Transcript
Chapter 5 Problem Solutions
511 .. The time derivative is related to the phasor form by jω ⇔ − jω µ H$ y ( z ) ⇔ − µ
∂ H y ( z, t ) ∂t
∂ . Hence ∂t
∂ Ex ( z, t ) . and − jω ε E$ x ( z ) ⇔ −ε ∂t
51 . .2 + − The time-domain equations are E x ( z , t ) = E m cos(ω t − β z ) + Em cos(ω t + β z ) and
H y ( z, t ) = gives −µ
+ Em
η
E− cos(ω t − β z ) − m cos(ω t + β z ) . Substituting into the first equation
η
∂ E x ( z, t ) + − = β Em sin(ω t − β z ) − β Em sin(ω t + β z ) and ∂z
∂ H y ( z, t ) ∂t
= µω
+ Em
η
sin(ω t − β z ) − µ ω
− Em
η
sin(ω t + β z ) . Matching the
corresponding coefficients of the sin terms gives the requirement that β = substituting β = ω
µ ε and η =
µω . But η
µ shows this to be true. Similar results are shown for ε
the second equation of Problem 5.1.1.
51 . .3 (a) pvc, (ε r = 35 .) β= v=
β=
vo
µ rε r
ω vo
= 16 . × 108
µ r ε r = 0.304
ω vo
µ r ε r = 0.392
µr rad , η = ηo = 202 Ω , εr m
v m , λ = = 16 m (b) Teflon, (ε r = 21 .) f s
rad µr , η = ηo = 260 Ω , v = εr m
vo
µ rε r
= 2.07 × 108
m , s
µr v rad ω = 20.7 m . (c) Mylar, (ε r = 5) β = µ r ε r = 0.468 , η = ηo = 169 Ω , f εr m vo vo m v , λ = = 13.4 m . (d) Polyurethane (ε r = 7) v= = 134 . × 108 f s µ rε r
λ=
5-1
β=
ω vo
λ=
µ r ε r = 0.554
rad µr , η = ηo = 142 Ω , v = εr m
vo
µ rε r
= 113 . × 108
m , s
v = 113 . m. f
51 . .4 A sketch is shown below. The phase constant is β =
ω
= 0105 .
vo
rad and the intrinsic m
impedance is η = η o = 377 Ω . The electric field intensity vector is given by . y $ = 10e j 0105 E a z or E = 10 cos 10π × 10 6 t + 0105 . y a z . In order for the power flow
(
)
E × H to be in the -y direction, the magnetic field intensity vector must be in the -x . y $ = −0.0265e j 0105 a x or H = −0.0265 cos 10π × 10 6 t + 0105 . y ax . direction so that H
(
)
z Ez Hx
y
x
51 . .5 v , the frequency of the wave is f
A sketch is shown below. Since the wavelength is λ = 800MHz. Also the velocity of propagation is v = constant is β =
(
ω v
=
2π
λ
= 251 .
vo
εr
. Hence ε r = 2.25 . The phase
rad . The electric field is m
)
E = 100 cos 16π × 108 t − 251 . x a z . The intrinsic impedance is η =
ηo = 251Ω . In εr
order that E × H be in the +x direction, the magnetic field must be directed in the -y
(
)
direction. Hence H = −0.398 cos 16π × 108 t − 251 . x ay
5-2
y
x
Ez Hy
z
51 . .6 The frequency of the wave is 40MHz. The phase constant is β = intrinsic impedance is η =
ω vo
ε r = 2.9
rad . The m
ηo = 109Ω . The problem is sketched below. The wave is εr
traveling in the +y direction. From this E must be in the z direction in order that E × H be in the +y direction. The magnitude of the electric field is the product of the magnitude of the magnetic field and the intrinsic impedance. Hence, the electric field is
(
)
E = 10.9 cos 8π × 10 7 t − 2.9 y a z . z Ez
y
x
Hx
5-3
51 . .7 The problem is sketched below. The phase constant is β =
ω vo
ε r µ r = 251
rad . The m
µr = 565Ω . The wave is traveling in the -z direction and εr
intrinsic impedance is η = η o
the magnetic field is in the y direction. Hence the magnetic field intensity vector is given
(
)
by H = 0.02 cos 4π × 109 t + 251z a y From this E must be in the -x direction in order that
E × H be in the -z direction. The magnitude of the electric field is the product of the magnitude of the magnetic field and the intrinsic impedance. Hence, the electric field is
(
)
E = −113 . cos 4π × 109 t + 251z a x .
x
z
Hy Ex
y
51 . .8 Since λ =
vo f εr
we have that
εr =
λo = 2 . Hence ε r = 4 . λ
5.21 . The frequency is 500MHz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j10π × 108 × 4π × 10 −7 × 41 + j10π × 108 × × 10 −9 × 36 36π
This evaluates to γ$ = 149∠67.5o = 57.2 + j138 . Hence we identify α = 57.2 and rad β = 138 . The intrinsic impedance is m
5-4
η$ =
jω µ o µ r j10π × 108 × 4π × 10 −7 × 4 . This evaluates to = (σ + jω ε oε r ) 1 + j10π × 108 × 1 × 10 −9 × 36 36π
η$ = 106∠22.5o . The problem is sketched below. The wave is traveling in the +x direction and the electric field is in the z direction. From this H must be in the -y direction in order that E × H be in the +x direction. Hence the magnetic field vector is
(
)
H = −0.946e −57.2 x cos 10π × 108 t − 138 x − 22.5o a y . y
x
Ez Hy
z
5.2.2 (a) 60Hz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j120π × 4π × 10 −7 0.01 + j120π × × 10 −9 × 15 36π
. × 10 −3 ∠45o = 154 . × 10 −3 + j154 . × 10 −3 . Hence we identify This evaluates to γ$ = 218 rad α = 154 . × 10 −3 and β = 154 . × 10 −3 . The velocity of propagation is m m ω v = = 2.45 × 105 . The intrinsic impedance is s β η$ =
jω µ o µ r j120π × 4π × 10 −7 = (σ + jω ε oε r ) 0.01 + j120π × 1 × 10 −9 × 15 36π
η$ = 0.22∠45o .
5-5
which evaluates to
(b) 1MHz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j 2π × 10 6 × 4π × 10 −7 0.01 + j 2π × 10 6 × × 10 −9 × 15 36π
This evaluates to γ$ = 0.281∠47.4o = 019 . + j 0.21 . Hence we identify α = 019 . and rad m ω β = 0.21 . The velocity of propagation is v = = 303 . × 107 . The intrinsic s β m impedance is η$ =
jω µ o µ r j 2π × 10 6 × 4π × 10 −7 .which evaluates to = (σ + jω ε oε r ) 0.01 + j 2π × 106 × 1 × 10 −9 × 15 36π
η$ = 28.05∠42.62 o . (c) 100MHz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j 2π × 108 × 4π × 10 −7 0.01 + j 2π × 108 × × 10 −9 × 15 36π
This evaluates to γ$ = 814 . ∠86.58o = 0.49 + j813 . . Hence we identify α = 0.49 and rad m ω β = 813 . . The velocity of propagation is v = = 7.73 × 107 . The intrinsic s β m impedance is η$ =
jω µ o µ r j 2π × 108 × 4π × 10 −7 .which evaluates to = (σ + jω ε oε r ) 0.01 + j 2π × 108 × 1 × 10 −9 × 15 36π
η$ = 96.99∠3.42 o . (d) 10GHz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j 2π × 1010 × 4π × 10 −7 0.01 + j 2π × 1010 × × 10 −9 × 15 36π
This evaluates to γ$ = 81116 . ∠89.97 o = 0.49 + j81116 . . Hence we identify α = 0.49 and ω m rad β = 8112 . . The velocity of propagation is v = = 7.75 × 10 7 . The intrinsic s β m impedance is η$ =
jω µ o µ r j 2π × 1010 × 4π × 10 −7 .which evaluates = (σ + jω ε oε r ) 0.01 + j 2π × 1010 × 1 × 10 −9 × 15 36π
to η$ = 97.34∠0.03o .
5-6
5.2.3 The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j 2π × 109 × 4π × 10 −7 × 16 2 + j 2π × 109 × × 10 −9 × 9 36π
This evaluates to γ$ = 510.33∠52.02 o = 314.06 + j 402.25 . Hence we identify α = 314.06 ω m rad and β = 402.25 . The velocity of propagation is v = = 156 . × 10 7 . The intrinsic s β m impedance is η$ =
jω µ o µ r j 2π × 109 × 4π × 10 −7 × 16 .which evaluates to = (σ + jω ε oε r ) 2 + j 2π × 109 × 1 × 10 −9 × 9 36π
η$ = 247.55∠37.98 o . 5.2.4 The frequency is 10GHz. Forming the propagation constant as γ$ = 200 + j300 =
1 j 2π × 1010 × 4π × 10 −7 σ + j 2π × 1010 × × 10 −9 × ε r . Squaring this 36π
16 gives γ$ 2 = −5 × 10 4 + j12 × 10 4 = − π 2 × 10 4 ε r + j 2π × 1010 × 4π × 10 −7 σ . Solving gives 36
S ε r = 114 . and σ = 152 . . The intrinsic impedance can now be computed from m η$ =
jω µ o µ r j 2π × 1010 × 4π × 10 −7 .which evaluates to = 1 (σ + jω ε oε r ) 152 . + j 2π × 1010 × . × 10 −9 × 114 36π
η$ = 218.98∠33.69 o . The problem is sketched below. The wave is traveling in the +y direction and the magnetic field is in the x direction. From this E must be in the +z direction in order that E × H be in the +y direction. Hence the electric field vector is
(
)
E = 219 . e −200 y cos 2π × 1010 t − 300 y + 33.69 o a z . z Ez
y
x
Hx
5-7
5.31 . The surface is sketched below and has a surface area of (3 − ( −1)) × ( 2 − ( −1)) = 12m 2 . The intrinsic impedance of the medium is η =
ηo
5
= 168.6Ω . Hence the average power
2
1 (10) W = 0.3 2 . The wave is propagating perpendicular to the surface and is density is 2 168.6 m uniform over it so that the total average power crossing the surface is PAV = ∫ S AV • ds = 356W . . s
x (3, –1, 2) m
(3, 2, 2) m
z (–1, –1, 2) m y
(–1, 2, 2) m
5.3.2 The surface is sketched below and has a surface area of 3 × 5 = 15m 2 . The intrinsic impedance of the medium is η =
ηo
9
intensity is E = ( 0.2) × 125.66 = 2513 .
= 125.66Ω . The magnitude of the electric field
V . Hence the average power density is m
2
. ) W 1 (2513 = 2.51 2 . The wave is propagating perpendicular to the surface and is 2 125.66 m uniform over it so that the total average power crossing the surface is PAV = ∫ S AV • ds = 37.7W . s
5-8
x (3, 0, 0) m
(3, 5, 0) m
(0, 0, 0) z
(0, 5, 0) m y
5.3.3 The frequency is 500MHz. The propagation constant is
γ$ =
jω µ o µ r (σ + jω ε oε r ) =
1 j10π × 108 × 4π × 10 −7 × 41 + j10π × 108 × × 10 −9 × 36 36π
This evaluates to γ$ = 149∠67.5o = 57.2 + j138 . Hence we identify α = 57.2 and rad . The intrinsic impedance is β = 138 m η$ =
jω µ o µ r j10π × 108 × 4π × 10 −7 × 4 . This evaluates to = (σ + jω ε oε r ) 1 + j10π × 108 × 1 × 10 −9 × 36 36π
η$ = 106∠22.5o . The problem is sketched below. The wave is propagating in the +x direction and is perpendicular to the side of area 2m × 3m = 6m 2 . The average power 2
1 (100) −2 × 57.2 x e cos 22.5o a y = 436 . e −114.4 x a y density is S AV = 2 106
(
(
)
)
power dissipated is 43.6 1 − e −114.4 × 20 mm × 6m 2 = 235W .
5-9
W m2
. Hence the
y (0, 2, 0) m
(20 mm, 2 m, 0)
3m (20 mm, 0, 0)
(0, 2 m, 3 m)
x 2m
(20 mm, 0, 3 m) 20 mm z
5.3.4 The propagation constant is
γ$ =
1 j 2π × 109 × 4π × 10 −7 × 16 2 + j 2π × 109 × × 10 −9 × 9 36π
jω µ o µ r (σ + jω ε oε r ) =
This evaluates to γ$ = 510.33∠52.02 o = 314.06 + j 402.25 . Hence we identify α = 314.06 rad . The intrinsic impedance is and β = 402.25 m η$ =
jω µ o µ r j 2π × 109 × 4π × 10 −7 × 16 .which evaluates to = (σ + jω ε oε r ) 2 + j 2π × 109 × 1 × 10 −9 × 9 36π
η$ = 247.55∠37.98 o . The wave is perpendicular to the surface of area 100cm 2 = 0.01m 2 . Hence the average power density is 2
1 (1) . z . × 10 −3 e −62812 S AV = e −2 × 314.06 z cos 37.98 o = 159 2 247.55
(
(
)
)
W m2
. Hence the power
. × 5mm dissipated is 159 . × 10 −3 1 − e −62812 × 0.01m 2 = 15.2µ W .
5.41 . We need to compute the attenuation constant from
γ$ = α + jβ =
jω µ o µ r (σ + jω ε oε r ) = −ω µ o µ r ε oε r 1 − j
5-10
σ . The ω ε oε r
attenuation of the amplitude varies as e −αd . In dB this is
(
)
20 log10 e −αd = −20α d log10 ( e ) = −8.69α d . For an attenuation of 80dB, d =
(a) 1kHz.
σ 0.889 × 109 = = 8.89 × 105 . Good conductor so f ω ε rε o
80 . 8.69α
α ≅ π fµ r µ oσ = 397 . × 10 −3 f = 0126 . . Hence d=73.3m. σ 0.889 × 109 = = 8.89 × 10 4 and is a good conductor. Thus α = 0.397 f ω ε rε o
(b) 10kHz.
and d=23.2m. (c) 100kHz.
σ 0.889 × 109 . = = 8.89 × 103 and is a good conductor. Thus α = 126 f ω ε rε o
and d=7.33m. (d) 1MHz.
σ 0.889 × 109 . = = 8.89 × 10 2 and is a good conductor. Thus α = 397 f ω ε rε o
and d=2.32m. (e) 10MHz.
σ 0.889 × 109 = = 88.9 and is a good conductor. Thus α = 12.6 and f ω ε rε o
d=0.733m. (f) 100MHz.
σ 0.889 × 109 = = 8.89 and is a good conductor. Thus α = 39.7 and f ω ε rε o
d=23.2cm. 5.4.2 The attenuation of the amplitude varies as e −αd . In dB this is
(
)
20 log10 e −αd = −20α d log10 ( e ) = −8.69α d . For an attenuation of 20dB, d =
20 . 8.69α
The attenuation constant was calculated for these cases in Problem 5.2.2. (a) 60Hz
α = 154 . × 10 −3 . Hence, d=1.49km. (b) 1MHz. α = 019 . . Hence, d=12.1m. (c) 100MHz. α = 0.49 . Hence, d=4.7m. (d) 10GHz. α = 0.49 . Hence, d=4.7m.
5-11
5.4.3 At these frequencies, sea water may be considered a good conductor. The attenuation constant was calculated in Problem 5.4.1 as α = 397 . × 10 −3 f . Hence the skin depth is 1 252 . The intrinsic impedance is δ= = α f 2 f jω µ o µ r 2 ≅ ∠45o = ∠45o = 14 . × 10 −3 f ∠45o . The average 252σ (σ + jω ε oε r ) σ δ
η$ =
power dissipated is d 2 −2 . × 103 1 (1) δ Area = 253 1 − e −2 × 5m 2 = 109 PAV = cos ∠θ n 1 − e W . (a) 1kHz. 2 η f 1 f 424 3 0.865
[
]
The average power dissipated is 34.5W. (b) 10kHz. The average power dissipated is 10.9W. (c) 100kHz. The average power dissipated is 3.45W. 5.4.4 The propagation constant is γ$ = α + jβ =
jω µ (σ + jω ε ) = −ω 2 µ ε (1 − j tan φ ) .
Squaring both sides and equating real and imaginary parts yields α 2 − β 2 = −ω 2 µ ε and 2α β = ω 2 µ ε tan φ . Solving gives the desired result.
5.5.1 The intrinsic impedances of each region are η 1 = η o
η
2
= ηo
µ r1 1 = η = 188Ω and ε r1 2 o
µ r2 2 = η = 251Ω . The reflection and transmission coefficients are ε r2 3 o
2 1 4 ηo − ηo ηo η 2 1 8 2 3 2 3 Γ= = = and T = = = . Observe as a 7 η 2 +η1 2η + 1 η η 2 +η1 2η + 1 η 7 o o o o 3 2 3 2
η 2 −η1
check that 1 + Γ = T . Since the phase constant in medium 1 is
β 1 = ω µ r1µ oε r1ε 0 =
ω µ r1ε r1 vo
= 6π the frequency of the wave is 450MHz. Hence
the phase constant in medium 2 is β 2 = ω µ r 2 µ oε r 2ε 0 =
(
ω µ r 2ε r 2 vo
)
electric fields can now be written as E i = 100 cos 9π × 108 t − 6π z a x ,
5-12
= 18π . The
Er =
(
)
(
)
100 800 cos 9π × 108 t + 6π z a x , E t = cos 9π × 108 t − 18π z a x . The magnetic 7 7
fields can be found by dividing the electric fields by the intrinsic impedance of the appropriate medium and ensuring that the sign is such that E × H is in the correct 100 direction for the particular wave. Hence, H i = cos 9π × 108 t − 6π z a y , 188 100 800 Hr = − cos 9π × 108 t + 6π z a y , H t = cos 9π × 108 t − 18π z a y . The 7 × 188 7 × 251
(
(
)
)
(
)
average power transmitted through a 2m 2 area of the surface is 2
t 1 E PAV,trans = × 2m 2 = 52W . 2 η2
5.5.2 The intrinsic impedances of each region are η 1 = η o
η
2 = ηo
µ r1 = 2η o = 754Ω and ε r1
µ r2 1 = η = 126Ω . The reflection and transmission coefficients are ε r2 3 o
1 2 η o − 2η o ηo 2 η 5 2 2 3 Γ= = 3 = − and T = = = . Observe as a η 2 + η 1 1 η + 2η η 2 + η 1 2η + 1 η 7 7 o o o o 3 3
η 2 −η1
check that 1 + Γ = T . Since the phase constant in medium 1 is
β 1 = ω µ r1µ oε r1ε 0 =
ω µ r1ε r1 vo
=
8π the frequency of the wave is 50MHz. Hence 3
the phase constant in medium 2 is β 2 = ω µ r 2 µ oε r 2ε 0 =
ω µ r 2ε r 2 vo
= π . The electric
8π z a x , fields can now be written as E i = 10 cos10π × 10 7 t − 3
Er = −
(
)
20 50 8π cos10π × 10 7 t + z a x , Et = cos 10π × 10 7 t − π z a x . The magnetic 7 3 7
fields can be found by dividing the electric fields by the intrinsic impedance of the appropriate medium and ensuring that the sign is such that E × H is in the correct 10 8π cos10π × 107 t − z a y , direction for the particular wave. Hence, H i = 754 3
5-13
Hr =
(
)
50 8π 20 cos10π × 10 7 t + z a y , H t = cos 10π × 107 t − π z a y . The 7 × 754 3 7 × 126
average power transmitted through a 5m 2 area of the surface is 2
t 1 E PAV,trans = × 5m 2 = 162mW . 2 η2
5.5.3 The intrinsic impedances of each region are η 1 = η o
η
µ r1 1 = η = 126Ω and ε r1 3 o
µ r2 1 = η = 188Ω . The reflection and transmission coefficients are ε r2 2 o
2 = ηo
1 1 ηo − ηo 2η 2 ηo 6 1 3 Γ= = 2 = = . Observe as a = and T = 1 1 1 1 η 2 +η1 η 2 + η1 ηo + ηo 5 ηo + ηo 5 2 3 2 3
η 2 −η1
check that 1 + Γ = T . Since the phase constant in medium 1 is
β 1 = ω µ r1µ oε r1ε 0 =
ω µ r1ε r1 vo
= 2π the frequency of the wave is 100MHz. Hence
the phase constant in medium 2 is β 2 = ω µ r 2 µ oε r 2ε 0 =
(
ω µ r 2ε r 2 vo
)
=
16π . The 3
electric fields can now be written as E i = 5 cos 2π × 108 t − 2π z a y ,
(
)
16π E r = 1 cos 2π × 108 t + 2π z a y , E t = 6 cos 2π × 108 t − z a y . The magnetic fields 3 can be found by dividing the electric fields by the intrinsic impedance of the appropriate medium and ensuring that the sign is such that E × H is in the correct direction for the 5 cos 2π × 108 t − 2π z a x , particular wave. Hence, H i = − 126 1 6 16π Hr = cos 2π × 108 t + 2π z a x , H t = − cos 2π × 108 t − z a x . The average 188 3 126
(
)
(
)
power transmitted through a 5m 2 area of the surface is 2
t 1 E PAV,trans = × 4m 2 = 383mW . 2 η2
5-14
5.5.4
µ r1 2 = η = 251Ω and ε r1 3 o
The intrinsic impedances of each region are η 1 = η o
η
2 = ηo
µ r2 = 4η o = 1508Ω . The reflection and transmission coefficients are ε r2
2 4η o − η o 2η 2 8η o 12 5 3 Γ= = = = . Observe as a = and T = η 2 + η 1 4η + 2 η 7 η 2 + η 1 4η + 2 η 7 o o o o 3 3
η 2 −η1
check that 1 + Γ = T . Since the phase constant in medium 1 is
β 1 = ω µ r1µ oε r1ε 0 =
ω µ r1ε r1 vo
= 8π the frequency of the wave is 200MHz. Hence
the phase constant in medium 2 is β 2 = ω µ r 2 µ oε r 2ε 0 =
ω µ r 2ε r 2
(
vo
=
)
16π . The 3
. cos 4π × 108 t − 8π z a y , electric fields can now be written as E i = −2513
(
)
16π E r = −17.95 cos 4π × 108 t + 8π z a y , E t = −4308 . cos 4π × 108 t − z a y . The 3 magnetic fields can be found by dividing the electric fields by the intrinsic impedance of the appropriate medium and ensuring that the sign is such that E × H is in the correct
(
)
. cos 4π × 108 t − 8π z a x , direction for the particular wave. Hence, H i = 01
Hr = −
(
)
5 2 16π × 01 . cos 4π × 108 t + 8π z a x , H t = × 01 . cos 4π × 108 t − z a x . The 7 3 7
average power transmitted through a 3m 2 area of the surface is 2
t 1 E PAV,trans = × 3m 2 = 185 . W. 2 η2
5.5.5 The intrinsic impedance and phase constant of the first medium are η1 = η o = 120π and β = β o = ω v = 0.063 . Medium 2 is a good conductor as evidenced by o
σ2 = 15 . × 10 6 . Hence the attenuation and phase constants can be computed as ω ε oε r 2 α
2 =
β2=
1
δ2
= π fµ 2σ 2 = 108.83 . The intrinsic impedance is
5-15
η$
=
2
Γ$ =
2
. ∠45o . The reflection and transmission coefficients are ∠45o = 0154
σ 2δ 2 η$ 2 − η o
η$ 2 + η o
= 0.9999∠179.97 o ≅ −1 and T$ =
(
2η$ 2 = 816 . × 10 −4 ∠45o . The fields η$ 2 + η o
)
(
)
in medium 1 are E i = 10 cos 6π × 10 6 t − 0.063z a x , E r = −10 cos 6π × 10 6 t + 0.063z a x ,
Hi =
(
)
(
)
10 10 cos 6π × 10 6 t − 0.063z a y , H r = cos 6π × 106 t + 0.063z a y . The fields 120π 120π
(
)
. × 10 −3 e −108.83z cos 6π × 10 6 t − 108.83z + 45o a x and in medium 2 are E t = 816
Ht =
816 . × 10 −3 −108.83z cos 6π × 10 6 t − 108.83z + 45o − 45o a y . Hence the average e 0154 .
(
)
power dissipated in the volume is $2 i 1 T E PAV = 2 η$ 2
2
( )
−3 cos θ η 2 1 − e −2α 2 ×10 × 2m 2 = 59.9µ W .
5.5.6 The intrinsic impedance and phase constant of the first medium are η1 = η o = 120π and β = β o = ω v = 20.94 . Medium 2 (stainless steel) is not a good conductor as evidenced o
by
σ2 = 0.36 . Hence we must compute the attenuation and phase constants directly ω ε oε r 2
as
γ$
2 =
jω µ o µ r (σ + jω ε oε r ) =
1 j 2π × 109 × 4π × 10 −7 × 500 0.02 + j 2π × 109 × × 10 −9 36π
This evaluates to γ$ 2 = 482.81∠801 . o = 83 + j 475.62 . Hence we identify α 2 = 83 and rad β 2 = 475.62 . The intrinsic impedance is m η$
2 =
jω µ o µ r j 2π × 109 × 4π × 10 −7 × 500 .which evaluates to = (σ + jω ε oε r ) 0.02 + j 2π × 109 × 1 × 10 −9 36π
η$ 2 = 8176.83∠9.9 o . The reflection and transmission coefficients are Γ$ =
η$ 2 − η o η$ 2 + η o
(
= 0.91∠0.91o and T$ =
)
2η$ 2 = 191 . ∠0.43o . The fields in medium 1 are η$ 2 + η o
(
)
E i = 100 cos 2π × 109 t − 20.94 z a x , E r = 91 cos 2π × 109 t + 20.94 z + 0.91o a x ,
5-16
Hi =
(
)
(
)
91 100 cos 2π × 109 t − 20.94 z a y , H r = − cos 2π × 109 t + 20.94 z + 0.91o a y . 120π 120π
(
)
The fields in medium 2 are E t = 191e −83z cos 2π × 109 t − 475.62 z + 0.43o a x and
(
)
191 e −83z cos 2π × 10 9 t − 475.62 z + 0.43o − 9.9 o a y . In the stainless steel a 8176.83 1 skin depth is δ = = 5.03mm . Hence the average power dissipated in the π fµ o µ r σ
Ht =
$2 i 1 T E volume is PAV = 2 η$ 2
2
( )
−3 cos θ η 2 1 − e −2α 2 × 5.03×10 × 2m 2 = 2.49 W .
5.5.7 The intrinsic impedance and phase constant of the first medium are η1 = η o and β =
ω
µr = 40π εr
µ r ε r = 314 . . Medium 2 is a good conductor as evidenced by
vo
σ2 = 720 . Hence the attenuation and phase constants can be computed as ω ε oε r 2 α
2 =
η$
2
Γ$ =
=
β2= 2
1
= π fµ 2σ 2 = 198.69 . The intrinsic impedance is
δ2
∠45o = 14.05∠45o . The reflection and transmission coefficients are
σ 2δ 2 η$ 2 − η o
η$ 2 + η o
= 0.854∠170.9 o and T$ =
(
)
2η$ 2 = 0.207∠40.81o . The fields in medium $ η 2 + ηo
(
)
. z a x , E r = 4.27 cos 10π × 108 t + 314 . z + 170.9o a x , 1 are E i = 5 cos 10π × 108 t − 314
Hi =
(
)
(
)
4.27 5 cos 10π × 108 t − 314 . z a y , Hr = − cos 10π × 108 t + 314 . z + 170.9o a y . 40π 40π
(
)
The fields in medium 2 are E t = 104 . e −198.69 z cos 10π × 108 t − 198.69 z + 40.81o a x and
Ht =
(
)
104 . e −198.69 z cos 10π × 108 t − 198.69 z + 40.81o − 45o a y . Hence the average 14.05
power dissipated in the volume is $2 i 1 T E PAV = 2 η$ 2
2
( )
−2 cos θ η 2 1 − e −2α 2 ×10 × 10 −4 m 2 = 2.64µ W .
5-17
5.5.8 The ocean is not a good conductor as evidenced by
σ = 0127 . . Hence we must ω ε oε r
calculate the intrinsic impedance directly as η$
2
=
jω µ o µ r j 2π × 7 × 109 × 4π × 10 −7 = 4172 . ∠362 . o . Hence the = (σ + jω ε oε r ) 4 + j 2π × 7 × 109 × 1 × 10 −9 × 81 36π
reflection coefficient is Γ$ =
η$ 2 − η o η$ 2 + η o
= 0.801∠17919 . o . The reflected power is
proportional to the square of the magnitude of the reflection coefficient. Hence the portion of the incident power that is reflected is 64.2% and the incident power that is dissipated in the ocean is 35.8%.
5.5.9 The total electric field is approximately zero at a distance of one-half wavelength from the surface of a good conductor. Hence λ o = 2m . Therefore the lowest possible frequency v of the wave is f = o = 150 MHz . 2
5.61 . The problem solution is sketched below. From Snell’s law sin θ t n sin θ t 1 1 sin ψ sin ψ = sin(90 − θ ) = cos θ . Similarly, = . Thus = 1. 1 sin ψ sin(90 − θ ) sin ψ n n Thus the direction of the beam as it exits the material is the same as the incident beam. t t The angle φ = 90 − θ − ψ and r = . Hence d = r sin φ = cos(θ + ψ ) . We cos ψ cos ψ need to eliminate ψ from this expression by writing it in terms of θ . We have the cos θ . identity cos(θ + ψ ) = cos θ cos ψ − sin θ sin ψ . Also we have Snell’s law sin ψ = n sin θ cos θ Substituting gives d = t cos θ − t . n 2 − cos 2 θ
5-18
n
d
r
t
90 – t
5.6.2 The critical angle is given by sin θ c =
1 1 . o . Hence, the ray strikes = giving θ c = 4181 . n 15
the back face with an angle of incidence of 45o which is greater than the critical angle and hence the ray is completely reflected. According to Snell’s law it will be reflected also with an angle of 45o and will strike the bottom face normal to it. The transmission 2 2η n = 0.8 . At the bottom face, the coefficient at the front face is T1 = = 1 ηo + η n +1 2η o 2 transmission coefficient is T1 = . . Hence the net transmission = = 12 ηo + η 1 + 1 n 2
Ei coefficient is T1T2 = 0.96 . The incident power density is S = and the transmitted 2η o i
2
T 2T 2 E i power density is S = 1 2 . Hence the ratio of the transmitted to incident power 2η o t
2
densities is (T1T2 ) = 0.922 .
5-19
5.6.3
The problem is sketched below. Snell’s law requires that sin θ t = n sin θ i . Hence 6 1 tan(90 − θ t ) = = so that θ t = 733 . o and θ i = 39.7 o . Thus 20 tan θ t D = 10 tan θ i = 8.3 ft . Thus the fish is at a distance of 28.3 ft from the boat.
t
6 ft.
10 ft.
n = 1.5
20 ft.
5-20
i
D
