Static behaviour and dynamic behaviour of a system
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I have a system with the static behavior:
$y(t)=a+by_{0}(t)$
where $y(t)$ is the output and $y_{0}(t)$ is the input.
The dynamic behavior of this system is:
$G(s)=frac{K}{1+Ts}$
To this equation corresponds a dynamic behavior expressed in time domain using a derivative.
If i put the derivative to 0 , should i get the linear equation that represents the static behavior of my system ?
ordinary-differential-equations dynamical-systems control-theory
asked Jan 12 at 19:02
JhdoeJhdoe
11
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add a comment |
$begingroup$
I have a system with the static behavior:
$y(t)=a+by_{0}(t)$
where $y(t)$ is the output and $y_{0}(t)$ is the input.
The dynamic behavior of this system is:
$G(s)=frac{K}{1+Ts}$
To this equation corresponds a dynamic behavior expressed in time domain using a derivative.
If i put the derivative to 0 , should i get the linear equation that represents the static behavior of my system ?
ordinary-differential-equations dynamical-systems control-theory
asked Jan 12 at 19:02
JhdoeJhdoe
11
$endgroup$
1
$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
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– SampleTime
Jan 13 at 14:55
add a comment |
$begingroup$
I have a system with the static behavior:
$y(t)=a+by_{0}(t)$
where $y(t)$ is the output and $y_{0}(t)$ is the input.
The dynamic behavior of this system is:
$G(s)=frac{K}{1+Ts}$
To this equation corresponds a dynamic behavior expressed in time domain using a derivative.
If i put the derivative to 0 , should i get the linear equation that represents the static behavior of my system ?
ordinary-differential-equations dynamical-systems control-theory
asked Jan 12 at 19:02
JhdoeJhdoe
11
$endgroup$
I have a system with the static behavior:
$y(t)=a+by_{0}(t)$
where $y(t)$ is the output and $y_{0}(t)$ is the input.
The dynamic behavior of this system is:
$G(s)=frac{K}{1+Ts}$
To this equation corresponds a dynamic behavior expressed in time domain using a derivative.
If i put the derivative to 0 , should i get the linear equation that represents the static behavior of my system ?
ordinary-differential-equations dynamical-systems control-theory
ordinary-differential-equations dynamical-systems control-theory
asked Jan 12 at 19:02
JhdoeJhdoe
11
asked Jan 12 at 19:02
JhdoeJhdoe
11
asked Jan 12 at 19:02
JhdoeJhdoe
11
asked Jan 12 at 19:02
JhdoeJhdoe
11
asked Jan 12 at 19:02
JhdoeJhdoe
11
11
1
$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
$endgroup$
– SampleTime
Jan 13 at 14:55
add a comment |
1
$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
$endgroup$
– SampleTime
Jan 13 at 14:55
1
1
$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
$endgroup$
– SampleTime
Jan 13 at 14:55
$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
$endgroup$
– SampleTime
Jan 13 at 14:55
add a comment |
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$begingroup$
Your equation doesn't correspond to a dynamic transfer function since it doesn't contain any derivatives. I guess you mean something like $dot{x}(t) = (K u(t) - x(t))/T$ with output $y(t) = x(t)$? That would be the correct differential (and output) equation of your transfer function $G(s)$. I didn't downvote, however, for me it is unclear what you are asking?
$endgroup$
– SampleTime
Jan 13 at 14:55