Let's solve each part step by step. --- # **Problem 1 Solution** --- ## **a) Distance to First Current Minimum** **Given:** - Reflection coefficient: \( \Gamma = .7\angle 45^\circ \) - Wavelength: \( \lambda = 20\,\text{cm} \) ### **Step 1: Find Distance to First Minimum** The distance from the load to the first current minimum is given by: \[ d = \frac{\theta_\Gamma + 180^\circ}{360^\circ} \lambda/2 \] where \( \theta_\Gamma \) is the phase angle of \( \Gamma \). Given \( \theta_\Gamma = 45^\circ \), \( \lambda = 20\,\text{cm} \): \[ d = \frac{45^\circ + 180^\circ}{360^\circ} \times \frac{20}{2} \] \[ d = \frac{225^\circ}{360^\circ} \times 10\,\text{cm} \] \[ d = .625 \times 10\,\text{cm} = 6.25\,\text{cm} \] ### **Step 2: Is There a Current Maximum Between Minimum and Load?** - **At the load**, the current is not necessarily a minimum or maximum unless \( |\Gamma| = 1 \). - **Current minima/maxima** alternate every \( \lambda/4 \). - Since the first minimum is at \( 6.25\,\text{cm} \) from the load, **there is no maximum between minimum and load** (the first maximum is \( \lambda/4 = 5\,\text{cm} \) from the minimum, i.e., \( 6.25 - 5 = 1.25\,\text{cm} \) _behind_ the load, which is not physically on the line). --- ## **b) Incident Power and Load Resistance** **Given:** - \( Z_ = 200\,\Omega \) - 25% of incident power reflected ### **Step 1: Reflection Coefficient (Magnitude)** \[ \text{Reflected Power} = |\Gamma|^2 \times \text{Incident Power} = .25 \implies |\Gamma|^2 = .25 \implies |\Gamma| = .5 \] ### **Step 2: Calculate Load Resistance** For a pure resistance: \[ |\Gamma| = \left| \frac{R_L - Z_}{R_L + Z_} \right| = .5 \] Let’s solve for \( R_L \): \[ \left| \frac{R_L - 200}{R_L + 200} \right| = .5 \] Two cases: **Case 1: \( R_L > Z_ \)** \[ \frac{R_L - 200}{R_L + 200} = .5 \] \[ R_L - 200 = .5(R_L + 200) \] \[ R_L - 200 = .5R_L + 100 \] \[ R_L - .5R_L = 100 + 200 \] \[ .5R_L = 300 \implies R_L = 600\,\Omega \] **Case 2: \( R_L < Z_ \)** \[ \frac{R_L - 200}{R_L + 200} = -.5 \] \[ R_L - 200 = -.5(R_L + 200) \] \[ R_L - 200 = -.5R_L - 100 \] \[ R_L + .5R_L = 200 - 100 \] \[ 1.5R_L = 100 \implies R_L = \frac{100}{1.5} \approx 66.67\,\Omega \] **Final Answers:** - \( R_L = 600\,\Omega \) - \( R_L \approx 66.7\,\Omega \) --- ## **c) SWR from Return Loss** ### **Definitions** Return Loss (\( RL \)): \[ RL = -20 \log_{10} |\Gamma| \] \[ |\Gamma| = 10^{-RL/20} \] SWR (\( S \)): \[ S = \frac{1 + |\Gamma|}{1 - |\Gamma|} \] --- ### **For RL = 6 dB:** \[ |\Gamma| = 10^{-6/20} = 10^{-.3} \approx .501 \] \[ S = \frac{1 + .501}{1 - .501} = \frac{1.501}{.499} \approx 3.01 \] --- ### **For RL = 20 dB:** \[ |\Gamma| = 10^{-20/20} = 10^{-1} = .1 \] \[ S = \frac{1 + .1}{1 - .1} = \frac{1.1}{.9} \approx 1.222 \] --- ## **Summary Table** | Part | Final Answer | |------|--------------| | a | Distance to first current minimum: **6.25 cm**. No maximum between minimum and load. | | b | Load resistances: **600 Ω** or **66.7 Ω** | | c | SWR for RL = 6 dB: **3.01**<br>SWR for RL = 20 dB: **1.22** | --- Let me know if you need further clarification!