Let's suppose we have:

G(s) = (s+1)/(s^2-2s+1)

how can we find the state space representation of the transfer function:

x_dot = x2

x2_dot = 2*x2-x1+u

where u is an arbitrary input.

I am very new to this topic, so a detailed answere would be great ! :)

  • $\begingroup$ Hint. The laplace operator roughly speaking performs derivatives. If you have $y/u=G(s)$, how can you substitute the "s" with derivatives? $\endgroup$ – LJSilver Jan 24 '17 at 9:23
  • $\begingroup$ @LjSilver Hmmm... I am sorry, could you help me a bit... $\endgroup$ – james Jan 24 '17 at 10:38
  • $\begingroup$ @LJSilver see my comment above. Help is needed. $\endgroup$ – james Jan 24 '17 at 15:35

So, assuming that the transfer function is between the output $Y(s)$ and the input $U(s)$, namely $$ \dfrac{Y(s)}{U(s)} = G(s)\qquad\qquad(1) $$ multiplying (1) by $U(s)\cdot(s^2-2s+1))$ yields $$ s^2 Y(s) -2s Y(s) + Y(s) = sU(s) + U(s) $$ Going back in the time domain we obtain $$ \ddot y -2\dot y + y = \dot u + u\qquad (2) $$ Now, we look for a realization of the kind \begin{align*} \dot x &= Ax+Bu\qquad (3)\\ y &=Cx \end{align*} with $x=(x_1,x_2)$ and \begin{align*} A &= \begin{pmatrix}0 & 1\\a_1 & a_2\end{pmatrix} & B&=\begin{pmatrix}0\\1\end{pmatrix}, & C&=\begin{pmatrix}c_1&c_2\end{pmatrix}\end{align*} The next step is to find the values of $(a_1,a_2,c_1,c_2)$ for which (3) has the same input-output behaviour of (2). From (3) we have \begin{align*} y&=c_1x_1+c_2x_2\\ \dot y&= c_1\dot x_1 + c_2\dot x_2 = c_1x_2 + c_2a_1x_1 + c_2a_2 x_2 + c_2u\\ \ddot y&= c_1\ddot x_1 + c_2\ddot x_2 = c_1a_1x_1+c_1a_2x_2 +c_1u+ c_2a_1x_2+ c_2a_2a_1x_1+c_2a_2^2 x_2 + c_2a_2u + c_2\dot u \end{align*} substituting into (2) yields \begin{align*} 0&=(c_1+c_1a_1+c_2a_2a_1-2c_2a_1)x_1 + (c_2+c_1a_2+c_2a_1+c_2a_2^2-2c_1-2c_2a_2)x_2\\ &+(c_1+a_2c_2-2c_2-1)u + (c_2-1)\dot u \end{align*} since that equality must hold for all $(x_1,x_2,u,\dot u)$ that' equivalent to ask \begin{align*} c_1+c_1a_1+c_2a_2a_1-2c_2a_1&=0\\c_2+c_1a_2+c_2a_1+c_2a_2^2-2c_1-2c_2a_2&=0\\ c_1+a_2c_2-2c_2-1&=0\\c_2-1&=0 \end{align*} and a solution is $$ (a_1,a_2,c_1,c_2)=(-1,2,1,1) $$ Therefore \begin{align*} \begin{pmatrix}\dot x_1\\\dot x_2\end{pmatrix} &=\begin{pmatrix}0&1\\-1&2\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix}\\y&=\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix} \end{align*}

  • $\begingroup$ Thanks a lot ! That was very clear. My prof. sometimes uses Y/U = Y/X*X/U, then says (s+1)/(s^2-2s+1) = (s+1)*1/(s^2-2s+1), hence Y/X=s+1 and X/U = s^2-2s+1, he then transforms it back to Time domain and hence finds the state space representation. Do you see why he can do it like that ? $\endgroup$ – james Jan 24 '17 at 16:51
  • 1
    $\begingroup$ Yes, you can proceed also in that way. In that way you obtain two models, the first with input u and output x, the second with input x and output y. Then you just have to consider their interconnectio $\endgroup$ – LJSilver Jan 24 '17 at 20:19
  • 2
    $\begingroup$ @totyped You also have to keep in mind that a state space model of a given transfer function is not unique (there are infinity many state space models which correspond to the same transfer function). $\endgroup$ – fibonatic Jan 25 '17 at 11:21

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.