Continuous vs. Discrete State-Space Model
$begingroup$
Time-invariant continuous model:
$dot{x}(t) = Ax(t)+Bu(t)$
$y(t) = Cx(t)+Du(t)$
Time-invariant discrete model:
$x_{k+1} = Ax_{k}+Bu_{k}$
$y_{k} = Cx_{k}+Du_{k}$
Why does the continuous model result in a rate of change $dot{x}$, while the discrete model results in a new state $x_{k+1}$?
ordinary-differential-equations control-theory
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
$endgroup$
add a comment |
$begingroup$
Time-invariant continuous model:
$dot{x}(t) = Ax(t)+Bu(t)$
$y(t) = Cx(t)+Du(t)$
Time-invariant discrete model:
$x_{k+1} = Ax_{k}+Bu_{k}$
$y_{k} = Cx_{k}+Du_{k}$
Why does the continuous model result in a rate of change $dot{x}$, while the discrete model results in a new state $x_{k+1}$?
ordinary-differential-equations control-theory
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
$endgroup$
add a comment |
$begingroup$
Time-invariant continuous model:
$dot{x}(t) = Ax(t)+Bu(t)$
$y(t) = Cx(t)+Du(t)$
Time-invariant discrete model:
$x_{k+1} = Ax_{k}+Bu_{k}$
$y_{k} = Cx_{k}+Du_{k}$
Why does the continuous model result in a rate of change $dot{x}$, while the discrete model results in a new state $x_{k+1}$?
ordinary-differential-equations control-theory
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
$endgroup$
Time-invariant continuous model:
$dot{x}(t) = Ax(t)+Bu(t)$
$y(t) = Cx(t)+Du(t)$
Time-invariant discrete model:
$x_{k+1} = Ax_{k}+Bu_{k}$
$y_{k} = Cx_{k}+Du_{k}$
Why does the continuous model result in a rate of change $dot{x}$, while the discrete model results in a new state $x_{k+1}$?
ordinary-differential-equations control-theory
ordinary-differential-equations control-theory
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
asked Dec 24 '18 at 8:30
Johannes KlausJohannes Klaus
225
225
add a comment |
add a comment |
1 Answer
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$begingroup$
It is not how the discretization is normally done, but see what happens if you replace the derivative by the forward difference for small $h>0$
$$
Ax_k+Bu_k=dot x(kh)approxfrac{x(kh+h)-x(kh)}{h}=frac{x_{k+1}-x_k}{h}.
$$
Thus, the rate of change is approximately proportional to the discrete state difference, and one can write
$$
x_{k+1}approx (Ah+I)x_k+Bhu_k.
$$
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
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add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
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active
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votes
$begingroup$
It is not how the discretization is normally done, but see what happens if you replace the derivative by the forward difference for small $h>0$
$$
Ax_k+Bu_k=dot x(kh)approxfrac{x(kh+h)-x(kh)}{h}=frac{x_{k+1}-x_k}{h}.
$$
Thus, the rate of change is approximately proportional to the discrete state difference, and one can write
$$
x_{k+1}approx (Ah+I)x_k+Bhu_k.
$$
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
It is not how the discretization is normally done, but see what happens if you replace the derivative by the forward difference for small $h>0$
$$
Ax_k+Bu_k=dot x(kh)approxfrac{x(kh+h)-x(kh)}{h}=frac{x_{k+1}-x_k}{h}.
$$
Thus, the rate of change is approximately proportional to the discrete state difference, and one can write
$$
x_{k+1}approx (Ah+I)x_k+Bhu_k.
$$
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
$endgroup$
add a comment |
$begingroup$
It is not how the discretization is normally done, but see what happens if you replace the derivative by the forward difference for small $h>0$
$$
Ax_k+Bu_k=dot x(kh)approxfrac{x(kh+h)-x(kh)}{h}=frac{x_{k+1}-x_k}{h}.
$$
Thus, the rate of change is approximately proportional to the discrete state difference, and one can write
$$
x_{k+1}approx (Ah+I)x_k+Bhu_k.
$$
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
$endgroup$
It is not how the discretization is normally done, but see what happens if you replace the derivative by the forward difference for small $h>0$
$$
Ax_k+Bu_k=dot x(kh)approxfrac{x(kh+h)-x(kh)}{h}=frac{x_{k+1}-x_k}{h}.
$$
Thus, the rate of change is approximately proportional to the discrete state difference, and one can write
$$
x_{k+1}approx (Ah+I)x_k+Bhu_k.
$$
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
answered Dec 24 '18 at 9:26
A.Γ.A.Γ.
22.9k32656
22.9k32656
add a comment |
add a comment |
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