Let's solve the problem step by step. --- ### **Given:** - **Loop gain function:** \[ T(jf) = \frac{5432 \times \beta}{\left(1 + j\frac{f}{2 \times 10^3}\right)\left(1 + j\frac{f}{3 \times 10^4}\right)\left(1 + j\frac{f}{4 \times 10^5}\right)} \] - \(\beta = 38.66 \times 10^{-3}\) We are to **determine the phase margin and the gain margin**. --- ## **Step 1: Substitute \(\beta\)** \[ 5432 \times \beta = 5432 \times 38.66 \times 10^{-3} \] \[ = 5432 \times 0.03866 \approx 209.94 \] So, \[ T(jf) = \frac{209.94}{\left(1 + j\frac{f}{2 \times 10^3}\right)\left(1 + j\frac{f}{3 \times 10^4}\right)\left(1 + j\frac{f}{4 \times 10^5}\right)} \] --- ## **Step 2: Find the **Unity Gain Frequency** (\(f_{0\,\text{dB}}\))** Set \(|T(jf)| = 1\): \[ |T(jf)| = \frac{209.94}{\sqrt{1 + \left(\frac{f}{2 \times 10^3}\right)^2} \sqrt{1 + \left(\frac{f}{3 \times 10^4}\right)^2} \sqrt{1 + \left(\frac{f}{4 \times 10^5}\right)^2}} \] Set \(|T(jf)| = 1\): \[ \sqrt{1 + \left(\frac{f}{2 \times 10^3}\right)^2} \sqrt{1 + \left(\frac{f}{3 \times 10^4}\right)^2} \sqrt{1 + \left(\frac{f}{4 \times 10^5}\right)^2} = 209.94 \] Square both sides: \[ \left[1 + \left(\frac{f}{2 \times 10^3}\right)^2\right]\left[1 + \left(\frac{f}{3 \times 10^4}\right)^2\right]\left[1 + \left(\frac{f}{4 \times 10^5}\right)^2\right] = (209.94)^2 \] \[ = 44079 \] --- ## **Step 3: **Approximate \(f\) at Unity Gain** For high frequency, \(f \gg\) all the corner frequencies, so: \[ 1 + \left(\frac{f}{f_c}\right)^2 \approx \left(\frac{f}{f_c}\right)^2 \] So, \[ \left(\frac{f}{2 \times 10^3}\right)^2 \left(\frac{f}{3 \times 10^4}\right)^2 \left(\frac{f}{4 \times 10^5}\right)^2 = 44079 \] \[ \left(\frac{f^6}{(2 \times 10^3)^2 (3 \times 10^4)^2 (4 \times 10^5)^2}\right) = 44079 \] \[ f^6 = 44079 \times (2 \times 10^3)^2 (3 \times 10^4)^2 (4 \times 10^5)^2 \] Calculate the denominator: - \(2 \times 10^3 = 2000\), squared: \(4 \times 10^6\) - \(3 \times 10^4 = 30000\), squared: \(9 \times 10^8\) - \(4 \times 10^5 = 400000\), squared: \(16 \times 10^{10}\) Multiply: \[ 4 \times 10^6 \times 9 \times 10^8 \times 16 \times 10^{10} = (4 \times 9 \times 16) \times 10^{6+8+10} = 576 \times 10^{24} \] So, \[ f^6 = 44079 \times 576 \times 10^{24} \] \[ = 25,399,104 \times 10^{24} \] \[ f = [25,399,104 \times 10^{24}]^{1/6} \] Calculate \(25,399,104^{1/6}\): Take log: \[ \log_{10}(25,399,104) \approx 7.4, \quad \frac{7.4}{6} \approx 1.23, \quad 10^{1.23} \approx 17 \] \(10^{24/6} = 10^4\) So, \[ f \approx 17 \times 10^4 = 1.7 \times 10^5~\text{Hz} = 170~\text{kHz} \] --- ## **Step 4: **Phase Margin** The phase of each pole at frequency \(f\): \[ \phi(f) = -\tan^{-1}\left(\frac{f}{2 \times 10^3}\right) - \tan^{-1}\left(\frac{f}{3 \times 10^4}\right) - \tan^{-1}\left(\frac{f}{4 \times 10^5}\right) \] At \(f = 1.7 \times 10^5\): - \(\frac{f}{2 \times 10^3} = \frac{1.7 \times 10^5}{2 \times 10^3} = 85\) - \(\frac{f}{3 \times 10^4} = \frac{1.7 \times 10^5}{3 \times 10^4} \approx 5.67\) - \(\frac{f}{4 \times 10^5} = \frac{1.7 \times 10^5}{4 \times 10^5} \approx 0.425\) Calculate phase shifts: - \(\tan^{-1}(85) \approx 89.3^\circ\) - \(\tan^{-1}(5.67) \approx 80^\circ\) - \(\tan^{-1}(0.425) \approx 23^\circ\) Total phase lag: \[ \phi = -(89.3 + 80 + 23) = -192.3^\circ \] Phase margin: \[ \text{PM} = 180^\circ + \phi = 180^\circ - 192.3^\circ = -12.3^\circ \] --- ## **Step 5: **Gain Margin** The gain margin is the reciprocal of the gain at the frequency where the phase is \(-180^\circ\). But, as seen, the phase crosses \(-180^\circ\) at approximately the unity gain frequency. So the system is **not stable** (since phase margin is negative). --- ## **Step 6: **Conclusion** ### **Summary:** - **Phase Margin:** \(\boxed{-12.3^\circ}\) - **Gain Margin:** The gain crosses unity at the phase crossover, so the gain margin is approximately 0 dB. - **Stability:** The system is **unstable** (\(\text{PM} < 0\)). --- If you need a more precise calculation, let me know!