Laplace transform - differential equation
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I have the following differential equation:
$$ddotθ + frac{b}{h}dotθ + frac{a}{h}cos θ = frac{1}{h}u$$
and I want to extract the transfer function (the equation actually describes a control system). Since $θ$ is a function of time ($t$), how do I find the laplace transform of $cos θ$ ? I think the classical Laplace transform of a cosine $frac{s}{s^2 + 1} $ does not apply here.
ordinary-differential-equations laplace-transform nonlinear-system
edited Dec 11 '18 at 15:20
Harry49
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
$endgroup$
add a comment |
$begingroup$
I have the following differential equation:
$$ddotθ + frac{b}{h}dotθ + frac{a}{h}cos θ = frac{1}{h}u$$
and I want to extract the transfer function (the equation actually describes a control system). Since $θ$ is a function of time ($t$), how do I find the laplace transform of $cos θ$ ? I think the classical Laplace transform of a cosine $frac{s}{s^2 + 1} $ does not apply here.
ordinary-differential-equations laplace-transform nonlinear-system
edited Dec 11 '18 at 15:20
Harry49
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
$endgroup$
1
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
1
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22
add a comment |
$begingroup$
I have the following differential equation:
$$ddotθ + frac{b}{h}dotθ + frac{a}{h}cos θ = frac{1}{h}u$$
and I want to extract the transfer function (the equation actually describes a control system). Since $θ$ is a function of time ($t$), how do I find the laplace transform of $cos θ$ ? I think the classical Laplace transform of a cosine $frac{s}{s^2 + 1} $ does not apply here.
ordinary-differential-equations laplace-transform nonlinear-system
edited Dec 11 '18 at 15:20
Harry49
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
$endgroup$
I have the following differential equation:
$$ddotθ + frac{b}{h}dotθ + frac{a}{h}cos θ = frac{1}{h}u$$
and I want to extract the transfer function (the equation actually describes a control system). Since $θ$ is a function of time ($t$), how do I find the laplace transform of $cos θ$ ? I think the classical Laplace transform of a cosine $frac{s}{s^2 + 1} $ does not apply here.
ordinary-differential-equations laplace-transform nonlinear-system
ordinary-differential-equations laplace-transform nonlinear-system
edited Dec 11 '18 at 15:20
Harry49
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
edited Dec 11 '18 at 15:20
Harry49
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
edited Dec 11 '18 at 15:20
Harry49
7,26831240
edited Dec 11 '18 at 15:20
Harry49
7,26831240
edited Dec 11 '18 at 15:20
Harry49
7,26831240
7,26831240
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
asked Dec 11 '18 at 11:49
Teo ProtoulisTeo Protoulis
94
94
1
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
1
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22
add a comment |
1
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
1
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22
1
1
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
1
1
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22
add a comment |
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1
$begingroup$
We cannot build a transfer function for $theta(t)$ because the cited DE is not linear in $theta(t)$
$endgroup$
– Cesareo
Dec 11 '18 at 12:39
1
$begingroup$
If $theta$ stays close to a given value, then $costheta$ can be linearized using its Taylor series approximation. Then, the transfer function of the linearized oscillator can be obtained by Laplace transform.
$endgroup$
– Harry49
Dec 11 '18 at 15:22