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State space analysis; TF of second-order model

Short answer

Advantages: handles MIMO, nonlinear, time-varying systems, includes initial conditions, reveals internal behaviour, supports modern design. Limitations: less intuitive, state model not unique, frequency-domain specs not directly visible. For the given model, G(s)=1/(s^2+3s+2)=1/[(s+1)(s+2)], poles at -1, -2, stable.

Discuss the advantages and limitations of state space analysis of control systems. Find the transfer function for the system represented by following state space model: ẋ = [−3 1; −2 0]x + [0; 1]u, y = [1 0]x.

Advantages of state space analysis

State space analysis describes a system by a set of first-order differential equations in terms of state variables, x˙=Ax+Bu\dot{x} = Ax + Bu, y=Cx+Duy = Cx + Du.

  1. Works for MIMO systems as easily as for single-input single-output systems.
  2. Applies to non-linear and time-varying systems; the transfer function method does not.
  3. Takes non-zero initial conditions into account.
  4. Gives the internal behaviour (all states), not only the input-output relation; controllability and observability can be tested.
  5. Suited to computer solution and to modern design methods (pole placement, optimal control, observers).
  6. Works in the time domain directly.

Limitations

  1. Needs more mathematics (matrices, eigenvalues); less physical insight for simple systems.
  2. The state model is not unique: different choices of states give different AA, BB, CC.
  3. Frequency-domain specifications (gain margin, phase margin, bandwidth) are not seen directly.
  4. Some states may not be measurable, so observers are needed for feedback.

Transfer function of the given model

A=[−31−20],  B=[01],  C=[10],  D=0A = \begin{bmatrix}-3 & 1\\ -2 & 0\end{bmatrix},\; B = \begin{bmatrix}0\\ 1\end{bmatrix},\; C = \begin{bmatrix}1 & 0\end{bmatrix},\; D = 0 G(s)=C(sI−A)−1B+DG(s) = C(sI - A)^{-1}B + D

Step 1:

sI−A=[s+3−12s],∣sI−A∣=s(s+3)+2=s2+3s+2sI - A = \begin{bmatrix}s+3 & -1\\ 2 & s\end{bmatrix}, \qquad |sI - A| = s(s+3) + 2 = s^2 + 3s + 2

Step 2:

(sI−A)−1=1s2+3s+2[s1−2s+3](sI - A)^{-1} = \frac{1}{s^2 + 3s + 2}\begin{bmatrix}s & 1\\ -2 & s+3\end{bmatrix}

Step 3:

(sI−A)−1B=1s2+3s+2[1s+3]G(s)=C(sI−A)−1B=[10]1s2+3s+2[1s+3]=1s2+3s+2\begin{aligned} (sI - A)^{-1}B &= \frac{1}{s^2 + 3s + 2}\begin{bmatrix}1\\ s+3\end{bmatrix} \\ G(s) = C(sI - A)^{-1}B &= \begin{bmatrix}1 & 0\end{bmatrix}\frac{1}{s^2 + 3s + 2}\begin{bmatrix}1\\ s+3\end{bmatrix} = \frac{1}{s^2 + 3s + 2} \end{aligned}

Answer:

Y(s)U(s)=1s2+3s+2=1(s+1)(s+2)\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 3s + 2} = \frac{1}{(s+1)(s+2)}

The poles are at s=−1s = -1 and s=−2s = -2, so the system is stable and overdamped.

✓ Answer checked by Claude Opus 5.5 (high effort)

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