how to obtain state space diagram and state space model for transfer functionHow to obtain a possible state space representation of this 2nd order transfer function?Two state space models represent the same transfer functionTransfer function from a block diagram with a loopDivision of two vectors? State-space model to transfer functionHow do I create a augmented plant of a state space model / Transfer function?Transfer function matrix to state space model?Numerical computations for transfer functions / state space models?Transform discrete state space into discrete transfer function?Minimal realization of a state space model?State-Space Realization of a Matrix of Transfer Functions
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how to obtain state space diagram and state space model for transfer function
How to obtain a possible state space representation of this 2nd order transfer function?Two state space models represent the same transfer functionTransfer function from a block diagram with a loopDivision of two vectors? State-space model to transfer functionHow do I create a augmented plant of a state space model / Transfer function?Transfer function matrix to state space model?Numerical computations for transfer functions / state space models?Transform discrete state space into discrete transfer function?Minimal realization of a state space model?State-Space Realization of a Matrix of Transfer Functions
$begingroup$
How do we obtain a state-space realization and a block diagram of a given transfer function?
Consider the transfer function
$$fracC(s)R(s)=frac5s3s^2+3s+1$$
Steps for solution are
$$fracC(s)R(s)=frac5s3s^2+3s+1fracQ(s)Q(s)\$$
$$C(s)=5(s^-1)Q(s)\$$
$$Rightarrow R(s)=(3+3s^-1+s^-2)Q(s)$$
$$Rightarrow R(s)=3Q(s)+3(s^-1)Q(s)+s^-2Q(s)$$
$$Rightarrow 3Q(s)=R(s)-3(s^-1)Q(s)-s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-s^-1Q(s)-frac13s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-Q(s)left[s^-1+frac13s^-2right]$$
the graph which is plotted in the book is of last equation of above solution.
I do not know how to post the graph here on Stack Exchange, but what I want to understand is:
How is the graph of this equation plotted?
control-theory
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
$endgroup$
add a comment |
$begingroup$
How do we obtain a state-space realization and a block diagram of a given transfer function?
Consider the transfer function
$$fracC(s)R(s)=frac5s3s^2+3s+1$$
Steps for solution are
$$fracC(s)R(s)=frac5s3s^2+3s+1fracQ(s)Q(s)\$$
$$C(s)=5(s^-1)Q(s)\$$
$$Rightarrow R(s)=(3+3s^-1+s^-2)Q(s)$$
$$Rightarrow R(s)=3Q(s)+3(s^-1)Q(s)+s^-2Q(s)$$
$$Rightarrow 3Q(s)=R(s)-3(s^-1)Q(s)-s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-s^-1Q(s)-frac13s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-Q(s)left[s^-1+frac13s^-2right]$$
the graph which is plotted in the book is of last equation of above solution.
I do not know how to post the graph here on Stack Exchange, but what I want to understand is:
How is the graph of this equation plotted?
control-theory
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
$endgroup$
$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
2
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53
add a comment |
$begingroup$
How do we obtain a state-space realization and a block diagram of a given transfer function?
Consider the transfer function
$$fracC(s)R(s)=frac5s3s^2+3s+1$$
Steps for solution are
$$fracC(s)R(s)=frac5s3s^2+3s+1fracQ(s)Q(s)\$$
$$C(s)=5(s^-1)Q(s)\$$
$$Rightarrow R(s)=(3+3s^-1+s^-2)Q(s)$$
$$Rightarrow R(s)=3Q(s)+3(s^-1)Q(s)+s^-2Q(s)$$
$$Rightarrow 3Q(s)=R(s)-3(s^-1)Q(s)-s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-s^-1Q(s)-frac13s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-Q(s)left[s^-1+frac13s^-2right]$$
the graph which is plotted in the book is of last equation of above solution.
I do not know how to post the graph here on Stack Exchange, but what I want to understand is:
How is the graph of this equation plotted?
control-theory
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
$endgroup$
How do we obtain a state-space realization and a block diagram of a given transfer function?
Consider the transfer function
$$fracC(s)R(s)=frac5s3s^2+3s+1$$
Steps for solution are
$$fracC(s)R(s)=frac5s3s^2+3s+1fracQ(s)Q(s)\$$
$$C(s)=5(s^-1)Q(s)\$$
$$Rightarrow R(s)=(3+3s^-1+s^-2)Q(s)$$
$$Rightarrow R(s)=3Q(s)+3(s^-1)Q(s)+s^-2Q(s)$$
$$Rightarrow 3Q(s)=R(s)-3(s^-1)Q(s)-s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-s^-1Q(s)-frac13s^-2Q(s)$$
$$Rightarrow Q(s) = frac13R(s)-Q(s)left[s^-1+frac13s^-2right]$$
the graph which is plotted in the book is of last equation of above solution.
I do not know how to post the graph here on Stack Exchange, but what I want to understand is:
How is the graph of this equation plotted?
control-theory
control-theory
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
edited Jun 1 '16 at 0:40
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
asked Dec 4 '12 at 9:26
Registered UserRegistered User
1113
1113
$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
2
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53
add a comment |
$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
2
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53
$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
2
2
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Only some parts of your question make sense. In particular, given the transfer function
$$
H(s) = frac5s3s^2+3s+1
$$
you invert the Laplace transforms to obtain the ode
$$3y''(t) + 3y'(t) + y = 5u'(t).$$
Define $X_1 = y$, $X_2 = y'$, $U_1 = u$, $U_2 = u'$. The state equations may therefore be written as
$$
beginaligned
X_1' &= X_2\
X_2' &= -X_2 - frac13X_1 + frac53U_2\
endaligned
$$
and thus
$$
beginbmatrix
X_1\
X_2\
endbmatrix'
= beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
beginbmatrix
X_1\
X_2\
endbmatrix
+
beginbmatrix
0 & 0\
0 &
frac53
endbmatrix
beginbmatrix
U_1\
U_2\
endbmatrix,
$$
corresponding to the partial
realization
$$
A = beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
, B = beginbmatrix
0 & 0\
0 &
frac53
endbmatrix.
$$
This is the only interpretation I can think of for "state space model". The only thing I can imagine is meant by "state variable diagram" (other than a block diagram which there is no "plotting" involved in making) is the $X_1$ vs. $X_2$ contour plot which you get by plotting the vector field of the RHS minus the controller parametrically. I have no idea what the calculation you show is trying to do and without any reference to where you got it from have no idea where to start with that.
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Only some parts of your question make sense. In particular, given the transfer function
$$
H(s) = frac5s3s^2+3s+1
$$
you invert the Laplace transforms to obtain the ode
$$3y''(t) + 3y'(t) + y = 5u'(t).$$
Define $X_1 = y$, $X_2 = y'$, $U_1 = u$, $U_2 = u'$. The state equations may therefore be written as
$$
beginaligned
X_1' &= X_2\
X_2' &= -X_2 - frac13X_1 + frac53U_2\
endaligned
$$
and thus
$$
beginbmatrix
X_1\
X_2\
endbmatrix'
= beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
beginbmatrix
X_1\
X_2\
endbmatrix
+
beginbmatrix
0 & 0\
0 &
frac53
endbmatrix
beginbmatrix
U_1\
U_2\
endbmatrix,
$$
corresponding to the partial
realization
$$
A = beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
, B = beginbmatrix
0 & 0\
0 &
frac53
endbmatrix.
$$
This is the only interpretation I can think of for "state space model". The only thing I can imagine is meant by "state variable diagram" (other than a block diagram which there is no "plotting" involved in making) is the $X_1$ vs. $X_2$ contour plot which you get by plotting the vector field of the RHS minus the controller parametrically. I have no idea what the calculation you show is trying to do and without any reference to where you got it from have no idea where to start with that.
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
$endgroup$
add a comment |
$begingroup$
Only some parts of your question make sense. In particular, given the transfer function
$$
H(s) = frac5s3s^2+3s+1
$$
you invert the Laplace transforms to obtain the ode
$$3y''(t) + 3y'(t) + y = 5u'(t).$$
Define $X_1 = y$, $X_2 = y'$, $U_1 = u$, $U_2 = u'$. The state equations may therefore be written as
$$
beginaligned
X_1' &= X_2\
X_2' &= -X_2 - frac13X_1 + frac53U_2\
endaligned
$$
and thus
$$
beginbmatrix
X_1\
X_2\
endbmatrix'
= beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
beginbmatrix
X_1\
X_2\
endbmatrix
+
beginbmatrix
0 & 0\
0 &
frac53
endbmatrix
beginbmatrix
U_1\
U_2\
endbmatrix,
$$
corresponding to the partial
realization
$$
A = beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
, B = beginbmatrix
0 & 0\
0 &
frac53
endbmatrix.
$$
This is the only interpretation I can think of for "state space model". The only thing I can imagine is meant by "state variable diagram" (other than a block diagram which there is no "plotting" involved in making) is the $X_1$ vs. $X_2$ contour plot which you get by plotting the vector field of the RHS minus the controller parametrically. I have no idea what the calculation you show is trying to do and without any reference to where you got it from have no idea where to start with that.
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
$endgroup$
add a comment |
$begingroup$
Only some parts of your question make sense. In particular, given the transfer function
$$
H(s) = frac5s3s^2+3s+1
$$
you invert the Laplace transforms to obtain the ode
$$3y''(t) + 3y'(t) + y = 5u'(t).$$
Define $X_1 = y$, $X_2 = y'$, $U_1 = u$, $U_2 = u'$. The state equations may therefore be written as
$$
beginaligned
X_1' &= X_2\
X_2' &= -X_2 - frac13X_1 + frac53U_2\
endaligned
$$
and thus
$$
beginbmatrix
X_1\
X_2\
endbmatrix'
= beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
beginbmatrix
X_1\
X_2\
endbmatrix
+
beginbmatrix
0 & 0\
0 &
frac53
endbmatrix
beginbmatrix
U_1\
U_2\
endbmatrix,
$$
corresponding to the partial
realization
$$
A = beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
, B = beginbmatrix
0 & 0\
0 &
frac53
endbmatrix.
$$
This is the only interpretation I can think of for "state space model". The only thing I can imagine is meant by "state variable diagram" (other than a block diagram which there is no "plotting" involved in making) is the $X_1$ vs. $X_2$ contour plot which you get by plotting the vector field of the RHS minus the controller parametrically. I have no idea what the calculation you show is trying to do and without any reference to where you got it from have no idea where to start with that.
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
$endgroup$
Only some parts of your question make sense. In particular, given the transfer function
$$
H(s) = frac5s3s^2+3s+1
$$
you invert the Laplace transforms to obtain the ode
$$3y''(t) + 3y'(t) + y = 5u'(t).$$
Define $X_1 = y$, $X_2 = y'$, $U_1 = u$, $U_2 = u'$. The state equations may therefore be written as
$$
beginaligned
X_1' &= X_2\
X_2' &= -X_2 - frac13X_1 + frac53U_2\
endaligned
$$
and thus
$$
beginbmatrix
X_1\
X_2\
endbmatrix'
= beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
beginbmatrix
X_1\
X_2\
endbmatrix
+
beginbmatrix
0 & 0\
0 &
frac53
endbmatrix
beginbmatrix
U_1\
U_2\
endbmatrix,
$$
corresponding to the partial
realization
$$
A = beginbmatrix
0 & 1\
-frac13 & -1
endbmatrix
, B = beginbmatrix
0 & 0\
0 &
frac53
endbmatrix.
$$
This is the only interpretation I can think of for "state space model". The only thing I can imagine is meant by "state variable diagram" (other than a block diagram which there is no "plotting" involved in making) is the $X_1$ vs. $X_2$ contour plot which you get by plotting the vector field of the RHS minus the controller parametrically. I have no idea what the calculation you show is trying to do and without any reference to where you got it from have no idea where to start with that.
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
answered Jun 1 '16 at 0:22
SZNSZN
2,718720
2,718720
add a comment |
add a comment |
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$begingroup$
Consider using LaTeX for formatting.
$endgroup$
– Epictetus
Dec 4 '12 at 9:56
$begingroup$
Yes, please, please learn some LaTeX for posting huge formulas like this. It was very difficult to read as it was.
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:05
2
$begingroup$
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post.
$endgroup$
– Julian Kuelshammer
Dec 4 '12 at 10:10
$begingroup$
Yes, that is the other problem! Cheers Julian. :)
$endgroup$
– Simon Hayward
Dec 4 '12 at 10:37
$begingroup$
yes the edits done are correct what is latex I am hearing it first time the edits are correct
$endgroup$
– Registered User
Dec 4 '12 at 10:53