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Please convert the below content to rendered equations without any errors .Alright, let’s go **clean and step-by-step** and treat this like a control-systems exam problem ? --- ## **Given system (rotational mechanical system)** From the diagram: * **Input:** Step torque ( T(t) ) * **Output:** Angular displacement ( \theta_2(t) ) ### **System parameters** * Moment of inertia: [ J = 1 \ \text{kg·m}^2 ] * Viscous damping coefficient: [ B = 1 \ \text{N·m·s/rad} ] * Torsional spring constant: [ K = 1 \ \text{N·m/rad} ] This is a **standard rotational mass–spring–damper system**. --- # **(a) Transfer function ( G(s) = \dfrac{\Theta_2(s)}{T(s)} )** --- ## **Step 1: Write the equation of motion** Sum of torques = inertial torque [ T(t) = J\ddot{\theta}_2(t) + B\dot{\theta}_2(t) + K\theta_2(t) ] Substitute values: [ T(t) = \ddot{\theta}_2(t) + \dot{\theta}_2(t) + \theta_2(t) ] --- ## **Step 2: Take Laplace transform (zero initial conditions)** [ T(s) = (s^2 + s + 1)\Theta_2(s) ] --- ## **Step 3: Transfer function** [ \boxed{ G(s) = \frac{\Theta_2(s)}{T(s)} = \frac{1}{s^2 + s + 1} } ] --- # **(b) Percent overshoot, settling time, and peak time** --- ## **Step 1: Compare with standard second-order system** Standard form: [ G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} ] Given: [ s^2 + s + 1 ] ### Match coefficients: * ( \omega_n^2 = 1 \Rightarrow \omega_n = 1 \ \text{rad/s} ) * ( 2\zeta\omega_n = 1 ) [ 2\zeta(1) = 1 \Rightarrow \zeta = 0.5 ] --- ## **System parameters** [ \boxed{ \omega_n = 1 \ \text{rad/s}, \quad \zeta = 0.5 } ] --- ## **Step 2: Percent Overshoot** [ %OS = e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}} \times 100 ] [ = e^{-\frac{0.5\pi}{\sqrt{1-0.25}}} \times 100 ] [ = e^{-1.813} \times 100 ] [ \boxed{%OS \approx 16.3%} ] --- ## **Step 3: Peak Time** [ t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} ] [ = \frac{\pi}{1 \cdot \sqrt{0.75}} ] [ \boxed{t_p \approx 3.63 \ \text{s}} ] --- ## **Step 4: Settling Time (2% criterion)** [ t_s = \frac{4}{\zeta\omega_n} ] [ = \frac{4}{0.5 \times 1} ] [ \boxed{t_s = 8 \ \text{s}} ] --- # **Final Answers (Exam-Ready)** ### **Transfer Function** [ \boxed{ G(s) = \frac{1}{s^2 + s + 1} } ] ### **Time-domain specifications** | Quantity | Value | | ------------------------------ | ------------------ | | Natural frequency ( \omega_n ) | (1 \ \text{rad/s}) | | Damping ratio ( \zeta ) | (0.5) | | Percent overshoot | **16.3%** | | Peak time | **3.63 s** | | Settling time (2%) | **8 s** | --- If you want, I can also: * Sketch the **step response** * Explain physically **why overshoot occurs** * Solve this using **state-space** * Convert this into a **perfect 10/10 exam solution** Just say the word ?

Question:

Please convert the below content to rendered equations without any errors .Alright, let’s go **clean and step-by-step** and treat this like a control-systems exam problem ? --- ## **Given system (rotational mechanical system)** From the diagram: * **Input:** Step torque ( T(t) ) * **Output:** Angular displacement ( \theta_2(t) ) ### **System parameters** * Moment of inertia: [ J = 1 \ \text{kg·m}^2 ] * Viscous damping coefficient: [ B = 1 \ \text{N·m·s/rad} ] * Torsional spring constant: [ K = 1 \ \text{N·m/rad} ] This is a **standard rotational mass–spring–damper system**. --- # **(a) Transfer function ( G(s) = \dfrac{\Theta_2(s)}{T(s)} )** --- ## **Step 1: Write the equation of motion** Sum of torques = inertial torque [ T(t) = J\ddot{\theta}_2(t) + B\dot{\theta}_2(t) + K\theta_2(t) ] Substitute values: [ T(t) = \ddot{\theta}_2(t) + \dot{\theta}_2(t) + \theta_2(t) ] --- ## **Step 2: Take Laplace transform (zero initial conditions)** [ T(s) = (s^2 + s + 1)\Theta_2(s) ] --- ## **Step 3: Transfer function** [ \boxed{ G(s) = \frac{\Theta_2(s)}{T(s)} = \frac{1}{s^2 + s + 1} } ] --- # **(b) Percent overshoot, settling time, and peak time** --- ## **Step 1: Compare with standard second-order system** Standard form: [ G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} ] Given: [ s^2 + s + 1 ] ### Match coefficients: * ( \omega_n^2 = 1 \Rightarrow \omega_n = 1 \ \text{rad/s} ) * ( 2\zeta\omega_n = 1 ) [ 2\zeta(1) = 1 \Rightarrow \zeta = 0.5 ] --- ## **System parameters** [ \boxed{ \omega_n = 1 \ \text{rad/s}, \quad \zeta = 0.5 } ] --- ## **Step 2: Percent Overshoot** [ %OS = e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}} \times 100 ] [ = e^{-\frac{0.5\pi}{\sqrt{1-0.25}}} \times 100 ] [ = e^{-1.813} \times 100 ] [ \boxed{%OS \approx 16.3%} ] --- ## **Step 3: Peak Time** [ t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} ] [ = \frac{\pi}{1 \cdot \sqrt{0.75}} ] [ \boxed{t_p \approx 3.63 \ \text{s}} ] --- ## **Step 4: Settling Time (2% criterion)** [ t_s = \frac{4}{\zeta\omega_n} ] [ = \frac{4}{0.5 \times 1} ] [ \boxed{t_s = 8 \ \text{s}} ] --- # **Final Answers (Exam-Ready)** ### **Transfer Function** [ \boxed{ G(s) = \frac{1}{s^2 + s + 1} } ] ### **Time-domain specifications** | Quantity | Value | | ------------------------------ | ------------------ | | Natural frequency ( \omega_n ) | (1 \ \text{rad/s}) | | Damping ratio ( \zeta ) | (0.5) | | Percent overshoot | **16.3%** | | Peak time | **3.63 s** | | Settling time (2%) | **8 s** | --- If you want, I can also: * Sketch the **step response** * Explain physically **why overshoot occurs** * Solve this using **state-space** * Convert this into a **perfect 10/10 exam solution** Just say the word ?

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Asked by: jack
Created at: 2026-02-17 10:44:39

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# Given System (Rotational Mechanical System) **Input:** Step torque \( T) \) **Output:** Angular displacement \theta_2(t) \) ## System Parameters - Moment of inertia: \( J = 1~\text{kg·m}^2 \) - Viscous damping coefficient: \( B = 1~\text{N·m·s/rad} \) - Torsional spring constant: \( K = 1~\text{N·m/rad} \) This is a standard rotational mass–spring–damper system. --- # (a) Transfer Function \( G(s) = \dfrac{\Theta_2(s)}{T(s)} \) ## Step 1: Equation of Motion Sum of torques equals inertial torque: \[ T(t) = J\ddot{\theta}_2(t) + B\dot{\theta}_2(t) + K\theta_2(t) \] Substitute given values: \[ T(t) = \ddot{\theta}_2(t) + \dot{\theta}_2(t) + \theta_2(t) \] ## Step 2: Laplace Transform (Zero Initial Conditions) \[ T(s) = (s^2 + s + 1)\Theta_2(s) \] ## Step 3: Transfer Function \[ \boxed{ G(s) = \frac{\Theta_2(s)}{T(s)} = \frac{1}{s^2 + s + 1} } \] --- # (b) Percent Overshoot, Settling Time, Peak Time ## Step 1: Compare with Standard Second-Order System Standard form: \[ G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Given denominator: \( s^2 + s + 1 \) **Match coefficients:** - \( \omega_n^2 = 1 \implies \omega_n = 1~\text{rad/s} \) - \( 2\zeta\omega_n = 1 \) \[ 2\zeta(1) = 1 \implies \zeta = .5 \] ## System Parameters \[ \boxed{ \omega_n = 1~\text{rad/s}, \quad \zeta = .5 } \] ## Step 2: Percent Overshoot \[ \%OS = e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}} \times 100 \] \[ = e^{-\frac{.5\pi}{\sqrt{1-.25}}} \times 100 \] \[ = e^{-1.813} \times 100 \] \[ \boxed{\%OS \approx 16.3\%} \] ## Step 3: Peak Time \[ t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} \] \[ = \frac{\pi}{1 \cdot \sqrt{.75}} \] \[ \boxed{t_p \approx 3.63~\text{s}} \] ## Step 4: Settling Time (2% Criterion) \[ t_s = \frac{4}{\zeta\omega_n} \] \[ = \frac{4}{.5 \times 1} \] \[ \boxed{t_s = 8~\text{s}} \] --- # Final Answers (Exam-Ready) ## Transfer Function \[ \boxed{ G(s) = \frac{1}{s^2 + s + 1} } \] ## Time-Domain Specifications | Quantity | Value | | ------------------------------ | ------------------- | | Natural frequency \( \omega_n \) | \( 1~\text{rad/s} \) | | Damping ratio \( \zeta \) | \( .5 \) | | Percent overshoot | **16.3%** | | Peak time | **3.63 s** | | Settling time (2%) | **8 s** |

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