Gebruiker:Madyno/Kladblok/Complex rekenen

Belaste lijn

Is de lijn belast met een belasting $Z_{D}$, dan is:

$Z_{in}={\frac {Z_{D}+Z_{0}\tanh(\gamma D)}{Z_{0}+Z_{D}\tanh(\gamma D)}}={\frac {Z_{D}+j\,Z_{0}\tan(\varphi )}{Z_{0}+j\,Z_{D}\tan(\varphi )}}$.

dus

${\frac {Z_{0}}{Z_{in}}}={\frac {1+j{\frac {Z_{D}}{Z_{0}}}\tan(\varphi )}{{\frac {Z_{D}}{Z_{0}}}+j\,\tan(\varphi )}}={\frac {\cos(\varphi )+j{\frac {Z_{D}}{Z_{0}}}\sin(\varphi )}{{\frac {Z_{D}}{Z_{0}}}\cos(\varphi )+j\,\sin(\varphi )}}={\frac {\cos(\varphi )+jQ\sin(\varphi )}{Q\cos(\varphi )+j\,\sin(\varphi )}}$.

${\frac {u_{out}}{u_{0}}}=\cos(\varphi )-j{\frac {Z_{0}}{Z_{in}}}\sin(\varphi )=\cos(\varphi )-j\left({\frac {\cos(\varphi )+jQ\sin(\varphi )}{Q\cos(\varphi )+j\,\sin(\varphi )}}\right)\sin(\varphi )=$,

$={\frac {Q\cos ^{2}(\varphi )+j\,\sin(\varphi )\cos(\varphi )-j\,\cos(\varphi )\sin(\varphi )+Q\sin ^{2}(\varphi )}{Q\cos(\varphi )+j\,\sin(\varphi )}}={\frac {Q}{Q\cos(\varphi )+j\,\sin(\varphi )}}$,

Voor ohmse belasting:

$|{\frac {u_{out}}{u_{0}}}|=|{\frac {1}{\cos(\varphi )+j{\frac {1}{Q}}\sin(\varphi )}}|={\frac {1}{\sqrt {\cos ^{2}(\varphi )+{\frac {1}{Q^{2}}}\sin ^{2}(\varphi )}}}={\frac {1}{\sqrt {1+({\frac {1}{Q^{2}}}-1)\sin ^{2}(\varphi )}}}$,

Als

$Q^{2}>1$,

dus:

$Z_{D}>Z_{0}$,

dan

$|{\frac {u_{out}}{u_{0}}}|>1$.

Voor inductieve belasting $L_{o}$:

${\frac {u_{out}}{u_{0}}}={\frac {Q}{Q\cos(\varphi )+j\,\sin(\varphi )}}={\frac {j\omega L_{o}}{j\omega L_{o}\cos(\varphi )+j\,Z_{0}\sin(\varphi )}}={\frac {1}{\cos(\varphi )+{\frac {Z_{0}}{\omega L_{o}}}\,\sin(\varphi )}}={\frac {\cos(\alpha )}{\cos(\alpha -\varphi )}}$,

waarin

$\tan(\alpha )={\frac {Z_{0}}{\omega L_{o}}}$.

Voor $0<\alpha <{\tfrac {1}{2}}\pi$ en $\varphi <\alpha$ is

${\frac {u_{out}}{u_{0}}}={\frac {\cos(\alpha )}{\cos(\alpha -\varphi )}}<1$.