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I was supposed to find state space representation and its matrices of this system:

enter image description here

and I have no idea, how to do this. We were told not to transfer the system to time domain, but I can only do state space representation from time domain schemas.

When I tried to solve this, I got matrices $$A = \left( \begin{array}{ccc@{\ }r} -a & k \\ -b & -p \\ \end{array} \right)$$

$$B = \left( \begin{array}{ccc@{\ }r} 0 \\ b \\ \end{array} \right)$$

$$C = \left( \begin{array}{ccc@{\ }r} 1 & 0 \\ \end{array} \right)$$

$$D = \left( \begin{array}{ccc@{\ }r} 0 \end{array} \right)$$


I went like: $$X_2(s) = (U(s)-X_1(s)) \cdot \frac {b}{s+p} $$ $$X_1(s) = X_2(s) \cdot \frac {k}{s+a}$$

That could mean: $$sX_2(s) + pX_2(s) = bU(s) - bX_1(s) \to \dot x_2(t) = bu(t) - bx_1(t) - px_2(t)$$ $$sX_1(s) + aX_1(s) = kX_2(s) \to \dot x_1(t) = -ax_1(t) + kx_2(t)$$

and output sould be: $$y(t) = x_1(t)$$

that would lead to matrices I wrote. But I don't know, if I can do that this way, or if that is what was the task, cause we were told not to transfer to time domain, but I can't imagine how to do it without transfer I did.

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    \$\begingroup\$ Can you show us your calculations in addition to the result you get? \$\endgroup\$ – user17592 Apr 14 '13 at 12:12
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From what I see what you did is correct.

I also couldn't think of a good way to do this without going to the time domain. Do note that from transfer function to state space, there are infinite possibilities. But some of the realizations are called canonical forms. See this wiki page for Canonical realizations.

Maybe what you can do is to get the complete transfer function for $$\frac{y}{u} = \frac{bk}{s^2 + (p+a)s + bk + pa}$$ first. And then just use that canonical transformation to plug in the numbers into the matrices by observation.

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  • \$\begingroup\$ should the matrices be the same, if I compute with Canonical realization, as these matrices I wrote before? I'm trying to learn how to use Canonical realization, but I can't get the same result. \$\endgroup\$ – Bo20 Apr 15 '13 at 8:16
  • \$\begingroup\$ I tried to put matrices in the formula of definition and both fit. Thanks \$\endgroup\$ – Bo20 Apr 15 '13 at 9:23
  • \$\begingroup\$ You should know, there are two types of canonical realizations: controllable canonical form and observer canonical form (both shown in the wiki page). You will learn that both are equivalent and both are so called minimal realization. It depends on how you choose your state variable. Normally, if you choose x1 = y, x2 = x1_dot, x3 = x2_dot. You will have controllable canonical form. \$\endgroup\$ – foresightyj Apr 16 '13 at 4:02
  • \$\begingroup\$ Following previous comment. So the answer to your question is No. Because in your case, your x2 != x1_dot. In your case, your x2 = x1_dot/k + (a/k)*x1. (Note, x1_dot means differentiation of x1 with respect to time, i.e. d(x1)/d(t)) \$\endgroup\$ – foresightyj Apr 16 '13 at 4:05

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