Let's break down the problem step by step, as required: --- # 1. **Open-Loop Transfer Function** Given system (excluding controller \( G_c \)): \[ G(s) = \frac{1}{(10s + 1)(s + 1)(2s + 1)} \] If the controller gain is \( K_c \), the open-loop transfer function is: \[ L(s) = K_c G(s) = \frac{K_c}{(10s + 1)(s + 1)(2s + 1)} \] --- # 2. **Bode Plot Asymptotic Analysis** - **Poles at:** \( s = -.1, -1, -.5 \) - **Break frequencies:** \( .1, .5, 1 \) rad/s ### **Magnitude Plot (Asymptotic)** - Slope starts at dB/dec. - At each pole, slope decreases by -20 dB/dec. ### **Phase Plot (Asymptotic)** - Each pole introduces -90° phase lag, starting about a decade before the pole frequency. --- # 3. **Ultimate Gain (\( K_u \)) and Period (\( P_u \)) via Bode Plot** ## **Step 1: Find Gain Crossover for -180° Phase** The Ziegler-Nichols method requires: - The gain at which the **open-loop system** has a **phase lag of -180°** and **|L(jω)| = 1 ( dB)**. - The frequency where this happens is called the **ultimate frequency** (\( \omega_u \)), and the gain is \( K_u \). ### **Open Loop Phase Equation** For \( G(s) \) at \( s = j\omega \): \[ \angle G(j\omega) = -\tan^{-1}(10\omega) - \tan^{-1}(\omega) - \tan^{-1}(2\omega) \] Set total phase lag to -180° (or -π radians): \[ \tan^{-1}(10\omega_u) + \tan^{-1}(\omega_u) + \tan^{-1}(2\omega_u) = 180° \] Or, since each arctan is positive for ω > , their sum should be 180°: \[ \tan^{-1}(10\omega_u) + \tan^{-1}(\omega_u) + \tan^{-1}(2\omega_u) = 180° \] But each arctan maxes out at 90°, so let's check possible ω numerically: #### Try ω = 1: - \( \tan^{-1}(10) \approx 84.3° \) - \( \tan^{-1}(1) = 45° \) - \( \tan^{-1}(2) \approx 63.4° \) - **Sum = 84.3 + 45 + 63.4 ≈ 192.7°** (too high) #### Try ω = .5: - \( \tan^{-1}(5) \approx 78.7° \) - \( \tan^{-1}(.5) \approx 26.6° \) - \( \tan^{-1}(1) = 45° \) - **Sum = 78.7 + 26.6 + 45 ≈ 150.3°** (too low) #### Try ω = .7: - \( \tan^{-1}(7) \approx 81.9° \) - \( \tan^{-1}(.7) \approx 34.99° \) - \( \tan^{-1}(1.4) \approx 54.5° \) - **Sum = 81.9 + 35 + 54.5 ≈ 171.4°** #### Try ω = .8: - \( \tan^{-1}(8) \approx 82.9° \) - \( \tan^{-1}(.8) \approx 38.7° \) - \( \tan^{-1}(1.6) \approx 58.° \) - **Sum = 82.9 + 38.7 + 58 ≈ 179.6°** So, **at ω ≈ .8 rad/s, phase lag is about -180°.** --- # 4. **Find Ultimate Gain (\( K_u \))** At \( \omega_u = .8 \): \[ |G(j\omega_u)| = \left| \frac{1}{(j8+1)(j.8+1)(j1.6+1)} \right| \] Compute each term's magnitude: - \( |j8+1| = \sqrt{8^2+1^2} = \sqrt{65} \approx 8.06 \) - \( |j.8+1| = \sqrt{.8^2+1^2} = \sqrt{1.64} \approx 1.28 \) - \( |j1.6+1| = \sqrt{1.6^2+1^2} = \sqrt{3.56} \approx 1.89 \) So, \[ |G(j.8)| = \frac{1}{8.06 \times 1.28 \times 1.89} \approx \frac{1}{19.51} \approx .0512 \] For critical gain (|L(jω)| = 1): \[ K_u \cdot .0512 = 1 \implies K_u = \frac{1}{.0512} \approx 19.53 \] --- # 5. **Ultimate Period (\( P_u \))** \[ P_u = \frac{2\pi}{\omega_u} = \frac{2\pi}{.8} \approx 7.85 \] --- # 6. **Ziegler-Nichols Tuning Formulas** | Controller | \( K_c \) | \( \tau_I \) (Integral Time) | \( \tau_D \) (Derivative Time) | |------------|-------------------|------------------------------|---------------------------------| | P | .5 \( K_u \) | — | — | | PI | .45 \( K_u \) | \( P_u / 1.2 \) | — | | PID | .6 \( K_u \) | \( P_u / 2 \) | \( P_u / 8 \) | ### **Substitute values:** - \( K_u \approx 19.53 \) - \( P_u \approx 7.85 \) #### **P Controller:** - \( K_c = .5 \times 19.53 = 9.77 \) #### **PI Controller:** - \( K_c = .45 \times 19.53 = 8.79 \) - \( \tau_I = P_u / 1.2 = 7.85 / 1.2 = 6.54 \) #### **PID Controller:** - \( K_c = .6 \times 19.53 = 11.72 \) - \( \tau_I = 7.85 / 2 = 3.93 \) - \( \tau_D = 7.85 / 8 = .98 \) --- # 7. **Summary Table** | Controller | \( K_c \) | \( \tau_I \) | \( \tau_D \) | |------------|-----------|--------------|--------------| | P | 9.77 | — | — | | PI | 8.79 | 6.54 | — | | PID | 11.72 | 3.93 | .98 | --- # 8. **Bode Plot (Asymptotic)** - **Magnitude:** Start at dB, slope decreases by -20 dB/dec at each break frequency (.1, .5, 1 rad/s). - **Phase:** Each pole introduces -90° lag, total -270° for high frequencies. - **Critical frequency:** At ω ≈ .8 rad/s, phase lag ≈ -180°. --- ## **Final Answers** **Proportional Controller:** - \( K_c = 9.77 \) **PI Controller:** - \( K_c = 8.79, \quad \tau_I = 6.54 \) **PID Controller:** - \( K_c = 11.72, \quad \tau_I = 3.93, \quad \tau_D = .98 \) --- ### **If you need a sketch of the Bode diagram, let me know!**