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Update AK105.md

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Ralf Mahler
2026-05-26 17:41:24 +02:00
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@@ -16,7 +16,7 @@ $d = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E}$
# Einsetzen # Einsetzen
$d_{{40}\degree} = \frac{\sqrt{30 \Omega \cdot P_{{40}\degree}}}{E} = \frac{\sqrt{30 \Omega \cdot 0,25 \cdot P_{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{30 \Omega \cdot P_{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{30 \Omega \cdot P_{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{30 \Omega \cdot P_{EIRP}}}{E} = 0,5 \cdot d$ $d_{{40}\degree} = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{{40}\degree}}}{E} = \frac{\sqrt{\qty{30}{\ohm} \cdot 0,25 \cdot P_{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = 0,5 \cdot d$
# Interpretation der Formel # Interpretation der Formel