Let's solve the problem step-by-step. --- # Given Data: Transfer function: \[ H(s) = \frac{s + 3}{(s + 2)^2} \] --- # Step 1: Find the frequency response \( H(j\omega) \) Replace s \) with \( j\omega \): \[ H(j\omega) = \frac{j\omega + 3}{(j\omega + 2)^2} \] --- # Step 2: Find magnitude \( |H(j\omega)| \) and phase \( \angle H(j\omega) \) ### Magnitude: \[ |H(j\omega)| = \frac{|j\omega + 3|}{|(j\omega + 2)^2|} = \frac{\sqrt{\omega^2 + 3^2}}{\left|j\omega + 2\right|^2} \] Calculate numerator: \[ |j\omega + 3| = \sqrt{\omega^2 + 3^2} = \sqrt{\omega^2 + 9} \] Calculate denominator: \[ |j\omega + 2| = \sqrt{\omega^2 + 2^2} = \sqrt{\omega^2 + 4} \] \[ |(j\omega + 2)^2| = (\sqrt{\omega^2 + 4})^2 = \omega^2 + 4 \] **Thus:** \[ |H(j\omega)| = \frac{\sqrt{\omega^2 + 9}}{\omega^2 + 4} \] --- ### Phase: \[ \angle H(j\omega) = \angle (j\omega + 3) - 2 \times \angle (j\omega + 2) \] Calculate: \[ \angle (j\omega + 3) = \arctan \left( \frac{\omega}{3} \right) \] \[ \angle (j\omega + 2) = \arctan \left( \frac{\omega}{2} \right) \] So: \[ \boxed{ \angle H(j\omega) = \arctan \left( \frac{\omega}{3} \right) - 2 \arctan \left( \frac{\omega}{2} \right) } \] --- # Step 3: Find the steady-state response for each input --- ## (a) Input: \( u(t) \) **Frequency response at \( \omega = \):** \[ |H(j)| = \frac{\sqrt{ + 9}}{ + 4} = \frac{3}{4} = .75 \] \[ \angle H(j) = \arctan() - 2 \arctan() = \] **Steady-state response:** \[ y(t) = |H(j)| \cdot \sin( \cdot t + ) = .75 \times \sin(t) \] --- ## (b) Input: \( \cos(2t + 60^\circ) u(t) \) Frequency: \( \omega = 2 \) Calculate magnitude: \[ |H(j2)| = \frac{\sqrt{4 + 9}}{4 + 4} = \frac{\sqrt{13}}{8} \approx \frac{3.606}{8} \approx .45075 \] Calculate phase: \[ \angle H(j2) = \arctan \left( \frac{2}{3} \right) - 2 \arctan \left( 1 \right) \] \[ = .588 - 2 \times \frac{\pi}{4} = .588 - \frac{\pi}{2} \approx .588 - 1.571 = -.983 \text{ radians} \] which is approximately **-56.3 degrees**. Input phase: \( 60^\circ = \pi/3 \) radians. **Steady-state response:** \[ y(t) = |H(j2)| \times \cos(2t + 60^\circ + \text{phase shift}) \] \[ = .45075 \times \cos \left( 2t + 60^\circ - 56.3^\circ \right) = .45075 \times \cos (2t + 3.7^\circ) \] or in radians: \[ \approx .45075 \times \cos (2t + .0646) \] --- ## (c) Input: \( \sin(3t - 45^\circ) u(t) \) Frequency: \( \omega = 3 \) Calculate magnitude: \[ |H(j3)| = \frac{\sqrt{9 + 9}}{9 + 4} = \frac{\sqrt{18}}{13} \approx \frac{4.243}{13} \approx .3264 \] Calculate phase: \[ \angle H(j3) = \arctan \left( \frac{3}{3} \right) - 2 \arctan \left( \frac{3}{2} \right) \] \[ = \frac{\pi}{4} - 2 \times .9828 \approx .785 - 1.9656 = -1.18 \text{ radians} \] which is approximately **-67.6 degrees**. Input phase: \( -45^\circ \). **Steady-state response:** \[ y(t) = |H(j3)| \times \sin \left( 3t - 45^\circ + 67.6^\circ \right) = .3264 \times \sin (3t + 22.6^\circ) \] or in radians: \[ \approx .3264 \times \sin (3t + .394) \] --- ## (d) Input: \( e^{j 3 t} u(t) \) This is a complex exponential input at \( \omega = 3 \). **The steady-state response:** \[ Y(j3) = H(j3) \times e^{j3t} \] \[ = |H(j3)| e^{j \angle H(j3)} \times e^{j3t} \] From previous calculations: \[ |H(j3)| \approx .3264 \] \[ \angle H(j3) \approx -1.18 \text{ radians} \] **Time domain response:** \[ y(t) = |H(j3)| \times e^{j(3t + \angle H(j3))} = .3264 \times e^{j(3t - 1.18)} \] In real form: \[ \boxed{ y(t) = .3264 \times \cos(3t - 1.18) } \] --- # **Summary:** | Input | Magnitude | Phase (radians) | Steady-state response | |---------|--------------|------------------|------------------------| | \( u(t) \) | .75 | | \( .75 \sin t \) | | \( \cos(2t + 60^\circ) u(t) \) | ~.45075 | ~ -.983 | \( .45075 \cos(2t + 3.7^\circ) \) | | \( \sin(3t - 45^\circ) u(t) \) | ~.3264 | ~ -1.18 | \( .3264 \sin(3t + 22.6^\circ) \) | | \( e^{j3t} u(t) \) | ~.3264 | ~ -1.18 | \( .3264 \cos(3t - 1.18) \) | Let me know if you'd like further clarification!