Fdtxjr

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Controllability of LTI Networks

Controllability of a linear time invariant system, whose A matrix is formed by Jordan BlocksSay whether an LTI system is controllable or not, without the controllability matrixFeedforward Control for Underactuated Systems: Linear Control TheoryCascade of state space models for linear systemsControllability of a convex polytope of matrices of LTI systemfrom multi input to single input in linear dynamical systems0-controllability of three simple systems.Adaptive Pole Placement and Canonical formsHow to divide an uncontrollable LTI system into controllable and uncontrollable parts?Controllability of cascade connection of two systems

0

$begingroup$

Let us assume a 4-node network, described by $dot x = A x + B u $, where

$$ A=beginpmatrix
0 & 0 & 0 & 0 \ b & 0 & 0 & 0 \ c & 0 & 0 & e \ d & 0 & 0 & 0
endpmatrix. $$

What is the intuitive difference from an engineering point of view between choosing the input to be a matrix

$$B=beginpmatrix
b_1 & 0 & 0 & 0 \
0 & b_2 & 0 & 0 \
0 & 0 & b_3 & 0 \
0 & 0 & 0 & b_4
endpmatrix
$$
and the input to be a vector
$$ B=beginpmatrix
b_1 \
b_2 \
b_3 \
b_4
endpmatrix?
$$

dynamical-systems control-theory linear-control

share|cite|improve this question

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your first choice of $B$ implies that $u$ has 4 components, while your second choice of $B$ implies that $u$ is a scalar. Those are very different situations. Do you have an independent control on each component of the dynamic (first choice)? or do all the components share the same one control (second choice)? Also the form for $A$ you provided seems irrelevant.
  $endgroup$
  – jnez71
  Mar 10 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is what I am interested in understanding. You said that in the first choice, we have an independent control on each component while in the second one, all the components share the same control. Does this have to do with the time I am applying control? For example, in the first case, am I applying consecutive stimulations to the components one after the other, while in the second case I am applying stimulations simultaneously?
  $endgroup$
  – Alex
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, in both cases you are simulating all the states simultaneously. For example, you could use Euler's method to update the state as $ x(t+Delta t) = x(t) + big(Ax(t) + Bu(t)big)Delta t$. You cannot decouple the simulations component-wise because there are off-diagonal terms in your $A$ matrix.
  $endgroup$
  – jnez71
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let us assume a 4-node network, described by $dot x = A x + B u $, where

$$ A=beginpmatrix
0 & 0 & 0 & 0 \ b & 0 & 0 & 0 \ c & 0 & 0 & e \ d & 0 & 0 & 0
endpmatrix. $$

What is the intuitive difference from an engineering point of view between choosing the input to be a matrix

$$B=beginpmatrix
b_1 & 0 & 0 & 0 \
0 & b_2 & 0 & 0 \
0 & 0 & b_3 & 0 \
0 & 0 & 0 & b_4
endpmatrix
$$
and the input to be a vector
$$ B=beginpmatrix
b_1 \
b_2 \
b_3 \
b_4
endpmatrix?
$$

dynamical-systems control-theory linear-control

share|cite|improve this question

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your first choice of $B$ implies that $u$ has 4 components, while your second choice of $B$ implies that $u$ is a scalar. Those are very different situations. Do you have an independent control on each component of the dynamic (first choice)? or do all the components share the same one control (second choice)? Also the form for $A$ you provided seems irrelevant.
  $endgroup$
  – jnez71
  Mar 10 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is what I am interested in understanding. You said that in the first choice, we have an independent control on each component while in the second one, all the components share the same control. Does this have to do with the time I am applying control? For example, in the first case, am I applying consecutive stimulations to the components one after the other, while in the second case I am applying stimulations simultaneously?
  $endgroup$
  – Alex
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, in both cases you are simulating all the states simultaneously. For example, you could use Euler's method to update the state as $ x(t+Delta t) = x(t) + big(Ax(t) + Bu(t)big)Delta t$. You cannot decouple the simulations component-wise because there are off-diagonal terms in your $A$ matrix.
  $endgroup$
  – jnez71
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let us assume a 4-node network, described by $dot x = A x + B u $, where

$$ A=beginpmatrix
0 & 0 & 0 & 0 \ b & 0 & 0 & 0 \ c & 0 & 0 & e \ d & 0 & 0 & 0
endpmatrix. $$

What is the intuitive difference from an engineering point of view between choosing the input to be a matrix

$$B=beginpmatrix
b_1 & 0 & 0 & 0 \
0 & b_2 & 0 & 0 \
0 & 0 & b_3 & 0 \
0 & 0 & 0 & b_4
endpmatrix
$$
and the input to be a vector
$$ B=beginpmatrix
b_1 \
b_2 \
b_3 \
b_4
endpmatrix?
$$

dynamical-systems control-theory linear-control

share|cite|improve this question

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 10 at 19:40

Mars Plastic

1,337121

edited Mar 10 at 19:40

Mars Plastic

1,337121

asked Mar 10 at 19:29

AlexAlex

11

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 19:29

AlexAlex

11

1 Answer
1

active

oldest

votes

0

$begingroup$

As @jnez71 proposed the first $boldsymbolB$ does give you more control authority, because you input vector $boldsymbolu=[u_1,u_2,u_3,u_4]^T$ has four components instead of a scalar component $u$ for the vector $boldsymbolb$. Because of this, you have a higher degree of freedom for the control of your system.

In order to see this, we can look at a simplified system

$$dotx_1=x_1 + u_1$$
$$dotx_2=x_2 - u_2$$

we can quickly see that $u_1=-x_1-k_1x_1$ and $u_2=x_2+k_2x_2$ will drive the system to the origin if $k_1$ and $k_2$ are positive.

If we look at the same system but with only one input we will have

$$dotx_1=x_1 + u$$
$$dotx_2=x_2 - u.$$

If we apply $u=-x_1-k_1x_1$, we can stabilize the first equation for $k_1>0$. The problem is now that the second equation will now be directly affected by this choice of $u$.

$$dotx_2=x_2+x_1+k_1x_1$$

Hence, we cannot freely choose $u$ without constraining the values of $k_1$ which will lead to an asymptotic behavior of the system. It turns out that for this example we will not be able to stabilize the system with such a control law.

I hope you can see by the example of how we lost control authority by having a vector instead of a matrix as the input matrix.

share|cite|improve this answer

edited 2 hours ago

answered yesterday

MachineLearnerMachineLearner

5847

$endgroup$

add a comment | 

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0

$begingroup$

As @jnez71 proposed the first $boldsymbolB$ does give you more control authority, because you input vector $boldsymbolu=[u_1,u_2,u_3,u_4]^T$ has four components instead of a scalar component $u$ for the vector $boldsymbolb$. Because of this, you have a higher degree of freedom for the control of your system.

In order to see this, we can look at a simplified system

$$dotx_1=x_1 + u_1$$
$$dotx_2=x_2 - u_2$$

we can quickly see that $u_1=-x_1-k_1x_1$ and $u_2=x_2+k_2x_2$ will drive the system to the origin if $k_1$ and $k_2$ are positive.

If we look at the same system but with only one input we will have

$$dotx_1=x_1 + u$$
$$dotx_2=x_2 - u.$$

If we apply $u=-x_1-k_1x_1$, we can stabilize the first equation for $k_1>0$. The problem is now that the second equation will now be directly affected by this choice of $u$.

$$dotx_2=x_2+x_1+k_1x_1$$

Hence, we cannot freely choose $u$ without constraining the values of $k_1$ which will lead to an asymptotic behavior of the system. It turns out that for this example we will not be able to stabilize the system with such a control law.

I hope you can see by the example of how we lost control authority by having a vector instead of a matrix as the input matrix.

share|cite|improve this answer

edited 2 hours ago

answered yesterday

MachineLearnerMachineLearner

5847

$endgroup$

add a comment | 

0

$begingroup$

As @jnez71 proposed the first $boldsymbolB$ does give you more control authority, because you input vector $boldsymbolu=[u_1,u_2,u_3,u_4]^T$ has four components instead of a scalar component $u$ for the vector $boldsymbolb$. Because of this, you have a higher degree of freedom for the control of your system.

In order to see this, we can look at a simplified system

$$dotx_1=x_1 + u_1$$
$$dotx_2=x_2 - u_2$$

we can quickly see that $u_1=-x_1-k_1x_1$ and $u_2=x_2+k_2x_2$ will drive the system to the origin if $k_1$ and $k_2$ are positive.

If we look at the same system but with only one input we will have

$$dotx_1=x_1 + u$$
$$dotx_2=x_2 - u.$$

If we apply $u=-x_1-k_1x_1$, we can stabilize the first equation for $k_1>0$. The problem is now that the second equation will now be directly affected by this choice of $u$.

$$dotx_2=x_2+x_1+k_1x_1$$

Hence, we cannot freely choose $u$ without constraining the values of $k_1$ which will lead to an asymptotic behavior of the system. It turns out that for this example we will not be able to stabilize the system with such a control law.

I hope you can see by the example of how we lost control authority by having a vector instead of a matrix as the input matrix.

share|cite|improve this answer

edited 2 hours ago

answered yesterday

MachineLearnerMachineLearner

5847

$endgroup$

add a comment | 

$begingroup$

As @jnez71 proposed the first $boldsymbolB$ does give you more control authority, because you input vector $boldsymbolu=[u_1,u_2,u_3,u_4]^T$ has four components instead of a scalar component $u$ for the vector $boldsymbolb$. Because of this, you have a higher degree of freedom for the control of your system.

In order to see this, we can look at a simplified system

$$dotx_1=x_1 + u_1$$
$$dotx_2=x_2 - u_2$$

we can quickly see that $u_1=-x_1-k_1x_1$ and $u_2=x_2+k_2x_2$ will drive the system to the origin if $k_1$ and $k_2$ are positive.

If we look at the same system but with only one input we will have

$$dotx_1=x_1 + u$$
$$dotx_2=x_2 - u.$$

If we apply $u=-x_1-k_1x_1$, we can stabilize the first equation for $k_1>0$. The problem is now that the second equation will now be directly affected by this choice of $u$.

$$dotx_2=x_2+x_1+k_1x_1$$

Hence, we cannot freely choose $u$ without constraining the values of $k_1$ which will lead to an asymptotic behavior of the system. It turns out that for this example we will not be able to stabilize the system with such a control law.

I hope you can see by the example of how we lost control authority by having a vector instead of a matrix as the input matrix.

share|cite|improve this answer

edited 2 hours ago

answered yesterday

MachineLearnerMachineLearner

5847

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered yesterday

MachineLearnerMachineLearner

5847

answered yesterday

MachineLearnerMachineLearner

5847

answered yesterday

MachineLearnerMachineLearner

5847