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I have this 2nd order transfer function: $$G(s) = \frac{2}{s} + \frac{1}{s+2}$$

And I need to find a possible state space representation in the form of:

$$ \frac{dx}{dt} = Ax + bu $$ $$y = c^Tx$$

Matrix A
Matrix A is the system matrix, and relates how the current state affects the state change x' . If the state change is not dependent on the current state, A will be the zero matrix. The exponential of the state matrix, eAt is called the state transition matrix.

Matrix B
Matrix B is the control matrix, and determines how the system input affects the state change. If the state change is not dependent on the system input, then B will be the zero matrix.

Matrix C
Matrix C is the output matrix, and determines the relationship between the system state and the system output.

I can see the eigenvalues, they are $s_1 = 0$ and $s_2 = -2$.
So I can write down a diagonal matrix like that I think: $$ A = \begin{pmatrix} 0 & 0 \\0 & -2 \end{pmatrix} $$ But know I am stuck.
Is there some sort of trick I can use?

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Do you know how to obtain minimal realization $G(s) = C(sI-A)^{-1}B+D$ ? – user13838 Oct 5 '11 at 11:55
1  
As one with control background, I know it is indeed a very basic problem in control theory. In fact, you can google it to get good answers. See here lpsa.swarthmore.edu/Representations/SysRepTransformations/… for example. – Shiyu Oct 5 '11 at 15:22
up vote 2 down vote accepted

The main tool to use in such exercises is to consider a dummy variable (which turns out to be the state of the system) and describe the input and output in terms of that variable. Let us call it $x(t)$ and its Laplace transformed version $X(s)$.

Then we use the following: $$ G(s) := \frac{Y(s)}{U(s)} = \frac{N(s)}{D(s)}\frac{X(s)}{X(s)} $$ where $N,D$ are the numerator and the denominator polynomials of the Laplace indeterminate $s$. such that $G = N/D$. Then apply this simple idea to your transfer function. $$ \frac{Y(s)}{U(s)} = \frac{(3s+4)X(s)}{(s^2+2s)X(s)} $$ From this (and from the assumed zero initial conditions) we have, $$ Y(s) = (3s+4) X(s) \implies y(t) = 3\dot x(t) + 4x(t) $$ and also $$ U(s) = (s^2+2s) X(s) \implies u(t) = \ddot x(t) + 2\dot x(t) $$ Then, rename the variables as $x_1=x,x_2=\dot x$ and fill in the matrices of $A,B,C$ $$\begin{align} \pmatrix{\dot x\\\ddot x} &= \pmatrix{0 &1\\0 &-2}\pmatrix{x\\ \dot x}+\pmatrix{0\\ 1}u\\ y &= \pmatrix{4&3}\pmatrix{x\\ \dot x} + 0 u \end{align} $$ Note that State-Space realizations are non-unique, you can obtain the $A$ matrix you have proposed by some state transformation. Hence it is not necessarily a mistake. For example, by noticing directly that $$ y(t) = \int^t_0 u(\tau)d\tau + e^{-2t}u(t) $$ You can identify the state space matrix entries by inspection e.g. ($\dot x_1 = u$) etc. but then you would probably obtain an $A$ matrix which is $3\times 3$. Therefore, I recommend you to read about minimal realizations of state space systems, which involves controllability and observability for a very good reason.

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Can you tell me where you get the "1" in your A matrix and the $\pmatrix{0\\ 1}u$ from? I only get the A matrix I wrote in my question. – madmax Oct 10 '11 at 12:46
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@madmax If you write $\ddot x = -2\dot x + u$. Then you only have the input on the bottom row. – user13838 Oct 10 '11 at 13:56
    
Ok, I understand that. But from where is the top row then? $ Y(s) = 3\dot x(t) + 4x(t) $ does not get me to the top row. – madmax Oct 10 '11 at 15:13
    
@madmax the top row is $\dot x_1 = \dot x_1$ – user13838 Oct 10 '11 at 15:18
1  
@madmax Just multiply all the matrices with the vectors and compare it with the equations I have written up. Also if you just search transfer function to state space you have tons of results on google. If your textbook does not have this, change your textbook. It should have it. – user13838 Oct 10 '11 at 15:46

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