Let's solve the problem step by step for **case (a)** first. We'll go through the method, and you can apply the same process to (b) and (c). ### **Given:** We want the **step response** of the closed-loop system to have a **10% overshoot**. #### The transfer function: \[ G(s) = \frac{K}{s(s+5)(s+10)} \] #### Unity feedback: \[ T(s) = \frac{G(s)}{1 + G(s)} \] --- ### **Step 1: Find Phase Margin for 10% Overshoot** The % overshoot (\( M_p \)) is related to the damping ratio (\( \zeta \)) by: \[ M_p = e^{-\frac{\pi \zeta}{\sqrt{1-\zeta^2}}} \times 100\% \] Set \( M_p = 10\% \): \[ 0.1 = e^{-\frac{\pi \zeta}{\sqrt{1-\zeta^2}}} \] Take natural log on both sides: \[ \ln(0.1) = -\frac{\pi \zeta}{\sqrt{1-\zeta^2}} \implies -2.3026 = -\frac{\pi \zeta}{\sqrt{1-\zeta^2}} \] \[ 2.3026 = \frac{\pi \zeta}{\sqrt{1-\zeta^2}} \] \[ \frac{\zeta}{\sqrt{1-\zeta^2}} = \frac{2.3026}{\pi} = 0.733 \] Let \( x = \zeta \): \[ \frac{x}{\sqrt{1-x^2}} = 0.733 \implies x = 0.733 \sqrt{1-x^2} \implies x^2 = 0.537 \cdot (1-x^2) \implies x^2 = 0.537 - 0.537 x^2 \implies x^2 + 0.537 x^2 = 0.537 \implies x^2 (1+0.537) = 0.537 \implies x^2 = \frac{0.537}{1.537} \implies x^2 = 0.349 \implies x = 0.591 \] So, **damping ratio \( \zeta = 0.591 \)**. #### Phase Margin (\( \phi_m \)): \[ \phi_m = \arccos(\zeta) \] \[ \phi_m = \arccos(0.591) = 53.8^\circ \] But actually, for phase margin, the formula is: \[ \phi_m = \arccos(\zeta) \] But usually, phase margin for a desired overshoot is: \[ \phi_m \approx \arccos(\zeta) \times \frac{180}{\pi} \] So, \[ \phi_m = \arccos(0.591) \times \frac{180}{\pi} \approx 53.8^\circ \] --- ### **Step 2: Find Gain Crossover Frequency (\( \omega_{gc} \))** At the gain crossover frequency, the phase of \( G(j\omega) \) is: \[ \angle G(j\omega_{gc}) = -180^\circ + \phi_m \] So, \[ \angle G(j\omega_{gc}) = -180^\circ + 53.8^\circ = -126.2^\circ \] --- ### **Step 3: Compute Phase at \( G(j\omega) \)** For \( G(s) = \frac{K}{s(s+5)(s+10)} \): \[ G(j\omega) = \frac{K}{j\omega (j\omega + 5)(j\omega + 10)} \] The phase is: - \( -90^\circ \) from \( j\omega \) - \( -\tan^{-1} \left( \frac{\omega}{5} \right) \) from \( j\omega + 5 \) - \( -\tan^{-1} \left( \frac{\omega}{10} \right) \) from \( j\omega + 10 \) So, \[ \angle G(j\omega) = -90^\circ - \tan^{-1} \left( \frac{\omega}{5} \right) - \tan^{-1} \left( \frac{\omega}{10} \right) \] Set this equal to \( -126.2^\circ \): \[ -90^\circ - \tan^{-1}\left(\frac{\omega_{gc}}{5}\right) - \tan^{-1}\left(\frac{\omega_{gc}}{10}\right) = -126.2^\circ \] \[ \tan^{-1}\left(\frac{\omega_{gc}}{5}\right) + \tan^{-1}\left(\frac{\omega_{gc}}{10}\right) = 36.2^\circ \] Let \( x = \omega_{gc} \): Let's try values for \( x \): Let \( x = 5 \): \[ \tan^{-1}(1) + \tan^{-1}(0.5) = 45^\circ + 26.6^\circ = 71.6^\circ \] Let \( x = 2 \): \[ \tan^{-1}(0.4) + \tan^{-1}(0.2) = 21.8^\circ + 11.3^\circ = 33.1^\circ \] Let \( x = 3 \): \[ \tan^{-1}(0.6) + \tan^{-1}(0.3) = 30.96^\circ + 16.7^\circ = 47.66^\circ \] Let \( x = 2.5 \): \[ \tan^{-1}(0.5) + \tan^{-1}(0.25) = 26.6^\circ + 14.0^\circ = 40.6^\circ \] Let \( x = 2.2 \): \[ \tan^{-1}(0.44) + \tan^{-1}(0.22) = 23.75^\circ + 12.5^\circ = 36.25^\circ \] So, at **\( \omega_{gc} \approx 2.2 \) rad/s**. --- ### **Step 4: Find the Value of \( K \) for Unity Gain at \( \omega_{gc} \)** At gain crossover frequency, the magnitude of the open-loop system is 1: \[ |G(j\omega_{gc})| = 1 \] \[ |G(j\omega_{gc})| = \frac{K}{|\omega_{gc}||j\omega_{gc} + 5||j\omega_{gc} + 10|} \] Calculate: - \( |\omega_{gc}| = 2.2 \) - \( |j2.2 + 5| = \sqrt{2.2^2 + 5^2} = \sqrt{4.84 + 25} = \sqrt{29.84} = 5.46 \) - \( |j2.2 + 10| = \sqrt{2.2^2 + 10^2} = \sqrt{4.84 + 100} = \sqrt{104.84} = 10.24 \) So, \[ |G(j2.2)| = \frac{K}{2.2 \times 5.46 \times 10.24} \] \[ 2.2 \times 5.46 = 12.012 \] \[ 12.012 \times 10.24 = 122.98 \] Set magnitude to 1: \[ \frac{K}{122.98} = 1 \implies K = 122.98 \] --- ## **Final Answer for (a):** \[ \boxed{K = 123} \] (Rounded to nearest integer) --- ### **You can use a similar process for (b) and (c):** - Calculate the phase equation for that transfer function. - Find \( \omega_{gc} \) such that phase equals \( -180^\circ + 53.8^\circ \). - Use the magnitude equation to solve for \( K \). If you need detailed solutions for (b) and (c), let me know!