How do we obtain a state-space realization and a block diagram of a given transfer function?

Consider the transfer function


Steps for solution are


$$C(s)=5(s^{-1})Q(s)\\$$ $$\Rightarrow R(s)=(3+3s^{-1}+s^{-2})Q(s)$$ $$\Rightarrow R(s)=3Q(s)+3(s^{-1})Q(s)+s^{-2}Q(s)$$ $$\Rightarrow 3Q(s)=R(s)-3(s^{-1})Q(s)-s^{-2}Q(s)$$

$$\Rightarrow Q(s) = \frac13R(s)-s^{-1}Q(s)-\frac13s^{-2}Q(s)$$

$$\Rightarrow Q(s) = \frac13R(s)-Q(s)\left[s^{-1}+\frac13s^{-2}\right]$$
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?

  • $\begingroup$ 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

Only some parts of your question make sense. In particular, given the transfer function

$$ H(s) = \frac{5s}{3s^2+3s+1} $$

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 $$ \begin{aligned} X_1' &= X_2\\ X_2' &= -X_2 - \frac{1}{3}X_1 + \frac{5}{3}U_2\\ \end{aligned} $$ and thus $$ \begin{bmatrix} X_1\\ X_2\\ \end{bmatrix}' = \begin{bmatrix} 0 & 1\\ -\frac{1}{3} & -1 \end{bmatrix} \begin{bmatrix} X_1\\ X_2\\ \end{bmatrix} + \begin{bmatrix} 0 & 0\\ 0 & \frac{5}{3} \end{bmatrix} \begin{bmatrix} U_1\\ U_2\\ \end{bmatrix}, $$ corresponding to the partial realization $$ A = \begin{bmatrix} 0 & 1\\ -\frac{1}{3} & -1 \end{bmatrix} ,\ B = \begin{bmatrix} 0 & 0\\ 0 & \frac{5}{3} \end{bmatrix}. $$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.