State space analysis; TF and stability (−4, 10)

Advantages of state-space analysis

1. Applies to MIMO (multi-input multi-output) systems as easily as to single-input single-output systems.
2. Can handle non-linear and time-varying systems; the transfer function needs linear time-invariant systems.
3. Includes initial conditions; the transfer function assumes zero initial conditions.
4. Gives the internal behaviour (all state variables), not only the input–output relation.
5. Allows tests of controllability and observability, and hidden (cancelled) modes are not lost.
6. Suited to computer solution and simulation, since it uses first-order matrix equations.
7. Forms the basis of modern and optimal control design: state feedback (pole placement), observers, Kalman filter.

Limitations

1. Needs more computation (matrix algebra, $n$ first-order equations); physical insight is less direct than with a transfer function or Bode plot.
2. The choice of state variables is not unique, so different models describe the same system.
3. State variables may not all be measurable; an observer may be needed for state feedback.
4. Classical specifications (gain margin, phase margin, bandwidth) are not read directly.

Transfer function

$$A = \begin{bmatrix} -4 & -1 \\ 10 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}$$ $$sI - A = \begin{bmatrix} s+4 & 1 \\ -10 & s \end{bmatrix},\qquad |sI-A| = s(s+4) + 10 = s^2 + 4s + 10$$ $$(sI-A)^{-1} = \frac{1}{s^2+4s+10}\begin{bmatrix} s & -1 \\ 10 & s+4 \end{bmatrix},\qquad (sI-A)^{-1}B = \frac{1}{s^2+4s+10}\begin{bmatrix} s \\ 10 \end{bmatrix}$$ $$\frac{Y(s)}{U(s)} = \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+4s+10}\begin{bmatrix} s \\ 10 \end{bmatrix} = \frac{s}{s^2 + 4s + 10}$$

Stability

Characteristic equation $s^2 + 4s + 10 = 0$:

$$s = \frac{-4 \pm \sqrt{16 - 40}}{2} = -2 \pm j2.449$$

Both poles have negative real parts, so the system is stable (underdamped: $\omega_n = \sqrt{10} = 3.16$ rad/s, $\zeta = 4/(2\sqrt{10}) = 0.632$). There is a zero at $s = 0$.