Inverse Laplace transform of rational function
$begingroup$
Is it true that the inverse Laplace transform of a proper rational function (where all the coefficients are real and positive),
begin{equation}
s mapsto frac{a_1 , s^{m-1} + a_2 , s^{m-2} + dotsb + a_{m}}{s^m + b_1 s^{m-1} + dotsb + b_{m}},
end{equation}
can be expressed as $langle b, e^{A t} b rangle$, for a matrix $A in mathbb R^{n times n}$ and a vector $b in mathbb R^n$ and $n geq m$?
I would say that it is, by partial fraction decomposition, inverse Laplace transform of the partial fractions, and using an ansatz of $A$ in normal Jordan form, but would like confirmation or a reference.
laplace-transform
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
$endgroup$
|
show 1 more comment
$begingroup$
Is it true that the inverse Laplace transform of a proper rational function (where all the coefficients are real and positive),
begin{equation}
s mapsto frac{a_1 , s^{m-1} + a_2 , s^{m-2} + dotsb + a_{m}}{s^m + b_1 s^{m-1} + dotsb + b_{m}},
end{equation}
can be expressed as $langle b, e^{A t} b rangle$, for a matrix $A in mathbb R^{n times n}$ and a vector $b in mathbb R^n$ and $n geq m$?
I would say that it is, by partial fraction decomposition, inverse Laplace transform of the partial fractions, and using an ansatz of $A$ in normal Jordan form, but would like confirmation or a reference.
laplace-transform
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
$endgroup$
$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23
|
show 1 more comment
$begingroup$
Is it true that the inverse Laplace transform of a proper rational function (where all the coefficients are real and positive),
begin{equation}
s mapsto frac{a_1 , s^{m-1} + a_2 , s^{m-2} + dotsb + a_{m}}{s^m + b_1 s^{m-1} + dotsb + b_{m}},
end{equation}
can be expressed as $langle b, e^{A t} b rangle$, for a matrix $A in mathbb R^{n times n}$ and a vector $b in mathbb R^n$ and $n geq m$?
I would say that it is, by partial fraction decomposition, inverse Laplace transform of the partial fractions, and using an ansatz of $A$ in normal Jordan form, but would like confirmation or a reference.
laplace-transform
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
$endgroup$
Is it true that the inverse Laplace transform of a proper rational function (where all the coefficients are real and positive),
begin{equation}
s mapsto frac{a_1 , s^{m-1} + a_2 , s^{m-2} + dotsb + a_{m}}{s^m + b_1 s^{m-1} + dotsb + b_{m}},
end{equation}
can be expressed as $langle b, e^{A t} b rangle$, for a matrix $A in mathbb R^{n times n}$ and a vector $b in mathbb R^n$ and $n geq m$?
I would say that it is, by partial fraction decomposition, inverse Laplace transform of the partial fractions, and using an ansatz of $A$ in normal Jordan form, but would like confirmation or a reference.
laplace-transform
laplace-transform
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
edited Jan 12 at 23:00
Roberto Rastapopoulos
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
asked Jan 12 at 20:07
Roberto RastapopoulosRoberto Rastapopoulos
955425
955425
$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23
|
show 1 more comment
$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23
$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23
|
show 1 more comment
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$begingroup$
We can always fill up $A$ and $b$ with as many zeros as we wish.
$endgroup$
– A.Γ.
Jan 12 at 20:24
$begingroup$
Yes, but I don't see your point?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:25
$begingroup$
We can make $n$ as large as we want. Isn't it the question?
$endgroup$
– A.Γ.
Jan 12 at 20:27
$begingroup$
The question is: Are there $A, b, n$ (with $n$ possibly larger than $m$), such that the Laplace transform of $<b, e^{-At} b>$ is the rational function given?
$endgroup$
– Roberto Rastapopoulos
Jan 12 at 20:32
$begingroup$
How do you think to write say $e^{-t}-e^{-2t}$ as $langle b,e^{At}brangle$? At $t=0$ it gives $|b|=0$. Do I miss something here?
$endgroup$
– A.Γ.
Jan 12 at 21:23