A state-space representation of an integro-differential equation implies a false statement
$begingroup$
I would like to convert the equation $ddot{y}+int_0^t y(tau)dtau=0$ to state-space representation. Below, I present my attempt, which seems to be contradicting, and then ask my question at the end.
Coversion
Let $x_1=y$ and $x_2=dot{y}$. Also, let $x=left[begin{array}{c}x_1\x_2end{array}right]$ and so:
$$dot{x}=
left[begin{array}{c}x_2\-int_0^t x_1 dtau end{array}right]
$$
Take Laplace transform, assuming 0 initial conditions:
$$
sX=left[begin{array}{c}X_2\ -frac{X_1}{s} end{array}right]=
left[begin{array}{cc}0 & 1\-frac{1}{s} & 0end{array}right]X
$$
Inverse Laplace transform:
$$
dot{x}=
left[begin{array}{cc}0 & delta(t)\-1 & 0end{array}right]x
$$
where $delta(t)$ is the delta-dirac function (infinity at 0, and 0 elsewhere).
Question
The last equation implies $dot{y}=delta(t)dot{y}$ and this implies $1=delta(t)$, a false statement.
Please let me know the error in my logic.
Comments
(1) I know I can model the original equation using another state assignments without running into such problem of contradicting statements. For example, I can use the states $x_1=int_0^t y(tau)dtau$, $x_2=y$, and $x_3=dot{y}$. This state assignment will not result into a problem like the former one. However, this assignment results in 3-by-3 system, whereas the former results in 2-by-2 system.
(2) I also know I can differentiate the original ODE to get rid of the integral, but this will also result in a 3-by-3 system.
(3) The bottomline here: The main objective of this question is to uncover the error in my first attempt that used Laplace transform.
dynamical-systems laplace-transform control-theory inverselaplace
asked Dec 7 '18 at 18:39
user8396743user8396743
134
$endgroup$
|
show 4 more comments
$begingroup$
I would like to convert the equation $ddot{y}+int_0^t y(tau)dtau=0$ to state-space representation. Below, I present my attempt, which seems to be contradicting, and then ask my question at the end.
Coversion
Let $x_1=y$ and $x_2=dot{y}$. Also, let $x=left[begin{array}{c}x_1\x_2end{array}right]$ and so:
$$dot{x}=
left[begin{array}{c}x_2\-int_0^t x_1 dtau end{array}right]
$$
Take Laplace transform, assuming 0 initial conditions:
$$
sX=left[begin{array}{c}X_2\ -frac{X_1}{s} end{array}right]=
left[begin{array}{cc}0 & 1\-frac{1}{s} & 0end{array}right]X
$$
Inverse Laplace transform:
$$
dot{x}=
left[begin{array}{cc}0 & delta(t)\-1 & 0end{array}right]x
$$
where $delta(t)$ is the delta-dirac function (infinity at 0, and 0 elsewhere).
Question
The last equation implies $dot{y}=delta(t)dot{y}$ and this implies $1=delta(t)$, a false statement.
Please let me know the error in my logic.
Comments
(1) I know I can model the original equation using another state assignments without running into such problem of contradicting statements. For example, I can use the states $x_1=int_0^t y(tau)dtau$, $x_2=y$, and $x_3=dot{y}$. This state assignment will not result into a problem like the former one. However, this assignment results in 3-by-3 system, whereas the former results in 2-by-2 system.
(2) I also know I can differentiate the original ODE to get rid of the integral, but this will also result in a 3-by-3 system.
(3) The bottomline here: The main objective of this question is to uncover the error in my first attempt that used Laplace transform.
dynamical-systems laplace-transform control-theory inverselaplace
asked Dec 7 '18 at 18:39
user8396743user8396743
134
$endgroup$
$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
1
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46
|
show 4 more comments
$begingroup$
I would like to convert the equation $ddot{y}+int_0^t y(tau)dtau=0$ to state-space representation. Below, I present my attempt, which seems to be contradicting, and then ask my question at the end.
Coversion
Let $x_1=y$ and $x_2=dot{y}$. Also, let $x=left[begin{array}{c}x_1\x_2end{array}right]$ and so:
$$dot{x}=
left[begin{array}{c}x_2\-int_0^t x_1 dtau end{array}right]
$$
Take Laplace transform, assuming 0 initial conditions:
$$
sX=left[begin{array}{c}X_2\ -frac{X_1}{s} end{array}right]=
left[begin{array}{cc}0 & 1\-frac{1}{s} & 0end{array}right]X
$$
Inverse Laplace transform:
$$
dot{x}=
left[begin{array}{cc}0 & delta(t)\-1 & 0end{array}right]x
$$
where $delta(t)$ is the delta-dirac function (infinity at 0, and 0 elsewhere).
Question
The last equation implies $dot{y}=delta(t)dot{y}$ and this implies $1=delta(t)$, a false statement.
Please let me know the error in my logic.
Comments
(1) I know I can model the original equation using another state assignments without running into such problem of contradicting statements. For example, I can use the states $x_1=int_0^t y(tau)dtau$, $x_2=y$, and $x_3=dot{y}$. This state assignment will not result into a problem like the former one. However, this assignment results in 3-by-3 system, whereas the former results in 2-by-2 system.
(2) I also know I can differentiate the original ODE to get rid of the integral, but this will also result in a 3-by-3 system.
(3) The bottomline here: The main objective of this question is to uncover the error in my first attempt that used Laplace transform.
dynamical-systems laplace-transform control-theory inverselaplace
asked Dec 7 '18 at 18:39
user8396743user8396743
134
$endgroup$
I would like to convert the equation $ddot{y}+int_0^t y(tau)dtau=0$ to state-space representation. Below, I present my attempt, which seems to be contradicting, and then ask my question at the end.
Coversion
Let $x_1=y$ and $x_2=dot{y}$. Also, let $x=left[begin{array}{c}x_1\x_2end{array}right]$ and so:
$$dot{x}=
left[begin{array}{c}x_2\-int_0^t x_1 dtau end{array}right]
$$
Take Laplace transform, assuming 0 initial conditions:
$$
sX=left[begin{array}{c}X_2\ -frac{X_1}{s} end{array}right]=
left[begin{array}{cc}0 & 1\-frac{1}{s} & 0end{array}right]X
$$
Inverse Laplace transform:
$$
dot{x}=
left[begin{array}{cc}0 & delta(t)\-1 & 0end{array}right]x
$$
where $delta(t)$ is the delta-dirac function (infinity at 0, and 0 elsewhere).
Question
The last equation implies $dot{y}=delta(t)dot{y}$ and this implies $1=delta(t)$, a false statement.
Please let me know the error in my logic.
Comments
(1) I know I can model the original equation using another state assignments without running into such problem of contradicting statements. For example, I can use the states $x_1=int_0^t y(tau)dtau$, $x_2=y$, and $x_3=dot{y}$. This state assignment will not result into a problem like the former one. However, this assignment results in 3-by-3 system, whereas the former results in 2-by-2 system.
(2) I also know I can differentiate the original ODE to get rid of the integral, but this will also result in a 3-by-3 system.
(3) The bottomline here: The main objective of this question is to uncover the error in my first attempt that used Laplace transform.
dynamical-systems laplace-transform control-theory inverselaplace
dynamical-systems laplace-transform control-theory inverselaplace
asked Dec 7 '18 at 18:39
user8396743user8396743
134
asked Dec 7 '18 at 18:39
user8396743user8396743
134
edited Dec 7 '18 at 18:47
user8396743
asked Dec 7 '18 at 18:39
user8396743user8396743
134
asked Dec 7 '18 at 18:39
user8396743user8396743
134
asked Dec 7 '18 at 18:39
user8396743user8396743
134
134
$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
1
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46
|
show 4 more comments
$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
1
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46
$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
1
1
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46
|
show 4 more comments
1 Answer
1
active
oldest
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$begingroup$
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
$endgroup$
add a comment |
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$begingroup$
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
$endgroup$
add a comment |
$begingroup$
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
$endgroup$
add a comment |
$begingroup$
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
$endgroup$
A very short and sketchy answer, because I don't have enought time right now, sorry.
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x′=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing.
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
answered Dec 7 '18 at 19:22
FedericoFederico
4,919514
4,919514
add a comment |
add a comment |
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$begingroup$
The problem is that the transform of a product isn't the product of the transforms
$endgroup$
– Federico
Dec 7 '18 at 18:58
$begingroup$
You end up with the convolution of $delta$ and $x$, which is $x$. Mystery solved
$endgroup$
– Federico
Dec 7 '18 at 18:59
$begingroup$
@Federico I can't follow. Can you please elaborate?
$endgroup$
– user8396743
Dec 7 '18 at 19:07
$begingroup$
You wrote $sX=AX$ for some matrix $A$. Then you said that the inverse transforms brings you to $x'=Bx$ for some other matrix $B$. That is not correct. To transform a product ($AX$) you get a convolution appearing
$endgroup$
– Federico
Dec 7 '18 at 19:09
1
$begingroup$
I guess you're right @obareey . However, I thought maybe somehow you can embed the third state into the structure of the 2-by-2 A matrix, which may no longer be a constant matrix. I guess this can be done; however, is it really useful? For example, can I get the eigenvalues of the non-constant A using the conventional method, i.e., $det (A-lambda I)=0$. I guess not, right? In other words, I lost the advantages of the nice LTI state-space form.
$endgroup$
– user8396743
Dec 8 '18 at 8:46