# State Space Question

Discussion in 'Homework Help' started by Kayne, May 15, 2010.

1. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Hi All,

Just wondering if I have done this correctly. To write the system below into a state space form with a unit step.
100(s+5)/ (s+2)(s+3)

100s+500/s^2+5s+6

d^2y/dt^2 + 5dy/dt + 6y = 100du/dt+500u

Y(s) = 100s^-1+500s^-2/ 1+5s^-1+6s^2
Y(s) = (100s^-1+500s^-2)E(s)
E(s)= U(s)/1+5s^-1+6s^-2 therefore
E(s) = U(s) - 5s^-1E(s)-6s^-2E(s)

dx1/dt=x2
dx2/dt2=x3
dx3/dt3= -6x3-5x2+u

y= 500x1 +100x2

d/dt
:x1: :0, 1, 0: :x1: :0:
:x2:=:0, 0, 1: :x2:+:0: u(t)
:x3: :0,-6,-5: :x3: :1:

y=:500, 100, 0:

I would also like to know where I can find the information to write the above in correct equations so it make more sence to the reader.

Thanks

Last edited: May 15, 2010
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I think it should be ....

$\frac{Y(s)}{U(s)}=\frac{100(s+5)}{s^2+5s+6}$

$\frac{Y(s)}{U(s)}=\frac{100(s^{-1}+5s^{-2})}{1+5s^{-1}+6s^{-2}}$

$E(s)=\frac{U(s)}{1+5s^{-1}+6s^{-2}$

$E(s)=U(s)-E(s)(5s^{-1}+6s^{-2})$

and so on ....

Dec 5, 2009
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4. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
I will re do the question using the link for correct formatting and re submit.

5. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thanks for the heads up on the code I have managed to write the answer so it is more understandable

$\frac {Y(s)}{U(s)}=\frac {100s+500}{s^{2}+5s+6}$

$Y(s)=\frac {100s^{-1}+500s^{-2}}{1+5s^{-1}+6s^{-2}}$

$Y(s)=(100s^{-1}+500s^{-2})E(s)$

$E(s)=\frac {U(s)}{1+5s^{-1}+6s^{-2}}$

$E(s)=U(s)-5s^{-1}E(s)-6s^{-2}E(s)$

$\frac {dx_{1}}{dt}= x_{2}$

$\frac {dx_{2}}{dt}= x_{3}$

$\frac {dx_{3}}{dt}= -6x_{2}-5x_{3}$

$y = 500x_{1}+100x_{2}$

$\frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\\x_{3}\end{array})=(\begin{array}{cc}0&1&0\\0&0&1\\-6&-5&0\end{array})(\begin{array}{cc}x_{1}\\x_{2}\\x_{3}\end{array})+(\begin{array}{cc}0\\0\\1\end{array})u(t)$

$y=(\begin{array}{cc}500&100&0\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\\x_{3}\end{array})$

Hopefully what I have done is correct, thanks for your help

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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You've obtained what appears to be a 3rd order representation of a 2nd order system. Is this how you were taught to do it?

The normal representation would be

d(X(t))/dt=A.X(t)+B.u(t)
y(t)= C.X(t)+D.u(t)

In a 2nd order system there would be two state variables, so A would be a 2x2 matrix .....

7. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
This is why i may be having trouble with the the rest of the question. I found a question that is similar and followed the example. Will go back and have another look

Thanks for the info

8. ### Ethan New Member

Feb 7, 2009
6
1
t_n_k is correct. You need to select a minimal set of variables that completely describe the system. In your case you only need 2 not 3 because it is a second order system.

9. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Have had another look at the question and found another example that is similar so hopefully this may be correct.

$\frac {Y(s)}{U(s)}=\frac {100s+500}{s^{2}+5s+6}$

$Y(s)=\frac {100s^{-1}+500s^{-2}}{1+5s^{-1}+6s^{-2}}$

$Y(s)=(100s^{-1}+500s^{-2})E(s)$

$E(s)=\frac {U(s)}{1+5s^{-1}+6s^{-2}}$

$E(s)=U(s)-5s^{-1}E(s)-6s^{-2}E(s)$

$\frac {dx_{1}}{dt}= x_{2}$

$\frac {dx_{2}}{dt}= -6x_{1}-5x_{2}$

$y = 500x_{1}+100x_{2}$

$\frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\0\end{array})u(t)$

$y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})$

thanks for the help

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Probably an oversight - You need a '1' in the first state space equation B matrix (the equation second from the bottom) to include u(t) in the function.

11. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
your correct, Looking back in my notes i did miss that when typing that in thanks TNK

12. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thanks for pointing that out TNK. You are correct it was an oversight.

$\frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\1\end{array})u(t)$

$y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})$

To continue on from this to a discrete time system in the form

$X(k+1)=\Phi X(k)=\Gamma u(k)$ ,
$y=CX(k)$

I have done the following

$A= (\begin{array}{cc}0&1\\-6&-5\end{array}) B=(\begin{array}{cr}0\\1\end{array})$

$\Phi(t)= L^-1[(\begin{array}{cc}s&0\\0&s\end{array})- (\begin{array}{cr}0&1\\-6&-5\end{array})]^-1$

$= L^-1[(\begin{array}{cc}s&-1\\6&s+5\end{array})]^-1$

$= L^-1[(\begin{array}{cc}\frac{1}{s}&\frac{-1}{s^2}\\6&\frac{1}{s+5}\end{array})]$

$\Phi(t)=[(\begin{array}{cc}1&-t\\6&e^-5t)\end{array})]$

I would like to know if what I have worked out for $\Phi(t)$ is correct I am not sure about changing $\frac {1}{s+5}$ that is all and this is required to work out $\Gamma$

Thanks for any help

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Keeping in mind that ....

$A_{d}=e^{AT}$

Where Ad is the discretized form of continuous matrix A. You called this $\Phi(t)$ rather than $A_{d}$ which is just a matter of convention.

Are you confident that ...

$A_{d}=e^{AT}=L^{-1}\left\{(sI-A)^{-1}\right\}_{t=T}$

You might show how you derived

$(sI-A)^{-1}$

which leads to Ad or $\Phi(t)$ as you have it in the post.

14. ### Ethan New Member

Feb 7, 2009
6
1
$image=http://forum.allaboutcircuits.com/mimetex.cgi?=%20L^-1[(\begin{array}{cc}\frac{1}{s}&\frac{-1}{s^2}\\6&\frac{1}{s+5}\end{array})]&hash=caebff6389c5f6c8d4f2074fc39e8bd9$ I believe this is not accurate. You need to do partial fraction expansion.

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I agree.

You can do the expansion of

$e^{AT}$

using Taylor's expansion or do the inversion of the matrix [sI-A] and apply the inverse Laplace transform at t=T.

16. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Have taken on board what has been said and have work though the question again.

$\frac {d}{dt}=(\begin{array}{cc}x_{1}\\x_{2}\end{array})=(\begin{array}{cc}0&1\\-6&-5\end{array})(\begin{array}{cc}x_{1}\\x_{2}\end{array})+(\begin{array}{cc}0\\1\end{array})u(t)$

$y=(\begin{array}{cc}500&100\end{array}) (\begin{array}{cr}x_{1}\\x_{2}\end{array})$

To continue on from this to a discrete time system in the form

$X(k+1)=\Phi X(k)=\Gamma u(k)$ ,
$y=CX(k)$

I have done the following

$A= (\begin{array}{cc}0&1\\-6&-5\end{array}) B=(\begin{array}{cr}0\\1\end{array})$

$\Phi(t)= L^-1[(\begin{array}{cc}s&0\\0&s\end{array})- (\begin{array}{cr}0&1\\-6&-5\end{array})]^-1$

$= L^-1[(\begin{array}{cc}s&-1\\6&s+5\end{array})]^-1$

$=L^-1\frac{1}{s^2+5s+6} (\begin{array}{cc}\ s+5&1\\-6&s\end{array})$

$=L^-1[(\begin{array}{cc}\frac{s+5}{s^2+5s+6}&\frac{1}{s^2+5s+6}\\\frac{-6}{s^2+5s+6}&\frac{s}{s^2+5s+6}\end{array})]$

$=L^-1[(\begin{array}{cc}\frac{3}{(s+2)}-\frac{2}{(s+3)}&\frac{1}{(s+2)}-\frac{1}{(s+3)}\\\frac{-6}{(s+2)}+\frac{6}{(s+3)}&\frac{-2}{(s+2)}+\frac{3}{(s+3)}\end{array})]$

$\Phi=[(\begin{array}{cc}\3e^-2T-2e^-3T&e^-2T-e^-3T\\-6e^-2T+6e^-3T&-2e^-2T+3e^-3T\end{array})]$

$\Gamma=\int^{T}_{0}e^Aq Bdq=\int^{T}_{0}[(\begin{array}{cc}\3e^-2q-2e^-3q&e^-2q-e^-3q\\-6e^-2q+6e^-3q&-2e^-2q+3e^-3q\end{array})][(\begin{array}{cc}\0\\1\end{array})]dq$

$=[(\begin{array}{cc}\\\frac{1}{2}-e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]$

So the original equation
$X(k+1)=\Phi X(k)=\Gamma u(k)$ ,
$y=CX(k)$

is then worked out to be

$X(k+1)=[(\begin{array}{cc}\3e^-2T-2e^-3T&e^-2T-e^-3T\\-6e^-2T+6e^-3T&-2e^-2T+3e^-3T\end{array})][(\begin{array}{cc}\\x_{1}(k)\\x_{2}(k)\end{array})] +[(\begin{array}{cc}\\\frac{1}{2}-e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]u(k)$

$y=(\begin{array}{cc}\500&100\end{array})[(\begin{array}{cc}\\x_{1}(k)\\x_{2}(k)\end{array})]$

This looks better than the first attempt so hopefully this method is correct.Thanks for the help given so far

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Re-check this.

18. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
$\Gamma=\int^{T}_{0}e^Aq Bdq=\int^{T}_{0}[(\begin{array}{cc}\3e^-2q-2e^-3q&e^-2q-e^-3q\\-6e^-2q+6e^-3q&-2e^-2q+3e^-3q\end{array})][(\begin{array}{cc}\0\\1\end{array})]dq$

$=[(\begin{array}{cc}\\\frac{-1}{3}e^-2T+\frac{1}{2}e^-3T\\e^-2T-e^-3T\end{array})]$

I think that I may have got the derivatives incorrect last post, hopefully this is the correct answer for $\Gamma$..

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
This is my solution ....

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20. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
Thanks TNK, I thought that the 1/6 had to be subtracted from the equation not just added into the equation.