State Space Analysis

For a state model $\dot{x} = Ax + Bu$, $y = Cx + Du$, the transfer function is

$$T(s) = \frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D$$

Here

$$A = \begin{bmatrix} 0 & 1 \\ -5 & -6 \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix},\quad D = 0$$

Step 1: $(sI - A)$ and its determinant

$$sI - A = \begin{bmatrix} s & -1 \\ 5 & s+6 \end{bmatrix}, \qquad |sI - A| = s(s+6) + 5 = s^2 + 6s + 5$$

Step 2: Inverse

$$(sI - A)^{-1} = \frac{\operatorname{adj}(sI-A)}{|sI-A|} = \frac{1}{s^2+6s+5}\begin{bmatrix} s+6 & 1 \\ -5 & s \end{bmatrix}$$

Step 3: Transfer function

$$\begin{aligned} (sI-A)^{-1}B &= \frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} \\ T(s) &= \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} = \frac{1}{s^2+6s+5} \end{aligned}$$ $$\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 6s + 5} = \frac{1}{(s+1)(s+5)}$$

Step 4: Stability

Characteristic equation: $|sI - A| = s^2 + 6s + 5 = 0$, so the eigenvalues of $A$ (poles) are

$$s = -1, \qquad s = -5$$

Both are real and negative (in the left half of the s-plane), so the system is stable. It is overdamped: comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$, $\omega_n = \sqrt5 = 2.24$ rad/s and $\zeta = 6/(2\sqrt5) = 1.34$. Its unit step response settles to $1/5 = 0.2$ without oscillation.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s^2+6s+5}$; poles at $-1$ and $-5$, so the system is stable.

The characteristic equation of a state model is $|sI - A| = 0$; its roots are the eigenvalues of $A$, which are the poles of the system.

$$A = \begin{bmatrix} -5 & -1 \\ 3 & -1 \end{bmatrix},\quad B = \begin{bmatrix} 2 \\ 5 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 2 \end{bmatrix}$$

Step 1: Form $sI - A$

$$sI - A = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -5 & -1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} s+5 & 1 \\ -3 & s+1 \end{bmatrix}$$

Step 2: Determinant

$$\begin{aligned} |sI - A| &= (s+5)(s+1) - (1)(-3) \\ &= s^2 + 6s + 5 + 3 \\ &= s^2 + 6s + 8 \end{aligned}$$

Characteristic equation:

$$s^2 + 6s + 8 = 0 \quad\Rightarrow\quad (s+2)(s+4) = 0$$

Roots (eigenvalues): $s = -2$ and $s = -4$. Both are in the left half of the s-plane, so the system is stable. Note that $B$ and $C$ do not affect the characteristic equation.

Check through the transfer function

$\operatorname{adj}(sI-A) = \begin{bmatrix} s+1 & -1 \\ 3 & s+5 \end{bmatrix}$, so

$$\begin{aligned} \operatorname{adj}(sI-A)B &= \begin{bmatrix} 2(s+1) - 5 \\ 6 + 5(s+5) \end{bmatrix} = \begin{bmatrix} 2s - 3 \\ 5s + 31 \end{bmatrix} \\ \frac{Y(s)}{U(s)} &= \frac{(2s-3) + 2(5s+31)}{s^2+6s+8} = \frac{12s + 59}{s^2 + 6s + 8} \end{aligned}$$

The denominator of the transfer function is the same characteristic polynomial, $s^2 + 6s + 8$.

Answer: $s^2 + 6s + 8 = 0$, roots $-2$ and $-4$.

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

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

Step 1: $sI - A$ and its determinant

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 0 & s & -1 \\ 1 & 2 & s+3 \end{bmatrix}$$ $$\begin{aligned} |sI - A| &= s\left[s(s+3) + 2\right] - (-1)\left[0\cdot(s+3) - (-1)(1)\right] + 0 \\ &= s^3 + 3s^2 + 2s + 1 \end{aligned}$$

Step 2: Required part of the adjoint

Because $B$ has only its first element non-zero and $C$ picks only the first row, only the (1,1) element of the adjoint is needed:

$$\operatorname{adj}(sI-A)_{11} = \begin{vmatrix} s & -1 \\ 2 & s+3 \end{vmatrix} = s^2 + 3s + 2$$

Full first column of the adjoint: $\begin{bmatrix} s^2+3s+2 \\ -1 \\ -s \end{bmatrix}$, so

$$\operatorname{adj}(sI-A)B = 10\begin{bmatrix} s^2+3s+2 \\ -1 \\ -s \end{bmatrix}$$

Step 3: Transfer function

$$C\,\operatorname{adj}(sI-A)B = 10(s^2 + 3s + 2)$$ $$\frac{Y(s)}{U(s)} = \frac{10(s^2+3s+2)}{s^3+3s^2+2s+1} = \frac{10(s+1)(s+2)}{s^3+3s^2+2s+1}$$

Stability check

Routh array for $s^3 + 3s^2 + 2s + 1$:

No sign change, so the system is stable. The poles are $-2.325$ and $-0.338 \pm j0.562$; the zeros are $-1$ and $-2$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{10(s^2+3s+2)}{s^3+3s^2+2s+1}$.

Choose the energy-storing elements' variables as states: capacitor voltages and inductor current.

• $x_1 = v_1$: voltage across the 1 F capacitor (node A)
• $x_2 = i_L$: current in the 1 H inductor (from A to B)
• $x_3 = v_2$: voltage across the 2 F capacitor (node B) $= V_o$

  Vs  1 ohm   A   1 H      B
 (+)-/\/\/--+--UUUU--+------+---o Vo
  |         |   iL   |      |
  |        === 1F   1 ohm  === 2F
  |         |        |      |
 (-)--------+--------+------+---o

KCL at node A

Current from the source through 1 Ω $= (V_s - v_1)/1$:

$$1\cdot\frac{dv_1}{dt} = (V_s - v_1) - i_L \;\Rightarrow\; \dot{x}_1 = -x_1 - x_2 + V_s$$

KVL around the inductor

$$1\cdot\frac{di_L}{dt} = v_1 - v_2 \;\Rightarrow\; \dot{x}_2 = x_1 - x_3$$

KCL at node B

$$2\frac{dv_2}{dt} = i_L - \frac{v_2}{1} \;\Rightarrow\; \dot{x}_3 = 0.5x_2 - 0.5x_3$$

State and output equations

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} -1 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 0.5 & -0.5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}V_s$$ $$V_o = y = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

Check: from this model $\dfrac{V_o(s)}{V_s(s)} = \dfrac{1}{2s^3 + 3s^2 + 4s + 2}$, the same as found directly by nodal analysis with impedances ($1/s$ for 1 F, $s$ for 1 H, $1/(2s)$ for 2 F).

$$\frac{Y(s)}{U(s)} = \frac{20(4s+2)}{s^3+5s^2+8s+2} = \frac{80s + 40}{s^3 + 5s^2 + 8s + 2}$$

Since the numerator has $s$ terms, use the phase-variable (controllable canonical) form with an intermediate variable $X(s)$.

Step 1: Split the transfer function

$$\frac{Y(s)}{U(s)} = \frac{X(s)}{U(s)}\cdot\frac{Y(s)}{X(s)},\qquad \frac{X(s)}{U(s)} = \frac{1}{s^3+5s^2+8s+2},\qquad \frac{Y(s)}{X(s)} = 80s + 40$$

So $\dddot{x} + 5\ddot{x} + 8\dot{x} + 2x = u$ and $y = 80\dot{x} + 40x$.

Step 2: Choose state variables

$x_1 = x$, $x_2 = \dot{x}$, $x_3 = \ddot{x}$:

$$\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= x_3 \\ \dot{x}_3 &= -2x_1 - 8x_2 - 5x_3 + u \\ y &= 40x_1 + 80x_2 \end{aligned}$$

Step 3: Matrix form

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -8 & -5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u$$ $$y = \begin{bmatrix} 40 & 80 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$$

Step 4: Characteristic equation from the matrix

The characteristic equation is $|sI - A| = 0$:

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 0 & s & -1 \\ 2 & 8 & s+5 \end{bmatrix}$$ $$\begin{aligned} |sI-A| &= s\left[s(s+5) + 8\right] + 1\left[0 + 2\right] \\ &= s^3 + 5s^2 + 8s + 2 = 0 \end{aligned}$$

This equals the denominator of the transfer function, as it must. Its roots (eigenvalues of $A$) are $s = -0.304$ and $-2.348 \pm j1.029$, all in the left half-plane, so the system is stable.

Answer: $A = \begin{bmatrix} 0&1&0\\0&0&1\\-2&-8&-5 \end{bmatrix}$, $B = \begin{bmatrix}0\\0\\1\end{bmatrix}$, $C = \begin{bmatrix}40&80&0\end{bmatrix}$; characteristic equation $s^3 + 5s^2 + 8s + 2 = 0$.

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

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 0 & -1 \\ -2 & -3 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 2 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}$$

Step 1: $(sI-A)$ and determinant

$$sI - A = \begin{bmatrix} s & 1 \\ 2 & s+3 \end{bmatrix},\qquad |sI-A| = s(s+3) - 2 = s^2 + 3s - 2$$

Step 2: Inverse

$$(sI-A)^{-1} = \frac{1}{s^2+3s-2}\begin{bmatrix} s+3 & -1 \\ -2 & s \end{bmatrix}$$

Step 3: Transfer function

$$\begin{aligned} (sI-A)^{-1}B &= \frac{1}{s^2+3s-2}\begin{bmatrix} (s+3) - 2 \\ -2 + 2s \end{bmatrix} = \frac{1}{s^2+3s-2}\begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix} \\ \frac{Y(s)}{U(s)} &= \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+3s-2}\begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix} = \frac{s+1}{s^2+3s-2} \end{aligned}$$

Stability

Characteristic equation $s^2 + 3s - 2 = 0$:

$$s = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm 4.123}{2} = 0.562,\ -3.562$$

One pole lies in the right half of the s-plane ($s = +0.562$). Also the coefficients of the characteristic polynomial do not all have the same sign, which already fails the necessary condition for stability. So the system is unstable; its response contains a growing term $e^{0.562t}$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{s+1}{s^2+3s-2}$; unstable.

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

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 1 \end{bmatrix}$$

Step 1: $(sI-A)$ and determinant

$$sI - A = \begin{bmatrix} s-1 & 0 \\ -1 & s-1 \end{bmatrix},\qquad |sI-A| = (s-1)^2 = s^2 - 2s + 1$$

Step 2: Inverse

$$(sI-A)^{-1} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 & 0 \\ 1 & s-1 \end{bmatrix}$$

Step 3: Transfer function

$$\begin{aligned} (sI-A)^{-1}B &= \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ 1 + s - 1 \end{bmatrix} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} \\ \frac{Y(s)}{U(s)} &= \begin{bmatrix} 1 & 1 \end{bmatrix}\frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} = \frac{2s-1}{(s-1)^2} = \frac{2s-1}{s^2-2s+1} \end{aligned}$$

Stability

Characteristic equation: $(s-1)^2 = 0$, so there is a repeated pole at $s = +1$ in the right half of the s-plane (also visible because $A$ is lower triangular with diagonal entries 1, 1).

The impulse response contains terms $e^{t}$ and $te^{t}$, which grow without bound. The system is unstable.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{2s-1}{(s-1)^2}$; poles at $s = 1, 1$, zero at $s = 0.5$; unstable.

The mass is pulled by $F(t)$, held back by the spring $K$ (left wall) and the damper $B$ (right wall). Both oppose the motion.

Differential equation

Newton's law (free-body diagram of $M$):

$$M\frac{d^2x}{dt^2} + B\frac{dx}{dt} + Kx = F(t)$$

State variables

Displacement and velocity of the mass (energy in the spring and the mass):

$$x_1 = x, \qquad x_2 = \dot{x}$$ $$\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -\frac{K}{M}x_1 - \frac{B}{M}x_2 + \frac{1}{M}F(t) \end{aligned}$$

State-space model

Taking the displacement $x$ as the output:

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\dfrac{K}{M} & -\dfrac{B}{M} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ \dfrac{1}{M} \end{bmatrix}F(t)$$ $$y = \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

(If velocity is wanted as output, use $C = [0 \;\; 1]$.) The corresponding transfer function is $\dfrac{X(s)}{F(s)} = \dfrac{1}{Ms^2 + Bs + K}$.

Take the capacitor voltages as state variables:

$$x_1 = v_2 \ (\text{voltage across } C_1), \qquad x_2 = v_o \ (\text{voltage across } C_2)$$

 Vi   R1     v2    R2     Vo
 o--/\/\/--+--/\/\/--+----o
   i1 ->   |  i2 ->  |
          === C1    === C2
           |         |
 o---------+---------+----o

Circuit equations

Mesh currents: $i_1 = \dfrac{V_i - v_2}{R_1}$ through $R_1$, $i_2 = \dfrac{v_2 - v_o}{R_2}$ through $R_2$.

KCL at node $v_2$ (current in $C_1$ is $i_1 - i_2$):

$$C_1\frac{dv_2}{dt} = \frac{V_i - v_2}{R_1} - \frac{v_2 - v_o}{R_2}$$

KCL at the output node (current in $C_2$ is $i_2$):

$$C_2\frac{dv_o}{dt} = \frac{v_2 - v_o}{R_2}$$

State equations

$$\begin{aligned} \dot{x}_1 &= -\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1}\right)x_1 + \frac{1}{R_2C_1}x_2 + \frac{1}{R_1C_1}V_i \\ \dot{x}_2 &= \frac{1}{R_2C_2}x_1 - \frac{1}{R_2C_2}x_2 \end{aligned}$$

Matrix form

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} -\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1}\right) & \dfrac{1}{R_2C_1} \\ \dfrac{1}{R_2C_2} & -\dfrac{1}{R_2C_2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} \dfrac{1}{R_1C_1} \\ 0 \end{bmatrix}V_i$$ $$V_o = y = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

This gives $\dfrac{V_o(s)}{V_i(s)} = \dfrac{1}{R_1C_1R_2C_2s^2 + (R_1C_1 + R_2C_2 + R_1C_2)s + 1}$, the known result for a two-stage RC ladder.

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B$$

with

$$A = \begin{bmatrix} 0 & 1 \\ -\dfrac{K}{M} & -\dfrac{B}{M} \end{bmatrix},\quad B_m = \begin{bmatrix} 0 \\ \dfrac{1}{M} \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}$$

(The input matrix is written $B_m$ to avoid confusion with the damping coefficient $B$.)

Step 1: $(sI - A)$ and determinant

$$sI - A = \begin{bmatrix} s & -1 \\ \dfrac{K}{M} & s + \dfrac{B}{M} \end{bmatrix}$$ $$|sI - A| = s\left(s + \frac{B}{M}\right) + \frac{K}{M} = s^2 + \frac{B}{M}s + \frac{K}{M}$$

Step 2: Inverse

$$(sI - A)^{-1} = \frac{1}{s^2 + \frac{B}{M}s + \frac{K}{M}}\begin{bmatrix} s + \dfrac{B}{M} & 1 \\ -\dfrac{K}{M} & s \end{bmatrix}$$

Step 3: Transfer function

$$(sI-A)^{-1}B_m = \frac{1}{s^2 + \frac{B}{M}s + \frac{K}{M}}\begin{bmatrix} \dfrac{1}{M} \\ \dfrac{s}{M} \end{bmatrix}$$ $$\frac{Y(s)}{U(s)} = \begin{bmatrix} 1 & 0 \end{bmatrix}(sI-A)^{-1}B_m = \frac{1/M}{s^2 + \frac{B}{M}s + \frac{K}{M}} = \frac{1}{Ms^2 + Bs + K}$$

This is the transfer function of the mass–spring–damper system $M\ddot{y} + B\dot{y} + Ky = u$, with $\omega_n = \sqrt{K/M}$ and $\zeta = \dfrac{B}{2\sqrt{KM}}$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{Ms^2 + Bs + K}$.

Let $y_1$ be the downward displacement of $M_1$ and $y = y_2$ that of $M_2$, both measured from the static equilibrium position, so gravity and the static spring force cancel and do not appear.

   //////////  support
       |
       K  (spring)
       |
   +-------+
   |  M1   | <-- u(t), y1
   +-------+
       |
       B  (damper)
       |
   +-------+
   |  M2   |  y(t) = y2
   +-------+

Equations of motion

For $M_1$: spring force $Ky_1$ and damper force $B(\dot{y}_1 - \dot{y}_2)$ oppose its motion:

$$M_1\ddot{y}_1 + B(\dot{y}_1 - \dot{y}_2) + Ky_1 = u(t)$$

For $M_2$: only the damper acts on it:

$$M_2\ddot{y}_2 + B(\dot{y}_2 - \dot{y}_1) = 0$$

State variables

$$x_1 = y_1,\quad x_2 = \dot{y}_1,\quad x_3 = y_2,\quad x_4 = \dot{y}_2$$ $$\begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -\frac{K}{M_1}x_1 - \frac{B}{M_1}x_2 + \frac{B}{M_1}x_4 + \frac{1}{M_1}u \\ \dot{x}_3 &= x_4 \\ \dot{x}_4 &= \frac{B}{M_2}x_2 - \frac{B}{M_2}x_4 \end{aligned}$$

State-space representation

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \\ \dot{x}_4 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -\dfrac{K}{M_1} & -\dfrac{B}{M_1} & 0 & \dfrac{B}{M_1} \\ 0 & 0 & 0 & 1 \\ 0 & \dfrac{B}{M_2} & 0 & -\dfrac{B}{M_2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} + \begin{bmatrix} 0 \\ \dfrac{1}{M_1} \\ 0 \\ 0 \end{bmatrix}u(t)$$ $$y(t) = \begin{bmatrix} 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$$

$y_2$ itself is needed as a state only because it is the output; the dynamics depend on $y_1$, $\dot{y}_1$ and $\dot{y}_2$.

Choose the inductor currents and the capacitor voltage as states. The left loop current flows through 2 Ω and 2 H (in series); the right loop current flows through 1 H and 3 Ω (in series); the capacitor carries the difference.

• $x_1 = i_1$: current in the 2 H inductor (left loop, clockwise)
• $x_2 = i_2$: current in the 1 H inductor (right loop, from A towards $V_2$)
• $x_3 = v_C = v_A - v_B$: voltage on the 2 F capacitor

     2 ohm   A    1 H
  +--/\/\/--+---UUUU---+
  |   i1 -> |  i2 ->   |
 (V1)      === 2F     (V2)
  |         |          |
  +--UUUU---+--/\/\/---+
      2 H   B   3 ohm

Left loop (KVL)

$$V_1 = 2i_1 + v_C + 2\frac{di_1}{dt} \;\Rightarrow\; \dot{x}_1 = -x_1 - 0.5x_3 + 0.5V_1$$

Right loop (KVL)

From A through 1 H, $V_2$ and 3 Ω back to B:

$$v_C = 1\cdot\frac{di_2}{dt} + V_2 + 3i_2 \;\Rightarrow\; \dot{x}_2 = -3x_2 + x_3 - V_2$$

Capacitor (KCL at A)

$$2\frac{dv_C}{dt} = i_1 - i_2 \;\Rightarrow\; \dot{x}_3 = 0.5x_1 - 0.5x_2$$

State model

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} -1 & 0 & -0.5 \\ 0 & -3 & 1 \\ 0.5 & -0.5 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0.5 & 0 \\ 0 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

Output: no output is marked, so the capacitor voltage is taken as output:

$$y = v_C = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

(For the current in $V_2$ as output, use $C = [0 \;\; 1 \;\; 0]$.)

Take the two inductor currents and the capacitor voltage as states (assumed directions shown):

• $x_1 = i_1$: current in the 2 H inductor, from $V_1$ towards node A
• $x_2 = i_2$: current in the 1 H inductor, from node A towards $V_2$
• $x_3 = v_C$: voltage on the 2 F capacitor (node A to common line)

    1 ohm  2 H     A    1 H  2 ohm
  +-/\/\/-UUUU----+---UUUU-/\/\/-+
  |    i1 ->      |    i2 ->     |
 (V1)            === 2F         (V2)
  |               |              |
  +---------------+--------------+

Left loop (KVL)

$$V_1 = 1\cdot i_1 + 2\frac{di_1}{dt} + v_C \;\Rightarrow\; \dot{x}_1 = -0.5x_1 - 0.5x_3 + 0.5V_1$$

Right loop (KVL)

$$v_C = 1\cdot\frac{di_2}{dt} + 2i_2 + V_2 \;\Rightarrow\; \dot{x}_2 = -2x_2 + x_3 - V_2$$

KCL at node A

$$2\frac{dv_C}{dt} = i_1 - i_2 \;\Rightarrow\; \dot{x}_3 = 0.5x_1 - 0.5x_2$$

State equation

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} -0.5 & 0 & -0.5 \\ 0 & -2 & 1 \\ 0.5 & -0.5 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0.5 & 0 \\ 0 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

Output equation

No output is marked in the figure, so the capacitor voltage (node A voltage) is taken as the output:

$$y = v_C = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 & 0 \end{bmatrix}\begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

If, for example, the voltage across the 2 Ω resistor is the output, then $y = 2x_2$, i.e. $C = [0 \;\; 2 \;\; 0]$.

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

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 0 & 1 & 0 \\ -1 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$$

Step 1: $sI - A$ and determinant

$$sI - A = \begin{bmatrix} s & -1 & 0 \\ 1 & s+1 & 0 \\ -1 & 0 & s \end{bmatrix}$$

Expanding along the third column:

$$|sI - A| = s\begin{vmatrix} s & -1 \\ 1 & s+1 \end{vmatrix} = s\left[s(s+1) + 1\right] = s^3 + s^2 + s$$

Step 2: Needed adjoint elements

$C$ picks row 3 and $B$ picks column 2, so only $\operatorname{adj}(sI-A)_{32}$ is needed. It is the cofactor of element (2,3):

$$\operatorname{adj}_{32} = (-1)^{2+3}\begin{vmatrix} s & -1 \\ -1 & 0 \end{vmatrix} = -\left[(s)(0) - (-1)(-1)\right] = 1$$

(The full column $\operatorname{adj}(sI-A)B = [\,s,\; s^2,\; 1\,]^T$.)

Step 3: Transfer function

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

Poles and zeros

• Poles: $s(s^2 + s + 1) = 0$:

$$s = 0, \qquad s = \frac{-1 \pm \sqrt{1-4}}{2} = -0.5 \pm j0.866$$

• Zeros: the numerator is a constant, so there are no finite zeros (three zeros at infinity).

The pole at the origin makes the system marginally stable (type 1); the complex pair has $\omega_n = 1$ rad/s and $\zeta = 0.5$.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s(s^2+s+1)}$; poles $0,\ -0.5 \pm j0.866$; no finite zeros.