Attenuation of Copper Cables

© 2018 Institute for Communications Engineering, Technical University of Munich
Authors: Jimmy He, David Jobst & Tasnad Kernetzky


Blue parameter settings
$\alpha_0 =0\text{ dB/km}$
$\alpha_1 =0\text{ dB}/(\text{km}\cdot\text{MHz})$
$\alpha_2 =2.4\text{ dB}/(\text{km}\cdot\sqrt{\text{MHz}})$
$l_\text{blue} =5$ km
$\alpha_\text{K}(f_*) = 0.0$ dB
$|H_\text{K}(f_*)| =$ $1.00\cdot 10^{0}$
$|H_\text{E}(f_*)| =$ $1.00\cdot 10^{0}$
$|H_\text{E}(f_*)|^2 =$ $1.00\cdot 10^{0}$

  $10\text{ lg }\eta_{\rm K+E} =$ -40.0$\\ $ dB   $10\text{ lg }\eta_{\rm K+E} =$ -51.4$\\ $ dB  

Nyquist frequency of the cosine-rolloff filter
$f_\text{Nyq} =15$ MHz
Rolloff factor of the cosine-rolloff filter
$r =0$
Currently selected frequency
$f_* =0$ MHz
Axis scaling for $|H_\text{E}(f)|$ and $|H_\text{E}(f)|^2$
$H_0 = 10^{0}$

510152025102030405060708090
o+
$f$
$\alpha_\text{K}(f)$ in [dB]


Red parameter settings
$k_1 =4.4\text{ dB/km}$
$k_2 =10.8\text{ dB/km}$
$k_3 =0.6$
$l_\text{red} =1$ km
$\alpha_\text{K}(f_*) = 4.4$ dB
$|H_\text{K}(f_*)| =$ $0.60\cdot 10^{0}$
$|H_\text{E}(f_*)| =$ $1.66\cdot 10^{0}$
$|H_\text{E}(f_*)|^2 =$ $2.75\cdot 10^{0}$
Exercises
For the following exercises,
the term Blue refers to the coaxial cable function (marked blue in the applet) and
the term Red refers to the two-wired cable function (marked red in applet)