Let's solve each part step by step. --- ## **(a) Natural Frequency (\(\omega_n\)) and Damping Ratio (\(\zeta\)) for the Plant** The plant's transfer function is: \[ G(s) = \frac{1}{s^2 + 4s + 20} \] A standard second order system has the form: \[ G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Comparing coefficients: - \(2\zeta\omega_n = 4\) - \(\omega_n^2 = 20\) So, \[ \omega_n = \sqrt{20} = 4.472 \ \text{rad/s} \] \[ 2\zeta\omega_n = 4 \implies \zeta = \frac{4}{2 \times 4.472} = \frac{4}{8.944} = .447 \] **Final Answers:** - \(\omega_n = 4.472 \ \text{rad/s}\) - \(\zeta = .447\) --- ## **(b) Closed Loop Transfer Function \(\frac{X(s)}{X_{set}(s)}\)** Let's write the open loop transfer function: - Controller: \(C(s) = K_P + \frac{K_I}{s}\) - Plant: \(G(s) = \frac{1}{s^2 + 4s + 20}\) Open-loop transfer function: \[ C(s)G(s) = \left(K_P + \frac{K_I}{s}\right) \cdot \frac{1}{s^2 + 4s + 20} = \frac{K_P s + K_I}{s(s^2 + 4s + 20)} \] Closed-loop transfer function: \[ \frac{X(s)}{X_{set}(s)} = \frac{C(s)G(s)}{1 + C(s)G(s)} \] Substituting in: \[ \frac{X(s)}{X_{set}(s)} = \frac{\frac{K_P s + K_I}{s(s^2 + 4s + 20)}}{1 + \frac{K_P s + K_I}{s(s^2 + 4s + 20)}} \] \[ = \frac{K_P s + K_I}{s(s^2 + 4s + 20) + K_P s + K_I} \] \[ = \frac{K_P s + K_I}{s^3 + 4s^2 + 20s + K_P s + K_I} \] \[ = \frac{K_P s + K_I}{s^3 + 4s^2 + (20+K_P)s + K_I} \] --- ## **(c) \(K_I = \): Find \(K_P\) for \(\omega_n = 6\) rad/s** With \(K_I = \), the denominator is: \[ s^3 + 4s^2 + (20+K_P)s \] But since the numerator is \(K_P s\), this is a second order system (since numerator and denominator share a root at \(s=\)), so ignore the \(s=\) mode for natural frequency. The characteristic equation: \[ s^2 + 4s + (20+K_P) = \] We want the natural frequency \(\omega_n = 6\): \[ \omega_n^2 = 20 + K_P \implies 6^2 = 20 + K_P \implies 36 = 20 + K_P \implies K_P = 16 \] --- ## **(d) Step Input, \(K_P = 20, K_I = \): Steady-State Error** Step input: \(x_{set}(s) = \frac{3}{s}\) Transfer function (from above, \(K_P=20, K_I=\)): \[ \frac{20s}{s^3 + 4s^2 + 40s} \] Steady-state error for step input (Final Value Theorem): \[ e_{ss} = \lim_{t \to \infty} [x_{set}(t) - x(t)] = \lim_{s \to } s \left[X_{set}(s) - X(s)\right] \] But \[ X(s) = \frac{20s}{s^3 + 4s^2 + 40s} X_{set}(s) \] \[ e(s) = X_{set}(s) - X(s) = X_{set}(s) \left[ 1 - \frac{20s}{s^3 + 4s^2 + 40s} \right] \] \[ e_{ss} = \lim_{s \to } s \cdot \frac{3}{s} \left[1 - \frac{20s}{s(s^2 + 4s + 40)}\right] \] \[ = 3 \cdot \lim_{s \to } \left[1 - \frac{20}{s^2 + 4s + 40}\right] \] At \(s=\): \[ s^2 + 4s + 40 = 40 \] \[ 1 - \frac{20}{40} = 1 - .5 = .5 \] \[ e_{ss} = 3 \times .5 = 1.5 \] --- ## **(e) Repeat (d) for \(K_I = 1\)** Now, the transfer function numerator is \(20s + 1\): \[ \frac{20s + 1}{s^3 + 4s^2 + 40s + 1} \] Repeat the same procedure: \[ e_{ss} = \lim_{s \to } s \frac{3}{s} \left[1 - \frac{20s + 1}{s^3 + 4s^2 + 40s + 1} \right] \] At \(s = \): \[ 1 - \frac{1}{1} = \] \[ e_{ss} = 3 \times = \] --- ## **Summary Table** | Part | Final Answer | |------|-------------| | a | \(\omega_n = 4.472\) rad/s, \(\zeta = .447\) | | b | \(\frac{K_P s + K_I}{s^3 + 4s^2 + (20+K_P)s + K_I}\) | | c | \(K_P = 16\) | | d | \(e_{ss} = 1.5\) | | e | \(e_{ss} = \) | Let me know if you need further clarification or details for any part!