Linia długa.
: 15 mar 2018, o 23:41
Obliczyć impedancję \(\displaystyle{ Z _{we}}\)
Link do rysunku:
Dane:
\(\displaystyle{ Z_{o} = 75 [ \Omega]}\)
\(\displaystyle{ Z_{2}= Z_{L}= 75j [ \Omega]}\)
\(\displaystyle{ Z_{w}(z=- \frac{1}{4} \lambda) = Z_{o} \frac{Z_{L}+jZ_{o}tan( \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}) }{Z_{o}+jZ_{L}tan( \frac{2 \pi}{\lambda} \frac{\lambda}{4} )} = Z_{o} \frac{jZ_{o}}{jZ_{L}}= \frac{Z_{o} ^{2} }{Z_{L}} = -75j [ \Omega]}\)
\(\displaystyle{ Z_{L2}= \frac{(-75j)(75j)}{-75j+75j} = \infty \Rightarrow}\) rozwarcie
\(\displaystyle{ Z_{we} = Z_{w}(z=- \frac{\lambda}{8} ) = -jZ_{o}cot(\beta l ) = -j Z_{o}cot( \frac{2 \pi}{\lambda} \frac{\lambda}{8} ) = -j75 [ \Omega]}\)
Proszę uprzejmie o sprawdzenie moich obliczeń.
Link do rysunku:
Dane:
\(\displaystyle{ Z_{o} = 75 [ \Omega]}\)
\(\displaystyle{ Z_{2}= Z_{L}= 75j [ \Omega]}\)
\(\displaystyle{ Z_{w}(z=- \frac{1}{4} \lambda) = Z_{o} \frac{Z_{L}+jZ_{o}tan( \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}) }{Z_{o}+jZ_{L}tan( \frac{2 \pi}{\lambda} \frac{\lambda}{4} )} = Z_{o} \frac{jZ_{o}}{jZ_{L}}= \frac{Z_{o} ^{2} }{Z_{L}} = -75j [ \Omega]}\)
\(\displaystyle{ Z_{L2}= \frac{(-75j)(75j)}{-75j+75j} = \infty \Rightarrow}\) rozwarcie
\(\displaystyle{ Z_{we} = Z_{w}(z=- \frac{\lambda}{8} ) = -jZ_{o}cot(\beta l ) = -j Z_{o}cot( \frac{2 \pi}{\lambda} \frac{\lambda}{8} ) = -j75 [ \Omega]}\)
Proszę uprzejmie o sprawdzenie moich obliczeń.