Diplomska naloga (.pdf)
Diplomska naloga (.pdf)
Diplomska naloga (.pdf)
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2.3 Nekaj enostavnih primerov 16<br />
5<br />
0<br />
odbojnost (dB)<br />
−5<br />
−10<br />
−15<br />
−20<br />
−25<br />
FM2<br />
FM4<br />
FM6<br />
0 2 4 6 8 10 12 14 16 18<br />
frekvenca (GHz)<br />
Slika 2.8: Graf odbojnosti za enostavne absorberje z materiali FM2, FM4 in FM6<br />
pri debelini 10 mm.<br />
naslednje vrednosti:<br />
Dobimo:<br />
ν = 4 · 10 9 s −1 , ε = 10 − 0.1i, µ = 1.1258 − 1.144i.<br />
R =<br />
. √<br />
i<br />
√<br />
1.1258−1.144i<br />
i<br />
10−0.1i<br />
tan(d · (3.0913 · 10 2 − 1.3132 · 10 2 i)) − 1<br />
.<br />
1.1258−1.144i<br />
10−0.1i<br />
tan(d · (3.0913 · 10 2 − 1.3132 · 10 2 i)) + 1<br />
Ko vstavimo želeno debelino, dobimo α, ki je kompleksno število, zato tangens<br />
izračunamo s formulo:<br />
sin(2Re(α))<br />
tan(Re(α) − iIm(α)) =<br />
cos(2Re(α)) + cosh(2Im(α)) − i sinh(2Im(α))<br />
cos(2Re(α)) + cosh(2Im(α)) .<br />
Pri debelini d 1 = 2 · 10 −3 m je α 1<br />
. = 0.6183 − 0.2626i ter tan(α1 ) . = 0.6430 − 0.3742i,<br />
pri debelini d 2 = 5 · 10 −3 m je α 2<br />
. = 1.5457 − 0.6566i ter tan(α2 ) . = 0.0504 − 1.7336i in<br />
pri debelini d 3 = 10 · 10 −3 m je α 3<br />
. = 3.0913 − 1.3132i ter tan(α3 ) . = −0.0126 − 0.8656i.<br />
Dobimo vrednosti odbojnih koeficientov:<br />
R 1<br />
. =<br />
(0.1556 + 0.3714i)(0.6430 − 0.3742i) − 1<br />
(0.1556 + 0.3714i)(0.6430 − 0.3742i) + 1<br />
R 2<br />
. =<br />
(0.1556 + 0.3714i)(0.0504 − 1.7336i) − 1<br />
(0.1556 + 0.3714i)(0.0504 − 1.7336i) + 1<br />
R 3<br />
. =<br />
(0.1556 + 0.3714i)(−0.0126 − 0.8656i) − 1<br />
(0.1556 + 0.3714i)(−0.0126 − 0.8656i) + 1<br />
.<br />
= −0.5806 + 0.2304i,<br />
.<br />
= −0.1835 − 0.1799i,<br />
.<br />
= −0.4990 − 0.1583i.