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State model from 10(s²+2s)/(s³+5s²+8s+15)

Short answer

Phase-variable realization: state equation ̇x = [0 1 0; 0 0 1; -15 -8 -5]x + [0;0;1]u, output equation y = [0 20 10]x.

A system is described by the transfer function Y(s)/U(s) = 10(s² + 2s)/(s³ + 5s² + 8s + 15). Find its state and output equation in matrix form.

A state model can be written directly from the transfer function in phase variable (controllable canonical) form. The denominator gives the system matrix and the numerator gives the output matrix.

Given transfer function

Y(s)U(s)=10s2+20ss3+5s2+8s+15\frac{Y(s)}{U(s)} = \frac{10s^2 + 20s}{s^3 + 5s^2 + 8s + 15}

The numerator order (2) is less than the denominator order (3), so there is no direct term (D=0D = 0).

Step 1: Introduce an intermediate variable

Let

Y(s)U(s)=Y(s)X(s)⋅X(s)U(s),X(s)U(s)=1s3+5s2+8s+15,Y(s)X(s)=10s2+20s\frac{Y(s)}{U(s)} = \frac{Y(s)}{X(s)}\cdot\frac{X(s)}{U(s)}, \quad \frac{X(s)}{U(s)} = \frac{1}{s^3 + 5s^2 + 8s + 15}, \quad \frac{Y(s)}{X(s)} = 10s^2 + 20s

So

x...+5x¨+8x˙+15x=u,y=10x¨+20x˙\dddot{x} + 5\ddot{x} + 8\dot{x} + 15x = u, \qquad y = 10\ddot{x} + 20\dot{x}

Step 2: Choose state variables

x1=x,x2=x˙,x3=x¨x_1 = x, \quad x_2 = \dot{x}, \quad x_3 = \ddot{x}

Then

x˙1=x2x˙2=x3x˙3=−15x1−8x2−5x3+u\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= x_3 \\ \dot{x}_3 &= -15x_1 - 8x_2 - 5x_3 + u \end{aligned}

Step 3: Output equation

y=10x¨+20x˙=0⋅x1+20x2+10x3y = 10\ddot{x} + 20\dot{x} = 0\cdot x_1 + 20x_2 + 10x_3

State and output equations in matrix form

[x˙1x˙2x˙3]=[010001−15−8−5][x1x2x3]+[001]u\begin{bmatrix}\dot{x}_1\\ \dot{x}_2\\ \dot{x}_3\end{bmatrix} = \begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ -15 & -8 & -5\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix} + \begin{bmatrix}0\\ 0\\ 1\end{bmatrix} u y=[02010][x1x2x3]y = \begin{bmatrix}0 & 20 & 10\end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}

So

A=[010001−15−8−5],  B=[001],  C=[02010],  D=0A = \begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ -15 & -8 & -5\end{bmatrix},\; B = \begin{bmatrix}0\\ 0\\ 1\end{bmatrix},\; C = \begin{bmatrix}0 & 20 & 10\end{bmatrix},\; D = 0

Block diagram of the realization

u -->(+)--> 1/s --x3--> 1/s --x2--> 1/s --x1
      ^ -5x3 -8x2 -15x1 (fed back)
y = 20*x2 + 10*x3

Check

Using G(s)=C(sI−A)−1BG(s) = C(sI - A)^{-1}B, the last column of (sI−A)−1(sI-A)^{-1} is 1Δ(s)[1    s    s2]T\frac{1}{\Delta(s)}[1\;\; s\;\; s^2]^T with Δ(s)=s3+5s2+8s+15\Delta(s) = s^3 + 5s^2 + 8s + 15. Hence

G(s)=0⋅1+20s+10s2s3+5s2+8s+15=10(s2+2s)s3+5s2+8s+15G(s) = \frac{0\cdot 1 + 20s + 10s^2}{s^3 + 5s^2 + 8s + 15} = \frac{10(s^2 + 2s)}{s^3 + 5s^2 + 8s + 15}

which is the given transfer function.

Answer: x˙=Ax+Bu\dot{x} = Ax + Bu, y=Cxy = Cx with AA, BB, CC as above (phase variable form). The state model is not unique; other forms (observable canonical, Jordan) give the same transfer function.

✓ Answer checked by Claude Opus 5.5 (high effort)

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