OnPlanas en Matlab

H 

|ρ| =

n1 > n 2

1

0 < ρ < 1

||

n1 < n 2



|ρ| = 0

2 E  = 0

(

2



+ κ2 )E  = 0

κ

  

  

∂ 2 ∂ 2 ∂ 2 + 2 + 2 + κ2 E x = 0 2 ∂x ∂y ∂z ∂ 2 ∂ 2 ∂ 2 + 2 + 2 + κ2 E y = 0 2 ∂x ∂y ∂z ∂ 2 ∂ 2 ∂ 2 + + + κ2 E z = 0 2 2 2 ∂x ∂y ∂z

E 

x





∂ 2 ∂ 2 ∂ 2 + 2 + 2 + κ2 E x = 0 2 ∂x ∂y ∂z E 





∂ 2 ∂ 2 + 2 + κ2 E x = 0 2 ∂y ∂z z

E  =

E 0 (z )ax

d2 E x + κ2 E x = 0 2 dz

E x (z ) = C 1 e− jκz + C 2 e jκz

∂  E  ∂x x

=0

C 1

C 1,2 = C 1,2 e jϕ 1,2

| |

C 2 E (t, z )

= [ C 1 cos(ωt cos(ωt

− κz + ϕ ) + |C  | cos(ωt cos(ωt + κz + ϕ )]a |C  | cos(ωt cos(ωt − κz + ϕ ) |C  | cos(ωt cos(ωt + κz + ϕ )

| |

1

2

1

z

1

x

1

1

1

z ∆z/∆ z/ ∆t

∆t ) ω(t + ∆t

− κ(z + ∆z ∆z ) ν  p =

∆ ω ω = = = ∆t κ ω µε

√

ν  p = √ µ10 ε0 = c = 299792458 ∆z

κ=

ν  p ωt

− κz =

√1µε ν  p 2π λ κ∆ = κλ = 2π

2π λ

z = ctte.

H 

×E  − ωµ

H  = H 

= =

−



−

−

H  =

 

× 

1 ∂  ∂  ∂  ax + ay + az  jωµ  jω µ ∂x ∂y ∂z 1 (  jκ)  jκ ) C 1 e− jκz C 2 e jκz  jωµ  jω µ



(az

−

E )

 × µ ε

= az

E 

ay

× E η

η = µ/ε 120π 120π = 377 Ω

η = η0 =

C 1 e− jκ ax C 1 − jκ H  = e ay η E  =

z

 

µ0 /ε0 =

0

ε = ε β π/2

≤ ≤

−  jε = |ε|−

µ = µ

−  jµ  = |µ|e− κ = ±ω |ε|− |µ|e− − j sin κ = ±ω |ε||µ| cos κ = κ − jκ   jα

 jβ

         α+β

α+β

2

2

 jα

 jβ

0

=

α π/2 α β −  j + ε µe 2

≤ ≤   ± | || | ω

κ , κ > 0 







E x (z) = C 1 e−κ z e− jκ z + C 2 eκ z e jκ z

E (t, z)



= [ C 1 e−κ z cos(ωt

| |

κ z

− κz + ϕ ) + |C  |e 1

2

cos(ωt + κ z + ϕ1 )]ax κ

κ L





E (z) = κ ∆z L = ln E (z + ∆z)





E (z) = 20 log eκ L = 20 log E (z + ∆z)



∆z

= κ ∆z20loge δ

δ = (κ )−1 



C 1 e−κ z e− jκ z ax C 1 −κ z − jκ z H  = e e ay η E  =



η=

φ = (β

− α)/2

 | |  | | µ e− jβ = ε e− jα

|| −45◦ ≤ φ ≤ 45◦



µ e ε

||

= C 1 e−κ z cos(ωt C 1 −κ z cos(ωt e H(t, z) = η E (t, z)

| | | | ||





−j

(β −α) 2

= η e− jφη

||

− κz + ϕ )a − κz + ϕ + φ )a 1

1

x

η

y

µ

 × H  = J  + ωεE   × H   j 

ε∼ = ε

tg∆ =



0

= [ jω(ε

 jε  ) + σ]E  σ  j ε + ω

 −−  

= jω ε

= jωεE  ε = |ε|e−

ε = 0

µ

σ ε + ω ε

σ =0



E 





=

σ 



1, 1,



σ ωε

; . σ =0

=0  

  

   − −     −     ···      ≈ −     ≈

κ=ω =ω

µ (

  )

    

µ 1

1   = ω µ 1    + 2 8   ω µ 1    2 µ 

η

2

1 + 16

 

3

 1+  2

 = 0

  −     −   √ − ≈   − ≈ −

κ=ω = =

σ ω σ ω 2 µ    ω ωµ (ω  σ)  ωµσ µ   

ωµσ (1 2

(1 + x)n = 1 + nx +

n(n−1)

2!

x2 +

n(n−1)(n−2)

3!

x3 + . . .

+

 )

n

σ

 ω

η

κ=κ



  ≈

ωµ (1 + ) 2σ

− κ

η=



κ ω µ ω µ0  ω µ ω µ

κ ω µ ω µ0  0

ϕη

R + X  = |η|e

 √√   √√   R    X     √            √         ω

2

ωµσ

2

ω ω

µ  µ0 

µ  µ  ωµ 2σ

µ  ωµσ

2

ω ω 0



2

µ  µ0 

µ  ωµ 2σ

x z



Ae− jκz ax A − jκz H  = ay e η E  =



αi βi



αi βi

γ i

i



γ i

i

∈ {1, 2, 3}

∈ {1, 2, 3}

Ae− jκ(x cos γ 1 +y cos γ 2+z cos γ 3 ) (cos α1 ax + cos α2 ay cos α3 az ) A − jκ (x cos γ 1+y cos γ 2 +z cos γ 3) H  = (cos β1 ax + cos β2 ay + cos β3 az ) e η E  =

x cos γ 1 + y cos γ 2 + z cos γ 3 = ctte. κ κ

= κ(cos γ 1 ax + cos γ 2 ay + cos γ 3 az )

r

r

·r

= xax + y ay + z az z

E (t)

= F −1

{E (ω)}

E (ω) ∞

1  jωt E (ω)e dω E (t) = 2π −∞

ˆ 

E (t)

∆ω ω0

E (t) κ = κ(ω)

E (t, z)

E (t, z)

= Re

 ˆ 

ω0 + ∆2ω

1 π

ω0

− ∆2ω

z

E (ω

κ

0

−

dω

κ(ω) κ0 = κ(ω0 ) dκ κ(ω) = κ0 + dω

E (t, z)

 j [ωt κ(ω )z ]

− ω )e



= Re

 ˆ 

ω0 + ∆2ω

1 π

ω0

− ∆2ω

E (ω



(ω

−ω )+ 0

ω0

 

 j ωt

− ω )e 0

κ0 +

−

|

dκ (ω dω ω 0



−ω0 )+

z

dω



∆ω

E (t, z)

= Re = Re

 ˆ   ˆ  1 π 1 π

ω0 + ∆2ω

ω0

−

∆ω 2

ω0 + ∆2ω

ω0

− ∆2ω

E (ω

E (ω

F −1 X (ω)e− ωt0

{

ν gr E (t, z)

 j ωt

− ω )e 0

 jω (t

0

=

E 

t

|

dκ (ω dω ω 0

|

dκ z) dω ω 0

1

dωe



−ω0 )

− j



κ0

z

−

dω

|



dκ ω0 dω ω 0



z

} = x(t − t ) F − {X (ω − ω )} = x(t)e dκ/dω | ν  = dω/dκ 0

 −  −  E 

κ0 +

−

−

− ω )e ω0

= Re

 

t

z ν gr

z ν gr

 ω0 t

0

gr

 jω 0 (t

e

cos(ω0 t

− νgrz )e− j (κ0− νωgr0 )z

− κ z) 0





z ν gr

ν gr ν gr κ(ω)

(ε1 , µ1 )

(ε2 , µ2 )

1 E 1 H 1



2 E 2 H 2

= C 3 e− κ2 z ax C 3 = e− κ2 z ay η2 1

◦

E 

◦

H 

= C 1 e− κ1z ax C 1 − κ1 z = e ay η1

1 −

E 

−

H 

= C 2 e κ1 z ax C 2  κ1 z = e ay η1

−



= C 1 e− κ1z + C 2 e κ1 ax 1 = C 1 e− κ1 z C 2 e κ1 ay η1



−



2 = C 3 e− κ2z ax C 3 − κ2 z ay = e η2

+

E 

+

H 

E 1 (0) = E 2 (0) H 1 (0) = H 2 (0)

C 1 + C 2 = C 3 C 1 C 2 C 3 = η1 η1 η2

−

ρ

E − (0) C 2 = ρ= ◦ E  (0) C 1 E + (0) C 3 = τ  = ◦ E  (0) C 1

1 + ρ = τ  1 ρ τ  = η1 η1 η2

−

−

η2 η1 η1 + η2 2η2 τ  = η1 + η2 ρ=

1 E 1 H 1

 

 

= C 1 e− κ1 z + ρe κ1z C 1 − κ1 z = e ρe κ1z η1

2 E 2 H 2

−

= C 1 τ e− κ2 z ax C 1 τ  − κ2z ay = e η2

ax ay

τ 

C 1

2 2

1

1.5 1 0.5 ρ

0

1

τ

−0.5

η  / η 2

−1 0

5

10

15

ρ

1

η2 /η1

20

ρ

τ 

−1

η2 /η1

1

−1 ≤ ρ ≤

|ρ| = 1

0

τ 

1

1

|ρ| = 0

0< ρ n2 ϕ n12

ϕ

↑⇒ θ ↑

ϕc ϕc = sin−1

cos θ =

  ± − 1

ϕ > ϕc θ sin2 θ

   ±

cos θ =  j

n12

n1 n2



↑

sin θ > sin ϕ θ = 90◦

⇒θ↑

n2 n1

sin θ = (n1 /n2 )sin ϕ > 1

2

sin2 ϕ

− 1 = ± jα(ϕ)

ϕc

xy

xy

ϕ > ϕc

ϕ = ϕc

xy ϕ > ϕc

n1 > n 2

β(ϕ) = sin θ =

n1 n2

sin ϕ

± jα(ϕ) f + (y, z) = e−κ2 zα(ϕ) e− jκ2 yβ (ϕ)

f + n1 < n2 n1 < n2

n1 sin θ n1 = sin ϕ n2

n

1⇒

2

n1 n2

→0⇒θ→0

o

E 

o

H 

−

E 

−

H 

+

E 

+

H 

A B

= Be − jκ 1(y sin ψ+z cos ψ) ae B = e− jκ1 (y sin ψ+z cos ψ) ah η1

−

−

= Ce− jκ2 (y sin θ+z cos θ) ae+ C  = e− jκ 2 (y sin θ+z cos θ) ah+ η2

C  aeo,−,+

aho,−,+

aen ahn

n

= Ae− jκ 1(y sin ϕ+z cos ϕ) ae ◦ A = e− jκ1 (y sin ϕ+z cos ϕ) ah ◦ η1

= cos α1n = cos β1n

+ cos α2n n ax + cos β2 ax

+ cos αn3 n ay + cos β3 ay

az az

∈ {◦, −, +} o

E 

o

H 

= Ae− jκ 1 (y sin ϕ+z cos ϕ) ax A = e− jκ1 (y sin ϕ+z cos ϕ) (cos ϕay η1

− sin ϕa ) z

x

, ,+

y

◦ −

α1◦ α2◦ α3◦ β1◦ β2◦ β3◦

−

α− 1 α− 2 α− 3 − β1 β2− β3−

0◦ 90◦ 90◦ 90◦ ϕ ◦ 90 + ϕ

,

,+

◦ −

0◦ α1+ 90◦ α2+ 90◦ α3+ 90◦ β1+ 180◦ ϕ β2+ 270◦ ϕ β3+

0◦ 90◦ 90◦ 90◦ θ ◦ 90 + θ

− −

= Be − jκ1 (y sin ψ+z cos ψ) ax B − jκ 1 (y sin ψ+z cos ψ) = (cos ϕay + sin ϕaz ) e η1

E 

−

−

H 

+

E 

+

H 

= Ce− jκ 2(y sin θ+z cos θ) ax C  = e− jκ 2 (y sin θ+z cos θ) (cos θay η2

− sin θa )

E ◦ (y, 0) + E − (y, 0) = E + (y, 0) H y◦ (y, 0) + H y− (y, 0) = H y+ (y, 0)

1 (A η1

−

ρ⊥

A + B = C  1 B)cos ϕ = C cos θ η2 τ ⊥

E − (y, 0) B = ρ⊥ = ◦ E  (y, 0) A E + (y, 0) C  = τ ⊥ = ◦ E  (y, 0) A κ1 sin ϕ = κ1 sin ψ = κ2 sin θ

z

1 (1 η1

−

1 + ρ⊥ = τ ⊥ 1 ρ⊥ )cos ϕ = τ ⊥ cos θ η2

η2 cos ϕ η1 cos θ η2 cos ϕ + η1 cos θ 2η2 cos ϕ τ ⊥ = η2 cos ϕ + η1 cos θ

−

ρ⊥ =

sin ψ = sin ϕ cos ψ = cos ϕ

−

z0 E 2 H 2







 

= E + = Aτ ⊥ e− jκ2 (y sin θ+z cos θ) ax A = H + = τ ⊥ e− jκ 2(y sin θ+z cos θ) (cos θay η2

E 

− sin θa ) z

H ⊥

⊥

a

◦,−,+

e ◦ ◦ ,−,+ ,−,+ α1,2,3 β1,2,3

o

E 

o

H 

H 

ah◦,−,+

= Ae− jκ1 (y sin ϕ+z cos ϕ) (cos ϕay A − jκ1 (y sin ϕ+z cos ϕ) ax = e η1

−



−

x y

z

− sin ϕa ) z

−E 

⊥

x

, ,+

y

◦ −

α1◦ α2◦ α3◦ β1◦ β2◦ β3◦

−

90◦ ϕ ◦ 90 + ϕ 180◦ 90◦ 90◦

α− 1 α− 2 α− 3 − β1 β2− β3−

−

,+

90◦ α1+ 180◦ ϕ α2+ 270◦ ϕ α3+ 180◦ β1+ 90◦ β2+ 90◦ β3+

− −

 jκ 1 (y sin ψ +z cos ψ )

−Be− B = − e− η =

E 

,

◦ −

(cos ϕay + sin ϕaz )

 jκ 1 (y sin ψ +z cos ψ )

H 

90◦ θ ◦ 90 + θ 180◦ 90◦ 90◦

ax

1

+

E 

+

H 

= Ce− jκ2 (y sin θ+z cos θ) (cos θay C  − jκ 2 (y sin θ+z cos θ) ax = e η2

− sin θa )

−

E y◦ (y, 0) + E y− (y, 0) = E y+ (y, 0) H ◦ (y, 0) + H − (y, 0) = H + (y, 0)

(A

− B)cos ϕ = C cos θ

1 C  (A + B) = η1 η2 ρ

τ 

E − (y, 0) B = ρ = ◦ E  (y, 0) A + E  (y, 0) C  = τ  = ◦ E  (y, 0) A

−

κ1 sin ϕ = κ1 sin ψ = κ2 sin θ

z

η2 cos ϕ η1 cos θ η2 cos ϕ + η1 cos θ η2 cos θ η1 cos ϕ ρ = η2 cos θ + η1 cos ϕ

2η2 cos ϕ η2 cos ϕ + η1 cos θ 2η2 cos ϕ τ  = η2 cos θ + η1 cos ϕ

−

ρ⊥ =

τ ⊥ =

(1 + ρ )cos ϕ = τ  cos θ 1 1 (1 ρ ) = τ  η1 η2

−

η2 cos θ η1 cos ϕ η2 cos θ + η1 cos ϕ 2η2 cos ϕ τ  = η2 cos θ + η1 cos ϕ

−

ρ =

sin ψ = sin ϕ cos ψ = cos ϕ

−

z0 E 2 H 2

= E + = Aτ  e− jκ 2 (y sin θ+z cos θ) (cos θay A − jκ 2 (y sin θ+z cos θ) = H + = τ  e ax η2

− sin θa ) z

−

ϕB η2 cos θ = η1 cos ϕB

ρ = 0 n1 cos θ = n2 cos ϕB

n1

n2

sin θ = sin ϕB θ = π2 ϕB

n1 n2

sin θ cos θ = sin ϕB cos ϕB

−

n1 cos θ = n2 sin ϕB ρ = 0 n1 sin ϕB = n2 sin θ

sin2 θ + cos2 θ = 1 = sin 2 ϕB + cos2 ϕB sin2 θ

cos2 θ n21 sin θ = 2 sin ϕB n2 n22 2 cos θ = 2 cos ϕB n1 2

n21 n22 sin ϕB + 2 cos ϕB = sin2 ϕB + cos2 ϕB 2 n2 n1



n2 ϕB = tan−1 n1

1 η2

2

η1

σ2

→0 ρ⊥ = ρ = 

−1 −1

2 E  = H  =

κT  = κ1 cos ϕ

τ ⊥ = 0 τ  = 0 

1

− j2A sin(κ

− jκ y ax

T z)e



2A cos(κT z)e− jκ y cos ϕay + j sin(κT z)e− jκ  y sin ϕaz η1

κ = κ1 sin ϕ



→∞⇒

3.5

3

2.5

2

1.5

1

0.5

0 0

2

4

6

8

10

12

14

16

18

20

zy

E (y , z , t)

= 2A sin(κT z) sin(ωt κ y)ax 2A [cos(κT z)cos(ωt κ y)cos ϕay H(y , z , t) = η1 sin(κT z)sin(ωt κ y)sin ϕaz ]

− −

−

κT 

−

ΛT  =

κ

2π κT 

  −    −    − 

2π z sin ωt ΛT  2π H(y , z , t) = H y cos z cos ωt ΛT  2π H z sin z sin ωt ΛT  E (y , z , t)

= E x sin

−

Λ = 2π y Λ 2π y Λ 2π y Λ

2π κ

ax ay

az

S 1,2 sin Λ2πT  z

z

H

y

ϕ

x

κT,

ΛT,

cos Λ2πT  z

E  H

y

κT 

↑

ΛT 

↑

↓

ϕ 0 < ϕ <

π

2

y y

λT /2 y

ΛT  D =

n Λ2T 

D n = 1, 2, 3 . . .

E 

H

z

2λ

3.5

3

2.5

2

1.5

1

0.5

0 0

2

4

6

8

10

12

14

16

18

20

zy

κT D = nπ nπ κ1 cos ϕ = D nc cos ϕ = 2Df  n = 1, 2, . . . , N   N 

D = 2

◦

◦

◦

◦

f  = 40

◦

n = 1, 2, 3, 4, 5

N  = 5

2



1

−2A  j sin(κ z)e− 2 jA cos(κ z)e− H  = − η E  =

1

T 

T 

 jκ  y

 jκ  y

cos ϕay + cos(κT z)e− jκ  y sin ϕaz

ax



E (y , z , t)

= 2A [sin(κT z)sin(ωt κ y)cos ϕay cos(κT z)cos(ωt κ y)sin ϕaz ] 2A cos(κT z)sin(ωt κ y)ax H(y , z , t) = η1

−

−

−

−

λT /2 y

f + (y, z) = e− jκ (y sin θ+z cos θ) κ=

 

ωσµ

2

(1

− )

z

2λ

×10 ×10 ×10 ×10

κ0 sin ϕ = κ sin θ sin θ

7 7

κ0 sin ϕ

7 7

κ0 sin ϕ =

κ

sin θ

          

f + (y, z) =e− jκ (y sin θ+z cos θ)

− j =e



yκ0 sin ϕ+z

√



κ2 κ20 sin2 ϕ

−

=e−Bz z e− j(Ay y+Az z) Ay = κ0 sin ϕ Az = γ 

   − κ2

κ20 sin2



ϕ

Bz =

   − κ2

2 0

κ

  −   − κ2

κ2

tan γ  =

Ay Az

≈0

2

sin2 ϕ Ay

κ20 sin2 ϕ

A

z



sin ϕ

y γ  = tan−1 A Az

e−Bz z κ2

κ20

ϕ

≈ ±κ

κ20 sin2 ϕ

≈κ

γ 

ϕ

γ  ≈ 0◦ 1 E  =

E + (0)e− κC z at

H  = az

× ηE 

C 

κC  ηC  at

E + (0) = E t◦ (0)+E t− (0)

ε = ε0 µ = µ0

σ = 6,1 20◦ 80◦

τ 

τ ⊥

ρ

ρ⊥ 0,5 J 

    −

θ = arcsin sin ϕ

√

κ1 = ω µ0 ε0

κ2 =

ϕ 20◦ 80◦

ϕ 20◦ 80◦

ϕ 20◦ 80◦

ωµ0 σ2

2

(1

 j)

θ 3,926 10−4 (1 + j)◦ 1,13 10−3 (1 + j)◦

× ×

τ  4,006 10−5 (1 + j) (4,006 + j4,005) 10−5

×

− −

5

× 10−− − × 10 − 4

× ×

τ ⊥ 10−5 (1 + j) 10−6 (1 + j)

ρ⊥ 1 + j3,765 1 + j6,957

κ1 sin ϕ

sin θ κ2 cos θ = κ2

E  +

E 

× 10−− × 10

5 6

= E 0 e− jκ 2 (y sin θ+z cos θ) a+ e

+

κ2 sin θ

3,765 6,957

×

ρ 1 + j4,263 1 + j2,307

E 

κ2

κ1 κ2

= E 0 e−Bz z e− j(Ay y+Az z) a+ e

 − 1

sin2 θ

× 10

7

f  = 0,88

Ay = κ2 sin θ Az = Real(κ2 cos θ)

Bz = Im(κ2 cos θ) θ∗

Ay ay + Az az

az

θ∗ = arctan ϕ 20◦ 80◦

≈ ≈

 Ay Az

θ∗ 0[◦ ] 0[◦ ]

ϕ +

E 

= 0,5e−Bz z e− j (Ay y+Az z) a+ e +

+

H 

= az

× E η

2

J 

ϕ = 20◦ +

E 

= σE +

= 0,5e−460300z e− j (6,308y+460300z) a+ e

+

= 33,127(1 + j)e−460300z e− j(6,308y+460300z) a+ h

J 

= 3,05

H 

ϕ = 80◦ +

E 

7

460300z

× 10 e−

e− j (6,308y+460300z) a+ e

= 0,5e−460300z e− j(18,163y+460300z) a+ e

+

= 33,127(1 + j)e−460300z e− j (18,163y+460300z) a+ h

J 

= 3,05

H 

2

7

× 10 e−

460300z

e− j(18,163y+460300z) a+ e

2

κ = κ (f )

κ = κ (f ) κ = κ (f )

κ = κ (f )

κ = κ

ν  p = ν  p (f ) ∆ = ∆(f )

≈

 

κ

ωµ 0 σ

≈ ω√µ ε 0

κ

≈ ω√µ ε

2

ν gr = ν gr (f ) tan ∆ = tan ∆(f )

δ = δ(f )

H (ω) = e− κ(ω)∆z ∆z

κ(ω) E i (ω) E i (ω) = F  i (t)

E (t) ∆z E (t)

{E  }

i

i

o

1

E (t) = F − {E  (ω)H (ω)} } = F − {E  (ω)e− F { } o

1

[]

FFT

IFFT

i i

 κ(ω )∆z

[]

ε (ω)





E (t)

N 

i

FFT

F −1

{}

ε (ω)

0

ε

2ε

N 



H (ω) N 

IFFT



E (t) o

ε = ε (ω)

δ

ε = ε (ω)

ω1 E i (t)

H (ω1 ) = e − κ(ω1)δ

Ei (t) e

|E  (t)| = | | E (t) o

i

E (t) o

∆z

ω0

E (t) i

