Question:

This problem asks for the **$S_{11}$ parameter** (input reflection coefficient) for a circuit consisting of a capacitive load connected to a transmission line. Here is the step-by-step solution. --- ## 1. Calculate the Load Impedance ($Z_L$) The load is a capacitor with capacitance $C = 10.61 \text{ pF}$ and the operating frequency is $f = 300 \text{ MHz}$. The impedance of a capacitor is: $$Z_L = \frac{1}{j \omega C}$$ where $\omega = 2\pi f$. $$Z_L = \frac{1}{j (2\pi \cdot 300 \times 10^6 \text{ Hz}) \cdot (10.61 \times 10^{-12} \text{ F})}$$ $$Z_L = \frac{1}{j (2\pi \cdot 3.183 \times 10^{-3})}$$ $$Z_L = \frac{1}{j (19.999 \times 10^{-3})} \approx \frac{1}{j (0.02)}$$ $$Z_L = -j \frac{1}{0.02} = -j 50 \ \Omega$$ The load impedance is $\mathbf{Z_L = -j 50 \ \Omega}$. --- ## 2. Calculate the Normalized Load Impedance ($z_L$) The normalized load impedance is $z_L = Z_L / Z_0$. The system characteristic impedance is $Z_0 = 50 \ \Omega$. $$z_L = \frac{Z_L}{Z_0} = \frac{-j 50 \ \Omega}{50 \ \Omega} = -j 1$$ The normalized load impedance is $\mathbf{z_L = -j 1}$. --- ## 3. Calculate the Reflection Coefficient at the Load ($\Gamma_L$) The reflection coefficient at the load is given by: $$\Gamma_L = \frac{Z_L - Z_0}{Z_L + Z_0} \quad \text{or} \quad \Gamma_L = \frac{z_L - 1}{z_L + 1}$$ Using the normalized impedance: $$\Gamma_L = \frac{-j 1 - 1}{-j 1 + 1} = \frac{-(1 + j)}{(1 - j)}$$ To simplify, multiply the numerator and denominator by the conjugate of the denominator $(1 + j)$: $$\Gamma_L = \frac{-(1 + j)}{(1 - j)} \cdot \frac{(1 + j)}{(1 + j)} = - \frac{(1 + j)^2}{1^2 - j^2}$$ $$\Gamma_L = - \frac{(1 + j2 + j^2)}{1 - (-1)} = - \frac{(1 + j2 - 1)}{2}$$ $$\Gamma_L = - \frac{j 2}{2} = -j 1$$ The reflection coefficient at the load is $\mathbf{\Gamma_L = -j 1}$. --- ## 4. Calculate the Input Reflection Coefficient ($S_{11}$) The $S_{11}$ parameter is the reflection coefficient at the input of the transmission line, which is $\Gamma_{\text{in}}$. For a transmission line of length $l$ terminating in a load with reflection coefficient $\Gamma_L$, the input reflection coefficient is: $$S_{11} = \Gamma_{\text{in}} = \Gamma_L e^{-j 2 \beta l}$$ where $\beta$ is the phase constant. The length of the line is given as $l = 0.5 \lambda$ (half a wavelength). Since $\beta = 2\pi / \lambda$, the phase shift is: $$2 \beta l = 2 \left( \frac{2\pi}{\lambda} \right) (0.5 \lambda) = 2\pi$$ Substitute this into the $\Gamma_{\text{in}}$ equation: $$S_{11} = \Gamma_L e^{-j 2\pi}$$ Since $e^{-j 2\pi} = \cos(-2\pi) + j \sin(-2\pi) = 1 + j 0 = 1$, we have: $$S_{11} = \Gamma_L \cdot 1$$ $$S_{11} = \Gamma_L$$ Therefore, the input reflection coefficient is equal to the load reflection coefficient. $$S_{11} = -j 1$$ *** The $\mathbf{S_{11}}$ parameter at the location shown is $\mathbf{-j 1}$. ***