Solution:

$z_0=R_0=\sqrt{\frac{L}{C}}=50 \Omega \\ $

with constant $-k$ section low pass filter,

$\begin{aligned} \\ & W_C=\frac{2}{\sqrt{L C}}=2 \pi \mathrm{fC}_C \\\\ \therefore & 2 \pi \times 50 M H_2=\frac{2}{\sqrt{L C}} \Rightarrow 50 \pi \times 10^6=\frac{1}{\sqrt{L C}}\\ \\ \therefore \quad & \sqrt{L C}=\frac{1}{50 \pi \times 10^6} \quad \therefore L C=4.05 \times 10^{-17} \\ \end{aligned} \\ $

$\begin{aligned} \\ &\therefore \frac{L}{C}=2500 \quad \therefore L=2500 \mathrm{C} \\\\ &\therefore \quad 2500 \mathrm{C}^2=4.05 \times 10^{-17} \quad \therefore \quad C=0.127 \mathrm{nF} \\\\ &\therefore \quad L=0.318 \mathrm{MH} \\ \end{aligned} \\ $

m-derived low pass filter:

$\begin{aligned}\\ &m=\sqrt{1-\frac{f_c^2}{f_{\infty}^2}}=\sqrt{1-\frac{50^2}{52^2 \mathrm{HH}_2}}=0.274 \\\\ &L / 2=0.159 \mu \mathrm{H}, \quad C=0.127 \mathrm{nF} \\\\ &\frac{m L}{2}=0.043 \mu \mathrm{H}, \quad m C=0.034 \mathrm{nF} \\\\ &\left(\frac{1-\mathrm{m}^2}{4 \mathrm{~m}}\right)^2 L=0.268 \mu \mathrm{H} \\ \end{aligned}\\ $

1lp \& Olp matching sectious:

$\begin{array}{ll}\\ 0.3 \mathrm{~L}=0.095 \mu \mathrm{H} & 0.3 \mathrm{C}=0.038 \mathrm{nF} \\\\ \left.(8]_{15}\right) \mathrm{L}=0.1696 \mathrm{\mu H} \\ \end{array}\\ $