end{bmatrix}
u
end{gather}
$$
Where:
- x$_1$ is the horizontal position
- x$_2$ is the vertical position
- x$_3$ is the horizontal velocity
- x$_4$ is the vertical velocity
And u is the command, accelerations on respectively the horizontal and vertical coordinates.
My problem is the acceleration is unidirectional, and its horizontal and vertical components are correlated:
$$
begin{gather}
u
=
begin{bmatrix}
atimescos{theta} \
atimessin{theta} - g
end{bmatrix}
end{gather}
$$
$a$ is the acceleration provided by the thruster and $theta$ is the angle of the lander. $g$ is the gravity constant.
Since the system is linear, I could calculate the feedback control with the gain K, but I realize this K would consider the two input commands are independent.
I've been stuck on this for a few days, I've tried including $theta$ and $a$ in $x$ so $u$ could simply be an angular velocity and an acceleration derivative, which would be independent, but then the system becomes nonlinear and I have no idea how to solve it.
Maybe my approach is wrong, but I'm honestly lost here, could anyone explain to me what I did wrong or how I could take this input restriction into account in the feedback loop?
ADDENDUM 1: I'm considering for the sake of simplicity that I can freely and instantly change $theta$ and $a$.
control-theory
control-theory
asked Dec 11 '18 at 15:49
AlderiAlderi
134
asked Dec 11 '18 at 15:49
AlderiAlderi
134
edited Dec 11 '18 at 16:32
Alderi
asked Dec 11 '18 at 15:49
AlderiAlderi
134
asked Dec 11 '18 at 15:49
AlderiAlderi
134
asked Dec 11 '18 at 15:49
AlderiAlderi
134
134
$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29
add a comment |
$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29
$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $a$ and $theta$ are essentially polar coordinates. So if you have a state feedback control law $u=-K,x$ then you can solve for $a$ and $theta$ given each value for $u$. Namely
begin{align}
a &= sqrt{u_1^2 + (u_2 + g)^2} \
theta &= text{atan2}(u_2 + g, u_1)
end{align}
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
add a comment |
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1 Answer
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1 Answer
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oldest
votes
$begingroup$
Since $a$ and $theta$ are essentially polar coordinates. So if you have a state feedback control law $u=-K,x$ then you can solve for $a$ and $theta$ given each value for $u$. Namely
begin{align}
a &= sqrt{u_1^2 + (u_2 + g)^2} \
theta &= text{atan2}(u_2 + g, u_1)
end{align}
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
add a comment |
$begingroup$
Since $a$ and $theta$ are essentially polar coordinates. So if you have a state feedback control law $u=-K,x$ then you can solve for $a$ and $theta$ given each value for $u$. Namely
begin{align}
a &= sqrt{u_1^2 + (u_2 + g)^2} \
theta &= text{atan2}(u_2 + g, u_1)
end{align}
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
add a comment |
$begingroup$
Since $a$ and $theta$ are essentially polar coordinates. So if you have a state feedback control law $u=-K,x$ then you can solve for $a$ and $theta$ given each value for $u$. Namely
begin{align}
a &= sqrt{u_1^2 + (u_2 + g)^2} \
theta &= text{atan2}(u_2 + g, u_1)
end{align}
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
Since $a$ and $theta$ are essentially polar coordinates. So if you have a state feedback control law $u=-K,x$ then you can solve for $a$ and $theta$ given each value for $u$. Namely
begin{align}
a &= sqrt{u_1^2 + (u_2 + g)^2} \
theta &= text{atan2}(u_2 + g, u_1)
end{align}
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
answered Dec 11 '18 at 16:58
Kwin van der VeenKwin van der Veen
5,3952827
5,3952827
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
add a comment |
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
$begingroup$
Thanks a lot for your help :)
$endgroup$
– Alderi
Dec 11 '18 at 17:19
add a comment |
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$begingroup$
What would be the dynamics of $theta$, $dot{theta}=tau$ with $tau$ the torque input to the system?
$endgroup$
– Kwin van der Veen
Dec 11 '18 at 16:26
$begingroup$
Here I'm considering I can freely change $theta$ and $a$, in order to make things simple so I guess $dot{theta}$ would be a Dirac delta function? I'm not sure this is the correct term.
$endgroup$
– Alderi
Dec 11 '18 at 16:29