how to derive the canonical form of a transfer second order equation?
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How to derive the canonical form of the second order transfer function??
$$frac(omega_n)^2s^2+2zetaomega_ns + (omega_n)^2$$
differential-equations derivatives control-theory derived-functors
asked Jun 3 '14 at 5:44
user25778
160114
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How to derive the canonical form of the second order transfer function??
$$frac(omega_n)^2s^2+2zetaomega_ns + (omega_n)^2$$
differential-equations derivatives control-theory derived-functors
asked Jun 3 '14 at 5:44
user25778
160114
Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to derive the canonical form of the second order transfer function??
$$frac(omega_n)^2s^2+2zetaomega_ns + (omega_n)^2$$
differential-equations derivatives control-theory derived-functors
asked Jun 3 '14 at 5:44
user25778
160114
How to derive the canonical form of the second order transfer function??
$$frac(omega_n)^2s^2+2zetaomega_ns + (omega_n)^2$$
differential-equations derivatives control-theory derived-functors
differential-equations derivatives control-theory derived-functors
asked Jun 3 '14 at 5:44
user25778
160114
asked Jun 3 '14 at 5:44
user25778
160114
asked Jun 3 '14 at 5:44
user25778
160114
asked Jun 3 '14 at 5:44
user25778
160114
asked Jun 3 '14 at 5:44
user25778
160114
160114
Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11
add a comment |Â
Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11
Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11
add a comment |Â
2 Answers
2
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oldest
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0
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The canonical example for this is the spring and damper system you can read about at wikipedia. If a mass $m$ is attached to an ideal, massless, perfrectly elastic spring and displaced by a length $x$, the force on the spring is $-kx$ where $k$ is a constant. Thus, applying Newton's second law, we find:
$$mddotx = -kx$$
Now imagine a dissipative term proportional to the velocity $-gamma dotx$. This is essentially like including a viscous damper on the spring so the overall equation is
$$mddotx + gammadotx + kx = 0,$$
Which is the correct form for the characteristic equation. Now we need to get the coefficients right:
Recall that $s = jomega$ so the dimensions of $s$ are Hz. We can introduce the natural frequency as the parameter for which $ddotx = -omega_n^2x$, and a dimensionless number $zeta$ for which $gamma = omega_nzeta$ so that the units of the terms all match and the dissipative term can be analyzed as a ratio of the damping coefficient to the natural frequency. Since any transfer function is trivial for the free response (by definition) we simply need to add a coefficient for the forcing term, which is typically $K$ but also often written as $omega^2_n$ so that oscillations about the natural frequency are rejected in the closed loop by default. The latter form in particular is a bit nicer because it's clear the overall TF is dimensionless and no ambiguous gain, which might be interpreted to be from some $P$-controller, etc. is accidentally inferred.
answered Jun 1 '16 at 23:18
SZN
2,551620
add a comment |Â
up vote
0
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The transfer function you've posted has this form
$$
G(s) = fracY(s)U(s) = frac b_0 s^2 + b_1 s + b_2 s^2 + a_1 s + a_2
$$
Controllable canonical form can be utilized to represents the system as follows:
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-a_2 & -a_1
endbmatrix \
y&=
beginbmatrix
b_2-a_2b_0 & b_1-a_1b_0
endbmatrix
endalign
$$
where $b_0=0,b_1=0,b_2= omega^2_n ,a_1=2zeta omega_n,a_2=omega^2_n $. The state space representation of the system is
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-omega^2_n & -2zeta omega_n
endbmatrix \
y&=
beginbmatrix
omega^2_n & 0
endbmatrix
endalign
$$
answered Apr 16 at 3:59
CroCo
225218
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The canonical example for this is the spring and damper system you can read about at wikipedia. If a mass $m$ is attached to an ideal, massless, perfrectly elastic spring and displaced by a length $x$, the force on the spring is $-kx$ where $k$ is a constant. Thus, applying Newton's second law, we find:
$$mddotx = -kx$$
Now imagine a dissipative term proportional to the velocity $-gamma dotx$. This is essentially like including a viscous damper on the spring so the overall equation is
$$mddotx + gammadotx + kx = 0,$$
Which is the correct form for the characteristic equation. Now we need to get the coefficients right:
Recall that $s = jomega$ so the dimensions of $s$ are Hz. We can introduce the natural frequency as the parameter for which $ddotx = -omega_n^2x$, and a dimensionless number $zeta$ for which $gamma = omega_nzeta$ so that the units of the terms all match and the dissipative term can be analyzed as a ratio of the damping coefficient to the natural frequency. Since any transfer function is trivial for the free response (by definition) we simply need to add a coefficient for the forcing term, which is typically $K$ but also often written as $omega^2_n$ so that oscillations about the natural frequency are rejected in the closed loop by default. The latter form in particular is a bit nicer because it's clear the overall TF is dimensionless and no ambiguous gain, which might be interpreted to be from some $P$-controller, etc. is accidentally inferred.
answered Jun 1 '16 at 23:18
SZN
2,551620
add a comment |Â
up vote
0
down vote
The canonical example for this is the spring and damper system you can read about at wikipedia. If a mass $m$ is attached to an ideal, massless, perfrectly elastic spring and displaced by a length $x$, the force on the spring is $-kx$ where $k$ is a constant. Thus, applying Newton's second law, we find:
$$mddotx = -kx$$
Now imagine a dissipative term proportional to the velocity $-gamma dotx$. This is essentially like including a viscous damper on the spring so the overall equation is
$$mddotx + gammadotx + kx = 0,$$
Which is the correct form for the characteristic equation. Now we need to get the coefficients right:
Recall that $s = jomega$ so the dimensions of $s$ are Hz. We can introduce the natural frequency as the parameter for which $ddotx = -omega_n^2x$, and a dimensionless number $zeta$ for which $gamma = omega_nzeta$ so that the units of the terms all match and the dissipative term can be analyzed as a ratio of the damping coefficient to the natural frequency. Since any transfer function is trivial for the free response (by definition) we simply need to add a coefficient for the forcing term, which is typically $K$ but also often written as $omega^2_n$ so that oscillations about the natural frequency are rejected in the closed loop by default. The latter form in particular is a bit nicer because it's clear the overall TF is dimensionless and no ambiguous gain, which might be interpreted to be from some $P$-controller, etc. is accidentally inferred.
answered Jun 1 '16 at 23:18
SZN
2,551620
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The canonical example for this is the spring and damper system you can read about at wikipedia. If a mass $m$ is attached to an ideal, massless, perfrectly elastic spring and displaced by a length $x$, the force on the spring is $-kx$ where $k$ is a constant. Thus, applying Newton's second law, we find:
$$mddotx = -kx$$
Now imagine a dissipative term proportional to the velocity $-gamma dotx$. This is essentially like including a viscous damper on the spring so the overall equation is
$$mddotx + gammadotx + kx = 0,$$
Which is the correct form for the characteristic equation. Now we need to get the coefficients right:
Recall that $s = jomega$ so the dimensions of $s$ are Hz. We can introduce the natural frequency as the parameter for which $ddotx = -omega_n^2x$, and a dimensionless number $zeta$ for which $gamma = omega_nzeta$ so that the units of the terms all match and the dissipative term can be analyzed as a ratio of the damping coefficient to the natural frequency. Since any transfer function is trivial for the free response (by definition) we simply need to add a coefficient for the forcing term, which is typically $K$ but also often written as $omega^2_n$ so that oscillations about the natural frequency are rejected in the closed loop by default. The latter form in particular is a bit nicer because it's clear the overall TF is dimensionless and no ambiguous gain, which might be interpreted to be from some $P$-controller, etc. is accidentally inferred.
answered Jun 1 '16 at 23:18
SZN
2,551620
The canonical example for this is the spring and damper system you can read about at wikipedia. If a mass $m$ is attached to an ideal, massless, perfrectly elastic spring and displaced by a length $x$, the force on the spring is $-kx$ where $k$ is a constant. Thus, applying Newton's second law, we find:
$$mddotx = -kx$$
Now imagine a dissipative term proportional to the velocity $-gamma dotx$. This is essentially like including a viscous damper on the spring so the overall equation is
$$mddotx + gammadotx + kx = 0,$$
Which is the correct form for the characteristic equation. Now we need to get the coefficients right:
Recall that $s = jomega$ so the dimensions of $s$ are Hz. We can introduce the natural frequency as the parameter for which $ddotx = -omega_n^2x$, and a dimensionless number $zeta$ for which $gamma = omega_nzeta$ so that the units of the terms all match and the dissipative term can be analyzed as a ratio of the damping coefficient to the natural frequency. Since any transfer function is trivial for the free response (by definition) we simply need to add a coefficient for the forcing term, which is typically $K$ but also often written as $omega^2_n$ so that oscillations about the natural frequency are rejected in the closed loop by default. The latter form in particular is a bit nicer because it's clear the overall TF is dimensionless and no ambiguous gain, which might be interpreted to be from some $P$-controller, etc. is accidentally inferred.
answered Jun 1 '16 at 23:18
SZN
2,551620
answered Jun 1 '16 at 23:18
SZN
2,551620
answered Jun 1 '16 at 23:18
SZN
2,551620
answered Jun 1 '16 at 23:18
SZN
2,551620
2,551620
add a comment |Â
add a comment |Â
up vote
0
down vote
The transfer function you've posted has this form
$$
G(s) = fracY(s)U(s) = frac b_0 s^2 + b_1 s + b_2 s^2 + a_1 s + a_2
$$
Controllable canonical form can be utilized to represents the system as follows:
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-a_2 & -a_1
endbmatrix \
y&=
beginbmatrix
b_2-a_2b_0 & b_1-a_1b_0
endbmatrix
endalign
$$
where $b_0=0,b_1=0,b_2= omega^2_n ,a_1=2zeta omega_n,a_2=omega^2_n $. The state space representation of the system is
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-omega^2_n & -2zeta omega_n
endbmatrix \
y&=
beginbmatrix
omega^2_n & 0
endbmatrix
endalign
$$
answered Apr 16 at 3:59
CroCo
225218
add a comment |Â
up vote
0
down vote
The transfer function you've posted has this form
$$
G(s) = fracY(s)U(s) = frac b_0 s^2 + b_1 s + b_2 s^2 + a_1 s + a_2
$$
Controllable canonical form can be utilized to represents the system as follows:
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-a_2 & -a_1
endbmatrix \
y&=
beginbmatrix
b_2-a_2b_0 & b_1-a_1b_0
endbmatrix
endalign
$$
where $b_0=0,b_1=0,b_2= omega^2_n ,a_1=2zeta omega_n,a_2=omega^2_n $. The state space representation of the system is
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-omega^2_n & -2zeta omega_n
endbmatrix \
y&=
beginbmatrix
omega^2_n & 0
endbmatrix
endalign
$$
answered Apr 16 at 3:59
CroCo
225218
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The transfer function you've posted has this form
$$
G(s) = fracY(s)U(s) = frac b_0 s^2 + b_1 s + b_2 s^2 + a_1 s + a_2
$$
Controllable canonical form can be utilized to represents the system as follows:
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-a_2 & -a_1
endbmatrix \
y&=
beginbmatrix
b_2-a_2b_0 & b_1-a_1b_0
endbmatrix
endalign
$$
where $b_0=0,b_1=0,b_2= omega^2_n ,a_1=2zeta omega_n,a_2=omega^2_n $. The state space representation of the system is
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-omega^2_n & -2zeta omega_n
endbmatrix \
y&=
beginbmatrix
omega^2_n & 0
endbmatrix
endalign
$$
answered Apr 16 at 3:59
CroCo
225218
The transfer function you've posted has this form
$$
G(s) = fracY(s)U(s) = frac b_0 s^2 + b_1 s + b_2 s^2 + a_1 s + a_2
$$
Controllable canonical form can be utilized to represents the system as follows:
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-a_2 & -a_1
endbmatrix \
y&=
beginbmatrix
b_2-a_2b_0 & b_1-a_1b_0
endbmatrix
endalign
$$
where $b_0=0,b_1=0,b_2= omega^2_n ,a_1=2zeta omega_n,a_2=omega^2_n $. The state space representation of the system is
$$
beginalign
dotx &=
beginbmatrix
0 & 1\
-omega^2_n & -2zeta omega_n
endbmatrix \
y&=
beginbmatrix
omega^2_n & 0
endbmatrix
endalign
$$
answered Apr 16 at 3:59
CroCo
225218
answered Apr 16 at 3:59
CroCo
225218
answered Apr 16 at 3:59
CroCo
225218
answered Apr 16 at 3:59
CroCo
225218
225218
add a comment |Â
add a comment |Â
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Isn't it already in canonical form?
– tpb261
Jun 3 '14 at 5:55
This follows directly from the relevant differential equation.
– copper.hat
Jun 3 '14 at 5:56
The question is how this form is derived. I would think it's derived from some standard physical system, but how?
– user25778
Jun 3 '14 at 6:11