From 5202d7912e35116b3c1906cdcba9480a16ca8a7c Mon Sep 17 00:00:00 2001 From: Ralf Mahler Date: Tue, 26 May 2026 17:41:24 +0200 Subject: [PATCH] Update AK105.md --- contents/solutions/AK105.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/contents/solutions/AK105.md b/contents/solutions/AK105.md index a02ff01..37bae8b 100644 --- a/contents/solutions/AK105.md +++ b/contents/solutions/AK105.md @@ -16,7 +16,7 @@ $d = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E}$ # 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