cert_exam:tech_radio_op_1:b:freq:h-feeder_impedance
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| cert_exam:tech_radio_op_1:b:freq:h-feeder_impedance [2019/12/31 10:49] – 作成 fu | cert_exam:tech_radio_op_1:b:freq:h-feeder_impedance [2020/01/01 01:45] (current) – fu | ||
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| ======無損失給電線の負荷側を見たインピーダンス====== | ======無損失給電線の負荷側を見たインピーダンス====== | ||
| + | 伝送路の電圧節では、電圧が最小となり、電流が最大となるので、 | ||
| + | 入射波の電圧がVi, | ||
| + | \[ | ||
| + | V_{min} = V_i - V_r | ||
| + | \] | ||
| + | 電圧節での電流は最大となるので, | ||
| + | \[ | ||
| + | I_{max} = \frac{V_i + V_r}{Z_0} | ||
| + | \] | ||
| - | * | + | これより、電圧節から負荷側をみた場合のインピーダンスは |
| + | \[ | ||
| + | Z = \frac{V_{min}}{I_{max}} = \frac{V_i - V_r}{V_i + V_r} Z_0 | ||
| + | \] | ||
| + | |||
| + | H29/ | ||
| + | |||
| + | |||
| + | 定在波比がSの時、入射はが$V_i$, | ||
| + | \[ | ||
| + | S = \frac{V_{max}}{V_{min}}= \frac{V_i+V_r}{V_i-V_r} | ||
| + | \] | ||
| + | なので、電圧波節からみた負荷のインピーダンスは | ||
| + | \[ | ||
| + | Z = \frac{Z_0}{S} | ||
| + | \] | ||
| + | で計算できる。 | ||
| ===== 出題履歴 ===== | ===== 出題履歴 ===== | ||
cert_exam/tech_radio_op_1/b/freq/h-feeder_impedance.1577756966.txt.gz · Last modified: by fu
