SuperminarenandClaude Sonnet 5.5
f7fa8cfbc1
- Sections and slides rewritten in Swedish; German law and institutions replaced by Swedish/CEPT/ITU equivalents (PTS, SSA, PTSFS 2025:1, SSMFS 2008:18), unverified statements marked "% TODO VERIFY" - New toc: INSTEG, HAREC, UPPGRADERING; section idents are Swedish (see ident-map.tsv) - Exam-question tags, BNetzA question pool, solutions and the German SWL course removed - Static pages, snippets, applets, READMEs and image alt-texts translated - TikZ labels translated; SVG drawings regenerated (drawing 689 still fails to build upstream: it references a missing photo) - Changed from the original: not endorsed by DARC e.V., SSA or PTS Co-Authored-By: Claude Sonnet 5.5 <[email protected]>
92 lines
2.8 KiB
Markdown
92 lines
2.8 KiB
Markdown
## Fältvågimpedans och fältstyrkor
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* Fältvågimpedans i vakuum:
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* $Z_{F0} = \sqrt{\dfrac{\mu_0}{\varepsilon_0}}$
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* $\mu_0$ är den magnetiska fältkonstanten, $\varepsilon_0$ den elektriska fältkonstanten
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---
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* I ett medium (t.ex. luft) gäller:
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* $Z_{F} = \sqrt{\dfrac{\mu}{\varepsilon}}$
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* Elektrisk och magnetisk fältstyrka beror på mediets vågimpedans
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---
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### Effekt vid antennens matningspunkt
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* Antennens ingångseffekt fås ur sändareffekten minus kabeldämpningen
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* Kabelförluster beaktas som dämpningsfaktor (t.ex. $\qty{10}{\dB}$ → $\num{0,1}$)
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* Formel: $P_{Ant} = D \cdot P_{Sändare}$
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---
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## Maximal sändareffekt
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* Skyddsavståndet måste ligga i fjärrfältet:
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<fragment>
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$d > \dfrac{\lambda}{2\pi}$
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</fragment>
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--- style="font-size: smaller;"
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#### Räkneexempel 1
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<left>
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* givet: $g_d = \qty{6}{\dBd}$
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* givet: $E = \qty{28}{\volt\per\meter}$
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</left>
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<right>
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* givet: $d = \qty{5}{\meter}$
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* sökt: $P_S$
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</right>
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<fragment>
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<left>
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$\begin{split}E &= \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{d}\\ \Rightarrow P_{EIRP} &= \frac{(E \cdot d)^2}{\qty{30}{\ohm}}\\ &= \frac{(\qty{28}{\volt\per\meter} \cdot \qty{5}{\meter})^2}{\qty{30}{\ohm}}\\ &\approx \qty{653}{\watt}\end{split}$
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</left>
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</fragment>
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<fragment>
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<right>
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$\begin{split}P_{EIRP} &= P_S \cdot 10^{\frac{g_d - a + 2,15dB}{\qty{10}{\dB}}}\\ \Rightarrow P_S &= \frac{P_{EIRP}}{10^{\frac{g_d - a + 2,15dB}{\qty{10}{\dB}}}}\\ &= \frac{\qty{653}{\watt}}{10^{\frac{\qty{6}{\dBd} - \qty{0}{\dB} + 2,15dB}{\qty{10}{\dB}}}}\\ &\approx \frac{\qty{653}{\watt}}{6,53}\\ &= \qty{100}{\watt}\end{split}$
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</right>
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</fragment>
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---
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--- style="font-size: smaller;"
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#### Räkneexempel 2
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<left>
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* givet: $g_i = \qty{12,15}{\dBi}$
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* givet: $P_A = \qty{250}{\watt}$
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</left>
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<right>
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* givet: $d = \qty{30}{\meter}$
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* sökt: $E$
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</right>
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<fragment>
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<left>
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$\begin{split}G_i &= 10^{\frac{g_i}{\qty{10}{\dB}}}\\ &= 10^{\frac{\qty{12,15}{\dBi}}{\qty{10}{\dB}}}\\ &\approx 16,4\end{split}$
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</left>
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</fragment>
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<fragment>
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<right>
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$\begin{split}E &= \frac{\sqrt{\qty{30}{\ohm} \cdot P_A \cdot G_i}}{d}\\ &= \frac{\sqrt{\qty{30}{\ohm} \cdot \qty{250}{\watt} \cdot 16,4}}{\qty{30}{\meter}}\\ &\approx \frac{\qty{350}{\volt}}{\qty{30}{\meter}}\\ &\approx \qty{11,7}{\volt\per\meter}\end{split}$
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</right>
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</fragment>
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---
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--- style="font-size: smaller;"
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#### Räkneexempel 3
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* givet: $P_{ERP} = \qty{100}{\watt}$
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* givet: $d = \qty{100}{\meter}$
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* sökt: $E$
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<fragment>
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<left>
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$\begin{split}P_{EIRP} &= P_{ERP} \cdot 1,64\\ &= \qty{100}{\watt} \cdot 1,64\\ &= \qty{164}{\watt}\end{split}$
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</left>
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</fragment>
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<fragment>
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<right>
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$\begin{split}E &= \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{d}\\ &= \frac{\sqrt{\qty{30}{\ohm} \cdot \qty{164}{\watt}}}{\qty{100}{\meter}}\\ &\approx \qty{0,7}{\volt\per\meter}\end{split}$
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</right>
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</fragment>
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