Question:

Please give me the below content same as it is without any changes in rendered equations Nice — let’s solve each part step-by-step. --- # a) Distance from the load to the first current minimum (Γ = 0.7∠45°, λ = 20 cm) **Given** * Reflection coefficient at the load: (\Gamma_L = 0.7\angle 45^\circ). * Wavelength (\lambda = 20\ \text{cm} = 0.20\ \text{m}). * For a lossless line (\beta = \dfrac{2\pi}{\lambda}). **Current magnitude on a lossless line** [ |I(z)|^2 \propto 1 + |\Gamma|^2 - 2|\Gamma|\cos(2\beta z + \varphi) ] where (\varphi) is the phase of (\Gamma) (here (\varphi=45^\circ=\pi/4)) and (z) is measured from the load toward the generator. A **current minimum** occurs when the cosine term is **+1** (that gives the largest subtraction): [ \cos(2\beta z + \varphi)=1 \quad\Rightarrow\quad 2\beta z + \varphi = 2\pi n. ] Smallest positive (z) is for (n=1): [ z = \frac{2\pi - \varphi}{2\beta}. ] Substitute (\beta = \dfrac{2\pi}{\lambda}): [ z = \frac{2\pi - \varphi}{2\cdot(2\pi/\lambda)}=\lambda\frac{2\pi-\varphi}{4\pi}. ] With (\varphi=\pi/4) and (\lambda=0.20\ \text{m}): [ z = 0.20\cdot\frac{2\pi - \pi/4}{4\pi} = 0.20\cdot\frac{7}{16}=0.0875\ \text{m}=8.75\ \text{cm}. ] **Answer (a):** The first current minimum is **8.75 cm** from the load (toward the generator). **Is there a current maximum between the load and that minimum?** Current maxima occur when (\cos(2\beta z + \varphi) = -1), i.e. [ 2\beta z + \varphi = \pi + 2\pi n. ] Take (n=0) to get the smallest positive (z): [ z = \frac{\pi - \varphi}{2\beta} = \lambda\frac{\pi-\varphi}{4\pi}. ] With (\varphi=\pi/4): [ z = 0.20\cdot\frac{3}{16}=0.0375\ \text{m}=3.75\ \text{cm}. ] This lies between 0 and 8.75 cm, so **yes** — there is a current maximum **3.75 cm** from the load (between load and first minimum). --- # b) 200 Ω line terminated in a pure resistance; 25% of incident power is reflected Reflected power fraction (P_r/P_i = 0.25). For a reflection coefficient magnitude (|\Gamma|), [ \frac{P_r}{P_i}=|\Gamma|^2 \quad\Rightarrow\quad |\Gamma|=\sqrt{0.25}=0.5. ] For a purely resistive load (R_L) on line with characteristic impedance (Z_0 = 200\ \Omega), [ \Gamma = \frac{R_L - Z_0}{R_L + Z_0} ] This is real, so set (\Gamma = +0.5) and (\Gamma = -0.5) to find the two possibilities. **Case 1:** (\dfrac{R_L - 200}{R_L + 200}=+0.5) [ R_L - 200 = 0.5(R_L + 200)\Rightarrow 0.5R_L = 300\Rightarrow R_L = 600\ \Omega. ] **Case 2:** (\dfrac{R_L - 200}{R_L + 200}=-0.5) [ R_L - 200 = -0.5(R_L + 200)\Rightarrow 1.5R_L = 100\Rightarrow R_L = \frac{100}{1.5}=66.\overline{6}\ \Omega. ] **Answer (b):** The two possible load resistances are **600 Ω** and **66.67 Ω** (i.e. (200/3)). --- # c) SWR from return loss Return loss (dB) is defined as [ \text{Return Loss (dB)} = -20\log_{10}|\Gamma| \quad\Rightarrow\quad |\Gamma| = 10^{-\text{RL(dB)}/20}. ] SWR (standing wave ratio) is [ \text{SWR} = \frac{1+|\Gamma|}{1-|\Gamma|}. ] **For return loss = 6 dB:** [ |\Gamma| = 10^{-6/20}=10^{-0.3}\approx 0.5012\approx 0.5, ] [ \text{SWR}=\frac{1+0.5}{1-0.5}=\frac{1.5}{0.5}=3. ] **For return loss = 20 dB:** [ |\Gamma| = 10^{-20/20}=10^{-1}=0.1, ] [ \text{SWR}=\frac{1+0.1}{1-0.1}=\frac{1.1}{0.9}\approx 1.222\ (\approx 1.22). ] **Answer (c):** * For 6 dB return loss: **SWR = 3.0**. * For 20 dB return loss: **SWR ≈ 1.22**.