Eletromagnetismo Aplicado 7 Exemplo Incidencia Obliqua
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Transcript of Eletromagnetismo Aplicado 7 Exemplo Incidencia Obliqua
Uma onda incide do ar em silício com ângulo de 37o. Se a onda transmitida tem polarização circular (mão esquerda). Obtenha as expressões para os campos ( ),iE r t
( )
x
e e determine a polarização das ondas incidentes e refletidas.( )i( ),rE r t
37°0 0,ε μ
021 0
2 0377μμη ηε ε= = = =
z
0 0,ε μ
z
y0 011,56 ,ε μ0ημμy
( )
0022
2 0110,8811,56 3, 4
ημμη ε ε= = = =
( ), 1 V/mtE r t =
( )( )( )( )
1 211
r r tx ixk kR
k kμ μ−
=( )( )( )( )
1 211
r r tx ixk kR
k kε ε−
=
TE TM
( )( )1 21 r r tx ixk kμ μ+
( )( )2
1T
k k=
( )( )1 21 r r tx ixk kε ε+
( )( )2T =( )( )1 21 r r tx ixk kμ μ+ ( )( )1 21 r r tx ixk kε ε+
0 0
37 0 8ik c
k
ω ε μ ω= =0 011,56 3, 4tk cω ε μ ω= =
0 0
0 0
cos37 0,8
sen 37 0,6ix
iz
k c
k c
ω ε μ ω
ω ε μ ω
= =
= = ( ) ( )( )1 0
sen 37 sen0,6 3, 4 sen
i t t
t
k kc c
θ
ω ω θ
θ
=
=
11 56 3 4k ( )1 0sen 0,6 3, 4 10,16tθ−= =
( )( )
0 011,56 3, 4
3, 4 cos10,16 3,34
3 4 sen10 16 0 6
t
tx
k c
k c c
k c c
ω ε μ ω
ω ω
ω ω
= =
= =1 3,34 /1 cω
−( )3, 4 sen10,16 0,6tzk c cω ω= =
3,34 /10 8 /TE
cc
ωω
−
111,56 0,8 / 0, 4691 3,34 /111,56 0,8 /
TM cR cc
ωωω
−= =
+
2
2 1, 4693,34 /1
TMT cω= =+
0,8 / 0,6143,34 /10,8 /
TE cR cc
ωωω
= = −+
, ,
2 0,3863,34 /10,8 /
TET cc
ωω
= =+
111,56 0,8 / cω
+×
Campos Elétricos transmitidos Onda TEOnda TM
TMtE TM
tETE
tETE
tE TMtH
tE
TEtH
+2 110,88η =
ˆ ˆ0.17 0.98TMtE x z= + Circular da mão esquerda
ˆ0,009TMtH y= ( ) /2ˆ ˆ0.17 0.98TM j
tE x z e π−= +
( ) /2ˆ0,009TM jtH y e π−=
ˆ1TEE y=
( )0,009tH y e
ˆ1TEtE y=
ˆ ˆ0.0015 0.0088TEtH x z= − −
1tE y=ˆ ˆ0.0015 0.0088TE
tH x z= − −
t y
( ) /2ˆ0 009TM jH y e π−= ( ) ( )/2
/2ˆ0,009ˆ0 006
jTMTM jt y eHH y e
ππ
−−= = =( )0,009tH y e= ( )0,006
1, 469j
i TMH y eT
= = =
TME ( ) /2ˆ ˆ0,006 377 0,6 0,8TM jiE x z e π−= × × +
iE
TMiH
( )( ) /2ˆ ˆ1,35 1,8TM j
iE x z e π−= +
0 8ω
1 377η =
iH
( )1 ˆ ˆ0,8 0.6k x zcω
= − +
0,8ixk
cω−
=
0,6ik ω=
ˆ1TEtE y= ˆ1 ˆ2,59
0,386
TETE t
i TE
E yE yT
= = =
izkc
( ) /2ˆ ˆˆ1,35 1,8 2,59TM TE ji i iE E E x z e yπ−= + = + +
( ) ( ) ( ) ( ) ( )1 1 1 1ˆ ˆ ˆ ˆˆ ˆ ˆ, 1,35 1,8 cos / 2 2,59 cosi x z x zE r t x z t k x k z y t k x k zω π ω= + − − − + − −
( ) /2ˆ0 006TM jH y e π− ( ) ( )/2 /2ˆ ˆ0 469 0 006 0 0028TM TM TM j jH R H y e y eπ π− −= = =( )0,006 jiH y e= ( ) ( )0, 469 0,006 0,0028r iH R H y e y e= = =
TME( ) /2ˆ ˆ0,0028 377 0,6 0,8TM j
rE x z e π−= × × −TM
rE
TMH
( )( ) /2ˆ ˆ0,63 0,84TM j
rE x z e π−= −
0 8ωrH
( )ˆ ˆ0,8 0.6rk x zcω
= +
0,8rxk
cω
=
0,6k ω=
ˆ ˆ0,61 2,59 1,525TE TE TEr iE R E y y= = − × = −
rzkc
ˆ2,59TEiE y=
( ) /2ˆ ˆˆ0,63 0,84 1,525TM TE jr r rE E E x z e yπ−= + = − −
( ) ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆˆ ˆ ˆ, 0,63 0,84 cos / 2 1,525 cosr rx rz rx rzE r t x z t k x k z y t k x k zω π ω= − − − − − − −