Let's solve the problem step by step as requested: --- ### **Step 1: Write the Open-Loop Transfer Function and Find Unity Gain Frequency** Given: \[ T(jf) = \frac{6000 \times \beta}{\left(1 + j \frac{f}{2 \times 10^3}\right) \left(1 + j \frac{f}{3 \times 10^4}\right) \left(1 + j \frac{f}{4 \times 10^5}\right)} \] where \(\beta = 38.66 \times 10^{-3}\). Let's substitute \(\beta\): \[ T(jf) = \frac{6000 \times 38.66 \times 10^{-3}}{\left(1 + j \frac{f}{2 \times 10^3}\right) \left(1 + j \frac{f}{3 \times 10^4}\right) \left(1 + j \frac{f}{4 \times 10^5}\right)} \] \[ 6000 \times 38.66 \times 10^{-3} = 231.96 \] So, \[ T(jf) = \frac{231.96}{\left(1 + j \frac{f}{2 \times 10^3}\right) \left(1 + j \frac{f}{3 \times 10^4}\right) \left(1 + j \frac{f}{4 \times 10^5}\right)} \] To find the **unity gain frequency** (\(|T(jf_1)| = 1\)), set the magnitude equal to 1: \[ 1 = \frac{231.96}{\sqrt{1 + \left(\frac{f_1}{2 \times 10^3}\right)^2} \sqrt{1 + \left(\frac{f_1}{3 \times 10^4}\right)^2} \sqrt{1 + \left(\frac{f_1}{4 \times 10^5}\right)^2}} \] \[ \sqrt{1 + \left(\frac{f_1}{2 \times 10^3}\right)^2} \sqrt{1 + \left(\frac{f_1}{3 \times 10^4}\right)^2} \sqrt{1 + \left(\frac{f_1}{4 \times 10^5}\right)^2} = 231.96 \] **Explanation:** We've substituted the value of \(\beta\) and set up the magnitude equation for unity gain to solve for the frequency where loop gain equals 1. --- ### **Step 2: Solve for \(f_1\) (Unity Gain Frequency), Find Phase Margin and Gain Margin** Let’s make an approximation since \(231.96\) is much greater than 1, the unity gain frequency will be much larger than the first pole but smaller than the last pole. Let’s try \(f_1 \approx 3 \times 10^4\) Hz as a starting point. Calculate the magnitude at \(f_1\): \[ \sqrt{1 + \left(\frac{f_1}{2 \times 10^3}\right)^2} \approx \frac{f_1}{2 \times 10^3} \] \[ \sqrt{1 + \left(\frac{f_1}{3 \times 10^4}\right)^2} \approx \sqrt{2} \] \[ \sqrt{1 + \left(\frac{f_1}{4 \times 10^5}\right)^2} \approx 1 \] This gives: \[ \frac{231.96}{\frac{f_1}{2 \times 10^3} \times \sqrt{2} \times 1} = 1 \] \[ \frac{231.96}{\frac{f_1}{2 \times 10^3} \times 1.414} = 1 \] \[ 231.96 = 1.414 \frac{f_1}{2 \times 10^3} \] \[ 231.96 = 1.414 \frac{f_1}{2000} \] \[ 231.96 \times 2000 = 1.414 f_1 \] \[ 463920 = 1.414 f_1 \] \[ f_1 \approx 327,970 \text{ Hz} \] Now, calculate the **phase at \(f_1\):** \[ \phi(f_1) = -\tan^{-1}\left(\frac{f_1}{2 \times 10^3}\right) - \tan^{-1}\left(\frac{f_1}{3 \times 10^4}\right) - \tan^{-1}\left(\frac{f_1}{4 \times 10^5}\right) \] \[ \tan^{-1}\left(\frac{327970}{2000}\right) \approx \tan^{-1}(164) \approx 89.65^\circ \] \[ \tan^{-1}\left(\frac{327970}{30000}\right) \approx \tan^{-1}(10.93) \approx 84.77^\circ \] \[ \tan^{-1}\left(\frac{327970}{400000}\right) \approx \tan^{-1}(0.82) \approx 39.15^\circ \] Sum of phase lags: \[ \phi(f_1) = -(89.65^\circ + 84.77^\circ + 39.15^\circ) = -213.57^\circ \] **Phase margin** = \(180^\circ + \phi(f_1)\) \[ \text{Phase margin} = 180^\circ - 213.57^\circ = -33.57^\circ \] **Gain margin:** Since the phase margin is negative, the system is **unstable** and gain margin is undefined (or negative). --- ### **Final Answer** - **Phase Margin:** \(\boxed{-33.57^\circ}\) (negative, so the system is unstable) - **Gain Margin:** Not defined (system is unstable) **Explanation:** After finding the unity gain frequency, we calculated the phase margin and found it to be negative, indicating instability. Thus, the gain margin is not meaningful for this unstable system.