Let's solve the step-by-step--- ## Given: - **Capacitance \( C \) = 10.61 pF** - **Transmission line length \( l = .5 \lambda \)** - **Operating frequency \( f = 300\, \text{MHz} \)** - **Characteristic impedance \( Z_ = 50\, \Omega \)** - **Load impedance \( Z_L = Z_{in} \) (to be calculated)** --- ## Step 1: Find the wavelength (\( \lambda \)) \[ \lambda = \frac{v}{f} \] where \( v \) is the speed of light in the medium; assuming the transmission line is in free space or an environment where \( v \approx c = 3 \times 10^8\, \text{m/s} \). \[ \lambda = \frac{3 \times 10^{8}\, \text{m/s}}{300 \times 10^{6}\, \text{Hz}} = 1\, \text{m} \] --- ## Step 2: Find the length of the transmission line \( l \) Given \( l = .5 \lambda \): \[ l = .5 \times 1\, \text{m} = .5\, \text{m} \] --- ## Step 3: Calculate the electrical length in radians The electrical length \( \beta l \): \[ \beta = \frac{2\pi}{\lambda} \] \[ \beta = \frac{2\pi}{1\, \text{m}} = 2\pi\, \text{rad/m} \] Therefore, for \( l = .5\, \text{m} \): \[ \beta l = 2\pi \times .5 = \pi\, \text{rad} \] --- ## Step 4: Use the transmission line impedance transformation formula The input impedance \( Z_{in} \) for a lossless transmission line terminated with \( Z_{L} \): \[ Z_{in} = Z_ \frac{Z_L + j Z_ \tan(\beta l)}{Z_ + j Z_L \tan(\beta l)} \] Given \( Z_L \) (the load impedance) is purely reactive (since it is a capacitor in parallel with the transmission line), we need to find \( Z_L \). --- ## Step 5: Find the load impedance \( Z_{load} \) The load is a capacitor: \[ Z_C = \frac{1}{j \omega C} \] Where: \[ \omega = 2 \pi f = 2 \pi \times 300 \times 10^6 = 1.884 \times 10^9\, \text{rad/sec} \] Calculate: \[ Z_C = \frac{1}{j \times 1.884 \times 10^9 \times 10.61 \times 10^{-12}} \] \[ Z_C = \frac{1}{j \times 1.884 \times 10^9 \times 10.61 \times 10^{-12}} \] \[ Z_C = \frac{1}{j \times 20.02} \approx -j \times .0499\, \Omega \] --- ## Step 6: Determine \( Z_{in} \) Since the load is purely reactive: \[ Z_L = Z_C = -j .0499\, \Omega \] Using the impedance transformation formula: \[ Z_{in} = Z_ \frac{Z_L + j Z_ \tan(\beta l)}{Z_ + j Z_L \tan(\beta l)} \] Substitute the known values: \[ Z_{in} = 50 \frac{-j .0499 + j 50 \times \tan(\pi)}{50 + j (-j .0499) \times \tan(\pi)} \] Recall: \[ \tan(\pi) = \] So: \[ Z_{in} = 50 \times \frac{-j .0499 + }{50 + } = 50 \times \frac{-j .0499}{50} = -j .0499\, \Omega \] --- ## **Final Result:** \[ \boxed{ Z_{in} \approx -j .0499\, \Omega } \] This is the input impedance at the specified location. --- ## **Summary:** - The input impedance \( Z_{in} \) at the specified location is approximately **\(-j .0499\, \Omega\)**, primarily reactive and very small in magnitude. If further clarification or calculations are needed, please let me know!