Let's break down and solve the problems step by step as requested: --- ## **Given Data** | Parameter | Symbol | Value | |--------------------------|-------------|---------------| | Moment of inertia | \( J \) | .55 kg·m² | | Motor arm length | \( d \) | .225 m | | Damping coefficient | \( b \) | .02 N·m·s/rad | | Motor thrust gain | \( k_m \) | 1.2 N/V | | Motor time constant | \( T_m \) | .1 s | Motor-propeller transfer function: \[ \frac{F(s)}{V(s)} = \frac{k_m}{T_m s + 1} \] Control input (differential voltage): \[ u(t) = V_1(t) - V_2(t) \] --- ## **1. Differential Equation of Motion** The biocopter rotates about the pitch axis only. - Let the pitch angle be \( \theta(t) \). - The forces from the motors generate a torque about the pitch axis. ### **Torque Equation:** \[ J \ddot{\theta} + b \dot{\theta} = d [F_1(t) - F_2(t)] \] ### **Motor Forces:** Since each motor's force is: \[ F_i(s) = \frac{k_m}{T_m s + 1} V_i(s) \] So in time domain (using Laplace inverse, it's a first-order lag): \[ T_m \dot{F}_i + F_i = k_m V_i \] --- ## **2. State-Space Representation** Let the state variables be: - \( x_1 = \theta \) (pitch angle) - \( x_2 = \dot{\theta} \) (angular velocity) - \( x_3 = F_1 \) (force from motor 1) - \( x_4 = F_2 \) (force from motor 2) ### **State Equations:** - \( \dot{x}_1 = x_2 \) - \( J \dot{x}_2 + b x_2 = d(x_3 - x_4) \implies \dot{x}_2 = -\frac{b}{J} x_2 + \frac{d}{J}(x_3 - x_4) \) - \( T_m \dot{x}_3 + x_3 = k_m V_1 \implies \dot{x}_3 = -\frac{1}{T_m} x_3 + \frac{k_m}{T_m} V_1 \) - \( T_m \dot{x}_4 + x_4 = k_m V_2 \implies \dot{x}_4 = -\frac{1}{T_m} x_4 + \frac{k_m}{T_m} V_2 \) But, \( u = V_1 - V_2 \). Assume symmetric operation: \( V_1 = u/2 \), \( V_2 = -u/2 \). ### **State-Space Equations:** \[ \begin{cases} \dot{x}_1 = x_2 \\ \dot{x}_2 = -\frac{b}{J} x_2 + \frac{d}{J} (x_3 - x_4) \\ \dot{x}_3 = -\frac{1}{T_m} x_3 + \frac{k_m}{2T_m} u \\ \dot{x}_4 = -\frac{1}{T_m} x_4 - \frac{k_m}{2T_m} u \\ \end{cases} \] Output: \( y = x_1 = \theta \) --- ## **3. Transfer Function \( \theta(s)/U(s) \)** ### **Step 1:** Motor Transfer Function for Differential Force \[ F_1(s) = \frac{k_m}{T_m s + 1} V_1(s) \] \[ F_2(s) = \frac{k_m}{T_m s + 1} V_2(s) \] \[ F_1(s) - F_2(s) = \frac{k_m}{T_m s + 1} (V_1(s) - V_2(s)) = \frac{k_m}{T_m s + 1} u(s) \] ### **Step 2:** Torque Equation in Laplace Domain \[ J s^2 \Theta(s) + b s \Theta(s) = d \left[ F_1(s) - F_2(s) \right] \] \[ J s^2 \Theta(s) + b s \Theta(s) = d \frac{k_m}{T_m s + 1} u(s) \] \[ \Theta(s) \left[ J s^2 + b s \right] = d \frac{k_m}{T_m s + 1} u(s) \] \[ \Theta(s) = \frac{d k_m}{(J s^2 + b s)(T_m s + 1)} u(s) \] \[ \boxed{ \frac{\Theta(s)}{U(s)} = \frac{d k_m}{(J s^2 + b s)(T_m s + 1)} } \] --- ## **4. Step & Ramp Input: Steady-State Response** Given \( u(t) \) as input: ### **Step Input \( u(t) = U_ \cdot 1(t) \):** - \( U(s) = \frac{U_}{s} \) - Steady-state value (\( t \to \infty \)): use Final Value Theorem \[ \lim_{t \to \infty} \theta(t) = \lim_{s \to } s \Theta(s) \] \[ \Theta(s) = \frac{d k_m}{(J s^2 + b s)(T_m s + 1)} U(s) \] Plug in \( U(s) = \frac{U_}{s} \): \[ \Theta(s) = \frac{d k_m U_}{s (J s^2 + b s)(T_m s + 1)} \] \[ \lim_{s \to } s \Theta(s) = \lim_{s \to } \frac{d k_m U_}{(J s^2 + b s)(T_m s + 1)} \] As \( s \to \): - \( J s^2 \to \) - \( b s \to \) - \( T_m s \to \) - So denominator \( \to \) So, for a step input, the system is **Type 1** (1 integrator, due to the \( s \) in denominator), so the steady-state output for step input is **infinite** -- i.e., the system responds with a ramp to a step input (physically, this makes sense: a constant torque results in ever-increasing angular displacement). --- ### **Ramp Input \( u(t) = U_ t \):** - \( U(s) = \frac{U_}{s^2} \) So, \[ \Theta(s) = \frac{d k_m U_}{s^2 (J s^2 + b s)(T_m s + 1)} \] \[ \lim_{t \to \infty} \theta(t) = \lim_{s \to } s \Theta(s) = \lim_{s \to } \frac{d k_m U_}{s (J s^2 + b s)(T_m s + 1)} \] As \( s \to \), denominator still goes to zero faster than numerator, so steady-state output is infinite (system is Type 1 for input, so tracks ramp input with constant error). --- ## **Summary of Answers** ### **1. Differential Equation:** \[ J \ddot{\theta} + b \dot{\theta} = d (F_1 - F_2) \] \[ T_m \dot{F}_i + F_i = k_m V_i \qquad (i=1,2) \] --- ### **2. State-Space Equations:** Let \( x_1 = \theta, x_2 = \dot{\theta}, x_3 = F_1, x_4 = F_2 \), \( u = V_1 - V_2 \): \[ \begin{cases} \dot{x}_1 = x_2 \\ \dot{x}_2 = -\frac{b}{J} x_2 + \frac{d}{J}(x_3 - x_4) \\ \dot{x}_3 = -\frac{1}{T_m} x_3 + \frac{k_m}{2T_m} u \\ \dot{x}_4 = -\frac{1}{T_m} x_4 - \frac{k_m}{2T_m} u \\ \end{cases} \] --- ### **3. Transfer Function:** \[ \boxed{ \frac{\Theta(s)}{U(s)} = \frac{d k_m}{(J s^2 + b s)(T_m s + 1)} } \] --- ### **4. Steady-State Output:** - **Step input**: Output ramps to infinity (system is Type 1). - **Ramp input**: Output grows even faster (infinite steady-state output). ---