Solutions to selected problems from Exercise 5 · Prof. Rakhesh Singh Kshetrimayum Solutions to...
Transcript of Solutions to selected problems from Exercise 5 · Prof. Rakhesh Singh Kshetrimayum Solutions to...
Prof. Rakhesh Singh Kshetrimayum
Solutions to selected problems from
Exercise 5
Prof. Rakhesh Singh Kshetrimayum
3/23/20181Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum
Exercise 5.1
� Since
� the direction of wave propagation
ˆ ˆx yk β
+ =
r
( ) 2
0ˆ ˆ
x yj
E E x y eβ
+ −
= − +r
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� the corresponding magnetic field can be obtained from Maxwell’s curl equation
ˆ ˆ
2
x yk β
+ =
Exercise 5.1
0
0
0 0
2 2
ˆ ˆ ˆ
0
x y x yj j
x y z
EEE j H H
j j x y z
e eβ β
ωµωµ ωµ
+ + − −
∇× ∂ ∂ ∂∇× = − ⇒ = =
− − ∂ ∂ ∂
−
rr r r
( ) 2
0ˆ ˆ
x yj
E E x y eβ
+ −
= − +r
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2 2
2 2 2 2
0
0
0
0
0
ˆ ˆ ˆ
2
x y x y x y x yj j j j
j
e e
E e e e ex y z
j z z x y
E je
j
β β β β
β
ωµ
β
ωµ
+ + + + − − − −
−
−
∂ ∂ ∂ ∂
= − − + + − ∂ ∂ ∂ ∂
−=
−2 2 2 2
0 0
0 0
2 2ˆ ˆ ˆ
2
x y x y x y x yj j jj j
e z E e z E e zj
β β ββ β
ωµ η
+ + + + − − −
− − + = = −
Exercise 5.1
� time-average power flow per unit area
( ) ( )( )2
* 2 2 0
0 0
0 0
ˆ ˆ1 1 2ˆ ˆ ˆRe Re
2 2 2
x y x yj j
avg
y xES E H E x y e E e z
β β
η η
+ + − +
+
= × = − + × =
r r r
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Exercise 5.2
� For any function to be a solution of the wave equation, it must satisfy the following equation
( ) ( )22tzftzf ±∂
=±∂ ωβ
εµωβ
( )tzf ωβ ±
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( ) ( )2002
t
tzf
z
tzf
∂
±∂=
∂
±∂ ωβεµ
ωβ
( ) ( )
∂
±∂
∂
∂=
∂
±∂
∂
∂⇒
t
tzf
tz
tzf
z
ωβεµ
ωβ00
Exercise 5.2
( ) ( )( )
( ) ( )( )
( )( )
( ) ( )( )
±∂
±∂±
∂
∂=
±∂
±∂
∂
∂⇒
±∂
±∂
∂
±∂
∂
∂=
±∂
±∂
∂
±∂
∂
∂
00
00
tz
tzf
ttz
tzf
z
tz
tzf
t
tz
ttz
tzf
z
tz
z
ωβ
ωβωεµ
ωβ
ωββ
ωβ
ωβωβεµ
ωβ
ωβωβ
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( )( )
( )
( )( )
( ) ( )( )
±∂
±∂=
±∂
±∂⇒
±∂±
∂=
±∂∂⇒
2
22
002
22
00
tz
tzf
tz
tzf
tzttzz
ωβ
ωβωεµ
ωβ
ωββ
ωβωεµ
ωββ
Exercise 5.2
( ) ( )cv p =⇒=⇒=⇒
00
2
2
2
00
2 1
εµβ
ωωεµβ
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� This wave has phase velocity equal to speed of light
� Hence, any function f(βz±ωt) is a solution of the wave equation
Exercise 5.3
� Already solved in the class
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Exercise 5.4
� Note that zxy form a right handed system, putting in time dependence
( )0ˆ ˆ( )
j ya E E jx z e
β−= +r
( ) tjyjeexjzEE
ωβ−+= ˆˆr
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� Taking the real part of the wave( ) tjyj
eexjzEE += ˆˆ0
( )
+−+−= xytzytEE ˆ
2cosˆcos0
πβωβω
r
Exercise 5.4
� Observing at y=0
( ) ( )( )xtztEE ˆsinˆcos0 ωω −=r
x
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z
x
Time t=∆t
Time t=0
LHCP
Exercise 5.4
� Obviously EP
� Putting time dependence
( ) ( ){ }0ˆ ˆ( ) 2 1 3
j yb E E j x z j e
β−= + + +r
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� Taking real part
( ) tjyjj
jeexezeEE
ωβ−
+=
−−
ˆ5ˆ10 2
1tan
3tan
0
11r
( ) ( )( )xytzytEE ˆ6.26cos5ˆ6.71cos1000
0 +−++−= βωβωr
Exercise 5.4� Observing at y=0
( ) ( )( )xtztEE ˆ6.26cos5ˆ6.71cos1000
0 +++= ωωr
x
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z
x
Time t=∆t Time t=0
RHEP
Exercise 5.4
� In that case, yx (-z) forms a right handed system
� Putting in time dependence and taking real part
( )0ˆ ˆ( )
j zc H H x jy e
β+= −r
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� Observing at z=0
( )
++
−+= ztxyztHH βω
πβω cosˆˆ
2cos0
r
( ) ( )( )txytHH ωω cosˆˆsin0 +=r
Exercise 5.4
x Time t=0
( ) ( )( )txytHH ωω cosˆˆsin0 +=r
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y
x
Time t=∆t
Time t=0
LHCP
Exercise 5.5
� equation of the ellipse of polarization
( ) ( )
( ) ( ) 1sincos
sin52
cos5,cos3
2222
=+=+⇒
−=
+==
ttEE
ttEtE
zy
zy
ωω
ωπ
ωω
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� Maximum field magnitude equals 5 V/m
( ) ( ) 1sincos259
22 =+=+⇒ ttEE
zy ωω
Exercise 5.5
� yzx forms a right handed system
z
( ) ( )ztytE ˆsin5ˆcos3 ωω −=r
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y
z
Time t=∆t
Time t=0
LHEP
Exercise 5.9� (a) Loss tangent of a medium is given by � At frequency f=10 kHz,
� loss tangent is approximately equal to 899 which is greater than 100 and
� hence the solid ground acts as a good conductor
ωε
σ
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� hence the solid ground acts as a good conductor� At frequency f=100 MHz,
� loss tangent is approximately equal to 0.0899 which is closer to 0.01 and
� hence the solid ground may act as a bad conductor or good dielectric
Exercise 5.9
� For good conductor,
� and good dielectric
014049629.0== µσπα f
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� The average Poynting vector reduces at the rate of
� inside the solid ground
297901922.02
==ε
µσα
ze
α2−
Exercise 5.9� Hence, for depth into the ground
�where the average power of the EM wave has been reduced by 1%
� to that of the surface is given by
( ) ( )α
αα
2
01.0ln01.0ln201.0
2 −=⇒=−⇒=−zze
z
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� For f=10kHz,
� the required depth is approximately 163.89m
� whereas for f=100MHz,
� it is approximately 7.72934m
( )α
α2
01.0ln201.0 −=⇒=−⇒= zze
Exercise 5.9
� (c) In order to image an object at a depth of 50 m,
� for roundtrip of the radar signals we need a distance of 100m
� From the knowledge of the above part of the section (100 MHz won’t be able to get that far),
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(100 MHz won’t be able to get that far),
�we may assume that the solid ground should
� behave as a good conductor for the problem at hand
Exercise 5.9
� Then for a power reflected back from the object
�which has dropped to just 1% of the original power,
�we have,
( ) ( ){ }22201.0ln401.0ln201.0 =⇒=−⇒=−
zfzfez µσπµσπα
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( ) ( ){ }22201.0ln401.0ln201.0 =⇒=−⇒=−
zfzfez µσπµσπα
( ){ } kHzz
f 86.2601.0ln4
1 2
2≅=⇒
πµσ
Exercise 5.9
� Hence the wavelength at this frequency is approximately 1.117 km,
� therefore, we may not be able to image at any small object at this depth
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