Let's solve the problem step by step: --- ## **Step 1: Write the Transfer Function** From the block diagram: - Forward path: \(\frac{K}{Js} \cdot \frac{1}{s}\) - Feedback: \(k\) So the open-loop transfer function is: \[ G(s) = \frac{K}{Js^2} \] The feedback path has a gain \(k). --- ## **Step 2: Closed-loop Transfer Function** Let’s use unity feedback with a proportional feedback \(k\): \[ \text{Closed-loop transfer function:} \quad T(s) = \frac{G(s)}{1 + G(s)H(s)} \] Here, \(H(s) = k\): \[ T(s) = \frac{\frac{K}{Js^2}}{1 + \frac{K}{Js^2}k} \] \[ = \frac{\frac{K}{Js^2}}{1 + \frac{Kk}{Js^2}} \] \[ = \frac{K}{Js^2 + Kk} \] --- ## **Step 3: Standard 2nd Order System Form** The standard 2nd order transfer function is: \[ T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Compare with: \[ T(s) = \frac{K}{s^2 + Kk} \] (Here, \(J=1\)) So, the denominator is: \[ s^2 + Kk \] But it’s missing the damping term. To include the feedback path more precisely, let’s analyze the loop: After combining, the characteristic equation is: \[ Js^2 C(s) + Kk C(s) = K R(s) \] \[ s^2 C(s) + Kk C(s) = K R(s) \] \[ s^2 C(s) + k s C(s) + K C(s) = K R(s) \] So, the denominator is: \[ s^2 + k s + K \] The closed-loop transfer function is: \[ T(s) = \frac{K}{s^2 + k s + K} \] --- ## **Step 4: Relate to Standard Form** \[ s^2 + 2\zeta\omega_n s + \omega_n^2 \] So, - \(k = 2\zeta\omega_n\) - \(K = \omega_n^2\) --- ## **Step 5: Use Given Specifications** ### **Maximum Overshoot \(M_p = 25\%\):** \[ M_p = e^{-\frac{\zeta \pi}{\sqrt{1-\zeta^2}}} \] Set \(M_p = .25\): \[ .25 = e^{-\frac{\zeta \pi}{\sqrt{1-\zeta^2}}} \] Take natural log on both sides: \[ \ln(.25) = -\frac{\zeta \pi}{\sqrt{1-\zeta^2}} \] \[ -1.386 = -\frac{\zeta \pi}{\sqrt{1-\zeta^2}} \] \[ \frac{\zeta \pi}{\sqrt{1-\zeta^2}} = 1.386 \] \[ \frac{\zeta}{\sqrt{1-\zeta^2}} = \frac{1.386}{\pi} = .4414 \] Let \(x = \zeta\): \[ \frac{x}{\sqrt{1-x^2}} = .4414 \] \[ x = .4414 \sqrt{1-x^2} \] \[ x^2 = .1948 (1-x^2) \] \[ x^2 = .1948 - .1948 x^2 \] \[ x^2 + .1948 x^2 = .1948 \] \[ x^2 (1 + .1948) = .1948 \] \[ x^2 = \frac{.1948}{1.1948} = .163 \] \[ x = \sqrt{.163} = .404 \] So, \[ \zeta \approx .404 \] --- ### **Peak Time \(t_p = 1\) s:** \[ t_p = \frac{\pi}{\omega_n \sqrt{1-\zeta^2}} \] \[ 1 = \frac{\pi}{\omega_n \sqrt{1-.404^2}} \] \[ \sqrt{1-.163} = \sqrt{.837} = .915 \] \[ 1 = \frac{\pi}{\omega_n \times .915} \] \[ \omega_n = \frac{\pi}{.915} = 3.436 \] --- ## **Step 6: Find \(K\) and \(k\)** Recall: - \(K = \omega_n^2 = (3.436)^2 = 11.81\) - \(k = 2\zeta\omega_n = 2 \times .404 \times 3.436 = 2.778\) --- ## **Final Answers** \[ \boxed{ K = 11.81 } \] \[ \boxed{ k = 2.78 } \] --- ### **Summary Table** | Parameter | Value | |-----------|--------| | \(K\) | 11.81 | | \(k\) | 2.78 | --- If you have any questions about the steps, feel free to ask!