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I am given the following initial value problem \$ y^{(4)} - y = u \$ with \$ y^{(3)}(0)=0, \ddot{y}(0)=1, \dot{y}=1, y(0)=10 \$ and am supposed to calculate the state space in the usual form

\begin{equation} \dot{x} = f(x,y) \end{equation}

\begin{equation}y = h(x,y) \end{equation} with \begin{equation} x^T = \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix} := \begin{bmatrix} y & \dot{y} +\ddot{y} & \dot{y} -\ddot{y} & y^{(3)} \end{bmatrix} \end{equation}

The initial values are non-zero, so Laplace transform and calculating \$ G(s) \$ as \$ \frac{Y(s)}{U(s)} \$ and then deriving the state space from the transfer function \$ G(s) \$ wont work, since we will get:

\begin{equation} s^4Y(s)-s^3y(0)-s^2\dot{y}(0)-s^\ddot{y}(0)-y^{(3)}(0)-Y(s) = U(s) \end{equation}

\begin{equation} s^4Y(s)-10s^3-s^2-s-Y(s) = U(s) \end{equation}

\begin{equation} Y(s)(s^4-1)-10s^3-s^2-s = U(s) \end{equation}

And \$ -10s^3-s^2-s \$ is a problem here


So i did it the usual way and transformed the 4th order DE into a 1st order DE-system.

\begin{equation} \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \\ \dot{x}_4 \end{bmatrix} = \begin{bmatrix} 0 & 0.5 & 0.5 & 0 \\ 0 & 0.5 & -0.5 & 0 \\ 0 & -0.5 & 0.5 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} x_1 (=y) \\ x_2(=\dot{y}+\ddot{y}) \\ x_3(=\dot{y}-\ddot{y}) \\ x_4(=y^{(3)}) \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0\\ 1 \end{bmatrix} u \end{equation}

I think this should be right. Now, how do i implement the initial values in this state space?


Later i am supposed to execute a linear state transform to

\begin{equation} \dot{z} = \tilde{A}z+\tilde{B}u \end{equation} \begin{equation} y = \tilde{C}z+\tilde{D}u \end{equation} with \begin{equation} z^T = \begin{bmatrix} z_1 & z_2 & z_3 & z_4 \end{bmatrix} := \begin{bmatrix} y & \dot{y} & \ddot{y} & y^{(3)} \end{bmatrix} \end{equation}

but i think i can not execute the linear state transform just now, since the initial values are not represented in the state space above, as of now.

Here are a few theorems, Theorem 4 is just the linear state transformation, Theorem 5 is just matrix exponentials, Theorem 6 i sadly dont really get what its used for and Theorem 7 is just stating that the output response is invariant under a linear state transformation.

Does any of these Theorems, especially Theorem 6 have to do with initial conditions? enter image description here

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how do i implement the initial values in this state space?

they are literally the initial condition of the ODE, they won´t show up on the definition of \$\dot{x}\$ but on the definition of \$x(0)\$ (or whichever point you choose as the starting one of the trajectory.)

but i think i can not execute the linear state transform just now, since the initial values are not represented in the state space above, as of now.

If you have the system

$$ \begin{align} \dot{x} &= Ax +Bu \\ x(0) &= x_0\end{align} $$

and you transform \$x\$ into \$z = Tx\$, where \$T\$ is a square matrix of appropriate size, you can then find \$\dot{z}\$ which will be

$$ \begin{align} \dot{z} &= T\dot{x} = TAx + TBu \\ z(0) &= Tx(0)\end{align} $$ lastly, for an invertible \$T\$ we have that \$x = T^{-1}z\$.

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  • \$\begingroup\$ So you dont even include it in the state space itself but just put all initial conditions below? So i can just transform the state space from \$ x \$ to \$ z \$ at this point already? What about the zero-initial state and zero-input response? \begin{equation} x(t) = x_h(t) + x_p(t) \end{equation} I also have these weird formulas : \begin{equation} x(t) = e^{At}x_0 + \int_{0}^t e^{A(t)-\tau} Bu(\tau)d\tau \end{equation} \begin{equation} y(t) = Ce^{At}x_0 + \int_{0}^t Ce^{A(t)-\tau} Bu(\tau)d\tau+Du(t)\end{equation} I dont need them for this exercise? \$\endgroup\$ Commented Aug 20, 2023 at 22:17
  • \$\begingroup\$ I cant add photos in the comments so i will add a few theorems which confuse me and i thought i will need to use for this exercise. Thanks a lot! \$\endgroup\$ Commented Aug 20, 2023 at 22:21
  • \$\begingroup\$ Dont \$ A,B,C \$ and \$ D\$ need different multiplications with \$ T^{-1}\$ and \$ T\$ looking at Theorem 4 above? Thanks a lot, i really appreciate your help! \$\endgroup\$ Commented Aug 20, 2023 at 22:46
  • \$\begingroup\$ Well, if you read it carefully you will notice that I defined \$z = Tx\$ while the Theorems you later added define \$x = Tz\$, that is why all that is \$T\$ on my equations becomes \$T^{-1}\$ in your book. \$\endgroup\$
    – jDAQ
    Commented Aug 20, 2023 at 23:06
  • \$\begingroup\$ You are dealing with a linear ODE, that is why you can decouple the response to initial conditions and to the input \$u(t)\$ \$\endgroup\$
    – jDAQ
    Commented Aug 20, 2023 at 23:08

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