Construction of a linear system stratifying certain requirements
$begingroup$
Construct a linear system $$ begin{cases} dot x = Ax + Bu \ y=Cx
end{cases} $$ that satisfies all the following requirements:
- 4th order
- 1 input & 1 output
- Unstable
- Stabilizable & detectable
- Rank of controlability matrix = Rank of obeservability matrix = 2
$C(sI-A)^{-1}B$ is first order, i.e., can be written as $frac{alpha}{s+beta}$ for some scalar constants $alpha$ and
$beta$.
Attempt
Requirement 1&2:
$$
A_{4 times 4}=
begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24} \
a_{31} & a_{32} & a_{33} & a_{34} \
a_{41} & a_{42} & a_{43} & a_{44} \
end{bmatrix},
hspace{0.6cm}
B_{4 times 1}=
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
b_{4}
end{bmatrix},
hspace{0.6cm}
C_{1 times 4}=
begin{bmatrix}
c_{1} & c_{2} & c_{3} & c_{4} \
end{bmatrix},
hspace{0.6cm}
x_{4 times 1}=
begin{bmatrix}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{bmatrix},
hspace{0.6cm}
u_{1 times 1},
hspace{0.6cm}
y_{1 times 1},
$$
Requirement 3:
Real${ lambda_i }_{i=1,2,3,4}$ is zero or positive. Let $a$ be non-negative real number, and $b,c,d$ be any real numbers.
$$
det(A-lambda I)=(a-lambda)(b-lambda)(c-lambda)(d-lambda)=0
$$
Requirement 4:
Stabilizable if $$
Rank(
begin{bmatrix}
sI-A & B \
end{bmatrix}
)=4
$$
Detectable if $$
Rank(
begin{bmatrix}
sI-A \
C \
end{bmatrix}
)=4
$$
Requirement 5:
The controllability matrix is given by
$$
mathcal{C}=
begin{bmatrix}
B & AB & A^2B & A^3B
end{bmatrix}
$$
The obeservability matrix is given by
$$
mathcal{O}=
begin{bmatrix}
C \
CA \
CA^2 \
CA^3
end{bmatrix}
$$
$Rank(mathcal{C})=Rank(mathcal{O})=2$
Requirement 6:
The transfer function is given by
$$
frac{Y(s)}{U(s)}=frac{alpha}{s+beta}
$$
for some scalar constants $alpha$ and $beta$.
Question
I understand all the requirements and implications but I am not sure how to starting actually constructing. Any help?
control-theory
asked Dec 5 '18 at 18:11
LodLod
128113
$endgroup$
|
show 1 more comment
$begingroup$
Construct a linear system $$ begin{cases} dot x = Ax + Bu \ y=Cx
end{cases} $$ that satisfies all the following requirements:
- 4th order
- 1 input & 1 output
- Unstable
- Stabilizable & detectable
- Rank of controlability matrix = Rank of obeservability matrix = 2
$C(sI-A)^{-1}B$ is first order, i.e., can be written as $frac{alpha}{s+beta}$ for some scalar constants $alpha$ and
$beta$.
Attempt
Requirement 1&2:
$$
A_{4 times 4}=
begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24} \
a_{31} & a_{32} & a_{33} & a_{34} \
a_{41} & a_{42} & a_{43} & a_{44} \
end{bmatrix},
hspace{0.6cm}
B_{4 times 1}=
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
b_{4}
end{bmatrix},
hspace{0.6cm}
C_{1 times 4}=
begin{bmatrix}
c_{1} & c_{2} & c_{3} & c_{4} \
end{bmatrix},
hspace{0.6cm}
x_{4 times 1}=
begin{bmatrix}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{bmatrix},
hspace{0.6cm}
u_{1 times 1},
hspace{0.6cm}
y_{1 times 1},
$$
Requirement 3:
Real${ lambda_i }_{i=1,2,3,4}$ is zero or positive. Let $a$ be non-negative real number, and $b,c,d$ be any real numbers.
$$
det(A-lambda I)=(a-lambda)(b-lambda)(c-lambda)(d-lambda)=0
$$
Requirement 4:
Stabilizable if $$
Rank(
begin{bmatrix}
sI-A & B \
end{bmatrix}
)=4
$$
Detectable if $$
Rank(
begin{bmatrix}
sI-A \
C \
end{bmatrix}
)=4
$$
Requirement 5:
The controllability matrix is given by
$$
mathcal{C}=
begin{bmatrix}
B & AB & A^2B & A^3B
end{bmatrix}
$$
The obeservability matrix is given by
$$
mathcal{O}=
begin{bmatrix}
C \
CA \
CA^2 \
CA^3
end{bmatrix}
$$
$Rank(mathcal{C})=Rank(mathcal{O})=2$
Requirement 6:
The transfer function is given by
$$
frac{Y(s)}{U(s)}=frac{alpha}{s+beta}
$$
for some scalar constants $alpha$ and $beta$.
Question
I understand all the requirements and implications but I am not sure how to starting actually constructing. Any help?
control-theory
asked Dec 5 '18 at 18:11
LodLod
128113
$endgroup$
$begingroup$
each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
$endgroup$
– Lod
Dec 5 '18 at 19:46
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
$endgroup$
– obareey
Dec 6 '18 at 17:51
$begingroup$
I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36
|
show 1 more comment
$begingroup$
Construct a linear system $$ begin{cases} dot x = Ax + Bu \ y=Cx
end{cases} $$ that satisfies all the following requirements:
- 4th order
- 1 input & 1 output
- Unstable
- Stabilizable & detectable
- Rank of controlability matrix = Rank of obeservability matrix = 2
$C(sI-A)^{-1}B$ is first order, i.e., can be written as $frac{alpha}{s+beta}$ for some scalar constants $alpha$ and
$beta$.
Attempt
Requirement 1&2:
$$
A_{4 times 4}=
begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24} \
a_{31} & a_{32} & a_{33} & a_{34} \
a_{41} & a_{42} & a_{43} & a_{44} \
end{bmatrix},
hspace{0.6cm}
B_{4 times 1}=
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
b_{4}
end{bmatrix},
hspace{0.6cm}
C_{1 times 4}=
begin{bmatrix}
c_{1} & c_{2} & c_{3} & c_{4} \
end{bmatrix},
hspace{0.6cm}
x_{4 times 1}=
begin{bmatrix}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{bmatrix},
hspace{0.6cm}
u_{1 times 1},
hspace{0.6cm}
y_{1 times 1},
$$
Requirement 3:
Real${ lambda_i }_{i=1,2,3,4}$ is zero or positive. Let $a$ be non-negative real number, and $b,c,d$ be any real numbers.
$$
det(A-lambda I)=(a-lambda)(b-lambda)(c-lambda)(d-lambda)=0
$$
Requirement 4:
Stabilizable if $$
Rank(
begin{bmatrix}
sI-A & B \
end{bmatrix}
)=4
$$
Detectable if $$
Rank(
begin{bmatrix}
sI-A \
C \
end{bmatrix}
)=4
$$
Requirement 5:
The controllability matrix is given by
$$
mathcal{C}=
begin{bmatrix}
B & AB & A^2B & A^3B
end{bmatrix}
$$
The obeservability matrix is given by
$$
mathcal{O}=
begin{bmatrix}
C \
CA \
CA^2 \
CA^3
end{bmatrix}
$$
$Rank(mathcal{C})=Rank(mathcal{O})=2$
Requirement 6:
The transfer function is given by
$$
frac{Y(s)}{U(s)}=frac{alpha}{s+beta}
$$
for some scalar constants $alpha$ and $beta$.
Question
I understand all the requirements and implications but I am not sure how to starting actually constructing. Any help?
control-theory
asked Dec 5 '18 at 18:11
LodLod
128113
$endgroup$
Construct a linear system $$ begin{cases} dot x = Ax + Bu \ y=Cx
end{cases} $$ that satisfies all the following requirements:
- 4th order
- 1 input & 1 output
- Unstable
- Stabilizable & detectable
- Rank of controlability matrix = Rank of obeservability matrix = 2
$C(sI-A)^{-1}B$ is first order, i.e., can be written as $frac{alpha}{s+beta}$ for some scalar constants $alpha$ and
$beta$.
Attempt
Requirement 1&2:
$$
A_{4 times 4}=
begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \
a_{21} & a_{22} & a_{23} & a_{24} \
a_{31} & a_{32} & a_{33} & a_{34} \
a_{41} & a_{42} & a_{43} & a_{44} \
end{bmatrix},
hspace{0.6cm}
B_{4 times 1}=
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
b_{4}
end{bmatrix},
hspace{0.6cm}
C_{1 times 4}=
begin{bmatrix}
c_{1} & c_{2} & c_{3} & c_{4} \
end{bmatrix},
hspace{0.6cm}
x_{4 times 1}=
begin{bmatrix}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{bmatrix},
hspace{0.6cm}
u_{1 times 1},
hspace{0.6cm}
y_{1 times 1},
$$
Requirement 3:
Real${ lambda_i }_{i=1,2,3,4}$ is zero or positive. Let $a$ be non-negative real number, and $b,c,d$ be any real numbers.
$$
det(A-lambda I)=(a-lambda)(b-lambda)(c-lambda)(d-lambda)=0
$$
Requirement 4:
Stabilizable if $$
Rank(
begin{bmatrix}
sI-A & B \
end{bmatrix}
)=4
$$
Detectable if $$
Rank(
begin{bmatrix}
sI-A \
C \
end{bmatrix}
)=4
$$
Requirement 5:
The controllability matrix is given by
$$
mathcal{C}=
begin{bmatrix}
B & AB & A^2B & A^3B
end{bmatrix}
$$
The obeservability matrix is given by
$$
mathcal{O}=
begin{bmatrix}
C \
CA \
CA^2 \
CA^3
end{bmatrix}
$$
$Rank(mathcal{C})=Rank(mathcal{O})=2$
Requirement 6:
The transfer function is given by
$$
frac{Y(s)}{U(s)}=frac{alpha}{s+beta}
$$
for some scalar constants $alpha$ and $beta$.
Question
I understand all the requirements and implications but I am not sure how to starting actually constructing. Any help?
control-theory
control-theory
asked Dec 5 '18 at 18:11
LodLod
128113
asked Dec 5 '18 at 18:11
LodLod
128113
edited Dec 7 '18 at 16:34
Lod
asked Dec 5 '18 at 18:11
LodLod
128113
asked Dec 5 '18 at 18:11
LodLod
128113
asked Dec 5 '18 at 18:11
LodLod
128113
128113
$begingroup$
each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
$endgroup$
– Lod
Dec 5 '18 at 19:46
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
$endgroup$
– obareey
Dec 6 '18 at 17:51
$begingroup$
I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36
|
show 1 more comment
$begingroup$
each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
$endgroup$
– Lod
Dec 5 '18 at 19:46
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
$endgroup$
– obareey
Dec 6 '18 at 17:51
$begingroup$
I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36
$begingroup$
each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
$endgroup$
– Lod
Dec 5 '18 at 19:46
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
$endgroup$
– Lod
Dec 5 '18 at 19:46
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
$endgroup$
– obareey
Dec 6 '18 at 17:51
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
$endgroup$
– obareey
Dec 6 '18 at 17:51
$begingroup$
I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
$endgroup$
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
$begingroup$
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
$endgroup$
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
add a comment |
$begingroup$
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
$endgroup$
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
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With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
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– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
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– Eduardo Elael
Dec 7 '18 at 18:32
add a comment |
$begingroup$
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
$endgroup$
Some Hints:
Start attempting with a 4x4 diagonal matrix for $A$ for simplicity(Requirement 1&2). You know that you need a positive eigen value for instability (Requirement 3), I would start with $A = text{Diag}(1,-1, -1, -1)$ beacause it is the simplest. As $A² = I$ and $A³ = A$, you get your "Rank of controlability matrix = Rank of obeservability matrix = 2" for free (Requirement 5).
Now you can go to the apparent paradox of having a 1st degree TF with a 4th order system. That means some part of your state is not controllable/observable, e.g. with $C = B^t = [1~0~0~0]$ (Requirement 6). Because I have choose the $C$ and $B$ so that I can control and observe the unstable part of the state, it will be Stabilizable/Detectable (Requirement 4).
Ps: Not sure you are aware, but on your Stabilizable/Detectable definition, that must be true for those values of $s$ equal the unstable eigenvalues (greater or equal than zero).
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
answered Dec 6 '18 at 19:30
Eduardo ElaelEduardo Elael
30115
30115
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
add a comment |
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
I understood requirements 1,2,3,5, but i'm confused about 6 and 4. With your choice of $A$, which states are not controllable/observable? $x_1$ because of the positive eigenvalue? I also do not understand the last comment on Stabilizable/Detectable definitions. Can you please clarify that too?
$endgroup$
– Lod
Dec 7 '18 at 16:44
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
With my choice of A, $x_2,x_3,x_4$ are "steered together". And the last comment was just because in your definition of Stabilizable you said "Stabilizable if Rank ... =4".. But it only needs to be 4 for $s=1$ which is the unstable eigenvalue.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:06
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
So how to know which states are the unobservable/uncontrollable states? I thought since you chose $C$ to be $[1~ 0 ~0 ~0]$ then you are only observing $x_1$ since it is the unobservable state. No?
$endgroup$
– Lod
Dec 7 '18 at 18:19
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
$begingroup$
You are correct, by my choice of C I'm only observing $x_1$. As, in some sense, $x_2, x_3, x_4$ would be multiples of each other, I could not observe all of them (distinguish), but I could still observe at least one of them, if I have chosen $C=[1~1~0~0]$ for example.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 18:32
add a comment |
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each of these requirement are describing conditions on your matrices, e.g. 4th order = start with A as a 4x4 matrix.
$endgroup$
– Eduardo Elael
Dec 5 '18 at 18:16
$begingroup$
@EduardoElael Please refer to the updated post. I do get these requirement but I am still unable to actually construct the linear system.
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– Lod
Dec 5 '18 at 19:46
$begingroup$
In Req 3. positiveness of "at least" one of the eigenvalues is enough, you don't need all eigenvalues being positive. In Req. 4 you only need to check the unstable eigenvalues. Req 6. means only one eigenvalue is "both" controllable and observable.
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– obareey
Dec 6 '18 at 17:51
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I don't get your comments about requirement 4 and 6. Can you explain it further?
$endgroup$
– Lod
Dec 7 '18 at 16:32
$begingroup$
$sI-A$ is always full rank if $s$ is different from the eigenvalues of $A$, by definition. So for the rank condition you only need to check for $s=lambda_i(A)$, eigenvalues of $A$. But you don't need to check for the stable eigenvalues, because you are not interested in them being controllable or not, since they are already stable. So, for stabilizability, you only need to check the unstable eigenvalues being controllable.
$endgroup$
– obareey
Dec 8 '18 at 11:36