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Transfer function from state model [0 −1; −2 −3]

Short answer

Compute G(s)=C(sI-A)⁻¹B with A=[0 -1;-2 -3], B=[1;2], C=[1 0]. Result: Y(s)/U(s) = (s+1)/(s²+3s-2). The denominator has a root at +0.562, so the system is unstable.

Determine TF for the system whose state space representation is given by: [ẋ1; ẋ2] = [0 −1; −2 −3][x1; x2] + [1; 2]u, y = [1 0][x1; x2].

For x˙=Ax+Bu\dot{x} = Ax + Bu, y=Cx+Duy = Cx + Du:

Y(s)U(s)=C(sI−A)−1B+D=C adj⁡(sI−A) B∣sI−A∣+D\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D A=[0−1−2−3],B=[12],C=[10],D=0A = \begin{bmatrix} 0 & -1 \\ -2 & -3 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 2 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix},\quad D = 0

Step 1: (sI−A)(sI - A)

sI−A=[s00s]−[0−1−2−3]=[s12s+3]sI - A = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} s & 1 \\ 2 & s+3 \end{bmatrix}

Step 2: Determinant and adjoint

∣sI−A∣=s(s+3)−(1)(2)=s2+3s−2|sI-A| = s(s+3) - (1)(2) = s^2 + 3s - 2 adj⁡(sI−A)=[s+3−1−2s]\operatorname{adj}(sI-A) = \begin{bmatrix} s+3 & -1 \\ -2 & s \end{bmatrix}

Step 3: Multiply by BB and CC

adj⁡(sI−A)B=[(s+3)(1)+(−1)(2)(−2)(1)+s(2)]=[s+12s−2]\operatorname{adj}(sI-A)B = \begin{bmatrix} (s+3)(1) + (-1)(2) \\ (-2)(1) + s(2) \end{bmatrix} = \begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix} C adj⁡(sI−A)B=[10][s+12s−2]=s+1C\,\operatorname{adj}(sI-A)B = \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix} = s + 1

Step 4: Transfer function

Y(s)U(s)=s+1s2+3s−2\frac{Y(s)}{U(s)} = \frac{s+1}{s^2+3s-2}
  • Zero: s=−1s = -1
  • Poles: s=−3±172=0.562s = \dfrac{-3 \pm \sqrt{17}}{2} = 0.562 and −3.562-3.562

The pole at +0.562+0.562 lies in the right half-plane, so this system is unstable.

Answer: Y(s)U(s)=s+1s2+3s−2\dfrac{Y(s)}{U(s)} = \dfrac{s+1}{s^2+3s-2}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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