Persamaan MAxwell Okf

8/18/2019 Persamaan MAxwell Okf

1/43

FISIKA DASAR 2

PERSAMAAN MAXWELL


2/43

Energi dan Momentum gelombang elektromagnetik dibawaoleh medan listrik E dan medan magnet B yang menjalarmelalui vakum.

Sumber gelombangnya berupa muatan-muatan listrik yangberosilasi dalam atom, molekul, atau mungkin juga dalamsuatu antene pemancar radio.

Untuk medan listrik E dan medan magnet B yang berubahdengan waktu, keberadaan E selalu disertai B, dansebaliknya. eterkaitan antara E dan B dituangkan dalampersamaan Ma!well yang mendasari teori medan magnetik.

PENDAHULUAN


3/43

GELOMBANG ELEKTROMAGNETIK

PERSAMAAN MAXWELL PERSAMAAN HELMHOLTZ

PEMANTULAN DAN TRANSMISI

TRANSMISI MELALUI TIGA MEDIUM

PENYESUAIAN IMPEDANSI


4/43

PERSAMAAN MAXWELL

"ersamaan Ma!well dirumuskan dalam besaran medan

listrik E dan medan magnet B. Seluruh persamaan Ma!well

terdiri dari # persamaan medan, yang masing-masing dapatdipandang sebagai hubungan antara medan dan distribusisumber, baik sumber muatan ataupun sumber arus.


5/43

0H0HB

0E0E0D

H jtH

tBE

E jt

E

t

DH

=•∇→=µ•∇=•∇

=•∇→=ε•∇→=•∇

ωµ−=∂∂µ−=

∂∂−=×∇

ωε=∂∂

ε=∂∂

=×∇

ω

µ

ε

jt

H B

E D

=∂∂=

=

0B

0D

t

BE

t

DH

=•∇=•∇

∂∂

−=×∇∂

∂=×∇

PERSAMAAN MAXWELL


6/43

0B

D

t

BE

t

DJH

v

=•∇

ρ=•∇∂∂−=×∇

∂∂

+=×∇

0B

0D0

t

BE

t

DH0EJ0

v

=•∇

=•∇→=ρ∂∂−=×∇

∂∂

=×∇→=σ=→=σ

RUANG HAMPA


7/43

EE) j( j

)H( j

)H j()E(

2

εµω=ωεωµ−=

×∇ωµ−=ωµ−×∇=×∇×∇

0H

0E

H jE

E jH

=•∇=•∇

ωµ−=×∇

ωε=×∇

E0

EE)E(2

2

∇−=∇−•∇=×∇×∇

0EE 22 =εµω+∇

PERSAMAAN HELMHOLTZ


8/43

!

"

#

)!t( j

## $EE β−ω=

)!t( j

"" $HH

β−ω

=##%EE =

""%HH =


9/43

22

)!t( j

#

2)!t( j

#

0

0$E$E) j)( j(

εµω=β→=εµω+β−

=εµω+β−β− β−ωβ−ω

vv

f ω π

λ

π

β === 22

)!t( j### $E)!&t(EE β−ω==

2

2

2

2

2

22

!

A

"

A

#

AA

∂∂

+∂∂

+∂∂

=∇

'v εµ=

0EE #2

#

2 =εµω+∇

0E!

E#22

#

2

=εµω+∂∂

B%*+%* +$,-%*+

K$.$/%t%* +$,-%*+ 1#'0 ,34,3H'0#5

,3F'0#657

8

'2

−

−

π=µ

=ε


10/43

)!t( j

#

"$E j

!

H β−ωωε=

∂

∂−

)!t( j

""" $H)!&t(HH β−ω==

!#"

"!#

#

"! %

"

A

#

A%

#

A

!

A%

!

A

"

AA

∂

∂−

∂

∂+

∂

∂−

∂

∂+

∂

∂−

∂

∂=×∇

E jH ωε=×∇

!#"

"!#

#

"! %"

H

#

H%#

H

!

H%!

H

"

HH

∂∂−

∂∂+

∂∂−

∂∂+

∂∂−

∂∂=×∇

## %E jH ωε=×∇


11/43

Ω=π=

ε

µ=η 188'20

)!t( j

#

"$E j

!

H β−ωωε=∂∂

−

η="

#

H

E

#

#

#

" EEEH µ

ε

=εµω

ωε

=β

ωε

=

ε

µ=

"

#

H

EI,/$9%*4 *4t:*4; <

)!t( j

"" $HH β−ω=

)!t( j#

)!t( j" $E j$H) j( β−ωβ−ω ωε=β−−


12/43

PEMANTULAN DAN TRANSMISI

E & H

E: & H:

Et & Ht

! 0

η' η2

)!t( j

'

'

)!t( j

'

'

'

$A

H

$AE

β−ω

β−ω

η=

=

)!t( j

2

2t

)!t( j2t

2

2

$A

H

$AE

β−ω

β−ω

η=

=

)!t( j

'

':

)!t( j

':

'

'

$B

H

$BE

β+ω

β+ω

η−=

=

S"%:%t=4"%:%t -%t%4 /%9% ! 0 <

t:

t:

HHH

EEE

=+

=+


13/43

E & H

E: & H:

Et & Ht

! 0

η' η2 S"%:%t=4"%:%t -%t%4 /%9% ! 0 <

t:

t:

HHH

EEE

=+=+

2

2

'

'

'

'

2''

ABA

ABA

η=

η−

η

=+

'2

'2

'

'

A

B

η+η

η−η

=Γ =

'2

2

'

2 2T

A

A

η+η

η==

F%;t: R$>$;4

F%;t: T:%*4,4


14/43

( )

0B

0D

H jtBE

E jE jEt

DJH

=•∇

=•∇

ωµ−=∂∂−=×∇

ωε+σ=ωε+σ=∂∂

+=×∇

MEDIUM DIELEKTRIK 0

EJ σ=


15/43

0E) j( jE

E) j( jE0EE

E) j( j

)H( j

)H j()E(

2

22

=ωε+σωµ−∇ωε+σωµ−=∇−=∇−•∇

ωε+σωµ−=×∇ωµ−= ωµ−×∇=×∇×∇

( )

0E

H jE

E jH

=•∇ ωµ−=×∇

ωε+σ=×∇


16/43

) j( j

) j( j

0$E) j( j$E

0E) j( j

!

E

$EE

2

!

!

2

2

2

!

ωε+σωµ±=γ

ωε+σωµ=γ

=ωε+σωµ−γ

=ωε+σωµ−

∂

∂=

γ γ

γ

0E) j( jE2 =ωε+σωµ−∇

K*4t%*t% /$*j%%:%* ( propagation constant )


17/43

) j(

j

) j(

) j( j

) j(H

E

$E) j($H

$E) j(!

H

"

#

!

#

!

"

!

#

"

ωε+σωµ

=ωε+σ

ωε+σωµ=

ωε+σγ

==η

ωε+σ=γ

ωε+σ=∂

∂−

γ −γ −

γ −

"

!

"

!

#

!

#

!

%$H$HH

%$E$EE

γ −γ −

γ −γ −

==

==

!#"

"!#

#

"! %"

H

#

H%

#

H

!

H%

!

H

"

HH

E) j(H

∂∂

−∂∂

+

∂∂

−∂

∂+

∂∂

−∂∂

=×∇

ωε+σ=×∇


18/43

?*t@ S% 17'

S%t +$,-%*+ $$;t:,%+*$t; ,$*j%%: 9 9%%, 4%t

,$9,7 K*4t%*t% /$*j%%:%* 9%* ,/$9%*4 *t:*4;*"%

,%4*+=,%4*+ γ (0&26 j 2) ,='

9%* η (C00 j 86) Ω7Ht*+ µR & ε

R 9%* σ 9%: ,$9, t$:4$-t7

J%%- <

08&1

'0#5

'0#8&1,3H'0#8&1

)'0#60(2

86&'2'

86&'2'86&'2' j)86 jC00)(2 j26&0( j

86 jC00 j

j2 j26&0) j( j

8

C

R

C

C =

π

=

µ

µ=µ→=

π

=µ

=ωµ→=++=ωµ=γη

+=ωε+σ ωµ=η+=ωε+σωµ=γ

−

−−

'C2&''0#65&'0#26&'0'0#26&'0

)'0#60(2'0#21'&1

'0#21'&1,3,S2'&0

'0#)21'&1 j2'&0(86 jC00

2 j26&0 j

'2

'2

R '2

C

1

1

1

==εε=ε→=π=ε

=ωε=σ

+=++

=ωε+σ=ηγ

−

−

−

−

−

−


19/43

S$-%@ /$4%%t t$:-%*+ ,$*++*%;%* %t,$t$: :%9%: *t;,$*$*t;%**"% ;$t*++%**"% "%t 9$*+%* ,$,%*.%:;%*

+$,-%*+ $$;t:,%+*$t; -$:>:$;$*4 8 GH! ;$ -%%@ 9%*,$*$:,% +$,-%*+ "%*+ 9/%*t;%* $@ /$:,;%%* 9

-%%@*"%7 B% /$4%%t t$:-%*+ t$:4$-t -$:%9% 9 %t%4 /$:,;%%* %t 9$*+%* /$:,tvt%4 :$%t/ ε

R '&

/$:,$%-t%4 :$%t/ µR ' 9%* ;*9;tvt%4 4t:; σ 5S3,& @t*+ /$:4$*t%4$ -$:;:%*+*"% 9%"% %;-%t /$,%*t%*

/$:,;%%* %t7

?*t@ S% 172 <

ωε+σ

ωµ=η

j

j

P$t*j; < I,/$9%*4 4%t ,$9, t$:@%9%/

+$,-%*+ $$;t:,%+*$t; %9%%@ <


20/43

η

ηL

I

L

2

I

2

η=

η+η==

2

G

L

L

G

F

2

F

2

F

2

'

)IFR$(2

')

GFFR$(

2

')IIR$(

2

'P

η=η=

=η

=η=

LT' η=η→=α

T

P

P=α

Jawab : (Analogi Rangkaian Lis!ik"

8


21/43

2

1

2

2

F'0#'6C&11188

2

F

2

'2

F

2

'P −=

=η

=

1

F'0#)8&6 j5&(F)C2&2 jC'0&5'(188

)C2&2 jC'0&5'(FF −+=++ +=η+η η=

1

'0#)0'6&0 j5&2()C2&2 jC'0&5'(188

I −−=

++=

η+η=

2

1

11

F'0#82&''

F'0#)0'6&0 j5&2('0#F)8&6 j5&(2

'R$)F#I

2

'R$(P

−

−−

=

++==

08&16'00F'0#'6C&11

F'0#82&''

P

P

1

1

T ===α −

−

C2&2 jC'0&5'

)'0#65&)(')('0#8)(2( j5

)'0#5)(')('0#8)(2( j'2

8

+=

π+ππ

=η −−

η

ηL

I

? t @ S 1 1


22/43

?*t@ S% 171 <

S%t +$,-%*+ $$;t:,%+*$t; -$:>:$;$*4 '0 MH!&

,$*j%%: 9%%, %:%@ a! 9 9%%, ,$9, 9$*+%*

;*9;tvt%4 4t:; σ C& # '0=5 ,@3,& /$:,tvt%4 :$%t/ε

R 8&'1 9%* /$:,$%-t%4 :$%t/ µ

R 5&667 T$*t;%*

-$4%:*"% ,$9%* ,%+*$t; /%9% ! 60 ., 9%* t 6 *4 -%

9;$t%@ %,/t9% 9%: ,$9%* 4t:; "%*+ ,$*j%%: 9%%,%:%@ a# %9%%@ 4$-$4%: '00 3,7

P$t*j; < K*4t%*t% /$*j%%:%* +$,-%*+

$$;t:,%+*$t; 9 9%%, 4%t ,$9, %9%%@ <

ωµωε+σ=β+α=γ j) j( j

J -


23/43

J%%- <

06&0

'2C

8C

80&2115&268&28)'065&)('1&8)('0'0)(2(5

)'05)(66&5)('0'0)(2(

je j x x j

x x j

=+=+

= −−

π

π π η

'&''02&0

)'05)(66&5)(2()'065&)('1&8)('0'0(2'0&CK( 8'2C5

j

x j x x j x

+=

+= −−− π π π γ

m A

x xe

z t e E H

z

xo y

328&0

80&2

J06&0)6&0('&')'06)('0'0(2.4K'00

).4(

C)6&0)('02&0(

=

−−=

−=

−−

−

π η

β ω α

? @ S 1 5


24/43

?*t@ S% 175 <

T*j;;%* -%@% 9%: ;*4t%*t% /$*j%%:%* <

''2

2

−

ωεσ

+µε

ω=α

''2

2

+

ωεσ

+µε

ω=β

) j( j j ωε+σωµ=β+α=γ ;*4t%*t% %t$*%4 9%* ;*4t%*t% >%4% ,%4*+=,%4*+

9%/%t 9@t*+ 9%: /$:4%,%%*=/$:4%,%%* <

J -


25/43

J%%- <

05

02

22

j j2

) j)( j( j

222

225

2

2

2

222

=σµω−µεαω+α

=µεω+

αωµσ−α

αωµσ

=β→ωµσ=αβωµσ+µεω−=αβ+β−α

ωε+σωµ=β+α

2

22222522 σµω+εµω±µεω−

=α

22222522


26/43

222225222 σµω+εµω±µεω−=α

''

2

2''

2'

'2

'

2

'

''2

'2

2

2

2

22

2

2

22222

2

2

222

+

ωε

σ+

µεω=β

µεω+ ωεσ+µεω=β

µεω+

ωεσ

+µεω−±µεω−=µεω+α=β

−

ωεσ

+µε

ω=α

ωε

σ+µεω−±µεω−=α

? t @ S 1 6


27/43

?*t@ S% 176

S%t +$,-%*+ $$;t:,%+*$t; -$:>:$;$*4 200 MH!

9$*+%* %,/t9% ,$9%* 4t:; 4$-$4%: E '00 3, "%*+

4$9%*+ ,$*j%%: ,$*+$*% %/4%* /$:%; t/4 (εR '& µR '& C'&8 MS3,) 4$t$-% 9 6 ,7 Ht*+ %,/t9% ,$9%*

4t:; E1 4$t$%@ ,$*$,-4 %/4%* t/4 t$:4$-t7

J%%- <

E E' E2 E1

9

J -


28/43

J%%- <

µσπ≈

+ ωεσ+µεω=β

µσπ=µσπ

=ωεσµε

ω≈

−

ωεσ

+µε

ω=α

>>==ωεσ

−

> ''2

> 2

> 2

2''

2

''0#16)'0#65&)('0#200(

'0#8&C'

2

2

'2C

C

2

2

) j'(2

2

')564* j56(.4> 2

56$$'

j j

j

j

j

j

56 j0 j

+σωµ=+σωµ=ηµσπ=β=α

∠σωµ

=σωµ

=σωµ

=

σ

ωµ=

σ

ωε

ε

µ≈

+ωεσ

εµ

=

ωε+σ

ωµ=η

C8 )'08C')('05)('02(> ππσµπα −


29/43

Ω+=+=η

∠=

∠ππ

=∠σωµ

=η

−−

−

−

11

2

1

C

8

2

'0#)688&1 j688&1() j')(808&0('0#06&6

56'0#06&6

56'0#8&C'

)'0#5)('0#2(256

,3F'0#8&'

'0#)688&1 j688&1(188

)188(2'0#0&0

2EE 1

1

1

2'

'21

−−

− =

++

=

η+η

η=

6

C8

222

'0#208&2

)'0#8&C')('0#5)('0#2(>

=

ππ=σµπ=α

,3'0#C5&2E'0#)&' j&'(

'0#)688&1 j688&1(188

'0#)688&1 j688&1(2)'00(

2EE

1

'

1

1

1

2'

2'

−−

−

−

=→+=

+++

=η+η

η=

,3F'0#0&0$'0#C5&2$EE 1)'0#6)('0#208&2(1!

'2

C62 −−−α− ===

−

E E' E2 E1

9


30/43

P%9% ,$9, ' <

E ,%;4,, t$:j%9 -% E 9%* E: 4$>%4% <

J$$KEEEE

$$EE$EE

)!t( j)!t( j

:

j)!t( j

:

)!t( j

''

''

φ+β+ωβ−ω

φβ+ωβ−ω

Γ+=+=

Γ=ΓΓ==

π+φ

=β− *2

!,%;4E

'*2!! '' π+φ+β=β−

)!t( j

,%;4'$)'(EE β−ωΓ +=

#EL$M%AN# %ERDIRI


31/43

Γ−Γ+

=='

'

E

E4

,*

,%;4

E ,*,, t$:j%9 -% E 9%* E: -$:%%*%* >%4% <

*2!! '' π+π+φ+β=β− π+π

+φ

=β− *22

!,*

E'

)!t( j

,*'$)'(EE

β−ωΓ −=

St%*9*+ W%v$ R%t (SWR) <


32/43

Sal&!an T!ans'isi

η ηL

! =9 ! 0

η'

η*

βη+η βη+ηη=η 9t+ j9t+ j

L

L*

ZLZv

Z'

>

9Z'

*/t,/$9%*49! =η −=

#EL$M%AN# MELALUI TI#A MEDIUM


33/43

0&0 ,

ε' 20 /F3,µ' 2 µH3,

ε2 60 /F3,

µ2 2&6 µH3,ε1 ε' µ1 µ'

D 9%$:%@ ' (! 0) t$:9%/%t ,$9, 9$*+%* ε' 20 /F3, 9%* µ

' 2 µH3,7 D

9%$:%@ 2 (0 ! .,) t$:9%/%t ,$9, 9$*+%* ε' 60 /F3, 9%* µ

' 2&6 µH3,7

M$9, /%9% 9%$:%@ 1 (! .,) 4%,% 9$*+%* ,$9, /%9% 9%$:%@ '7B% ,$9%* 4t:; "%*+ 9%t%*+ 9%: ,$9, ' %9%%@ E '000 .4 (5#'0 π t = β

' !)

3,& t$*t;%* ,$9%* ,%+*$t; H "%*+ 9t$:4;%* ;$ ,$9, 17

Ω=== 2&1'C'

'1'

ε

µ η η

Ω== C&2212

22

ε µ η

sm xv 3'05&0'

22

2 ==ε µ

sm xvv 3'06'&''

''

1' ===

ε µ

Jawab :

ono) Soal *+, :


34/43

66&68 j01&'1)C16&'(2&1'C jC7221

)C16&'(C&221 j2&1'CC&221* −=

++

=η

C16&'9t+022&')0&0('0#5&0

'0#22

9

MH!200> 9v

> 29

v9

22

22

2

2

2

2

22

=β→=π

=β

=π

=ω

=β

βη+ηβη+η

η=η2212

22212*

9t+ j

9t+ j

η' η*

E Ht


35/43

η' η*

E Ht

6&C j

*

tt $&156C&0 j61&1

66&68 j01&'1

2'8&'55 j0&85EH =+=

−−

=η

=

2'8&'55 j0&85)'000(66&68 j01&'12&1'C

)66&68 j01&'1(2

E2

E *'

*t

−=−+

−=

η+ηη=

)!5&8t'0#5.4('000)!t.4(EE ' −π=β−ω=

,3:%95&8'0#6'&'

)'0#200(2

v

C

'

1'

=π=

ω=β=β

)6&C!5&8t'0#5.4(&1)!t.4(HH

1tt +−π=β−ω=


36/43

S$-%@ $*4% ;%.%,%t% -$:*9$;4 -%4 '&C2 9%/4 $@ %/4%* %*t :$>$;47L%/4%* /$*"$4% ,/$9%*4 * -%4%*"% 9:%*.%*+ *t; ,$*$:4;%* .%@%"%

9$*+%* /%*j%*+ +$,-%*+ 660 *, (;**+) "%*+ 9%t%*+ 9%: 9%:% ;%:$*% ,%t%,$,/*"% 4$*4tvt%4 ,%;4,, /%9% %:*% ;**+7 S$/$:t ;t% ;$t%@.%@%"% t%,/%; ,$,/*"% 9%$:%@ /%*j%*+ +$,-%*+ 500 *, 439 800 *,7Ht*+ /$:4$*t%4 -$:;:%*+*"% *t$*4t%4 .%@%"% "%*+ 9t$:4;%* /%9% %:*%*+ (500 *,) 9%* ,$:%@ (800 *,)7 A/% ;,$*t%: 4%9%:%

*' '&C2 *2*1 '

5

kuning λ

ono) Soal *+- :

Jawab :

'


37/43

oo

oo

v

cn

ε µ

µε

µε

ε µ ===

'

'

nnnn

o

o

o

o

o 188'2

===== η

ε

µ

ε

µ

ε

µ η

B%@%* **=,%+*$t; o µ µ =

o

nε

ε =

on ε ε 2=

Ω==Ω==Ω= 2&2C8&212

C2&'

188188 1'2'1 η η η η η


38/43

η1

η*

I

*' '&C2 *2*1 '

5

kuning λ

η1 η*


39/43

2

12

1

2

1

G

;**+ F'0#CC1&0)188(5

F

5

F

2

F

2

F)IFR$(P −==

η=

η==

η1

η*

I

'9t+

25

29

*,660

t22

2

2

22

1

=α→∞=β

π=λλ

π=β

=λ

1

1*

1*

2

I

2

η=

η+η==→η=η

a)a.a k&ning :


40/43

( )

88&''00#'0#085&'

'0#066&''0#085&'P

'0#066&''0#)218&0 j'6&0()066&0 j61&0(IR$P

'0#)218&0 j'6&0(8&2125&6 j&2C1

I

)066&0 j61&0(8&2125&6 j&2C1

5&6 j&2C1

5&6 j&2C1)58&'(188 j2&2C

)58&'(2&2C j1882&2C

9; t+ j

9; t+ j

58&'9t+)186&'(

25

660

500

29*,500

2

0

1

2

0

12

0

1

2

0

1

0

1

0

*+

0

10

'*

0

00

'*

*

2212

22212*

22221

=−

=∆

=−+==

+=++

=η+η

=

+=++ +=η+η η=

+=

−+−+

=

η+ηη+η

η=η

−=β→π

=π

=β→=λ

−

−−

−

a)a.a &ng& :


41/43

( )

8&0'00#'0#085&'

'0#0C8&''0#'085&0P

'0#0C8&''0#)'C2&0 j20&0()0188&0 j6'5&0(IR$P

'0#)'C2&0 j20&0(8&212&1C j&252

I

)0188&0 j6'5&0(8&212&1C j&252

&1C j&252

&1C j&252)6&2(188 j2&2C

)6&2(2&2C j1882&2C

6&29t+)8C&0(25

660

800

29*,800

2

0

1

2

0

12

0

1

2

0

1

0

1

0

,$:%@

0

10

'*

0

00'*

*

*

22221

=−

=∆

=++==

−=+−

=η+η

=

−=+−−

=η+ηη

=

−=

+

+=η

=β→π

=π

=β→=λ

−

−−

−

a)a.a '/!a) :


42/43

S$-%@ 4,-$: t$+%*+%* 4$-$4%: 0 9$*+%* >:$;$*4 500 ;H!,$,/*"% ,/$9%*4 t/t 4$-$4%: '0 @,7 S,-$: t$+%*+%*

* %;%* ,$*t:%*4,4;%* 9%"% ;$ 4$-%@ -$-%* 4$-$4%: C0 @,,$% 4%:%* t:%*4,4 -$:/% ;%-$ ;%;4% 4$/%*j%*+ 86 ,9$*+%* Z

50 @, 9%* v 2&6#'0 ,347 Ht*+ 9%"% "%*+

9t:%*4,4;%* $@ 4,-$: t$+%*+%* t$:4$-t7

'0 Ω

C0 Ω

Z 50 Ω v 2&6#'0 ,34

0 > 500 KH!

ono) Soal *+ :


43/43

[ ] W5C6&6)5C&0 j60&')(C2&5 j'6&C5(R$P

5C&0 j60&'

850&'6 j65&18

0I

C2&5 j'6&C5)0('0850&'6 j65&18

850&'6 j65&18

850&'6 j65&18)1&0(C0 j50

)1&0(50 jC050Z

1&0;9t+:%9865&0)86('0#6&2

'0#52

9v9

*

6

=−−=

+=

−

=

−=+−

−=

−=

++

=

=→=π

=ω

=β

Jawab :

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