Transfer function of differential equationWhat is the Laplace transform transfer function of affine expression $dot x = bu + c$?Transfer Function, Newtonian CoolingWhat is the Laplace transform transfer function of affine expression $dot x = bu + c$?How to write a transfer function (in Laplace domain) from a set of linear differential equations?Transfer function unity and output function poles are related?Second order differential equation with Heaviside functionGet transfer function of a nonlinear diff. equationLaplace Transforms/Differential Equations using MATLABLine Follower Transfer FunctionLaplace transform - differential equationTransfer Function, what does s Stand for
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Transfer function of differential equation
What is the Laplace transform transfer function of affine expression $dot x = bu + c$?Transfer Function, Newtonian CoolingWhat is the Laplace transform transfer function of affine expression $dot x = bu + c$?How to write a transfer function (in Laplace domain) from a set of linear differential equations?Transfer function unity and output function poles are related?Second order differential equation with Heaviside functionGet transfer function of a nonlinear diff. equationLaplace Transforms/Differential Equations using MATLABLine Follower Transfer FunctionLaplace transform - differential equationTransfer Function, what does s Stand for
$begingroup$
I'm trying to find out the transfer function of simple differential equation:
$$a_0dot y + a_1y=b_0x+b_1$$
The problem is i have no idea what to do with $b_1$.
If we apply the Laplace transform then we will have
$$a_0sY(s) + a_1Y(s)=b_0X(s)+b_1/s$$
The problem is that i have to reach $X(s)/Y(s)$ .
What is the Laplace transform transfer function of affine expression $dot x = bu + c$?
Based on this we do not have a solution?
laplace-transform control-theory
edited 16 hours ago
Ugo Mela
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
$endgroup$
add a comment |
$begingroup$
I'm trying to find out the transfer function of simple differential equation:
$$a_0dot y + a_1y=b_0x+b_1$$
The problem is i have no idea what to do with $b_1$.
If we apply the Laplace transform then we will have
$$a_0sY(s) + a_1Y(s)=b_0X(s)+b_1/s$$
The problem is that i have to reach $X(s)/Y(s)$ .
What is the Laplace transform transfer function of affine expression $dot x = bu + c$?
Based on this we do not have a solution?
laplace-transform control-theory
edited 16 hours ago
Ugo Mela
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
$endgroup$
$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38
add a comment |
$begingroup$
I'm trying to find out the transfer function of simple differential equation:
$$a_0dot y + a_1y=b_0x+b_1$$
The problem is i have no idea what to do with $b_1$.
If we apply the Laplace transform then we will have
$$a_0sY(s) + a_1Y(s)=b_0X(s)+b_1/s$$
The problem is that i have to reach $X(s)/Y(s)$ .
What is the Laplace transform transfer function of affine expression $dot x = bu + c$?
Based on this we do not have a solution?
laplace-transform control-theory
edited 16 hours ago
Ugo Mela
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
$endgroup$
I'm trying to find out the transfer function of simple differential equation:
$$a_0dot y + a_1y=b_0x+b_1$$
The problem is i have no idea what to do with $b_1$.
If we apply the Laplace transform then we will have
$$a_0sY(s) + a_1Y(s)=b_0X(s)+b_1/s$$
The problem is that i have to reach $X(s)/Y(s)$ .
What is the Laplace transform transfer function of affine expression $dot x = bu + c$?
Based on this we do not have a solution?
laplace-transform control-theory
laplace-transform control-theory
edited 16 hours ago
Ugo Mela
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
edited 16 hours ago
Ugo Mela
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
edited 16 hours ago
Ugo Mela
256
edited 16 hours ago
Ugo Mela
256
edited 16 hours ago
Ugo Mela
256
256
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
asked Mar 20 '15 at 12:30
KilyenOrsKilyenOrs
1162
1162
$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38
add a comment |
$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38
$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you could do is to introduce a new variable $barx(t) := x(t) + fracb_1b_0$. This would enable you to express the transfer function by means of $barX(s)$, at least:
$$ fracbarX(s)Y(s) = fraca_0s+a_1b_0. $$
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
What you could do is to introduce a new variable $barx(t) := x(t) + fracb_1b_0$. This would enable you to express the transfer function by means of $barX(s)$, at least:
$$ fracbarX(s)Y(s) = fraca_0s+a_1b_0. $$
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
$endgroup$
add a comment |
$begingroup$
What you could do is to introduce a new variable $barx(t) := x(t) + fracb_1b_0$. This would enable you to express the transfer function by means of $barX(s)$, at least:
$$ fracbarX(s)Y(s) = fraca_0s+a_1b_0. $$
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
$endgroup$
add a comment |
$begingroup$
What you could do is to introduce a new variable $barx(t) := x(t) + fracb_1b_0$. This would enable you to express the transfer function by means of $barX(s)$, at least:
$$ fracbarX(s)Y(s) = fraca_0s+a_1b_0. $$
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
$endgroup$
What you could do is to introduce a new variable $barx(t) := x(t) + fracb_1b_0$. This would enable you to express the transfer function by means of $barX(s)$, at least:
$$ fracbarX(s)Y(s) = fraca_0s+a_1b_0. $$
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
answered Apr 20 '15 at 11:05
Max HerrmannMax Herrmann
704418
704418
add a comment |
add a comment |
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$begingroup$
What would you do without the $b_1$. It is good if you show some effort. Do you know what a transfer function is?
$endgroup$
– mickep
Mar 20 '15 at 12:34
$begingroup$
without $b_1$ it would be $$b_0/(a_0*s+a_1)$$ i think
$endgroup$
– KilyenOrs
Mar 20 '15 at 12:38