Component Modeling

Assumptions: the masses have no friction with the ground; $x_1$ is the displacement of $M_1$, $x$ the displacement of $M_2$, and $x_3$ the displacement of the junction between damper $B$ and spring $K_3$ (B on the $M_1$ side).

Equations of motion (Laplace, zero initial conditions)

Mass $M_1$:

$$F(s) = (M_1s^2 + K_1)X_1 + K_2(X_1 - X) + Bs(X_1 - X_3)$$

Junction ($x_3$, massless): $\;Bs(X_3 - X_1) + K_3(X_3 - X) = 0$

Mass $M_2$: $\;M_2s^2X + K_2(X - X_1) + K_3(X - X_3) = 0$

Solving

The series $B$–$K_3$ branch acts as an equivalent spring $\dfrac{BsK_3}{Bs+K_3}$. Together with $K_2$ in parallel, the coupling between the masses is

$$K_c(s) = K_2 + \frac{BK_3s}{Bs+K_3} = \frac{N(s)}{Bs+K_3}, \quad N(s) = B(K_2+K_3)s + K_2K_3$$

The two mass equations become $(M_1s^2+K_1+K_c)X_1 - K_cX = F$ and $-K_cX_1 + (M_2s^2+K_c)X = 0$. So

$$\frac{X(s)}{F(s)} = \frac{K_c}{(M_1s^2+K_1+K_c)(M_2s^2+K_c) - K_c^2} = \frac{K_c}{(M_1s^2+K_1)(M_2s^2+K_c) + K_cM_2s^2}$$

Multiplying top and bottom by $(Bs+K_3)$ and expanding:

$$\frac{X(s)}{F(s)} = \frac{B(K_2+K_3)s + K_2K_3}{\Delta(s)}$$ $$\begin{aligned} \Delta(s) ={}& BM_1M_2s^5 + K_3M_1M_2s^4 + B\left[K_1M_2 + (K_2+K_3)(M_1+M_2)\right]s^3 \\ &+ K_3\left[K_1M_2 + K_2(M_1+M_2)\right]s^2 + BK_1(K_2+K_3)s + K_1K_2K_3 \end{aligned}$$

Force–voltage (F-V) analogy

Force → voltage, mass → inductance $L = M$, damper → resistance $R = B$, spring → capacitance $C = 1/K$, velocity → loop current. Mechanical parallel → electrical series and vice versa, so the coupling becomes $C_2$ in series with ($R \parallel C_3$).

  +--[L1]--[C1]--+--[L2]--+
  |              |        |
 (f) i1->       C2   <-i2 |
  |              |        |
  |           +--+--+     |
  |           R     C3    |
  |           +--+--+     |
  |              |        |
  +--------------+--------+
 L1=M1, C1=1/K1, C2=1/K2, R=B, C3=1/K3, L2=M2

Force–current (F-I) analogy

Force → current source, mass → capacitance $C = M$, damper → conductance ($R = 1/B$), spring → inductance $L = 1/K$, velocity → node voltage. Each displacement point becomes a node.

     v1                v3             v2
  +---o---+---[R]------o-----[L3]---+---o
  |   |   |                         |   |
  |   |   +----------[L2]-----------+   |
  |   |   |                             |
 (f)  C1  L1                            C2
  |   |   |                             |
  +---+---+-----------------------------+
            ground (reference)
 C1=M1, L1=1/K1, L2=1/K2, R=1/B, L3=1/K3, C2=M2

Node equation at $v_1$, for example: $f = M_1\dot v_1 + K_1\!\int v_1dt + K_2\!\int(v_1-v_2)dt + B(v_1-v_3)$, which matches the force equation of $M_1$ with $v = \dot x$.

Assumption: ideal op-amp (infinite gain and input impedance), so $V_- = V_+$ and no current enters the inputs.

        R1 (feedback)
     +---/\/\/---+
     |           |
Ei --+-/\/\/--(-)|\
        R1       | >--+-- Eo
Ei ---||--+---(+)|/
       C  |
         R2
          |
         GND

Non-inverting input

$C$ and $R_2$ form a voltage divider:

$$V_+ = E_i\frac{R_2}{R_2 + \dfrac{1}{sC}} = E_i\frac{sCR_2}{1 + sCR_2}$$

Inverting input (KCL)

Current through input $R_1$ = current through feedback $R_1$, with $V_- = V_+$:

$$\frac{E_i - V_+}{R_1} = \frac{V_+ - E_o}{R_1} \;\Rightarrow\; E_o = 2V_+ - E_i$$

Transfer function

$$\begin{aligned} E_o &= E_i\left(\frac{2sCR_2}{1+sCR_2} - 1\right) = E_i\,\frac{sCR_2 - 1}{sCR_2 + 1} \\ \frac{E_o(s)}{E_i(s)} &= \frac{sR_2C - 1}{sR_2C + 1} = -\frac{1 - sR_2C}{1 + sR_2C} \end{aligned}$$

This is an all-pass (phase-shift) network: $|E_o/E_i| = 1$ at all frequencies, and phase $\phi = 180^\circ - 2\tan^{-1}(\omega R_2C)$, going from $180^\circ$ at $\omega = 0$ to $0^\circ$ at high frequency (90° at $\omega = 1/R_2C$). It has a zero at $s = +1/R_2C$ (non-minimum phase) and a pole at $s = -1/R_2C$.

Assumptions: zero initial conditions; $\theta_3$ is the angle of the junction between damper $D$ and spring $K$ (D on the $J_1$ side). The series $D$–$K$ coupling is massless.

Equations (Laplace form)

Inertia $J_1$:

$$T(s) = (J_1s^2 + D_1s)\theta_1 + Ds(\theta_1 - \theta_3)$$

Junction: $\;Ds(\theta_3 - \theta_1) + K(\theta_3 - \theta_2) = 0$

Inertia $J_2$: $\;(J_2s^2 + D_2s)\theta_2 + K(\theta_2 - \theta_3) = 0$

Solving

Damper and spring in series transmit the same torque, so they act as one element with "stiffness"

$$K_e(s) = \frac{DsK}{Ds + K}$$

Then $(J_1s^2+D_1s+K_e)\theta_1 - K_e\theta_2 = T$ and $-K_e\theta_1 + (J_2s^2+D_2s+K_e)\theta_2 = 0$, giving

$$\frac{\theta_2}{T} = \frac{K_e}{(J_1s^2+D_1s)(J_2s^2+D_2s) + K_e\left[(J_1+J_2)s^2 + (D_1+D_2)s\right]}$$

Multiply top and bottom by $(Ds+K)$ and cancel the common $s$:

$$\frac{\theta_2(s)}{T(s)} = \frac{DK}{s\left[(Ds+K)(J_1s+D_1)(J_2s+D_2) + DK\{(J_1+J_2)s + D_1 + D_2\}\right]}$$

Expanded:

$$\frac{\theta_2(s)}{T(s)} = \frac{DK}{s\left[a_3s^3 + a_2s^2 + a_1s + a_0\right]}$$ $$\begin{aligned} a_3 &= DJ_1J_2, \quad a_2 = D(D_1J_2 + D_2J_1) + KJ_1J_2 \\ a_1 &= DD_1D_2 + K\left[D(J_1+J_2) + D_1J_2 + D_2J_1\right] \\ a_0 &= K\left[D(D_1+D_2) + D_1D_2\right] \end{aligned}$$

Torque–voltage (T-V) analogy

Torque → voltage, $J$ → $L$, $D$ → $R$, $K$ → $C = 1/K$, angular velocity → loop current. The series $D$–$K$ pair becomes $R \parallel C$ shared by the two loops.

  +--[L1]--[R1]--+-----+--[L2]--[R2]--+
  |              |     |              |
 (T)  i1 ->      R     C     <- i2    |
  |              |     |              |
  +--------------+-----+--------------+
 L1=J1, R1=D1, R=D, C=1/K, L2=J2, R2=D2

Torque–current (T-I) analogy

Torque → current source, $J$ → $C$, $D$ → conductance ($R = 1/D$), $K$ → $L = 1/K$, angular velocity → node voltage.

      ω1          ω3           ω2
  +----o----+--[R]--o--[L]--+----o
  |    |    |               |    |
 (T)   C1   R1              C2   R2
  |    |    |               |    |
  +----+----+---------------+----+
              ground
 C1=J1, R1=1/D1, R=1/D, L=1/K, C2=J2, R2=1/D2

Assumptions: $f_{c1}$, $f_{c2}$ are viscous friction coefficients of $m_1$, $m_2$ with the ground; $x_3$ is the displacement of the massless junction between $K_2$ and $B_1$ ($K_2$ on the $m_1$ side); zero initial conditions.

Equations of motion (Laplace form)

Mass $m_1$:

$$F(s) = (m_1s^2 + f_{c1}s + K_1)X_1 + K_2(X_1 - X_3)$$

Junction: $\;K_2(X_3 - X_1) + B_1s(X_3 - X_2) = 0$

Mass $m_2$:

$$0 = (m_2s^2 + f_{c2}s + B_2s)X_2 + B_1s(X_2 - X_3)$$

Solving

$K_2$ and $B_1$ in series carry the same force, so they act as one coupling element

$$Z_c(s) = \frac{K_2B_1s}{K_2 + B_1s}$$

Let $Z_1 = m_1s^2 + f_{c1}s + K_1$ and $Z_2 = m_2s^2 + (f_{c2}+B_2)s$. Then

$$\begin{bmatrix} Z_1 + Z_c & -Z_c \\ -Z_c & Z_2 + Z_c \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} F \\ 0 \end{bmatrix}$$ $$\frac{X_2}{F} = \frac{Z_c}{(Z_1+Z_c)(Z_2+Z_c) - Z_c^2} = \frac{Z_c}{Z_1Z_2 + Z_c(Z_1 + Z_2)}$$

Substituting $Z_c$, multiplying by $(K_2 + B_1s)$ and cancelling the common factor $s$:

$$\frac{X_2(s)}{F(s)} = \frac{K_2B_1}{(K_2 + B_1s)(m_1s^2 + f_{c1}s + K_1)(m_2s + f_{c2} + B_2) + K_2B_1\left[(m_1+m_2)s^2 + (f_{c1}+f_{c2}+B_2)s + K_1\right]}$$

The denominator is a fourth-order polynomial; its leading term is $B_1m_1m_2s^4$ and its constant term is $K_1K_2(B_1 + B_2 + f_{c2})$.

Force–voltage (F-V) analogy

Force → voltage, mass → $L$, friction/damper → $R$, spring → $C = 1/K$, velocity → loop current. Mechanical series ($K_2$, $B_1$) → electrical parallel ($C_2 \parallel R_{B1}$), shared by both loops.

  +--[L1]--[Rf1]--[C1]--+--[L2]--[Rf2]--[RB2]--+
  |                     |                      |
 (F)  i1 ->          +--+--+          <- i2    |
  |                  C2   RB1                  |
  |                  +--+--+                   |
  |                     |                      |
  +---------------------+----------------------+
 L1=m1, Rf1=fc1, C1=1/K1, C2=1/K2, RB1=B1
 L2=m2, Rf2=fc2, RB2=B2

Loop equations, with $i = \dot x$:

$$\begin{aligned} F &= L_1\frac{di_1}{dt} + R_{f1}i_1 + \frac{1}{C_1}\int i_1dt + v_p \\ 0 &= L_2\frac{di_2}{dt} + (R_{f2} + R_{B2})i_2 - v_p \end{aligned}$$

where $v_p$ is the voltage across $C_2 \parallel R_{B1}$ carrying $(i_1 - i_2)$ in total; these match the mechanical equations.

Assumptions: zero initial conditions; $D_1$, $D_2$ are viscous friction coefficients to the fixed frame.

Equations (Laplace form)

Inertia $J_1$:

$$T(s) = (J_1s^2 + D_1s + K)\theta_1 - K\theta_2$$

Inertia $J_2$:

$$0 = -K\theta_1 + (J_2s^2 + D_2s + K)\theta_2$$

Solving (Cramer's rule)

$$\Delta = (J_1s^2 + D_1s + K)(J_2s^2 + D_2s + K) - K^2$$ $$\theta_2 = \frac{K\,T(s)}{\Delta}$$

Expanding $\Delta$:

$$\Delta = J_1J_2s^4 + (J_1D_2 + J_2D_1)s^3 + \left[K(J_1+J_2) + D_1D_2\right]s^2 + K(D_1+D_2)s$$ $$\frac{\theta_2(s)}{T(s)} = \frac{K}{s\left[J_1J_2s^3 + (J_1D_2 + J_2D_1)s^2 + \{K(J_1+J_2) + D_1D_2\}s + K(D_1+D_2)\right]}$$

The factor $s$ in the denominator shows that a constant torque makes the whole system rotate continuously (angle grows with time).

Torque–current (T-I) analogy

Torque → current source, inertia $J$ → capacitance $C = J$, viscous friction $D$ → conductance ($R = 1/D$), stiffness $K$ → inductance $L = 1/K$, angular velocity $\omega$ → node voltage. Each inertia is a node.

      ω1                 ω2
  +----o----+---[L]---+----o
  |    |    |         |    |
 (T)   C1   R1        C2   R2
  |    |    |         |    |
  +----+----+---------+----+
          ground
 C1=J1, R1=1/D1, L=1/K, C2=J2, R2=1/D2

Node equations:

$$\begin{aligned} T &= C_1\frac{d\omega_1}{dt} + \frac{\omega_1}{R_1} + \frac{1}{L}\int(\omega_1 - \omega_2)dt \\ 0 &= C_2\frac{d\omega_2}{dt} + \frac{\omega_2}{R_2} + \frac{1}{L}\int(\omega_2 - \omega_1)dt \end{aligned}$$

which match the mechanical equations with $\omega = \dot\theta$.

Assumptions: damper $B_2$ and spring $K_3$ are in series and this pair is in parallel with $K_2$ between $M_1$ and $M_2$; $X_3$ is the junction between $B_2$ and $K_3$; zero initial conditions. Values: $M_1 = M_2 = 1$ kg, $K_1 = K_2 = 1$ N/m, $K_3 = 2$ N/m, $B_1 = 2$, $B_2 = 1$, $B_3 = 1$ N·s/m.

F-V analogous circuit

Force → voltage, mass → $L$, damper → $R$, spring → $C = 1/K$, velocity → loop current. The coupling ($K_2$ parallel to $B_2$–$K_3$ series) becomes $C_2$ in series with ($R_2 \parallel C_3$).

  +--[L1]--[R1]--[C1]--+--[L2]--[R3]--+
  |                    |              |
  |     <- i1         C2      i2 ->  (F)
  |                    |              |
  |                 +--+--+           |
  |                 R2    C3          |
  |                 +--+--+           |
  |                    |              |
  +--------------------+--------------+
 L1=M1=1 H, R1=B1=2 Ω, C1=1/K1=1 F
 C2=1/K2=1 F, R2=B2=1 Ω, C3=1/K3=0.5 F
 L2=M2=1 H, R3=B3=1 Ω, source = F(t)

Loop currents $i_1 = \dot x_1$, $i_2 = \dot x_2$; the shared branch carries $(i_2 - i_1)$.

Equations of motion

Equivalent of series $B_2$–$K_3$: $\dfrac{B_2sK_3}{B_2s+K_3} = \dfrac{2s}{s+2}$. Total coupling:

$$K_c(s) = 1 + \frac{2s}{s+2} = \frac{3s+2}{s+2}$$ $$\begin{aligned} M_1: &\quad (s^2 + 2s + 1)X_1 + K_c(X_1 - X_2) = 0 \\ M_2: &\quad (s^2 + s)X_2 + K_c(X_2 - X_1) = F(s) \end{aligned}$$

Transfer function $X_1/F$

From the first equation, $X_2 = \dfrac{s^2+2s+1+K_c}{K_c}X_1$. Substituting in the second:

$$\frac{X_1}{F} = \frac{K_c}{(s^2+2s+1)(s^2+s) + K_c(2s^2+3s+1)}$$

Using $(s^2+2s+1)(s^2+s) = s(s+1)^3$ and $2s^2+3s+1 = (2s+1)(s+1)$, then multiplying top and bottom by $(s+2)$:

$$\begin{aligned} \frac{X_1}{F} &= \frac{3s+2}{s(s+2)(s+1)^3 + (3s+2)(2s+1)(s+1)} \\ &= \frac{3s+2}{(s+1)\left[s(s+2)(s+1)^2 + (3s+2)(2s+1)\right]} \\ &= \frac{3s+2}{(s+1)(s^4 + 4s^3 + 11s^2 + 9s + 2)} \end{aligned}$$

Answer:

$$\frac{X_1(s)}{F(s)} = \frac{3s+2}{s^5 + 5s^4 + 15s^3 + 20s^2 + 11s + 2}$$

In an armature-controlled DC motor the field current $i_f$ is kept constant (constant $v_f$) and the speed/position is controlled by changing the armature voltage $v_a$.

Assumptions and symbols

• $R_a, L_a$: armature resistance and inductance; $i_a$: armature current
• $e_b$: back emf; $K_b$: back-emf constant
• $T_m$: motor torque; $K_t$: torque constant
• $J, B$: inertia and viscous friction of motor plus load; $\theta$: shaft angle
• Flux $\phi \propto i_f$ is constant, so torque and back emf are linear.

Governing equations

1. Armature circuit (KVL):

$$v_a = R_a i_a + L_a \frac{di_a}{dt} + e_b$$

2. Back emf: $e_b = K_b \dfrac{d\theta}{dt}$
3. Torque: $T_m = K_t\, i_a$
4. Load (Newton's law): $T_m = J\dfrac{d^2\theta}{dt^2} + B\dfrac{d\theta}{dt}$

Laplace transform (zero initial conditions)

$$\begin{aligned} I_a(s) &= \frac{V_a(s) - E_b(s)}{R_a + sL_a} \\ T_m(s) &= K_t I_a(s) \\ \theta(s) &= \frac{T_m(s)}{s(Js + B)} \\ E_b(s) &= K_b\, s\,\theta(s) \end{aligned}$$

Block diagram

 Va +    ┌──────┐ Ia ┌──┐ Tm ┌─────┐ ω  ┌───┐ θ
 ──►(Σ)─►│  1   ├───►│Kt├───►│  1  ├─┬─►│1/s├─►
    ▲ -  │Ra+sLa│    └──┘    │Js+B │ │  └───┘
    │    └──────┘            └─────┘ │
    │ Eb        ┌────┐               │
    └───────────┤ Kb │◄──────────────┘
                └────┘

The back emf forms an internal negative feedback loop on speed $\omega = s\theta$.

Transfer function

Reducing the inner loop (forward path $G = \dfrac{K_t}{(R_a+sL_a)(Js+B)}$, feedback $K_b$):

$$\frac{\omega(s)}{V_a(s)} = \frac{K_t}{(R_a + sL_a)(Js + B) + K_t K_b}$$

Multiplying by the integrator $1/s$:

$$\frac{\theta(s)}{V_a(s)} = \frac{K_t}{s\left[(R_a + sL_a)(Js + B) + K_t K_b\right]}$$

If $L_a$ is negligible, this becomes $\dfrac{\theta(s)}{V_a(s)} = \dfrac{K_m}{s(1 + s\tau_m)}$ with $K_m = \dfrac{K_t}{R_aB + K_tK_b}$ and $\tau_m = \dfrac{R_aJ}{R_aB + K_tK_b}$, a type-1 second-order system.

Take $x_1$ (mass) and $x_2$ (junction of $K_1$, $K_2$, $B_2$) as the two displacements. The junction has no mass.

Equations of motion

$$\begin{aligned} M\ddot{x}_1 + K_1(x_1 - x_2) &= f \\ K_1(x_2 - x_1) + B_2\dot{x}_2 + K_2 x_2 &= 0 \end{aligned}$$

F-V analogy (mesh/loop)

$f \to e$, $M \to L$, $B \to R$, $K \to 1/C$, $\dot x \to i$. Each displacement gives one mesh; $K_1$ is common to both.

$$\begin{aligned} L\frac{di_1}{dt} + \frac{1}{C_1}\int (i_1 - i_2)\,dt &= e \\ \frac{1}{C_1}\int (i_2 - i_1)\,dt + R_2 i_2 + \frac{1}{C_2}\int i_2\,dt &= 0 \end{aligned}$$

with $L = M$, $C_1 = 1/K_1$, $R_2 = B_2$, $C_2 = 1/K_2$.

      L = M               R2 = B2
  +---LLLL----+---------/\/\/----+
  |           |                  |
 (~) e      C1=1/K1           C2=1/K2
  |    i1     |       i2         |
  +-----------+------------------+

F-I analogy (node)

$f \to i$, $M \to C$, $B \to 1/R$, $K \to 1/L$, $\dot x \to v$. Each displacement gives one node; $K_1$ joins the nodes.

$$\begin{aligned} C\frac{dv_1}{dt} + \frac{1}{L_1}\int (v_1 - v_2)\,dt &= i \\ \frac{1}{L_1}\int (v_2 - v_1)\,dt + \frac{v_2}{R_2} + \frac{1}{L_2}\int v_2\,dt &= 0 \end{aligned}$$

with $C = M$, $L_1 = 1/K_1$, $R_2 = 1/B_2$, $L_2 = 1/K_2$.

    v1          L1=1/K1         v2
  +---+--------LLLL-------+--------+
  |   |                   |        |
 (i) C=M              R2=1/B2  L2=1/K2
  |   |                   |        |
  +---+-------------------+--------+  ground

Let $x_1$, $x_2$ be the displacements of the two masses ($M_1$, $M_2$) and $x_3$ the displacement of the massless junction between $K_2$ and $B_1$ (spring on the $M_1$ side).

Differential equations

$$\begin{aligned} M_1\ddot x_1 + f_{c1}\dot x_1 + K_1 x_1 + K_2(x_1 - x_3) &= f \\ K_2(x_3 - x_1) + B_1(\dot x_3 - \dot x_2) &= 0 \\ M_2\ddot x_2 + (f_{c2} + B_2)\dot x_2 + B_1(\dot x_2 - \dot x_3) &= 0 \end{aligned}$$

Eliminating the junction

From the second equation, $X_3 = \dfrac{K_2X_1 + B_1sX_2}{K_2 + B_1 s}$. The force through the series spring–damper is then

$$K_2(X_1 - X_3) = K_{eq}(X_1 - X_2), \qquad K_{eq}(s) = \frac{K_2 B_1 s}{K_2 + B_1 s}$$

So, writing $a = M_1s^2 + f_{c1}s + K_1$ and $b = M_2s^2 + (f_{c2}+B_2)s$:

$$\begin{aligned} (a + K_{eq})X_1 - K_{eq}X_2 &= F \\ -K_{eq}X_1 + (b + K_{eq})X_2 &= 0 \end{aligned}$$

Transfer function

$$\frac{X_2(s)}{F(s)} = \frac{K_{eq}}{(a+K_{eq})(b+K_{eq}) - K_{eq}^2} = \frac{K_{eq}}{ab + K_{eq}(a+b)}$$

Multiplying through by $(K_2 + B_1 s)$ and cancelling the common factor $s$:

$$\frac{X_2(s)}{F(s)} = \frac{B_1K_2}{(K_2 + B_1s)(M_1s^2 + f_{c1}s + K_1)(M_2s + f_{c2} + B_2) + B_1K_2\left[(M_1+M_2)s^2 + (f_{c1}+f_{c2}+B_2)s + K_1\right]}$$

The denominator is a 4th-order polynomial in $s$.

F-I analogy

$f \to i$, $M \to C$, friction/damper $B \to$ conductance $1/R$, $K \to 1/L$, velocity $\to$ node voltage. Nodes $v_1, v_3, v_2$ correspond to $\dot x_1, \dot x_3, \dot x_2$.

  v1               v3               v2
 +--+---+----+-LLLL-+--/\/\/--+--+----+----+
 |  |   |    | 1/K2   1/B1    |  |    |    |
(i) C1 R1   L1               C2 R2   R3    |
 |  =M1 =1/ =1/              =M2 =1/ =1/B2 |
 |  |  fc1  K1                |  fc2  |    |
 +--+---+----+----------------+--+----+----+ gnd

Node equation at node 1: $C_1\dfrac{dv_1}{dt} + \dfrac{v_1}{R_1} + \dfrac{1}{L_1}\int v_1dt + \dfrac{1}{L_2}\int (v_1 - v_3)dt = i$, and similarly for nodes 3 and 2 (matching the three mechanical equations).

The circuit is an inverting integrator (the input, through $R_1$, and the capacitor both go to the inverting input). $R_2$ only balances the input bias current; for an ideal op-amp it carries no current, so the non-inverting input is at 0 V.

Assumptions (ideal op-amp)

• Infinite input impedance: no current into the op-amp terminals.
• Virtual ground: $V_- = V_+ = 0$ (no current in $R_2$, so no drop across it).

Derivation

Current through $R_1$:

$$I_1(s) = \frac{E_{in}(s) - 0}{R_1}$$

This whole current flows through $C$ (impedance $1/sC$):

$$I_1(s) = \frac{0 - E_{out}(s)}{1/(sC)} = -sC\,E_{out}(s)$$

Equating:

$$\frac{E_{in}(s)}{R_1} = -sC\,E_{out}(s)$$

Transfer function

$$\frac{E_{out}(s)}{E_{in}(s)} = -\frac{1}{sR_1C}$$

In time domain, $e_{out}(t) = -\dfrac{1}{R_1C}\displaystyle\int e_{in}\,dt$. So the circuit is an integrator with time constant $R_1C$ and a sign inversion; it gives a pole at $s = 0$ and is used to realise the integral (I) action of a controller. Usually $R_2$ is chosen equal to $R_1$ to reduce output offset.

Reflect everything to one shaft. Mechanical impedances move across a gear pair by $(N_{dest}/N_{source})^2$, and torques by $N_{dest}/N_{source}$.

Step 1: Reflect the input torque to the $N_2$ shaft

$$T_2 = T\,\frac{N_2}{N_1} = T\cdot\frac{33}{11} = 3T$$

Step 2: Reflect the load damping to the $N_3$ shaft

$$D_{eq} = 0.04\left(\frac{N_3}{N_4}\right)^2 = 0.04\left(\frac{50}{10}\right)^2 = 1\ \text{N-m-s/rad}$$

Step 3: Equivalent system

Let $\theta_2$ = angle of the $N_2$ shaft (with $J = 1$, $D = 2$) and $\theta_3$ = angle of gear $N_3$ (end of the spring, $K = 3$, carrying the reflected damping 1):

 3T ─► [J=1, D=2] ──spring K=3── (θ3) ── D=1 ─► frame
        θ2

Equations (Laplace):

$$\begin{aligned} (s^2 + 2s + 3)\,\theta_2 - 3\,\theta_3 &= 3T \\ -3\,\theta_2 + (s + 3)\,\theta_3 &= 0 \end{aligned}$$

Step 4: Solve

From the second equation $\theta_2 = \dfrac{(s+3)}{3}\theta_3$. Substituting:

$$\begin{aligned} \left[\frac{(s^2+2s+3)(s+3)}{3} - 3\right]\theta_3 &= 3T \\ \left[(s^3 + 5s^2 + 9s + 9) - 9\right]\theta_3 &= 9T \\ \frac{\theta_3(s)}{T(s)} &= \frac{9}{s(s^2 + 5s + 9)} \end{aligned}$$

Step 5: Output shaft

$$\theta_L = \theta_3\,\frac{N_3}{N_4} = 5\,\theta_3$$

Answer:

$$\frac{\theta_L(s)}{T(s)} = \frac{45}{s(s^2 + 5s + 9)}$$

The pole at $s=0$ appears because no spring ties the system to the frame, so a constant torque gives a steadily increasing angle.

Let $x_1$ = displacement of the outer frame $M_1$ and $x_2$ = displacement of the inner mass $M_2$. Assumption: no friction (none is shown) between $M_1$ and ground or between $M_2$ and $M_1$.

Free body equations

• $M_1$: force $f$ acts; spring $k_2$ to the wall resists by $k_2x_1$; spring $k_1$ pulls by $k_1(x_1 - x_2)$.
• $M_2$: only spring $k_1$ acts, $k_1(x_2 - x_1)$.

$$\begin{aligned} M_1\ddot x_1 + k_1(x_1 - x_2) + k_2x_1 &= f \\ M_2\ddot x_2 + k_1(x_2 - x_1) &= 0 \end{aligned}$$

Laplace form

$$\begin{aligned} (M_1s^2 + k_1 + k_2)X_1 - k_1X_2 &= F \\ -k_1X_1 + (M_2s^2 + k_1)X_2 &= 0 \end{aligned}$$

From the second equation, $X_1 = \dfrac{M_2s^2 + k_1}{k_1}X_2$. Substituting:

$$\left[(M_1s^2 + k_1 + k_2)(M_2s^2 + k_1) - k_1^2\right]X_2 = k_1F$$

Transfer function

$$\frac{X_2(s)}{F(s)} = \frac{k_1}{M_1M_2s^4 + \left[M_1k_1 + M_2(k_1 + k_2)\right]s^2 + k_1k_2}$$

With no damping the poles lie on the $j\omega$ axis (undamped oscillation at two natural frequencies).

F-I analogous circuit

$f\to i$, $M\to C$, $k\to 1/L$, velocity $\to$ node voltage. Note that $M_2$ is referred to the ground reference (absolute velocity), so its capacitor goes to ground.

   v1          L1 = 1/k1        v2
  +----+---+-----LLLL-------+------+
  |    |   |                |      |
 (i)  C1  L2               C2      |
  |  =M1 =1/k2             =M2     |
  |    |   |                |      |
  +----+---+----------------+------+ gnd

$$\begin{aligned} C_1\frac{dv_1}{dt} + \frac{1}{L_2}\int v_1dt + \frac{1}{L_1}\int(v_1 - v_2)dt &= i \\ C_2\frac{dv_2}{dt} + \frac{1}{L_1}\int(v_2 - v_1)dt &= 0 \end{aligned}$$