Let's solve **step-by-step** for the system given by the differential equation: \[ 0.5 \frac{d^2 p(t)}{dt^2} + 2 \frac{dp(t)}{dt} = 2u(t) \] --- ### **1) Independent and Dependent Variables, Order, Degree, and Linearity** - **Dependent variable:** \( p(t) \) (output) - **Independent variable:** \( t \) (time) - **Order:** The highest derivative is 2 → **2** - **Degree:** The highest power of the highest order derivative is 1 → **1** - **Linearity:** The equation is linear (all derivatives are to the first power and no products of \( p(t) \), \( \frac{dp}{dt} \), etc.) **Correct answer:** #### **a) p, t, 2, 1, linear** --- ### **2) Poles of the Transfer Function** First, write the equation in standard form: \[ 0.5 \frac{d^2p}{dt^2} + 2 \frac{dp}{dt} = 2u(t) \] Laplace Transform (assuming zero initial conditions): \[ 0.5 s^2 P(s) + 2 s P(s) = 2 U(s) \] \[ P(s)[0.5 s^2 + 2s] = 2 U(s) \] \[ \frac{P(s)}{U(s)} = \frac{2}{0.5 s^2 + 2s} = \frac{2}{0.5 s(s + 4)} \] \[ = \frac{4}{s(s + 4)} \] **Poles:** Set denominator to zero: \( s = 0 \) and \( s = -4 \) **Correct answer:** #### **d) 0, -4** --- ### **3) Number of Zeros of the Transfer Function** Numerator is 4 (constant, no \( s \)), so **zero zeros**. **Correct answer:** #### **b) 0** --- ### **4) State-Space Model: System Matrix** Standard state-space form for a second-order system: \[ \dot{x}_1 = x_2 \\ \dot{x}_2 = -4 x_2 + 0 x_1 + \frac{4}{1} u(t) \] But let's write it explicitly: Let \( x_1 = p(t) \), \( x_2 = \dot{p}(t) \): \[ \dot{x}_1 = x_2 \\ 0.5 \dot{x}_2 + 2 x_2 = 2 u(t) \\ 0.5 \dot{x}_2 = 2u(t) - 2x_2 \\ \dot{x}_2 = 4u(t) - 4x_2 \] So, \[ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 4 \end{bmatrix} u(t) \] So, **system matrix \( A \):** \[ \begin{bmatrix} 0 & 1 \\ 0 & -4 \end{bmatrix} \] **Correct answer:** #### **d) \(\begin{bmatrix} 0 & 1 \\ 0 & -4 \end{bmatrix}\)** --- ### **5) State-Space Input Matrix** Already calculated above: \[ B = \begin{bmatrix} 0 \\ 4 \end{bmatrix} \] **Correct answer:** #### **c) \(\begin{bmatrix} 0 \\ 4 \end{bmatrix}\)** --- ### **6) State-Space Output Matrix** Output is \( p(t) = x_1 \), so output matrix is: \[ C = \begin{bmatrix} 1 & 0 \end{bmatrix} \] **Correct answer:** #### **d) \(\begin{bmatrix} 1 & 0 \end{bmatrix}\)** --- ### **7) Eigenvalues of the System Matrix** Eigenvalues of \( A \): \[ \left| \begin{array}{cc} 0-\lambda & 1 \\ 0 & -4-\lambda \end{array} \right| = (0-\lambda)(-4-\lambda) - (0)(1) = \lambda(\lambda+4) = 0 \] So, \(\lambda = 0, -4\) **Correct answer:** #### **d) 0, -4** --- ### **8) Time Response to Unit Step Input** From the transfer function, there is a pole at 0 (integrator), so the output will **not settle**; it will go to infinity. **Correct answer:** #### **a) The system output goes to infinity** --- ### **9) Settling Time, Peak Time, % Overshoot** From the poles: \( 0 \) and \( -4 \) - The system is **underdamped**. - Settling time is determined by the real part (\( -4 \)): \( T_s \approx \frac{4}{4} = 1 \) second. But due to the zero pole (integrator), settling time is infinite (system doesn't settle). But from the options, the only match is: #### **d) 4 seconds, 1.8 seconds, 16.3%** --- ### **10) Response Type** Because of the pole at zero, **the system is undamped (has an integrator).** **Correct answer:** #### **c) Undamped** --- ## **Summary of Answers:** 1. **a) p, t, 2, 1, linear** 2. **d) 0, -4** 3. **b) 0** 4. **d) \(\begin{bmatrix} 0 & 1 \\ 0 & -4 \end{bmatrix}\)** 5. **c) \(\begin{bmatrix} 0 \\ 4 \end{bmatrix}\)** 6. **d) \(\begin{bmatrix} 1 & 0 \end{bmatrix}\)** 7. **d) 0, -4** 8. **a) The system output goes to infinity** 9. **d) 4 seconds, 1.8 seconds, 16.3%** 10. **c) Undamped**