ページ3:

def
=
if o
,
0 則稱為
完全
perfect
非完全
介質
medium
math:
if σ=0, (D-ME) f(r,t) = 0
k
向
thm: general solution
量
W
flr,t) = f+(Å --wt) +
*(any)
f(kr+wt)
#k² = kk = wμG => k = W JME
<p> f is solution
a
3 f₁ = 3 + ( I . r - w t ) · f = -wf+
22
=f+= w²f" +
* dxf+ = x(x)+
y ǝy f+ = êy (ǝy α) ft
ƒ₁ = ( k· ŕ – wt) ƒ f = þƒ +) ada + = 2 (dad) ft
J³ƒ+ = v ·(v4) = 3 ⋅ (iff)
f+
=
J(α) ft
= k⋅ v(t + ) = k² ft"
→ 10-μ t. = { 1² - 4 ε w³ } t ; -o
同理f
22
31 ant_ = w³f", √`ƒ_ = k²f"
= 0
=)
<物理意義>
ff(È
=
= O
=wt):往坨向行進,以速率為傳播平面波
f+(R.F-wt) = C₁ 【時變 純量場
等值面方程:F.P= C2 + wt
t+dt
k
f+=C₁
3.

ページ4:

=
kz+wt Constant
k(z+dz) tw(t+ dt) = Constant
kdz + wdt=0
dz
W
FI
= √
=
=
=
dt
干
time-harmonic solution >
諧
Cos[wlt+T)] = coswt
sin [wlt+T)] = Sin wt
T=週期(S)
* frequency f (st or Hz)
W =
w
=
250
·角頻率 angular frequency
C₁ Cos(wt) + Cz Sin (wt)
C₁
C₂
Cos(wt) +
A
8038
- Sing
= A cos(wt+6)
=
system
Fi
Euler
I
j
eje = cos o+j sin
理
i
es = cose + i sine
Re{Ã ed
time-harmonic phasor
ejwt
eint
f(k-wt)
to e-jk-wt)
to e² (k-wt)
=
-jk.t
= 140 €
@Jwt
fo ekint
time-harmonic phasor solution
(補充) Complex表示
ZEC, XY GIR
'
Z =
}
x+2y
Pejo
y
(X,Y)
→ x
let à = A eju
x=pcosp
e: modulus
y=p sinc
4 = tan".
4: argument

ページ5:

實係規、線性, if phasor 是解,則Re {phasor }也是解
thm
ex: F = 主+从F=0
F:管+从
O
前=0
H)
Re (× E) + Re (μ ²²) = 0
√x(ke) + M₂(Re) = 0
非聯立,PDE
k = W JME
H(r,t).
= Re
ed(wt-I-F)
聯立,PDE(O=0)
(1) Ge = 0
Gm
3.- 0
聯立 complex 代規方程
-JR.Z...
-JR.A. -o
=
= 0
ǝt
- jhx ¥. + juu Ho
(2) F √x + M.
AM -6=0
ǝt
Rule
:
v = jk
= jw
A=H.
-¿x-jεĔ. =0
<P>:
C
› = jlwt - k. r)
» o. v. (ř. е ' } [é Ĕ) (é)····) é'
(1) 0 =
=) 0 =

ページ6:

(2)
0 =
def
def
[lè]从[e]
· [ ŵé '] × ². + μй.е² '
ǒ
· é ' - j ) × й + μj w й₁ e' ..
jxx
>> - jb x²² + jw μl to 0
1. 1. - 0
K. H. =
- 1 x
= O
+ who =
kx Ho+w€
[n] =
。
= 0
= 0
沉厝
%m
A/m
ko = wμl Ho
ko = wel
ñ =
=>
V
电
=
#
⇒ 波是橫波
磁
K
L
n: intrinsic impendence
= =2
A
本質
阻抗
WJME
"
[
A
真空。
=
wll
k
Me
Go
H₂ =
n
压器
点
ㄨˋ
=
=
k
WE
Mo
Jul.Go
=> n =
wu
M
JE
- = MoC = 45 × 10 7 x 3 x 10² = 2012 (12) = 377 (52)
>> = -nē-
. = nй.
êt x ke (Foed')
kel Ĥo edic ;=
ALF,t) =
ex x
(³)
(=)
E(rt) = -nêxxo

ページ7:

Elz,t)
[H
Ĥ(z, t)
E. =l Eledu.
Re
è In
IEol
edcwt-kz +40)
n
ex El
cos(wt-kz +60)
1.
2
freguancy 單位 time 的振動次规
B =w= =2xf
空:入=波長=
I wave
Cos(k(z+2)+
Cos (0+2) Cos
kλ= 25 k =
Ex (≥, t=to)
Fol
x
=)
e length (m)
= cos(kz+6)
Ho
|
a
V =
=
JME
T
25
wave
2
number(波) 單位長度的波之個
題型I,phasor 5=0
:
example uniform plane wave in Mr =1, Er=4
[ê. + Eye, +(2+6) és] ēd (2.3) (-0.3x+0.44) educ
0.44)
(½m)
(F) - [Hex + Hyêy+ H &₁]ed(2.1) (-0.3x + 0. by edu (A/m)
Ĥ = êx
Q: Find Ey, Hx, Hy, Hz
②等相面 eq.
Ⓡ f = ? λ = ?

ページ8:

<Sol> Ĕo • êx + Ey by + ( 2+ j6) êà
H₁ = Hx ex + Ey ey + H₂ là
3
2.3(-0.3,0.4,0)
2.3
=
(-0.6,0.8,0)
2
0ên 0 -0.640.8 Ey=0) Ey
= O
从
120元
n =
=
= 60To (n)
E
Er Eo
2
× ŷ
2
l
-0.6 0.8
°
Ho=
e₁₁ x
=
n
60%
1
3/42+36
2 -3x+ 4y = constant
(%)
2y j
= ex ( 16 + J48 ) + ( 12 +31 6 ) + 2 (−1·25) (/m)
6070
6070
③
2 = 27
=
2T
2.3
4k
47 ~ 5.5m
2.3
v
C
3×108
f.
=
=
=
2.7 × 10 Hz
入 Jerder
27
2.22%
題型Ⅱ.read
ho, E., = (free space
or Vacuum)
:
example uniform plane wave
in air
Ĕ(2,t) = êx 100 sin (wt-kz) + by zw cos(wt -kz) (Ym)
Ĥlz,t) = ?
Note:
=
10 - êx 200 cos(wt - kb) + êy lwo sin(ut - k
-ka)] (/m)

ページ9:

§ 5-2 Poynting theorem p
Rmk:靜磁學
1.体電流密度
j
Sm
体電荷密度
How to 体能量流 density
2. Charge Conservation
ǝt
=>
energy Conservation
a
-6-8-2 love
1. Poynting vector
ot
ev
IS
(二)体電能量密度
df de p = 0
=)
a.s
real = P(P, t) ; complex:
微
2.
Poynting theorem
稍
d
form
de
a.s
dz Utot = 0
P(F)
Poynting vector.
real
or instantaneous (瞬時)
P₁f,t) = E(Pt) x HP,t) ExB
σ=0
說明
性:①方向
② 因次
[] = [][] =
V
(c/s)
J
m
m
M² S

ページ10:

= P = Uem V
<p> P= EH = EHEH
E
H
78=MV (V.)
==√
=
=
Ev
EH = E(EE) = ε ȳ V = U e V
EH = H (MH) = — WH² √ = Um√
↳ B =
H
=> + B²
M
2/4
=) P = (Ue+ Um)
system 電 Toke
I
H
理
E
B
电
E
Tott
=
|
Juε
=
P
p = Ĕ + H
=V BĒ奶
Complex Poynting vector
~*
PCP) = ± ẾCF) × (P)
if σ = 0
33
ㄨ
(f) • . ****
=
=
=
***
=
2n
1.1
2
=
W/m²
W: watt

ページ11:

例:前+步+(2+6)全
n=60元
x = (-0.6, 0.8,0)
P(P) = ex n
識
To
= (-0.6,0.8,0)
[+(++(2+6)]
hote
time averaged
flt+ 2) = fit)
def : <f(t))ta =
性質:
(1) <1>t.a = 1
dt fit)
2π/w
(2) { etj zwt ta = 0
<pf> =
25
It=
#j2wt
左=
dt e
W/250
tjzw
= ( e²-45 - 1) = 0
(3)
31. < cos³ ( wt + α) > t.a = =
2. (Sin (wt)>t.a
=
Y is time- -independent.
<p>
1. 左= (
utte
(wt+)
=
jzwt
e
-jawt
• e-jay +2) > t.a
2
=
=
(0+0+2)
2. <sin² (wt+ )) ta = <|- Cos² (wt + 4)) = 1 - —=—- *
重要公式
| Re ( P (F,t)) = ( P (Pt)) = P(F)
t.a

ページ12:

<pf>
2=x+iy EC, X, YER
Z*=x-iy
Relz) = x = = ( 2 + 2 *) ; Im (2) = Y = — — - (2 - 2*)
22
=
+
头
=
+
· Re (± {2 \(D× *)}>
ㄨ
- Re ( ± ₤(P) × й*(F))
= Re P(P)
3*
+ ** H*(P) e
3
+ ECP) × HCD + [* * H(P) }
use
C.C
Z+2* = 2 Re (2)
題型:
e.g. 1: uniform plane wave (M = Mo, E=4ε0) phasor
Ĕ = [ êx 1 + êy — + ê₂ (2j+6) ] ¿ª³² ².1)(− 0.3% +0.44)
Pov = ?
(Sol>
|
|
(1/m)
Par = 1 - (-0.62 +0.89) 1 [1 + 5+ (2²+6)] (%)
—-|·|² = - -
êk
2n
120元
#
eg.2: real
uniform plane wave
in air
Ĕlz,t) = ex loosin (wt-kz) + êy 200 cos (wt-kz) (Y/m)
Find P(z,t) = ?
Pav = ?

ページ13:

<sol>
Plat)
=
=
no
4
10
120元
Par = ½ 10 (±² + 4 + ) %½³
Poynting :
120元
2
微分型式:
=
+
<pf>:
£ = - ³· p = - 7 · (Ĕ×H)
物理意義:
= [Sin` ( wt-k²) + 4 co³ (wt-k²)] (W/m²)
#
H+
=
- H. (×) + · (×H)
=
at
一、(一)+苣(菇+號)
B
=
•
+
+
at
at
24h
ale
2 Um
+
+
at
at
at
-Utot
(3)
a
= Ht
dz Utot = 0 ⇒ 總能守恆
(1)
(2)
d
dt
a.s
說明:
ds
dz
=
I
-----
P
at
#

ページ14:

(1)
(2)
:
ue =
a
at
0.6{號+..]
Um =
a
at
Um
=
DE
2μ
=
Ət
=
B. B
|
2M
2 B
ǝt
at
·B+B ·
.
ǝt
(3)
=
B.
u
at
=
ǝt
电力
ef
磁力
dz PfE + de jfx BdF
=
dw· F · dÌ » dw = F . 4 . ‹†
I duh
de dt
dun
dt
dt
呈 F V
=
dt
D
(Pƒ Ĕ + B). V
l+v
tes
磁力對带电粒子不作功
=
=
=
#
積分型式:
=
=
+
+
182
at
Joule 熱
+
+
器

ページ15:

常考題:(非EMwave 範例)
space
I
indep of
time
Z=0
Q:①証明:Joule定理
U:柱体,
Joc Js ·E - I'R
②証明:
Poynting
定理
-§ då
P
•E +
+
)
<sol>
①
=
© [de Is 6 - [ d e f g + + + d + (x²)
l
=
= 1) = I'R
( ƒ = GE³)
Ⓡ P = EXH
P<a
I
σra'
P
=
..
Moj
2
→ P = ± (±±² ± 2 × 6
I indep of space, time
I
a
=
20
-(-7-7-4-5-6-8)
- då p = ±±±±² (xal). I'R
xa' 2
B

ページ16:

[de(蒜8+電器+)
= I²R
故 Poynting thm 成立.
§5-3:均匀平面波(5=0)
M₁E,σ = 0
E
k = kēk
H
k = WJME
n =
② Polarization 偏振or偏極化
def: 固定一等相面,看电場的箭頭隨時間變化的軌跡。
real
10->
linear polarized
system
圆偏振
偏振
偏振
I
左旋] 右旋
理
右旋 左旋
左旋
右旋
左旋
右旋
for
for
左
右
Circular polarized
left-handed
right-handed
eliptical polarized{
left-handed
right-handed

ページ17:

左
右
線、圆偏振為橢圓之特例
題型I: real
uniform plane wave
(zt) = èx100 sin(wt-kz) +êy 200 cos(wt-kz)
polarization?
<Sol>
[固定一等相面 Z=0
(z=0,t) = exo sin(wt) + êy 200 cos(wt)
=êx Ex(t) + êy Ey(t)
參毯式:
Ex(t) = luo sin(vt)
Ey(t) = 200 cos(wt)
Ey
方程式:(x)+(
-} = 1
200
100
y
to
=)
橢圓
t-of
⇒左旋橢圓偏振
題型I: phasor ←外積
例:平面波
(z)=(êxA+&jAz) e
-jkz
其中A,A.是不全為零的real constant,討論其偏振。
<sol>
letz=0
Ĕ(t) = ke { Ễ (z = 0) edit } = ( êx A₁ + y ¿ A₁₂) (cos wt + j ≤n wt)
:
Re
ex A, Cos(wt) +ey (-A₂) Sin (wt)

ページ18:

ex Ex + y Ey
Ex(t) = A₁ Cos(wt)
Ey(t) = - A₂ Sin (wt)
Ex
=)
A₁
①線偏振
Ey
(參規式)
;+ (- -=1橢圓
(i) A₂ = 0 #A₁
-A₂
-A₁
A₁
(ii) A₂ +0 = A₁
A₂
E(z) = o êx ejk² (x- polarized)
=10led(不影響)
②圓偏振:1A1 = 1Az | ≠ 0
(i) A, A₂ > 0
k
A₁
x
(a) = ê dika ŷ-polarized
=(z)
(ii) AA₂ <0
t=0+
too
(--)
左旋
LHCP
too (++)
toot
右旋
RHCP
t=o+
(-+)
t=0
toot
too (+-)
-jk z
=
+
1) e'
= (êx + j ĝy) ¿¿(†k (ếx xêy) · ¥)
推廣
左
kz
ERHCP = (ex-jey) ejka
食 任意右手系
=
( x + (-)) e¯¯
右

ページ19:

ĔLHcp (ê+jê) e
LHCP =
③ 橢圓:1A1≠1A21≠0
(i) A₁ Az>o
y
-¿ [+ kê₁₁r]
CRHCP
= (₁- ¿é) - kê, 7]
(ii) A. A co
x
x
3
ELHEP (2x+ j)ē
(不一定 2:1)
練習:判斷下列平面波的偏振
初級: Ễ₁ = ( еy - jêx) eð ka
(1)
XHEP (2x-jey) eka
=
(sol>
(1)
中級:
(2)
Ĕ₁ = (-¿еz + y)ēk
-jkx
ê₁ = êy × - еx) = êz
eška = é³(-kéz-?)
(二)右旋圆偏振(RHCP)
(2) ê₁ = êy × (- éε) = - êx
-k(-ex).)
ejkx-
=
=) RHCP
=
-
x+2
S
(4) x = (-jéx + √³ x - ³) e*
(sol>
(3) ₁ =
x
₂ =
-
2
&-ê
35
=( − êx + y) =
tjk
4+532
2
ê+ex

ページ20:

=
=) LHCP
(4) ₁ =
e
=
+jk
√3 êy - z
2
4+532
=) LHCP
=
=
2
‚½₁₂ = - ex
√3 4 -
× (-x)
=
¸¯åk [+ ( √¹ à + êy ). ŕ ]
[+]
2
高級:
(5)
-jkx
Ĕ¸ = ( ĝ− êà + j êy + êà ) ¯¯¿kx
(6)
E6
=
'S
√3 - êx
√
2
*) edky
<sol>
(5)
ê₁ =
еy - z
:
= y + z
√
ex
ê‚ê‚×ê ê tê â
= =
2
ėjkx ĕjk [+ êx. F]
=
=> LHCP
(6) ê, 5 ê - êx z + ♫ êx
=
2
ê₁ = ê₁ × е₁₂ = −³ еy - êy
edby ēd (+ (-Ê;)-8)
=
4
2
=
-ey
=) ZHCP
[thm]:任意平面波(橢圓)必可分解兩個非零的左旋圖右旋圓之疊加
<p>:
=
+j A₂)
ELHCP = (êx + jêy) ēka

ページ21:

2 = (A,+A₂)
=)
ß =
(A₁-A₂)
ĔRN-(ê -jê) e
✓ =
-jk z
= 2 ELHCP + PERHCP
ex: A₁ = 2+ P
y = j A₂ = j (α-ß³)
AA2
2 2,3 0
§5-4 非均匀平面波(5≠0)
§ 1 σ = 0
U.S. §40+0
均匀平面波
非均匀平面波
̃ · Ĕ = σẺ · Ĕ ( Joule )
-
能量↓
非聯立 PDE
-ME-
H(r,t)
time-harmonic phasor solution.
= 0
ECF)
ejut
MCP,t)]
前
Ecr,t) →
H(Ft) → HCF)
E)
[V-Mε(jw) - Mσ (jw)]
= 0
let -ME(jw) - Mol(jw) =
→ (+k³) (r) = 0
Helmholtz
K²
if w=0
=
= 0
(Laplace equation)

ページ22:

the solution:
etȧ( k ên). Y
rule → ± j k êk
:
V² - K²
→
非聯立::wuE-jwuo
=
+ Wμσ
= WμE 1 +
= Will E
=
笔。E(1+ue)
jwE
jwE
les
Complex number
note:
jf
E
JD
jwEE
· (Ft)]
HCF,t)
K = w√ME
jwt
· Ĕ = ε ( 1 +
聯立 PDE
jwE
Ge
j. Ễli,t)
= 0
Gm
. (,t) = O
F
x) +
2
= °
at
AM
√ × π(³,t) - (6 +0) Ĕci,t) = 0·
>
聯立 complex 代方
( -¿ k)еk · Po
(-jk)êk · Ho
等償 C
ů → - j k êk
= 。 => 橫波
= 。 => 橫波
- j k ê x x Ễ + j w μ й₁ = 0
³ − j k ê x x π • − ( j w € + σ) Ễ₂ = 0
=0
→ jw
3
Ecf,t)
-
E.
> kêk

ページ23:

F = k Ĕo = wre Ĥo
=
Ho
=
W ll
k
K = w √ M Ě 1 € x
=
说
k
WE
=
M
E
ey
ejwt éjkz Re{},
X = w√ME, Ĕ = E ( 1 + 16)
ñ³
=
W
=
Lo = lol eve
Ho= 1 Hole³ H
jwE
-dz
Ĕla,t) ex | Fol e cos(ut-p²+ DE)
=
H(8,t) = еy | Hole cos (wt - p² + Pμ)
def j = ɣ =
PE
ão
PH
2+jß
傳播 constant L
constant phase Constant
(attenuation)
單位:
phy
math
[r]=[] = [p] = m²₁
ß
=
rad
m
Np
2 = Or
m
單位
phy
math(底0)
(Naper)
Np
Irad = 57.3°
發現log
場 e
dB
m
Bell
10
發明電話
r=d+ jp = jw√ ME √ 1 + 1√ √6
平方取Re
2²-1² = -wμE
絕對值平方
jwe
L'+P² = W'Mε | +(6)
22² = W M6 [ √1 + (~1~ 6 5 -]
dB 功率
(分貝)場
d: deci
⇒ 1 Np = 8.69 dB
(e¹)²
= 10
X= 8.69
告
=> -2 =
=-10

ページ24:

23² =
W ME
+
刂
ME
2 = W
|
2
WE
=>
ME
ß = w
1+3 +┃
2
从
ñ
E
def:
| +
|
jwe
loss tangent
angle
一
WE
|+
jwe
品
说
On
=
8
2
8
tan S
= |-
= 1+
WE
=
WE
=
WE
29

ページ25:

ME
α = W
√2
1+(7) I
ß = W
ň =
ME
√
√1+1565
1+(+)+1
→完全介質
特例1: if 5=0,則
WE
exp { } = ton" ( 5 ) }}
2=0,p=wJME
W
8 = 2 + j ß = j k = j w J M Ğ
=
座
WE
特例2: ①if0621,稱為良導体(good conductor)
②if000,稱為完全導体(perfect conductor)
(pec perfect electric conductor)
:
2 ~
def e
e-/
3
ds
Ss skin depth
2
集膚深度
who
=Jufo ß
WE
2
4W=2x5
= e

ページ26:

Ss=
I
69
WE
221
=> O₂ = ½ tan³ ( E ) =
TO
4
11
匹
✓ ~
CE-CH
u
E
WE
4
PE
σ
wu
=
(1+j)
Tulf
=
20
= Rs (1+j)
CH
Suface resistivity
(1+j)(表面電阻)
b ch
Ch6phy mean.
=> Ks &s
Sm
check:
|
=
The f
Tu fo
5/m
② 完全尊体5300
85
=
Jufo
πclef
183=0=⇒完全尊体不存在 Em wave
Rs =
→0
σ
考題:
σ = 2m, Mr = 1, Er = 36, f= 1 GHS
Find ①2 = ?, ß=?
© | | = ?, On = ?
2
2nf Eoεr 2
(sol>
σ
=
WE
①
2 = w
MoEoEr
2
2×2×9×109
'
109 × 36
=1 (475 = 9x109)
2TG +10 × 6
√2 × 3 ×10
8
1 + ( 105 )² - 1
- T
= 57.2 (w/m)

ページ27:

ß
= W
MoEoEr
2
=
2TG +10 × 6
√2 × 3 ×10
+1
+ 1
= 138.1 rad/
Mo
Eo Er
120元
6
=
52.8452
+
4√2
WE
On ton (E) = (rad)
181 2: : σ = 0.58 × 10° (5/m) Mr = 1 Ex = 1
①証銅為良導体
© εs = ? ; Rs = ?
③ 入。=?
<Sol>
σ
0.58×10 x 45 x
9×109
①
~10221
⇒ Cu為良導体
20x10
10
f = 10 GHz
②
|
Es
= 0.66 m)
=
Tox 4 x10x10 x 0.58 +10°
Rs =
=
2.6×102 (5)
So
0.66×10 6 x 0.88 ×108
③
C
3×108
20 =
=
3×10³m
= 3cm
10'0
Note: Es << lo
* Es Rs =
n
=)
=
r
± for w/ >>
I
I
8=jw√μ
¼ ju z jw ε (tf z
=
jw
Rs (1+8)
(1+j)
JWE(14
= Rs ds =
jwɛ
1
ň = √√
2/20

ページ28:

§ 5-5 (normal incidence)
補充:
Maxwell
交界條件
Ge
· B = ls
ĥ⋅ (D₁ - Ď₂) = 6f
Gm
J. B.
= 0
ĥ. (B₁ - B₂)
= 0
F
+
at
= 0
AM
=
at
=
。
ĥx (H₁-H₂) = dfs
ĥ, 2.
rule=
→n
field
OF
fields difference 1-2
vol source surface source
<pf>
= AM
=
da
+
at
dr
1. 上下压扁(Ch3)
+
床
=
上
ch了相同, 故略
→ ^ × ( H₁ - H₁ = Jss
dfs
+0
有限

ページ29:

20
1° 0₁ =0 102 300
2° σ₁ = 0 ± σ₂ =0
By 1° 0₁ =0 1 σ₂ 700
ex Ero - êd'Elo
x
C
>> 2
y
Eto
-ey Eso
2.
w
no
w'
①②
01=0
02-300
W = W'=W"
def: 反射係e卩 三
<sol>
Ero
Eso
a ñ · (Ď₁ - B₂) = σƒ equation
© ĥ (B - B₁₂ = 0
× CE₁- E) = 0
④ ^ × (₁₁ - H₁₂ ) = ƒ
by at 2:0
jwt
Ezoe
to edit
→ trivial (o-o=0)
1/4 (Q² continous)
equation
+ PE to ew't
0-
=) w' = wn Ecto + Eto = 0
1=-1(實際上為倒立)
Eto
T =
Eto
:
11700 = 0
= O
Of = 0 (by @)
Eio
ey
no
By
(-1) Elo
2°
3- - ê × [ê - -
=
2 Eio
by④
2°

ページ30:

note: 物理意義(類似虛像法)
177
2°
Elo
30
541:2
Eio
2 Eto
no
€
Eio
Eig
*Elo
σ = 0
σ = 0
Mo, E.
Mo, Eo
LHCP
← f
-jkz
Ĕ₁. E.(x+j)
x
-jw(-2)
⇒ ĔR = - Ĕ¿ = - E。 ( êx + jêy) ē¯¨³
=> KHCP
Eig
=) No EM wave
30
2°
ex Elo
ex Eto
ext Elo
WJME, W/ME.
WfMzEZ
0
2
Eio
Eio
ZELO
ey
n
2
n
σ₁ = 0
02=0
Mz Ez
M₁ε₁
几
層蛋

ページ31:

[ = ?, T = ?
<sol>=
at = 0
ĥ · (D₁ - D) = σf = 0 (trivial)
ĥ. (B₁-B₂) = 0 (trivial)
x (H₁-H) ==
2 = 1 + 1 =
0
⑥= ±(1-1) = ±²
:
72-71
2n2
14+ 4 = 2 + 1 = 6
公式記憶
if n₁ = n = 17 = 0, 2 = 1
② if σ₁ = 0, G.-00
,
n² =
= Rs (1+ j) = 0
=> [ = -1, Z = 0
各章節比較
章節
Chz:靜电
Ch3:靜磁
Chs:平面波
Ei
公式
E₁
3
q
q
Q"
q
=
=
E₁-82
E₁ + ε2
2ε₂
M. Mr
Et
ni
22
Y_I I_I
I'
Mr-Mi
Ex
2=-2
=
=
Mithle
Ei
n₁ + n
I'
2441
Et
E₁
ni+n

ページ32:

131
Ecl=1m
f=IGHE
1. 畫出(Eca)l v.sz圖
(1) Er = 1, Er 4
12) Er = 4, Er=l
<Sol>
(1) Er₁ = 1, k₁ =
n₁ = no = 120πG
3/0
n₂ =
=6070
√4
C =
22-21
2+2=
=
-1
x = 1+1
=
+3
| E(3)| =
Evi
Er₁
Mr₁ = 1
Mr₂ = 1
σ₁ = 0
02=0
2 × 109
3×108
=
20T
rad
3
=
=)
Irl=
= Tu
2
-jka
IzEto e
² | = |Ecol = — —
Z>0
,
-jk₁z
+jkız
Z<o
-jkiz
+jzkız
E+PI
= | | 1+11
22% ZM + To
(2k, Z+T),
(24+1)
=
Jo
even π
=> | E | m = 1 + 1 = —
1+
2元
22M = 25
=> 02m =
* def:
dielectric
介電質
medium
介質
σ = 0 ^ Mr = 1

ページ33:

2
2π
2x 2m + Tv = odd x To
2.
2,
Jo
= 25
=) 02m=
2,
2
-
hote
Cose
2T
|Cosel
TO
TU
26
-45
-30
-15
λ = 30cm
256
2
+ TG = 0
2
(2) Er₁ = 4, Er₂ = 1
K₁ = WJMO Eo Er₁ = W
21=15cm
=
2
=
C=l+=+
4
¾
3
-3
=
2% x 107
40元
2
2 =
3×108
3
-
=> 7 =
1 + 1/2
=
1
5
0μ=0

ページ34:

|Ez] =
z =
告
| Eiol | | + || el p+2kz) |
2.
2x 2 m + 0 =
even To =) | Elm = 1/144
21
2. 2h 23m to = odd =) (Elm = 123
AAA
-22.5
-15
-2.5
4/
3/3
33
言
M
§5-6傾斜入射(oblique incidence)
10波向量
反射 Law
:
折射 Law
(Snell Law)
=>臨界角 全內反射、表面波
垂直偏振:「乙」(Frenshel公式)
:
平行偏振:,,=⇒極化角 or Brewster angle
(偏振)
回沒向量
def: 入射面:买与所決定的平面(eg.以下為x-plane)
(kk, ky
X
(k", ky", ka")
Oi
y
0120
(kx, o, kz, M₁, ε,
√₂ =0
Mr, E.

ページ35:

=)
at z=o
- XẺ, XẺ,
{
+
=> kx = kx = kx (b)
- ky ký - kỷ 20
反射 Law :
=
①下躺在入射面上
© Sin Oi
=
kx
k
=
kx
k'
Sin Or O Or
def index of refraction (4) = n
MoEo
n = =
=
Mr Er
21
v
JUE
* π1, Mr = 1 = n=Jer
折射 Law
①"躺在入射面上
"
k kx
n₁
Sint = k"
=
=
Sindi
k" k
n₂
=>
n, Sindin₂ Sin Ot
Snell's Law
几大⇒光密
n 小⇒光疏
Jur, Eri
=
Jun En
=
n₁
n₂
edky

ページ36:

光疏→光密
光密→光疏
Di
Let
def
:
→ 2
ni zn₂
n₁ > n₂
-
臨界角: Critical angle De (only在密 疏成立)
O t = 90° = Oi = Oc
-1
n₁ Sinoc = n₂ Sin 90°
=) Oc sin
=
(
Oi
=
Oi < Oc
=
Z
Oi
n₁ Sin De = n₂ Sin Ot
na
01 > O c = Sin D ₂ > Sin O c = 1₁₁
n. Sin On Sin Ot > n₂
=> Sin Oε > 1
Z
ML
total internal reflection
Oi > Oc
x
x
全内
Sin (0+) >1
互射
At=90°
Oc > Oi
JOŁEC
k = k*cos Ot
cosBt 純虛
sin Otx
IR
Cost = 1-sin²de =+) sim²+-1=
(sim² -
-1
-kz
(2700)=0

ページ37:

常考題:
例1:step - index optical fiber 階梯拆射率光纖
air
n₂
ni
Oc
BE
m₂
n₁>n> air = 1
Q:Omax Sin (√n₁² - n² )
when h₁ = 1.6, n₂ = 1.4 =) Omax = ?
<sol>:
①
I sin Omax n, Sin O₂ = n, cos Oc
ni
Joc
= Jn₁²-n
Jni-n₂³
=) Omax = Sin" (Jn₁-n³)
© = Sin" ( √1.6 -1.4") = 50.8°
139 2:
©
TE:
find
Pr
?
A 45°
Pi
t₁E;
Pr
T₂TE
n = 2
45° Z₁Fi
Oc = Sin () = 30° = Oc
<sol>
cladding
Core()
ni
包饅
no
(air) |
45°入射)全內反射
n = Ju=
=
Ser
n
17₂z₁ Eil
2n
n
2
Pr
Pi
no
C₁ =
=
E
=
=(2221)²
70+2
n+1
16:12
=
no
4n
(n+1)²
240
2
C₂ =
=
no+n
1+
2n
h+i
Cove

ページ38:

n=21€'x
9
64
= 279%
81
1543:
30'
①= Oc = ?
② 在air中每經一波長衰減多少分?
(Sols:
②2
σ= 0
Mr= |
Ex = 81
*air
Oc = Sin" (√) = 6.38°
e
-22
2= "sin³ Di-1
,
200-120 x 8.69
·Sin 2-1 = -j 4.39 (P328)
-coot = -j √ √ ², sin 30 °² - 1
=) 2 x 4.39 x 8.69
= 240dB
-24
=) 10 CCI(衰減很快)
回場.
入射面二基準面
品
x
(1) 垂直偏振
TE
Hr
Snell √ Sin 30= |· Sin Ot =
Oi
10t
82
H₁ 22
名
Er
E
Et
7₁ =
Ei

ページ39:

(2) 平行偏振
=) TM
öö
x
Py
名
22 =
"
③:
Ht
一般:
M: Magnetic
74415
E: electric 電波
T: trensverse 橫向⇒相對食橫向,(不是B,F)
TM = H₂ = 0,
Ez to
TE: E₂ = 0, Hz to
TEM: E₂ =0, Hz= 0
for (1)
W
def: 横向阻抗
ZTE
=
=
HT Hcose
Cose
Ert Er
橫向互射係犯 PTE.
=
EST
Es
橫向透射係QTTE
EtT
=
=71
ECT
at Z=0
-
=)
Er
=
E¿
21 =
Ei
ESTET = Ett
HET + HrT
1+TE
=
- HtT
=
TTE
{1 (Elt-Em) = -1 EET » 1 (1 - Pr) = TE
}0
TE
TTE =
Z₁ - Z
Z₁ + Z₂
222
21 +22
=)
+
Coset CosBi
=
+
Coset
Cosoi
n
2
coset
14
n
+
Cosot
COSO:
=
n
nicus Di-n cos Ot
hi cosi + h₂ Cos Ot
271 COS Di
nicos Di + n₂ Cos Ot

ページ40:

for (2)
ET
ECOSO
def ZTM =
HT
HT
CTM =
Ert
Er Costc
EST
Ez Cos Di
EtT
TTM
=
=
EiT
Ett
E=T
=
Et cost
Cos Ot
=
=
E cos Di
COS Di
159 =
at Z=0
=)
ELT + EXT =
EST
Zi
{
"
{
=)
Er
=
Ett
Ett
ZL
1+1TM = TTM
(1-T) = TTM
CTM
TTM =
Cu
=
=
Z₁ - Z₁
Zz + Z,
2Z2
Z₁+22
Cost - Costc
n₂ cost +n₁ Cosoi
T₁ = (1+1)
=
Cosot
介
=
nicosbt-n₂ cos
nicoset+nCoSD:
2 Cosoi
n₁ Coso; +n₂ cosot
x
Q = 1 += ? C₁ = ?
= ? Z = ?
水
8=0,h=1
E1281
air

ページ41:

(sols: h₁ = 581 = 9
01-30°
n₁ = 1
Cost = -j 4.39
①
n, co son cos Ot
9.
-1(-j4.39)
1₁ =
=
= 1. / 58.8°
nicos OE +n₂ CoSO t
9.-1.- 4.39)
2111全反射
71 = 1 + 1 = 1+1/58.80
= 1.74/29.4°
Note: 透射無效
②
nicos Ot-n₂ cos Di
=
nicos Ot +Nz Cos Oi
Cosoi
√
2
91-34.39) - 1.
91-4.39)-1.5
T₁ = (1 + [₁₁) = (1 + 1/2.51°)
coset
= 0.39 / 88.740
√
-j4.39
Brewster
原理:
OB
0
83.90
f
n₁
no
def : [ = 0 = 0; = 0₁ = tan^ (^) (< 1 g)
①OB + O₂ = 90°
介电髓
def =0→ 日三日(不存在)
= 1/2.510
==>

ページ42:

<p>
ni cos Ot n₂ COSOB
0 = 1,, =
nicos Ot+n₂ COSDB
=) h₁ cos Ot
n₂ COS OB = 0
Snell, Sin OB = n₂ Sin Ot
2
| = cos² Ot + Sin² Ot = ( m², COSDB)² + (^_ SinDB) "
"
| = Cos² + Sin Op
= cos³ OB, let t tan OB
n₁
⇒ 1 + t² = ( — 3³ + m² ² = ²
t² =
T
» OB = tan
<p>
0=P₁ =
n
12
ni
n₂
n₂
2
=
ni
nicose-n₂ Cosot
= 0
nicos Op + n₂ Cosot
Snell, SinB = n₂ simot
| = Cos² Ot + Sin Ot = (1) Cos² OB + ( ^_^ Sin " OB
ni
=)0-
(→←-)
<pf①>
100
n₁
n, Sin On Sin Ot
m
n₁
n₂
Sin Ot =
nz Jnitni
=> OB + O t = 90°
=
M

ページ43:

綜合題
×
<sol>:
©
2755
k
20
30°
60°
真空 介質
n=53
= 46 (0)
Oi = 60°
Ostan" () = 60°
O i = 0₁ => [,, = 0
↳ OB +0 += 90°
polarization of E
ê₁ =
ex -Sier
2
x ey =
2
+ j ŷ ) e°
4 = (x+2)
(½m)
2
êu
2
Q = ①O₁ = ? 0₁ = ?
palarization of Ĕ = ?
②
=>
=) LHCP
Ex - (ên Eo + jê E)
5x A
0,
Z
© Ĕy = ?
polarisation ?
Et
polarization?
1×2 - 55×
S
=
nicos On₂ Cos Ot
√ ₁ =
=
nicos Di+n₂ Cosot
₁ = ŷ
=)
Er = еy jE. (-)
4(x-2)
(m)

ページ44:

polarization of Er = y-
= y-polarized
or perpendicular.
(3)
Ĕ₁ = E₁ ( ê,” T₁ + j êï 7) е° (%)
*(0)
=
Check: k"= √3 = 45
C₁₁ = (1 + 1).
COSOC
Cos Ot
=== {
No 1 : 1) #19
-j4(x+2)
(Ym)
-> Ĕt = E α ( √ ŷ - ε √ +jys) e
2
e₁
polarization. of Et
ê = ex
=
3
53-12
√3 2 +
x ey
2
2
=
e
=) ZHEP