Transcript
Chapter 7 Problem Solutions
7.11 . Using the complete equations in (7.1): (a) 10cm. At 100MHz a wavelength is 3m. Hence 10cm is
1 λ o so the fields are in the 30
near field of the dipole. The magnetic field is o A H$ φ = 2.468 × 10 −2 ∠30 o ( j 4.77 + 22.797)e − j12 = 0.575∠29.8 o . The electric fields are m E$θ o V o V $ $ Er = 2069.67∠ − 6017 . . ∠ − 59.64 and Eθ = 9914 . The ratios are = 0.479 m m E$ r
and
E$θ H$ φ
= 1724.65 Ω .
(b) 1m. At 100MHz a wavelength is 3m. Hence 1m is
1 λ o so the fields are in the border 3
between the near field and far field of the dipole. The magnetic field is V A × 10 −2 ∠ − 255 and H$ φ = 1306 . . o . The electric fields are E$ r = 4.701∠ − 1155 . o m m E$θ E$θ V = 0.858 and = 308.8 Ω . E$θ = 4.033∠ − 3174 . o . The ratios are m H$ φ E$ r
(c) 10m. At 100MHz a wavelength is 3m. Hence 10m is 3333 . λ o so the fields are in the A far field of the dipole. The magnetic field is H$ φ = 118 . × 10 −3 ∠ − 2.7o . The electric m V V and E$θ = 0.444∠ − 2.74o . The ratios are fields are E$ r = 4.247 × 10 −2 ∠ − 92.73o m m E$θ E$θ = 10.45 and = 376.08 Ω ≅ η o . Since 10m is in the far field of the dipole it is E$ r H$ φ expected that the fields should approach the far-field approximations.
7-1
7.1.2 2
2
2 dl Rrad = 80π = 8.773 × 10 −3 Ω , PAV, radiated = I$rms Rrad = 438.64mW where λo 10 ∠30 o . I$rms = 2
7.1.3 The field points are in the far field of the dipole and at electrical distances of 166.7λ o and e 1667λ o . Hence the electric field and magnetic field are E$θ , H$ φ ∝ the electric field at 1000m is 1/10 of the field at 100m or 10
r − j 2π λo
r
. Hence
mV . The magnetic field at m
100m is the ratio of the electric field and the intrinsic impedance of free space: E$ E$ A A H$ = = 2.65 × 10 −4 . The magnetic field at 1000m is H$ = = 2.65 × 10 −5 . The ηo ηo m m r . Hence the difference in phase angles between the two phase angle at a point is −2π
λo
points is
2π (1000 − 100) = 1500 × 2π = 0o . Hence the fields at 100m and 1000m are 0.6
exactly in phase. The fields at 1000m lag those at 100m because of the minus sign in the $2 1 E . exponential. The average power densities are those of plane waves: S AV = 2 ηo µW µW Hence the average power density at 100m is 13.26 2 and at 1000m is 01326 . . m2 m Observe that the electric and magnetic fields decay inversely with distance, whereas the power densities decay inversely with the square of the distance.
7.21 . E µA Im I mV F (θ ) = 60 m = 60 , Hφ = θ = 15915 . . The power ηo 2π r r m m $ 2 µW 1 Eθ , and the total average power radiated = 4.775 density in the wave is S AV = 2 ηo m2 F (θ ) = 1, Eθ = η o
2 is PAV, rad = I$m, rms Rrad = 365mW .
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7.2.2 The input impedance to the monopole is (36.5 + j 2125 . )Ω . Attaching the source, we 100 = 1123 determine the input current to the antenna as I$m = . ∠ − 1388 . oA. . 50 + 36.5 + j 2125 2 1 Hence the total average power radiated is PAV, rad = I$m 36.5 = 23W . Broadside to he 2 Im V = 0.674 antenna Eθ = 60 which is perpendicular to the ground plane (hence r m satisfying the boundary condition on the electric field at the surface of a perfect $ 2 1 Eθ mW . = 0.602 conductor). The average power density in the wave is S AV = 2 ηo m2
7.2.3 1 λ o monopole is ( 20 − j50)Ω . Attaching the source, we 5 100 = 1162 . ∠35.54o A . determine the input current to the antenna as I$m = 50 + 20 − j50 2 1 . W and the radiation Hence the total average power radiated is PAV, rad = I$m 20 = 1351 2 The input impedance to the
resistance for this antenna is the real part of its input impedance since it is assumed lossless.
7.2.4 1 λ o monopole is ( 4 − j180)Ω . Attaching the source, we 10 100 = 0.532∠733 . o A . Hence determine the input current to the antenna as I$m = 50 + 4 − j180 2 1 the total average power radiated is PAV, rad = I$m 4 = 0.566W and the radiation 2 The input impedance to the
resistance for this antenna is the real part of its input impedance since it is assumed lossless.
7.2.5 The load on the transmission line is the input impedance to the antenna,
Z$ L = 73 + j 42.5 Ω . The input impedance to the transmission line can be found using the methods of Chapter 6. The load reflection coefficient is 7-3
73 + j 42.5 − 50 = 0.3713∠42.52 o . The reflection coefficient at the input to the line Γ$ L = 73 + j 42.5 + 50 is Γ$ ( 0) = Γ$ L e
− j 4π
λ = 0.3713∠42.52 o × 1∠ − 216 o = 0.3713∠ − 173.48 o . The input
impedance is Z$in = ZC
[1 + Γ$ (0)] = 2309 . ∠ − 5585 . o = 22.98 − j 2.25 . Hence the input $ [1 − Γ(0)]
current to the line is I$input to line =
100 . ∠176 . o . Hence the power = 137 50 + 22.98 − j 2.25
2 . W . But since the line delivered to the line input is PAV, to line = I$input to line 22.98 = 431
is lossless this is also the power delivered to the antenna input.
7.31 . π d α Since E ∝ cos cos φ + , we can determine the location of maxima and minima by 2 λo
differentiating this expression with respect to φ and setting the result to zero: π d d π d α α π d cos φ + = − sin cos φ + × − sin φ = 0 . The sin φ = 0 cos dφ λ o 2 2 λo λo condition results in maxima or minima at φ = 0 and φ = 180 o . This condition is a direct result of the pattern being symmetrical with respect to a line through the two antennas. πd α The other condition yields cos φ + = 0,±180o ,±360 o , L . λo 2
7.3.2 π d π α π λ E ∝ cos cos φ + (a) d = o 2 , α = 90 o , E ∝ cos cos φ + . Nulls at 2 4 2 λo 1 π π π 3π 5π or cos φ = or 60 o . Maxima and minima at cos φ + = ± ,± ,± 2 2 4 2 2 2 π 1 π π π sin cos φ + = 0 or cos φ + = 0,±π or cos φ = − giving ±120o ,0,180 o . The 2 4 2 2 4
pattern is sketched below.
0.707
0.707
1 60°
120°
7-4
(b) d =
5λ o
π 5π o 45 , α = , E ∝ cos cos + φ . Nulls at 8 8 8
5π π π 3π 5π cos φ + = ± ,± ,± or cos φ = 0.6,−1 or φ = ±5313 . o ,180 o . Maxima and 8 8 2 2 2 π 5π π 5π cos φ + = 0,±π giving φ = 0,180 o ,±10154 . o. minima at sin cos φ + = 0 or 8 8 8 8 The pattern is sketched below.
0.707
0.92
1.0
53.13° 101.54°
π π π 3π 5π (c) d = λ o , α = 180 o , E ∝ cos π cos φ + . Nulls at π cos φ + = ± ,± ,± or 2 2 2 2 2 π cos φ = 0,−1,+1 or φ = 90 o ,270o ,180o ,0 . Maxima and minima at sin π cos φ + = 0 or 2 π cos φ +
π 2
= 0,±π giving φ = ±120 o ,±60 o . The pattern is sketched below.
1.0
1.0 120°
60°
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π π π π 3π 5π π λ (d) d = o 4 , α = 180 o , E ∝ cos cos φ + . Nulls at cos φ + = ± ,± ,± 4 2 4 2 2 2 2 π π or cos φ = 0,−4 or φ = ±90o . Maxima and minima at sin cos φ + = 0 or 4 2 π 4
cos φ +
π 2
= 0,±π giving φ = 0 o ,180 o . The pattern is sketched below.
0.707
0.707
7.3.3 d = 164ft = 50m , f = 1500 × 103 so that λ o = 200m . Therefore d =
λo 4
.
π d π π 3π 5π 3π 3π α π = ± ,± or ,± E ∝ cos cos φ + = cos cos φ + . Nulls at cos φ + 2 4 8 4 8 2 2 2 λo
3π π cos φ = 0.5,−35 . or φ = ±60 o . Maxima and minima at sin cos φ + = 0 or 4 8 π 3π = 0,±π giving φ = 0 o ,180 o . The pattern is sketched below. cos φ + 4 8
0.38
0.92
60°
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7.3.4 π d α π π π 3π 5π or E ∝ cos cos φ + = cos π cos φ + . Nulls at π cos φ + = ± ,± ,± 2 4 4 2 2 2 λo
cos φ = 0.25,−0.75 or φ = ±75.52 o ,±138.58 o . Maxima and minima at
π π sin π cos φ + = 0 or π cos φ + = 0,±π giving φ = 0 o ,180 o ,±104.48 o ,±4141 . o . The 4 4 pattern is sketched below.
0.707
41.41°
0.707
138.58°
0.707
1
75.52°
104.48°
7.41 . 1 S AV ∝ 2 , r 2
2
S AV,5000ft × (5000ft ) = S AV, max × (rmax ) ⇒ rmax = 3535 . × 106 ft = 670miles .
7.4.2 2
λ PR = PT G R GT o , GT = 12dB = 15.85 , 4π d PR = 10 −9 , PT = 01 . , λ o = 3, d = 3844 . × 108 m . 2
PR 4π d GR = . × 109 = 9214dB . . = 1636 PT GT λ o
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7.4.3 45dB ⇒ 31,622.78 , PR = 10 −3 , λ o = 01 . m, d = 30miles = 48,280m . 2
PR 4π d PT = = 36.81W . G R GT λ o
7.4.4 60 PT GT , PT = 5 × 10 −3 , GT = 12dB = 15.85, d = 2 miles = 3218.7m . d V E$ = 0.68 . m E$ =
7.4.5 10 = 7.68 × 10 −2 ∠ − 191 . o, 123 + j 42.5 2 P W 1 Prad = I$ant 73 = 0.216W , S rad = G rad2 = 2.82 × 10 −10 2 , 2 m 4π r
G = 215 . dB ⇒ 1.64 . I$ant =
$2 60 PT GT mV 1 E mV S rad = ⇒ E$ = 0.461 . E$ = = 0.461 . m 2 ηo m d η I mV By equation (7.14) E$ = o m F (θ ) = 4.608 . Equation (7.16) gives the same result. m 2π r { 1
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