$$\left\{\begin{matrix} v(z,t)- v(z+ \Delta z) &=L \Delta z \frac {\delta i}{\delta t} + R \Delta z i \\ i(z,t)- i(z+ \Delta z) &=C \Delta z \frac {\delta v}{\delta t} + g \Delta z v \end{matrix}\right.$$
$$\left\{\begin{matrix} \frac{v(z,t)- v(z+ \Delta z)}{\Delta z} =L \frac {\delta i}{\delta t} + R i \\ \frac{i(z,t)- i(z+ \Delta z)}{\Delta z} =C \frac {\delta v}{\delta t} + g v \end{matrix}\right.$$
z infinitisimaal klein maken:
$$\left\{\begin{matrix} \frac{\delta v}{\delta z} &=L \frac {\delta i}{\delta t} + R i \\ \frac{\delta i}{\delta z} &=C \frac {\delta v}{\delta t} + g v \end{matrix}\right.$$
We groeperen van stroom en spanning:
$$\left\{\begin{matrix} \frac{\delta v}{\delta z} &= (j \omega L + R ) i \\ \frac{\delta i}{\delta z} &= (j \omega C + g ) v \end{matrix}\right.$$
Beide vergelijkingen combineren geeft:
$$ \frac{\delta v}{\delta i} = \frac{(j \omega L + R ) i}{(j \omega C + g ) v }$$
We groeperen stroom en spanning:
$$ \frac{v \delta v}{i\delta i} = \frac{j \omega L + R }{j \omega C + g }$$
We schrijven het resultaat in de vorm van karakteristieke impedantie:
$$ Z_o^2 = \frac{j \omega L + R }{j \omega C + g }$$
$$ Z_o=\sqrt{\frac{j \omega L+R}{j \omega C+g}}$$
$$ Z_o \approx \sqrt{\frac{L}{C}}$$
Voor een typische coax-kabel met binnendiameter 1.2 mm en buiten diameter 2.8 mm bekomen we de volgende parameters:
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>
<IPython.core.display.Image object>