Please convert the below content to rendered equations without any errors.Let’s go cleanly, step by step, using standard **lossless transmission-line theory**. --- ## Given * Characteristic impedance: [ Z_0 = 400\ \Omega ] * Operating frequency: [ f = 200\ \text{MHz} ] * Input impedance: [ Z_{in} = 200 - j200\ \Omega ] * Dielectric: **air** ⇒ phase velocity [ v \approx c = 3\times 10^8\ \text{m/s} ] --- # (a) Standing Wave Ratio (SWR) ### Step 1: Reflection coefficient at the input For a **lossless line**, the magnitude of the reflection coefficient is the same everywhere on the line. [ \Gamma = \frac{Z_{in}-Z_0}{Z_{in}+Z_0} ] [ \Gamma = \frac{(200-j200)-400}{(200-j200)+400} = \frac{-200-j200}{600-j200} ] ### Step 2: Magnitude of (\Gamma) [ |\Gamma| = \frac{\sqrt{(-200)^2+(-200)^2}}{\sqrt{600^2+(-200)^2}} = \frac{282.84}{632.46} = 0.447 ] ### Step 3: Standing-wave ratio [ \boxed{ S = \frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1+0.447}{1-0.447} = 2.62 } ] --- # (b) Load impedance (Z_L) (line length = 1 m) ### Step 1: Wavelength and phase constant [ \lambda = \frac{c}{f} = \frac{3\times10^8}{200\times10^6} = 1.5\ \text{m} ] [ \beta = \frac{2\pi}{\lambda} = \frac{2\pi}{1.5} = 4.189\ \text{rad/m} ] [ \beta l = 4.189 \times 1 = 4.189\ \text{rad} = 240^\circ ] [ \tan(\beta l) = \tan(240^\circ) = \sqrt{3} = 1.732 ] --- ### Step 2: Transmission-line impedance equation [ Z_{in} = Z_0 \frac{Z_L + jZ_0\tan(\beta l)} {Z_0 + jZ_L\tan(\beta l)} ] Solve for (Z_L): [ Z_L = Z_0 \frac{jZ_0\tan(\beta l)-Z_{in}} {jZ_{in}\tan(\beta l)-Z_0} ] --- ### Step 3: Substitute values [ Z_L = 400 \frac{j(400)(1.732)-(200-j200)} {j(200-j200)(1.732)-400} ] After complex arithmetic: [ \boxed{ Z_L \approx 1041 + j70\ \Omega } ] --- # (c) Distance from the load to the nearest voltage maximum ### Step 1: Reflection coefficient at the load [ \Gamma_L = \frac{Z_L-Z_0}{Z_L+Z_0} ] [ \Gamma_L = \frac{(1041+j70)-400}{(1041+j70)+400} ] Phase of (\Gamma_L): [ \angle\Gamma_L \approx 3.43^\circ = 0.0599\ \text{rad} ] --- ### Step 2: Condition for voltage maximum Voltage maxima occur when: [ \angle\Gamma_L - 2\beta z = 0 ] [ z = \frac{\angle\Gamma_L}{2\beta} ] --- ### Step 3: Calculate distance [ z = \frac{0.0599}{2(4.189)} = 0.00715\ \text{m} ] [ \boxed{ z \approx 7.2\ \text{mm from the load} } ] --- ## ✅ Final Answers | Part | Result | | --------------------------------------- | ------------------------------------ | | (a) SWR | **2.62** | | (b) Load impedance | **(Z_L \approx 1041 + j70\ \Omega)** | | (c) Distance to nearest voltage maximum | **≈ 7.2 mm** | If you want, I can also: * Show this on a **Smith chart** * Verify results numerically * Explain the physical meaning of each result