Let's solve the problem **step-by-step** as requested. --- # 1. **Open Loop Transfer Function** The plant transfer function is: \[ G(s) = \frac{1}{(10s+1)(s+1)(2s+1)} \] --- # 2. **Ziegler-Nichols Tuning Method (Ultimate Gain Method)** ## **Step 1: Find Critical Gain (\(K_{cr}\)) and Critical Period (\(P_{cr}\))** To use the **Ziegler-Nichols** method, set the controller to **P-only** (i.e., \( G_c = K_c \)) and increase \(K_c\) until the system oscillates continuously (sustained oscillations). This is the **ultimate gain** (\(K_{cr}\)), and the oscillation period is the **ultimate period** (\(P_{cr}\)). The open-loop transfer function with proportional controller: \[ L(s) = K_c \cdot G(s) = K_c \cdot \frac{1}{(10s+1)(s+1)(2s+1)} \] The Bode plot is used to determine the **gain margin**. At the phase crossover frequency (\(\omega_{pc}\)), the phase is \(-180^\circ\). ## **Step 2: Find Phase Crossover Frequency (\(\omega_{pc}\))** Each factor contributes phase lag: - \((10s+1)\): \(-\tan^{-1}(10\omega)\) - \((s+1)\): \(-\tan^{-1}(\omega)\) - \((2s+1)\): \(-\tan^{-1}(2\omega)\) Total phase lag: \[ \text{Phase} = -\tan^{-1}(10\omega) - \tan^{-1}(\omega) - \tan^{-1}(2\omega) \] Set phase to \(-180^\circ\) (i.e., \(-\pi\) radians): \[ \tan^{-1}(10\omega) + \tan^{-1}(\omega) + \tan^{-1}(2\omega) = 180^\circ \] But all terms are positive, so sum to \(180^\circ\): \[ \tan^{-1}(10\omega) + \tan^{-1}(\omega) + \tan^{-1}(2\omega) = 180^\circ \] Let’s try values to find \(\omega_{pc}\): - For \(\omega = 1\): - \(\tan^{-1}(10) \approx 84^\circ\) - \(\tan^{-1}(1) = 45^\circ\) - \(\tan^{-1}(2) \approx 63^\circ\) - Total: \(84 + 45 + 63 = 192^\circ\) (too high) - For \(\omega = .5\): - \(\tan^{-1}(5) \approx 78.7^\circ\) - \(\tan^{-1}(.5) \approx 26.6^\circ\) - \(\tan^{-1}(1) = 45^\circ\) - Total: \(78.7 + 26.6 + 45 = 150.3^\circ\) (too low) - For \(\omega = .8\): - \(\tan^{-1}(8) \approx 82.9^\circ\) - \(\tan^{-1}(.8) \approx 38.7^\circ\) - \(\tan^{-1}(1.6) \approx 58.^\circ\) - Total: \(82.9 + 38.7 + 58. = 179.6^\circ\) (very close!) So, \(\omega_{pc} \approx .8\) rad/s. ## **Step 3: Find Ultimate Gain (\(K_{cr}\))** At phase crossover, magnitude needs to be 1 for sustained oscillations: \[ |L(j\omega_{pc})| = 1 \] So, \[ K_{cr} = \left| (10j\omega_{pc}+1)(j\omega_{pc}+1)(2j\omega_{pc}+1) \right| \] Plug in \(\omega_{pc} = .8\): - \(10j\cdot.8 + 1 = 1 + j8\), magnitude \(= \sqrt{1^2 + 8^2} = \sqrt{65} \approx 8.06\) - \(j\cdot.8 + 1 = 1 + j.8\), magnitude \(= \sqrt{1^2 + .8^2} = \sqrt{1.64} \approx 1.28\) - \(2j\cdot.8 + 1 = 1 + j1.6\), magnitude \(= \sqrt{1^2 + 1.6^2} = \sqrt{3.56} \approx 1.89\) Total magnitude: \[ |G(j\omega_{pc})| = \frac{1}{8.06 \times 1.28 \times 1.89} = \frac{1}{19.45} \approx .0514 \] So, to make the open loop gain 1: \[ K_{cr} = \frac{1}{.0514} \approx 19.45 \] ## **Step 4: Find Ultimate Period (\(P_{cr}\))** \[ P_{cr} = \frac{2\pi}{\omega_{pc}} = \frac{2\pi}{.8} \approx 7.85 \text{ sec} \] --- # 3. **Ziegler-Nichols Controller Settings** | Controller | \(K_c\) | \(T_i\) | \(T_d\) | |------------|----------------|--------------|--------------| | P | \(.5 K_{cr}\) | - | - | | PI | \(.45 K_{cr}\)| \(P_{cr}/1.2\)| - | | PID | \(.6 K_{cr}\) | \(.5 P_{cr}\)| \(.125 P_{cr}\)| --- ## **A. P Controller** \[ K_c = .5 \times 19.45 = 9.73 \] ## **B. PI Controller** \[ K_c = .45 \times 19.45 = 8.75 \] \[ T_i = \frac{P_{cr}}{1.2} = \frac{7.85}{1.2} = 6.54 \text{ sec} \] ## **C. PID Controller** \[ K_c = .6 \times 19.45 = 11.67 \] \[ T_i = .5 \times 7.85 = 3.93 \text{ sec} \] \[ T_d = .125 \times 7.85 = .98 \text{ sec} \] --- # 4. **Summary Table** | Controller | \(K_c\) | \(T_i\) (sec) | \(T_d\) (sec) | |------------|---------|---------------|---------------| | P | 9.73 | - | - | | PI | 8.75 | 6.54 | - | | PID | 11.67 | 3.93 | .98 | --- # 5. **Bode Plot (Asymptotic Sketch, Explanation)** - The open loop transfer function has three lag terms: break frequencies at \(.1\), \(.5\), and \(1\) rad/s. - Initial slope: dB/dec. - At each break frequency, slope decreases by -20 dB/dec. - After all breaks: -60 dB/dec. - Phase starts at °, drops by -45° at each break, reaching -135° at high frequency. - At \(\omega_{pc} = .8\), phase is about -180°. --- # **Final Answers** ### **P Controller:** - \(K_c = 9.73\) ### **PI Controller:** - \(K_c = 8.75\) - \(T_i = 6.54\) sec ### **PID Controller:** - \(K_c = 11.67\) - \(T_i = 3.93\) sec - \(T_d = .98\) sec --- ## **Bode Diagram (Asymptotic)** - **Magnitude:** - Flat until .1 rad/s, drops -20 dB/dec after each break at .1, .5, 1 rad/s. - **Phase:** - Drops -45° at each break, totaling nearly -180° at high frequencies. --- **If you need a hand-drawn Bode plot or want to see the equations for controller transfer functions, let me know!**