Mixing
Transfer function is $0$?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Given the continuous time state space model:
$dotx(t)=Ax(t)+Bu(t)$, $quad y(t)=Cx(t), quad tin R^+$
with:
$left[
beginarrayc
A & B \
hline
C & \
endarray
right]$ =
$ left[
beginarrayccc
0 & 1 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 \
hline
0 & 1 & 0 \
endarray
right]$
Using: $C(sI-A)^-1B$ yields the following transfer function:
$0$.
I'm used to seeing $s$-terms in the denominator, for instance: $frac1s+7$.
Which then provides the pole location(s) and thus the stability.
What does this zero say about stability?
linear-algebra control-theory linear-control
edited Aug 3 at 12:34
Sou
2,6192820
asked Aug 3 at 11:38
user463102
1157
add a comment |Â
up vote
2
down vote
favorite
Given the continuous time state space model:
$dotx(t)=Ax(t)+Bu(t)$, $quad y(t)=Cx(t), quad tin R^+$
with:
$left[
beginarrayc
A & B \
hline
C & \
endarray
right]$ =
$ left[
beginarrayccc
0 & 1 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 \
hline
0 & 1 & 0 \
endarray
right]$
Using: $C(sI-A)^-1B$ yields the following transfer function:
$0$.
I'm used to seeing $s$-terms in the denominator, for instance: $frac1s+7$.
Which then provides the pole location(s) and thus the stability.
What does this zero say about stability?
linear-algebra control-theory linear-control
edited Aug 3 at 12:34
Sou
2,6192820
asked Aug 3 at 11:38
user463102
1157
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the continuous time state space model:
$dotx(t)=Ax(t)+Bu(t)$, $quad y(t)=Cx(t), quad tin R^+$
with:
$left[
beginarrayc
A & B \
hline
C & \
endarray
right]$ =
$ left[
beginarrayccc
0 & 1 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 \
hline
0 & 1 & 0 \
endarray
right]$
Using: $C(sI-A)^-1B$ yields the following transfer function:
$0$.
I'm used to seeing $s$-terms in the denominator, for instance: $frac1s+7$.
Which then provides the pole location(s) and thus the stability.
What does this zero say about stability?
linear-algebra control-theory linear-control
edited Aug 3 at 12:34
Sou
2,6192820
asked Aug 3 at 11:38
user463102
1157
Given the continuous time state space model:
$dotx(t)=Ax(t)+Bu(t)$, $quad y(t)=Cx(t), quad tin R^+$
with:
$left[
beginarrayc
A & B \
hline
C & \
endarray
right]$ =
$ left[
beginarrayccc
0 & 1 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 \
hline
0 & 1 & 0 \
endarray
right]$
Using: $C(sI-A)^-1B$ yields the following transfer function:
$0$.
I'm used to seeing $s$-terms in the denominator, for instance: $frac1s+7$.
Which then provides the pole location(s) and thus the stability.
What does this zero say about stability?
linear-algebra control-theory linear-control
edited Aug 3 at 12:34
Sou
2,6192820
asked Aug 3 at 11:38
user463102
1157
edited Aug 3 at 12:34
Sou
2,6192820
edited Aug 3 at 12:34
Sou
2,6192820
edited Aug 3 at 12:34
Sou
2,6192820
2,6192820
asked Aug 3 at 11:38
user463102
1157
asked Aug 3 at 11:38
user463102
1157
asked Aug 3 at 11:38
user463102
1157
1157
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your equations give:
$$y = x_2\sx_2=x_3\sx_3=0$$
Which implies that:
$$s^2y=0$$
Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent:
$$exists n | A^n = mathbb0 $$
Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.
answered Aug 3 at 11:58
Niki Di Giano
650211
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your equations give:
$$y = x_2\sx_2=x_3\sx_3=0$$
Which implies that:
$$s^2y=0$$
Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent:
$$exists n | A^n = mathbb0 $$
Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.
answered Aug 3 at 11:58
Niki Di Giano
650211
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
add a comment |Â
up vote
1
down vote
accepted
Your equations give:
$$y = x_2\sx_2=x_3\sx_3=0$$
Which implies that:
$$s^2y=0$$
Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent:
$$exists n | A^n = mathbb0 $$
Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.
answered Aug 3 at 11:58
Niki Di Giano
650211
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your equations give:
$$y = x_2\sx_2=x_3\sx_3=0$$
Which implies that:
$$s^2y=0$$
Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent:
$$exists n | A^n = mathbb0 $$
Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.
answered Aug 3 at 11:58
Niki Di Giano
650211
Your equations give:
$$y = x_2\sx_2=x_3\sx_3=0$$
Which implies that:
$$s^2y=0$$
Thus the transfer function is indeed zero.
Such systems are called "finite-memory" (particularly for discrete-time systems) and their matrices are nilpotent:
$$exists n | A^n = mathbb0 $$
Finite-memory systems have null output in a finite amount of time (as opposed to the usual, asymptotic behaviour of stable systems) when input is also null.
answered Aug 3 at 11:58
Niki Di Giano
650211
answered Aug 3 at 11:58
Niki Di Giano
650211
answered Aug 3 at 11:58
Niki Di Giano
650211
answered Aug 3 at 11:58
Niki Di Giano
650211
650211
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
add a comment |Â
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
So does that mean this system is stable, since the output is always zero? It's not asymptotically stable right?
– user463102
Aug 3 at 12:02
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
Yes, it is simply stable. You can try this with a discrete-time system and check for yourself that the "free movement" of the system is going to drop and reach exactly zero after finite time.
– Niki Di Giano
Aug 3 at 12:07
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I've checked it with Matlab and you're right. This makes it a lot clearer. Thanks a lot Niki.
– user463102
Aug 3 at 12:13
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
I'm really glad I could help.
– Niki Di Giano
Aug 3 at 12:17
1
1
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
$s^2,y=0$ means that the time derivate of the output is constant. If that constant is nonzero then as time goes to infinity the output would go to infinity as well, so the system is (Lyapunov) unstable (but is BIBO stable).
– Kwin van der Veen
Aug 4 at 0:37
add a comment |Â
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