Exosystem for reference generation
$begingroup$
Recently I have read a paper where an LTI system of the form
$$
begin{align}
dot{x}_p &= A_p x_p + B_p u \
y &= C_p x_p
end{align}tag{1}
$$
for the control plant was considered. In order to analyze the system for different reference inputs, the author used the notation of what they called exosystem defined as:
$$
begin{align}
dot{x}_e &= A_e x_e \
r &= C_e x_e
end{align}tag{2}
$$
which serves as a reference generator. Assume a P controller with $u = K(r - y)$, then they combine $(1)$ and $(2)$ to
$$
begin{align}
dot{x} &= A x \
y &= C x
end{align} tag{3}
$$
with $x = [x_p^T , x_e^T]^T$ and $A, C$ the resulting matrices from the feedback interconnection of $(1)$ and $(2)$. So far so good.
But now it is stated that such an exosystem can be used to generate all of the common input functions (step, ramp, sine, etc.) However I cannot see how the system $(2)$ is able to generate even just a simple step for example as it by itself, does not have any input.
Question: How can such an exosystem used to bring a system like $(1)$ to a form like $(3)$ given a specific input (e.g. step)? And how is the exosystem supposed to generate such input functions?
ordinary-differential-equations dynamical-systems control-theory linear-control
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
$endgroup$
add a comment |
$begingroup$
Recently I have read a paper where an LTI system of the form
$$
begin{align}
dot{x}_p &= A_p x_p + B_p u \
y &= C_p x_p
end{align}tag{1}
$$
for the control plant was considered. In order to analyze the system for different reference inputs, the author used the notation of what they called exosystem defined as:
$$
begin{align}
dot{x}_e &= A_e x_e \
r &= C_e x_e
end{align}tag{2}
$$
which serves as a reference generator. Assume a P controller with $u = K(r - y)$, then they combine $(1)$ and $(2)$ to
$$
begin{align}
dot{x} &= A x \
y &= C x
end{align} tag{3}
$$
with $x = [x_p^T , x_e^T]^T$ and $A, C$ the resulting matrices from the feedback interconnection of $(1)$ and $(2)$. So far so good.
But now it is stated that such an exosystem can be used to generate all of the common input functions (step, ramp, sine, etc.) However I cannot see how the system $(2)$ is able to generate even just a simple step for example as it by itself, does not have any input.
Question: How can such an exosystem used to bring a system like $(1)$ to a form like $(3)$ given a specific input (e.g. step)? And how is the exosystem supposed to generate such input functions?
ordinary-differential-equations dynamical-systems control-theory linear-control
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
$endgroup$
add a comment |
$begingroup$
Recently I have read a paper where an LTI system of the form
$$
begin{align}
dot{x}_p &= A_p x_p + B_p u \
y &= C_p x_p
end{align}tag{1}
$$
for the control plant was considered. In order to analyze the system for different reference inputs, the author used the notation of what they called exosystem defined as:
$$
begin{align}
dot{x}_e &= A_e x_e \
r &= C_e x_e
end{align}tag{2}
$$
which serves as a reference generator. Assume a P controller with $u = K(r - y)$, then they combine $(1)$ and $(2)$ to
$$
begin{align}
dot{x} &= A x \
y &= C x
end{align} tag{3}
$$
with $x = [x_p^T , x_e^T]^T$ and $A, C$ the resulting matrices from the feedback interconnection of $(1)$ and $(2)$. So far so good.
But now it is stated that such an exosystem can be used to generate all of the common input functions (step, ramp, sine, etc.) However I cannot see how the system $(2)$ is able to generate even just a simple step for example as it by itself, does not have any input.
Question: How can such an exosystem used to bring a system like $(1)$ to a form like $(3)$ given a specific input (e.g. step)? And how is the exosystem supposed to generate such input functions?
ordinary-differential-equations dynamical-systems control-theory linear-control
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
$endgroup$
Recently I have read a paper where an LTI system of the form
$$
begin{align}
dot{x}_p &= A_p x_p + B_p u \
y &= C_p x_p
end{align}tag{1}
$$
for the control plant was considered. In order to analyze the system for different reference inputs, the author used the notation of what they called exosystem defined as:
$$
begin{align}
dot{x}_e &= A_e x_e \
r &= C_e x_e
end{align}tag{2}
$$
which serves as a reference generator. Assume a P controller with $u = K(r - y)$, then they combine $(1)$ and $(2)$ to
$$
begin{align}
dot{x} &= A x \
y &= C x
end{align} tag{3}
$$
with $x = [x_p^T , x_e^T]^T$ and $A, C$ the resulting matrices from the feedback interconnection of $(1)$ and $(2)$. So far so good.
But now it is stated that such an exosystem can be used to generate all of the common input functions (step, ramp, sine, etc.) However I cannot see how the system $(2)$ is able to generate even just a simple step for example as it by itself, does not have any input.
Question: How can such an exosystem used to bring a system like $(1)$ to a form like $(3)$ given a specific input (e.g. step)? And how is the exosystem supposed to generate such input functions?
ordinary-differential-equations dynamical-systems control-theory linear-control
ordinary-differential-equations dynamical-systems control-theory linear-control
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
edited Dec 10 '18 at 22:53
SampleTime
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
asked Dec 10 '18 at 22:35
SampleTimeSampleTime
53539
53539
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you set the initial conditions at $t=0$ of $x_p$ to zero, then system $(3)$ for $t>0$ behaves the same as when the input from $(2)$ would have been multiplied by a step function. It can be noted that the initial conditions of $x_e$ are not zero, unless a $r=0$ is desired. So for example to generate a step or a ramp then the reference should be $r=1$ or $r=t$ respectively which can be obtained by choosing appropriate $A_e$, $C_e$ and $x_e(0)$, namely after multiplying these with a step function you get the desired inputs.
If you want $r$ to be an $n$th order polynomial in time then you could for example choose $A_e$ to be a Jordan block of size $n+1$ with eigenvalue zero. For example when $n=2$ one gets
$$
x_e(t)=e^{A_e,t} x_e(0)=
begin{bmatrix}
1 & t & tfrac12 t^2 \
0 & 1 & t \
0 & 0 & 1
end{bmatrix}x_e(0)
$$
Now by choosing $x_e(0)$ and $C_e$ one can get any linear combination of $1$, $t$ and $t^2$. For polynomial times sinusoids you would need to use completely imaginary values or real Jordan blocks for the eigenvalues instead of zero.
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
add a comment |
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1 Answer
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$begingroup$
If you set the initial conditions at $t=0$ of $x_p$ to zero, then system $(3)$ for $t>0$ behaves the same as when the input from $(2)$ would have been multiplied by a step function. It can be noted that the initial conditions of $x_e$ are not zero, unless a $r=0$ is desired. So for example to generate a step or a ramp then the reference should be $r=1$ or $r=t$ respectively which can be obtained by choosing appropriate $A_e$, $C_e$ and $x_e(0)$, namely after multiplying these with a step function you get the desired inputs.
If you want $r$ to be an $n$th order polynomial in time then you could for example choose $A_e$ to be a Jordan block of size $n+1$ with eigenvalue zero. For example when $n=2$ one gets
$$
x_e(t)=e^{A_e,t} x_e(0)=
begin{bmatrix}
1 & t & tfrac12 t^2 \
0 & 1 & t \
0 & 0 & 1
end{bmatrix}x_e(0)
$$
Now by choosing $x_e(0)$ and $C_e$ one can get any linear combination of $1$, $t$ and $t^2$. For polynomial times sinusoids you would need to use completely imaginary values or real Jordan blocks for the eigenvalues instead of zero.
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
add a comment |
$begingroup$
If you set the initial conditions at $t=0$ of $x_p$ to zero, then system $(3)$ for $t>0$ behaves the same as when the input from $(2)$ would have been multiplied by a step function. It can be noted that the initial conditions of $x_e$ are not zero, unless a $r=0$ is desired. So for example to generate a step or a ramp then the reference should be $r=1$ or $r=t$ respectively which can be obtained by choosing appropriate $A_e$, $C_e$ and $x_e(0)$, namely after multiplying these with a step function you get the desired inputs.
If you want $r$ to be an $n$th order polynomial in time then you could for example choose $A_e$ to be a Jordan block of size $n+1$ with eigenvalue zero. For example when $n=2$ one gets
$$
x_e(t)=e^{A_e,t} x_e(0)=
begin{bmatrix}
1 & t & tfrac12 t^2 \
0 & 1 & t \
0 & 0 & 1
end{bmatrix}x_e(0)
$$
Now by choosing $x_e(0)$ and $C_e$ one can get any linear combination of $1$, $t$ and $t^2$. For polynomial times sinusoids you would need to use completely imaginary values or real Jordan blocks for the eigenvalues instead of zero.
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
add a comment |
$begingroup$
If you set the initial conditions at $t=0$ of $x_p$ to zero, then system $(3)$ for $t>0$ behaves the same as when the input from $(2)$ would have been multiplied by a step function. It can be noted that the initial conditions of $x_e$ are not zero, unless a $r=0$ is desired. So for example to generate a step or a ramp then the reference should be $r=1$ or $r=t$ respectively which can be obtained by choosing appropriate $A_e$, $C_e$ and $x_e(0)$, namely after multiplying these with a step function you get the desired inputs.
If you want $r$ to be an $n$th order polynomial in time then you could for example choose $A_e$ to be a Jordan block of size $n+1$ with eigenvalue zero. For example when $n=2$ one gets
$$
x_e(t)=e^{A_e,t} x_e(0)=
begin{bmatrix}
1 & t & tfrac12 t^2 \
0 & 1 & t \
0 & 0 & 1
end{bmatrix}x_e(0)
$$
Now by choosing $x_e(0)$ and $C_e$ one can get any linear combination of $1$, $t$ and $t^2$. For polynomial times sinusoids you would need to use completely imaginary values or real Jordan blocks for the eigenvalues instead of zero.
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
$endgroup$
If you set the initial conditions at $t=0$ of $x_p$ to zero, then system $(3)$ for $t>0$ behaves the same as when the input from $(2)$ would have been multiplied by a step function. It can be noted that the initial conditions of $x_e$ are not zero, unless a $r=0$ is desired. So for example to generate a step or a ramp then the reference should be $r=1$ or $r=t$ respectively which can be obtained by choosing appropriate $A_e$, $C_e$ and $x_e(0)$, namely after multiplying these with a step function you get the desired inputs.
If you want $r$ to be an $n$th order polynomial in time then you could for example choose $A_e$ to be a Jordan block of size $n+1$ with eigenvalue zero. For example when $n=2$ one gets
$$
x_e(t)=e^{A_e,t} x_e(0)=
begin{bmatrix}
1 & t & tfrac12 t^2 \
0 & 1 & t \
0 & 0 & 1
end{bmatrix}x_e(0)
$$
Now by choosing $x_e(0)$ and $C_e$ one can get any linear combination of $1$, $t$ and $t^2$. For polynomial times sinusoids you would need to use completely imaginary values or real Jordan blocks for the eigenvalues instead of zero.
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
edited Dec 12 '18 at 0:49
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
answered Dec 11 '18 at 4:49
Kwin van der VeenKwin van der Veen
5,3952827
5,3952827
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
add a comment |
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
I dont really understand how I can choose $A_e, C_e, x_e(0)$ such that $r$ is, for example a simple ramp signal $r = t$. If I take $A_e neq 0$, $r$ will decay exponentially to zero (or to $pm infty$ if $A_e$ is not Hurwitz). If I take $A_e = 0$, then I can only get a step at $t = 0$ with height depending on $x_e(0)$. Can you give an example how to choose $A_e, C_e, x_e(0)$ to generate $r = t$ with system $(2)$?
$endgroup$
– SampleTime
Dec 11 '18 at 18:32
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
$begingroup$
@SampleTime I have updated my answer.
$endgroup$
– Kwin van der Veen
Dec 12 '18 at 0:49
add a comment |
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