How to linearize a kinematic bicycle model? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the fundamental matrix of a system of ODEs?Derivation for state equation linearizationDetermining stability of equilibria of a nonlinear pendulum with torque ode systemDifferential equation with constant term to state spaceDerive state space modelHow do I find the Lyapunov function from a nonlinear state space model?How do I linearize this system to achieve the stated result?Use direct method of lyapunov to determine stability of a systemWhy is my state space model of a hydraulic servo system unstable?Augmented nonlinear state space model for nonlinear model predictive control?
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How to linearize a kinematic bicycle model?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the fundamental matrix of a system of ODEs?Derivation for state equation linearizationDetermining stability of equilibria of a nonlinear pendulum with torque ode systemDifferential equation with constant term to state spaceDerive state space modelHow do I find the Lyapunov function from a nonlinear state space model?How do I linearize this system to achieve the stated result?Use direct method of lyapunov to determine stability of a systemWhy is my state space model of a hydraulic servo system unstable?Augmented nonlinear state space model for nonlinear model predictive control?
$begingroup$
I have the following system:
$$beginaligned x(k+1) &= x(k) + T_svcos(phi(k) + beta(k)) \ y(k+1) &= y(k) + T_svsin(phi(k) + beta(k)) \ phi(k+1) &= phi(k) + fracT_svlsin(beta(k))endaligned$$
I am confused how should I put this in linear state-space representation?
- Clearly there is no equilibria when $vneq0$, does this mean that I cannot linearize the system or I can linearize it arbitrarily?
- I got the following ss form using the standard linearization method. Again, how to pick $x_e/u_e$? does it make sense and is there a better way to linearize this system?
$$
beginbmatrixdelta x(k+1) \ delta y(k+1) \ delta phi(k+1) \endbmatrix=
beginbmatrix1&0&-T_svsin(phi_e + beta_e) \ 0&1&T_svcos(phi_e + beta_e) \ 0&0&1 \endbmatrix
beginbmatrixdelta x(k) \ delta y(k) \ delta phi(k) \endbmatrix+
beginbmatrix-T_svsin(phi_e + beta_e) \ T_svcos(phi_e + beta_e) \ fracT_svlcos(beta_e) \endbmatrix
beginbmatrix delta beta(k) \ endbmatrix$$
where $ delta x(k) = x(k) - x_e(k) $
linear-algebra dynamical-systems control-theory jacobian
New contributor
$endgroup$
add a comment |
$begingroup$
I have the following system:
$$beginaligned x(k+1) &= x(k) + T_svcos(phi(k) + beta(k)) \ y(k+1) &= y(k) + T_svsin(phi(k) + beta(k)) \ phi(k+1) &= phi(k) + fracT_svlsin(beta(k))endaligned$$
I am confused how should I put this in linear state-space representation?
- Clearly there is no equilibria when $vneq0$, does this mean that I cannot linearize the system or I can linearize it arbitrarily?
- I got the following ss form using the standard linearization method. Again, how to pick $x_e/u_e$? does it make sense and is there a better way to linearize this system?
$$
beginbmatrixdelta x(k+1) \ delta y(k+1) \ delta phi(k+1) \endbmatrix=
beginbmatrix1&0&-T_svsin(phi_e + beta_e) \ 0&1&T_svcos(phi_e + beta_e) \ 0&0&1 \endbmatrix
beginbmatrixdelta x(k) \ delta y(k) \ delta phi(k) \endbmatrix+
beginbmatrix-T_svsin(phi_e + beta_e) \ T_svcos(phi_e + beta_e) \ fracT_svlcos(beta_e) \endbmatrix
beginbmatrix delta beta(k) \ endbmatrix$$
where $ delta x(k) = x(k) - x_e(k) $
linear-algebra dynamical-systems control-theory jacobian
New contributor
$endgroup$
1
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12
add a comment |
$begingroup$
I have the following system:
$$beginaligned x(k+1) &= x(k) + T_svcos(phi(k) + beta(k)) \ y(k+1) &= y(k) + T_svsin(phi(k) + beta(k)) \ phi(k+1) &= phi(k) + fracT_svlsin(beta(k))endaligned$$
I am confused how should I put this in linear state-space representation?
- Clearly there is no equilibria when $vneq0$, does this mean that I cannot linearize the system or I can linearize it arbitrarily?
- I got the following ss form using the standard linearization method. Again, how to pick $x_e/u_e$? does it make sense and is there a better way to linearize this system?
$$
beginbmatrixdelta x(k+1) \ delta y(k+1) \ delta phi(k+1) \endbmatrix=
beginbmatrix1&0&-T_svsin(phi_e + beta_e) \ 0&1&T_svcos(phi_e + beta_e) \ 0&0&1 \endbmatrix
beginbmatrixdelta x(k) \ delta y(k) \ delta phi(k) \endbmatrix+
beginbmatrix-T_svsin(phi_e + beta_e) \ T_svcos(phi_e + beta_e) \ fracT_svlcos(beta_e) \endbmatrix
beginbmatrix delta beta(k) \ endbmatrix$$
where $ delta x(k) = x(k) - x_e(k) $
linear-algebra dynamical-systems control-theory jacobian
New contributor
$endgroup$
I have the following system:
$$beginaligned x(k+1) &= x(k) + T_svcos(phi(k) + beta(k)) \ y(k+1) &= y(k) + T_svsin(phi(k) + beta(k)) \ phi(k+1) &= phi(k) + fracT_svlsin(beta(k))endaligned$$
I am confused how should I put this in linear state-space representation?
- Clearly there is no equilibria when $vneq0$, does this mean that I cannot linearize the system or I can linearize it arbitrarily?
- I got the following ss form using the standard linearization method. Again, how to pick $x_e/u_e$? does it make sense and is there a better way to linearize this system?
$$
beginbmatrixdelta x(k+1) \ delta y(k+1) \ delta phi(k+1) \endbmatrix=
beginbmatrix1&0&-T_svsin(phi_e + beta_e) \ 0&1&T_svcos(phi_e + beta_e) \ 0&0&1 \endbmatrix
beginbmatrixdelta x(k) \ delta y(k) \ delta phi(k) \endbmatrix+
beginbmatrix-T_svsin(phi_e + beta_e) \ T_svcos(phi_e + beta_e) \ fracT_svlcos(beta_e) \endbmatrix
beginbmatrix delta beta(k) \ endbmatrix$$
where $ delta x(k) = x(k) - x_e(k) $
linear-algebra dynamical-systems control-theory jacobian
linear-algebra dynamical-systems control-theory jacobian
New contributor
New contributor
edited Apr 8 at 6:22
Rodrigo de Azevedo
13.2k41962
13.2k41962
New contributor
asked Apr 6 at 22:47
n33n33
33
33
New contributor
New contributor
1
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12
add a comment |
1
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12
1
1
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You should always start any state space modelling effort by clearly defining (a) the state variables, (b) the inputs, and (c) the outputs. It looks like your state variables are $x,y,phi$. What physical quantities do those represent? It looks like $beta$ is an input. What physical quantity does that represent? What outputs are there?
You can still linearize the system. You just need to linearize about a nonstationary solution. Assume that $v neq 0$ and $beta = 0$ (note that it is common to linearize about the "free-response", e.g. with an input of zero.) Then it looks like $phi$ is constant and $x$ and $y$ have simple solutions. That looks like a good place to start. Remember that you will need to add in this solution when you want to go from the perturbative analysis back into the first-order approximation of the overall solution.
That looks reasonable for the $A$ and $B$ matrices. The reason you have uncertainty at this point is that you skipped the crucial step of defining the solution you are linearizing about before linearizing. $beta_e(t),x_e(t),x_e(t),phi_e(t)$ can be any solution you want. I suggested the solution consistent with $beta_e(t) = 0$, since this would be the most common choice, but you might have some other need. The procedure of linearization cannot tell you this--you as an engineer need to decide what is best for your application.
$endgroup$
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
add a comment |
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1 Answer
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$begingroup$
You should always start any state space modelling effort by clearly defining (a) the state variables, (b) the inputs, and (c) the outputs. It looks like your state variables are $x,y,phi$. What physical quantities do those represent? It looks like $beta$ is an input. What physical quantity does that represent? What outputs are there?
You can still linearize the system. You just need to linearize about a nonstationary solution. Assume that $v neq 0$ and $beta = 0$ (note that it is common to linearize about the "free-response", e.g. with an input of zero.) Then it looks like $phi$ is constant and $x$ and $y$ have simple solutions. That looks like a good place to start. Remember that you will need to add in this solution when you want to go from the perturbative analysis back into the first-order approximation of the overall solution.
That looks reasonable for the $A$ and $B$ matrices. The reason you have uncertainty at this point is that you skipped the crucial step of defining the solution you are linearizing about before linearizing. $beta_e(t),x_e(t),x_e(t),phi_e(t)$ can be any solution you want. I suggested the solution consistent with $beta_e(t) = 0$, since this would be the most common choice, but you might have some other need. The procedure of linearization cannot tell you this--you as an engineer need to decide what is best for your application.
$endgroup$
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
add a comment |
$begingroup$
You should always start any state space modelling effort by clearly defining (a) the state variables, (b) the inputs, and (c) the outputs. It looks like your state variables are $x,y,phi$. What physical quantities do those represent? It looks like $beta$ is an input. What physical quantity does that represent? What outputs are there?
You can still linearize the system. You just need to linearize about a nonstationary solution. Assume that $v neq 0$ and $beta = 0$ (note that it is common to linearize about the "free-response", e.g. with an input of zero.) Then it looks like $phi$ is constant and $x$ and $y$ have simple solutions. That looks like a good place to start. Remember that you will need to add in this solution when you want to go from the perturbative analysis back into the first-order approximation of the overall solution.
That looks reasonable for the $A$ and $B$ matrices. The reason you have uncertainty at this point is that you skipped the crucial step of defining the solution you are linearizing about before linearizing. $beta_e(t),x_e(t),x_e(t),phi_e(t)$ can be any solution you want. I suggested the solution consistent with $beta_e(t) = 0$, since this would be the most common choice, but you might have some other need. The procedure of linearization cannot tell you this--you as an engineer need to decide what is best for your application.
$endgroup$
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
add a comment |
$begingroup$
You should always start any state space modelling effort by clearly defining (a) the state variables, (b) the inputs, and (c) the outputs. It looks like your state variables are $x,y,phi$. What physical quantities do those represent? It looks like $beta$ is an input. What physical quantity does that represent? What outputs are there?
You can still linearize the system. You just need to linearize about a nonstationary solution. Assume that $v neq 0$ and $beta = 0$ (note that it is common to linearize about the "free-response", e.g. with an input of zero.) Then it looks like $phi$ is constant and $x$ and $y$ have simple solutions. That looks like a good place to start. Remember that you will need to add in this solution when you want to go from the perturbative analysis back into the first-order approximation of the overall solution.
That looks reasonable for the $A$ and $B$ matrices. The reason you have uncertainty at this point is that you skipped the crucial step of defining the solution you are linearizing about before linearizing. $beta_e(t),x_e(t),x_e(t),phi_e(t)$ can be any solution you want. I suggested the solution consistent with $beta_e(t) = 0$, since this would be the most common choice, but you might have some other need. The procedure of linearization cannot tell you this--you as an engineer need to decide what is best for your application.
$endgroup$
You should always start any state space modelling effort by clearly defining (a) the state variables, (b) the inputs, and (c) the outputs. It looks like your state variables are $x,y,phi$. What physical quantities do those represent? It looks like $beta$ is an input. What physical quantity does that represent? What outputs are there?
You can still linearize the system. You just need to linearize about a nonstationary solution. Assume that $v neq 0$ and $beta = 0$ (note that it is common to linearize about the "free-response", e.g. with an input of zero.) Then it looks like $phi$ is constant and $x$ and $y$ have simple solutions. That looks like a good place to start. Remember that you will need to add in this solution when you want to go from the perturbative analysis back into the first-order approximation of the overall solution.
That looks reasonable for the $A$ and $B$ matrices. The reason you have uncertainty at this point is that you skipped the crucial step of defining the solution you are linearizing about before linearizing. $beta_e(t),x_e(t),x_e(t),phi_e(t)$ can be any solution you want. I suggested the solution consistent with $beta_e(t) = 0$, since this would be the most common choice, but you might have some other need. The procedure of linearization cannot tell you this--you as an engineer need to decide what is best for your application.
answered Apr 8 at 15:02
SZNSZN
2,850720
2,850720
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
add a comment |
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
$begingroup$
I tried to make the question concise so I skipped some details. Nonlinear control theory states that only a small region around the equilibria shares stability with the linearized model, that's why I was confused which equilibrium I should use, but it turns out I can linearize at any point in this case. I simply use linearization as a tool so I never thought it thorough, thanks for your explanation now it makes more senses to me!
$endgroup$
– n33
Apr 8 at 20:04
add a comment |
n33 is a new contributor. Be nice, and check out our Code of Conduct.
n33 is a new contributor. Be nice, and check out our Code of Conduct.
n33 is a new contributor. Be nice, and check out our Code of Conduct.
n33 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
what do you know about linearization? What have you tried?
$endgroup$
– SZN
Apr 7 at 1:04
$begingroup$
how can variables that depend on the state be in the system matrix? Perhaps things might be clearer if you fixed an operating point. What is are reasonable values for the operating point for the state and the operating point of the control?
$endgroup$
– SZN
Apr 7 at 1:41
$begingroup$
I edited my question, hoping it is clearer now
$endgroup$
– n33
Apr 7 at 3:12