Question 2.6: Consider a copper coaxial cable with a Teflon (εr = 2) diele......
Consider a copper coaxial cable with a Teflon (\epsilon _r = 2) dielectric. Find the ratio b/a corresponding to the minimum conductor losses and evaluate the resulting impedance. Dimension the cable so that the maximum operating frequency is 50 GHz.
The attenuation is proportional to a function f(b/a) that can be minimized with respect to the line shape ratio by looking for a zero of its first derivative:
\begin{aligned} \alpha_c & \propto \frac{1+x}{\log x}=f(x), \quad x=\frac{b}{a}>1 \\ \frac{ d f(x)}{ d x} & =\frac{1}{\log x}-\frac{1+x}{(\log x)^2} \frac{1}{x}=\frac{x \log x-(1+x)}{x(\log x)^2}=0 . \end{aligned}
The zero corresponds to x = 3.5911 and it can be easily shown by inspection that this is a minimum of f and therefore of the attenuation. The corresponding impedance is, from (2.31),
Z_0=\frac{60}{\sqrt{\epsilon_r}} \log \left(\frac{b}{a}\right)\enspace \Omega,\hspace{30 pt} \text{(2.31)}Z_0 = 54.24 \Omega . We then have from (2.34):
\begin{aligned} f_{\max } & =\frac{c_0}{\lambda_c}=\frac{c_0}{\pi \sqrt{\epsilon_r} (a+b)}=\frac{c_0}{\pi a \sqrt{\epsilon_r} (1+b / a)} \rightarrow \\ a & =\frac{c_0}{\pi f_{\max } \sqrt{\epsilon_r} (1+b / a)}=\frac{3 \times 10^8}{\pi \cdot 50 \times 10^9 \sqrt{2} (1+3.5911)}=0.294 \text{ mm} \\ b & =3.5911 \cdot a=3.5911 \cdot 0.294=1.06 \text{ mm}. \end{aligned}
Thus the coaxial cable outer diameter is 2.12 mm while the inner diameter is ≈ 0.6 mm. The conductor attenuation at 50 GHz is from (2.33)