Let's solve the problem step by step: ## **Given:** Open-loop transfer function: \[ G(s) = \frac{5K}{s^3 + 6s^2 + 5s} \] We are to: 1. Obtain the Bode plot, 2. Compute the **phase margin (PM)** and **gain margin (GM)** for: - \( K = 2 \) - \( K = 20 \) --- ### **Step 1: Factorize the Denominator** \[ s^3 + 6s^2 + 5s = s(s^2 + 6s + 5) = s(s+1)(s+5) \] So, \[ G(s) = \frac{5K}{s(s+1)(s+5)} \] --- ### **Step 2: Express in Frequency Domain (\(s = j\omega\))** \[ G(j\omega) = \frac{5K}{(j\omega)((j\omega)+1)((j\omega)+5)} \] --- ### **Step 3: Bode Plot Construction** #### **Magnitude:** \[ |G(j\omega)| = \frac{5K}{\omega \sqrt{\omega^2+1}\sqrt{\omega^2+25}} \] #### **Phase:** \[ \angle G(j\omega) = -90^\circ - \tan^{-1}\left(\frac{\omega}{1}\right) - \tan^{-1}\left(\frac{\omega}{5}\right) \] --- ### **Step 4: Gain Margin and Phase Margin** #### **Definitions:** - **Gain Margin (GM):** Additional gain required to bring the system to the verge of instability (phase = -180°). - **Phase Margin (PM):** Additional phase lag required to bring the system to the verge of instability (gain = 1 or 0 dB). --- ## **A. For \(K = 2\):** ### **Magnitude (in dB):** \[ 20\log_{10}|G(j\omega)| = 20\log_{10}(10) - 20\log_{10}(\omega) - 10\log_{10}(\omega^2+1) - 10\log_{10}(\omega^2+25) \] \[ \Rightarrow 20\log_{10}(10) = 20 \] So, \[ 20 - 20\log_{10}(\omega) - 10\log_{10}(\omega^2+1) - 10\log_{10}(\omega^2+25) \] ### **Phase:** \[ \angle G(j\omega) = -90^\circ - \tan^{-1}(\omega) - \tan^{-1}\left(\frac{\omega}{5}\right) \] --- ### **Find Gain Crossover Frequency (\(\omega_{gc}\))** At gain crossover, \(|G(j\omega_{gc})| = 1\), or \(0\) dB. Set magnitude to 1 and solve for \(\omega_{gc}\): \[ \frac{10}{\omega \sqrt{\omega^2+1} \sqrt{\omega^2+25}} = 1 \] \[ 10 = \omega \sqrt{\omega^2+1} \sqrt{\omega^2+25} \] Square both sides: \[ 100 = \omega^2 (\omega^2 + 1)(\omega^2 + 25) \] Let \(x = \omega^2\): \[ 100 = x(x + 1)(x + 25) \] \[ 100 = x(x^2 + 26x + 25) \] \[ 100 = x^3 + 26x^2 + 25x \] \[ x^3 + 26x^2 + 25x - 100 = 0 \] This is a cubic equation. Let's estimate the root numerically: For \(x = 1\): \(1 + 26 + 25 - 100 = -48\) For \(x = 2\): \(8 + 104 + 50 - 100 = 62\) So, the root is between 1 and 2. Try \(x = 1.5\): \[ (1.5)^3 + 26(1.5)^2 + 25(1.5) - 100 = 3.375 + 58.5 + 37.5 - 100 = -0.625 \] Try \(x = 1.6\): \[ (1.6)^3 + 26(1.6)^2 + 25(1.6) - 100 = 4.096 + 66.56 + 40 - 100 = 10.656 \] So at \(x \approx 1.52\): \[ (1.52)^3 + 26(1.52)^2 + 25(1.52) - 100 \approx 3.51 + 60.07 + 38 - 100 = 1.58 \] So \(x \approx 1.5\), thus \(\omega_{gc} \approx \sqrt{1.5} \approx 1.22\) rad/s. --- ### **Phase at \(\omega_{gc}\):** \[ \theta = -90^\circ - \tan^{-1}(1.22) - \tan^{-1}(1.22/5) \] \[ \tan^{-1}(1.22) \approx 50.1^\circ \] \[ 1.22/5 = 0.244, \quad \tan^{-1}(0.244) \approx 13.7^\circ \] \[ \theta = -90^\circ - 50.1^\circ - 13.7^\circ = -153.8^\circ \] --- ### **Phase Margin (PM):** \[ \text{PM} = 180^\circ + (-153.8^\circ) = 26.2^\circ \] --- ### **Find Phase Crossover Frequency (\(\omega_{pc}\)):** Set phase to -180°: \[ -90^\circ - \tan^{-1}(\omega) - \tan^{-1}\left(\frac{\omega}{5}\right) = -180^\circ \] \[ \tan^{-1}(\omega) + \tan^{-1}\left(\frac{\omega}{5}\right) = 90^\circ \] Recall: \[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left( \frac{a+b}{1-ab} \right) \] Set this equal to 90°: \[ \frac{\omega + \frac{\omega}{5}}{1 - \omega \cdot \frac{\omega}{5}} \to \infty \] So denominator = 0: \[ 1 - \frac{\omega^2}{5} = 0 \implies \omega^2 = 5 \implies \omega_{pc} = \sqrt{5} \approx 2.24 \] --- ### **Gain at Phase Crossover:** \[ |G(j2.24)| = \frac{10}{2.24 \sqrt{2.24^2 + 1} \sqrt{2.24^2 + 25}} \] \[ 2.24^2 = 5.02; \quad 5.02 + 1 = 6.02 \implies \sqrt{6.02} \approx 2.45 \] \[ 5.02 + 25 = 30.02 \implies \sqrt{30.02} \approx 5.48 \] \[ |G(j2.24)| = \frac{10}{2.24 \times 2.45 \times 5.48} \approx \frac{10}{30.04} \approx 0.333 \] In dB: \(20\log_{10}(0.333) = -9.54\,\text{dB}\) #### **Gain Margin:** \[ \text{GM} = -(-9.54) = 9.54\,\text{dB} \] --- ## **B. For \(K = 20\):** Similar steps, but now \(5K = 100\): ### **Magnitude:** \[ |G(j\omega)| = \frac{100}{\omega \sqrt{\omega^2+1} \sqrt{\omega^2+25}} \] Set \( |G(j\omega_{gc})| = 1 \): \[ 100 = \omega \sqrt{\omega^2+1} \sqrt{\omega^2+25} \] \[ 10000 = \omega^2 (\omega^2+1)(\omega^2+25) \] Let \(x = \omega^2\): \[ 10000 = x(x+1)(x+25) \] \[ 10000 = x^3 + 26x^2 + 25x \] \[ x^3 + 26x^2 + 25x - 10000 = 0 \] Try \(x = 10\): \(1000 + 2600 + 250 - 10000 = -6150\) Try \(x = 20\): \(8000 + 10400 + 500 - 10000 = 8900\) Try \(x = 12\): \(1728 + 1872 + 300 - 10000 = -6100\) Try \(x = 18\): \(5832 + 8424 + 450 - 10000 = 2706\) Try \(x = 16\): \(4096 + 6656 + 400 - 10000 = 1152\) Try \(x = 14\): \(2744 + 5096 + 350 - 10000 = -1810\) So between 14 and 16. Try \(x = 15\): \[ (15)^3 + 26(15)^2 + 25(15) - 10000 = 3375 + 5850 + 375 - 10000 = -400 \] Try \(x = 15.5\): \[ (15.5)^3 + 26(15.5)^2 + 25(15.5) - 10000 = 3724 + 6251 + 387.5 - 10000 = 362.5 \] So root is approximately at \(x \approx 15.25\). Thus, \(\omega_{gc} \approx \sqrt{15.25} \approx 3.9\) rad/s. --- ### **Phase at \(\omega_{gc} \approx 3.9\):** \[ \tan^{-1}(3.9) \approx 75.4^\circ \] \[ 3.9/5 = 0.78, \tan^{-1}(0.78) \approx 38.2^\circ \] \[ \theta = -90^\circ - 75.4^\circ - 38.2^\circ = -203.6^\circ \] --- ### **Phase Margin:** \[ \text{PM} = 180^\circ + (-203.6^\circ) = -23.6^\circ \] (Negative phase margin, system is **unstable**.) --- ### **Phase Crossover Frequency (\(\omega_{pc}\))** Previously found as \(\omega_{pc} = \sqrt{5} \approx 2.24\) rad/s. --- ### **Gain at \(\omega_{pc} = 2.24\):** \[ |G(j2.24)| = \frac{100}{2.24 \times 2.45 \times 5.48} = \frac{100}{30.04} \approx 3.33 \] In dB: \(20\log_{10}(3.33) = 10.45\) dB #### **Gain Margin:** \[ \text{GM} = -10.45\, \text{dB} \] (Again, negative, system is unstable.) --- ## **Summary Table:** | \( K \) | Phase Margin (\(^\circ\)) | Gain Margin (dB) | |---------|--------------------------|------------------| | 2 | 26.2 | 9.54 | | 20 | -23.6 | -10.45 | --- ## **Final Answers:** ### **a. \(K = 2\):** - **Phase Margin:** \(26.2^\circ\) - **Gain Margin:** \(9.54\) dB ### **b. \(K = 20\):** - **Phase Margin:** \(-23.6^\circ\) (**unstable**) - **Gain Margin:** \(-10.45\) dB (**unstable**) --- If you need Bode plots, let me know and I can sketch or provide MATLAB code for them!

Certainly! Here's MATLAB code to plot the Bode plots for the open-loop transfer function \( G(s) = \frac{5K}{s(s+1)(s+5)} \) at \( K = 2 \) and \( K = 20 \). This code will generate the magnitude and phase plots, and also mark the gain and phase crossover points to help visualize gain margin and phase margin. --- ### MATLAB Code for Bode Plots and Margins ```matlab % Define transfer function parameters K_values = [2, 20]; % Loop over each K value for i = 1:length(K_values) K = K_values(i); % Define numerator and denominator num = 5 * K; den = [1 6 5 0]; % s^3 + 6s^2 + 5s, plus the 's' in numerator % Create transfer function G(s) = numerator / denominator G = tf(num, den); % Plot Bode plot figure; margin(G); title(['Bode Plot for K = ', num2str(K)]); % Display gain and phase margins [Gm, Pm, Wcg, Wcp] = margin(G); fprintf('K = %d:\n', K); fprintf('Gain Margin (dB): %.2f at %.2f rad/sec\n', 20*log10(Gm), Wcg); fprintf('Phase Margin (degrees): %.2f at %.2f rad/sec\n\n'); end ``` --- ### How to run this code: 1. Copy and paste the code into a new script or command window in MATLAB. 2. Run the script. 3. MATLAB will generate Bode plots for both \( K=2 \) and \( K=20 \), displaying gain and phase margins directly on the plots. 4. The command window will also print the numerical gain margin (in dB) and phase margin (in degrees) for each case. --- ### Additional notes: - The `margin` command in MATLAB automatically plots the Bode magnitude and phase plots, along with the gain and phase crossover points. - You can modify the frequency range by adding `'FrequencyRange', [lower, upper]` as an argument to `margin()` if desired. - To get a clearer view, you can also save the plots or customize axes. Let me know if you'd like the plots in a specific format or further analysis!