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Problem 8.18 For some types of glass, the index of refraction varies withwavelength. A prism made of a material with

n = 1 . 71 4

30 0 ( 0 in m) ,

where 0 is the wavelength in vacuum, was used to disperse white light as shown inFig. P8.18. The white light is incident at an angle of 50 , the wavelength 0 of redlight is 0 . 7 m, and that of violet light is 0 . 4 m. Determine the angular dispersionin degrees.

Solution:

50

60

R e d

G r e e n

V i o l e t

Angular dispersionA

B

C 4 3

2

Figure P8.18: Prism of Problem 8.18.

For violet,

n v = 1 . 71 430 0 . 4 = 1 . 66 , sin 2 = sin n v

= sin50

1 . 66 ,

or 2 = 27 . 48 .

From the geometry of triangle ABC ,

180 = 60 + ( 90 2) + ( 90 3) ,

or 3 = 60 2 = 60 27 . 48 = 32 . 52 ,

and sin 4 = n v sin 3 = 1 . 66sin32 . 52 = 0 . 89 ,

or 4 = 63 . 18 .

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For red,

n r = 1 . 71 430 0 . 7 = 1 . 62 ,

2 = sin 1 sin50

1 . 62 = 28 . 22 ,

3 = 60 28 . 22 = 31 . 78 , 4 = sin

1 [1 . 62sin31 . 78 ] = 58 . 56 .

Hence, angular dispersion = 63 . 18 58 . 56 = 4 . 62 .

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Problem 8.27 A plane wave in air with

E i = y 20e j(3 x+ 4 z) (V/m)

is incident upon the planar surface of a dielectric material, with r = 4, occupying thehalf-space z 0. Determine:

(a) The polarization of the incident wave.

(b) The angle of incidence.

(c) The time-domain expressions for the reected electric and magnetic elds.

(d) The time-domain expressions for the transmitted electric and magnetic elds.

(e) The average power density carried by the wave in the dielectric medium.

Solution:(a) E

i = y 20e j(3 x+ 4 z) V/m.Since E i is along y, which is perpendicular to the plane of incidence, the wave is

perpendicularly polarized.(b) From Eq. (8.48a), the argument of the exponential is

jk 1( xsin i + zcos i) = j(3 x+ 4 z) .Hence,

k 1 sin i = 3, k 1 cos i = 4,

from which we determine that

tan i = 34 or i = 36 . 87,

andk 1 = 32 + 42 = 5 (rad/m) .

Also, = upk = ck = 3 108 5 = 1. 5 109 (rad/s) .

(c)

1 = 0 = 377 ,

2 = 0 r

2

= 0

2 = 188 . 5 ,

t = sin1sin i r2 = sin

1 sin36 . 87 4 = 17 . 46, =

2 cos i 1 cos t 2 cos i + 1 cos t

= 0 . 41 ,

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= 1 + = 0. 59 .

In accordance with Eq. (8.49a), and using the relation E r0 = E i0 ,

E r = y 8 . 2 e j(3 x4 z) ,H r = (x cos i + z sin i)

8. 2 0

e j(3 x4 z) ,

where we used the fact that i = r and the z-direction has been reversed.

E r = Re [E re j t ] = y 8 . 2cos (1 . 5 109t 3 x+ 4 z) (V/m) ,H r = (x 17 . 4 + z 13 . 06) cos (1 . 5 109t 3 x+ 4 z) (mA/m) .

(d) In medium 2,

k 2 = k 1 2 1 = 5 4 = 20 (rad/m) ,and

t = sin1 1 2 sin i = sin1 12 sin36 . 87 = 17 . 46and the exponent of E t and H t is

jk 2( xsin t + zcos t) = j10( xsin17 . 46+ zcos17 . 46) = j(3 x+ 9 . 54 z) .

Hence,

E t = y 20 0 . 59 e j(3 x+ 9 . 54 z) ,H t = ( x cos t + z sin t)

20 0 . 59 2

e j(3 x+ 9 . 54 z) .

E t = Re [E te j t ] = y 11 . 8cos (1 . 5 109t 3 x9 . 54 z) (V/m) ,H t = ( x cos17 . 46+ z sin17 . 46)

11 . 8188 . 5

cos (1 . 5 109t 3 x9 . 54 z)= ( x 59 . 72 + z 18 . 78) cos(1 . 5 109t 3 x9 . 54 z) (mA/m) .

(e)

S tav = | E t0|

2

2 2= (11 . 8)

2

2 188 . 5 = 0 . 36 (W/m 2) .

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Problem 9.2 A 1-mlong dipole is excited by a 1-MHz current with an amplitudeof 12 A. What is the average power density radiated by the dipole at a distance of 5 km in a direction that is 45 from the dipole axis?

Solution: At 1 MHz, = c/ f = 3 108/ 106 = 300 m. Hence l/ = 1/ 300, andtherefore the antenna is a Hertzian dipole. From Eq. (9.12),

S ( R, ) = 0k 2 I 20 l

2

32 2 R2sin 2

= 120 (2 / 300 )2 122 12

32 2 (5 103)2 sin2 45 = 1.51 10 9 (W/m 2).

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Problem 9.12 Assuming the loss resistance of a half-wave dipole antenna to benegligibly small and ignoring the reactance component of its antenna impedance,calculate the standing-wave ratio on a 50- transmission line connected to the dipoleantenna.

Solution: According to Eq. (9.48), a half wave dipole has a radiation resistance of 73 . To the transmission line, this behaves as a load, so the reection coefcient is

= R rad Z 0 R rad + Z 0

=73 50 73 + 50

= 0 . 187 ,

and the standing wave ratio is

S =1 + | |

1 |

|

=1 + 0 . 187

1 0 . 187 = 1 . 46 .

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Problem 9.20 A 3-GHz line-of-sight microwave communication link consists of two lossless parabolic dish antennas, each 1 m in diameter. If the receive antennarequires 10 nW of receive power for good reception and the distance between theantennas is 40 km, how much power should be transmitted?

Solution: At f = 3 GHz, = c/ f = ( 3 108 m/s )/ (3 109 Hz) = 0.10 m. Solvingthe Friis transmission formula (Eq. (9.75)) for the transmitted power:

Pt = Prec 2 R2

tr At Ar

= 10 8 (0.100 m )2(40 103 m)2

1 1 ( 4 (1 m )2)( 4 (1 m )

2)= 25.9 10 2 W = 259 mW .

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Problem 8.19 The two prisms in Fig. P8.19 are made of glass with n = 1.5. Whatfraction of the power density carried by the ray incident upon the top prism emergesfrom the bottom prism? Neglect multiple internal reections.

45

45

45

45

90

90

S i

S t

65

4

3

21

Figure P8.19: Periscope prisms of Problem 8.19.

Solution: Using = 0/ n , at interfaces 1 and 4,

a = n 1 n 2n 1 + n 2

= 1 1.51 + 1.5

= 0.2.

At interfaces 3 and 6, b = a = 0.2.

At interfaces 2 and 5,

c = sin 1 1

n= sin 1

11.5

= 41.81 .

Hence, total internal reection takes place at those interfaces. At interfaces 1, 3, 4and 6, the ratio of power density transmitted to that incident is (1 2). Hence,

S t

S i = ( 1 2)4 = ( 1 (0.2)2)4 = 0.85 .

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Problem 8.22 Figure P8.22 depicts a beaker containing a block of glass on thebottom and water over it. The glass block contains a small air bubble at an unknowndepth below the water surface. When viewed from above at an angle of 60 , the airbubble appears at a depth of 6.81 cm. What is the true depth of the air bubble?

60

10 cm

2

d a

d 2

2

3

x2

x

x1

d t

Figure P8.22: Apparent position of the air bubble inProblem 8.22.

Solution: Let

d a = 6 . 81 cm = apparent depth ,

d t = true depth .

2 = sin 1 n 1

n 2sin i = sin

1 11 . 33

sin60 = 40 . 6 ,

3 = sin 1 n 1

n 3sin i = sin

1 11 . 6

sin60 = 32 . 77 ,

x1 = ( 10 cm ) tan40 . 6 = 8 . 58 cm ,

x = d a cot30 = 6 . 81cot30 = 11 . 8 cm .

Hence, x2 = x x1 = 11 . 8 8 . 58 = 3 . 22 cm ,

andd 2 = x 2 cot32 . 77

= ( 3 . 22 cm ) cot32 . 77 = 5 cm .Hence, d t = ( 10 + 5) = 15 cm.

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Problem 8.28 Repeat Problem 8.27 for a wave in air with

H i = y 2 102e j(8 x+ 6 z) (A/m)incident upon the planar boundary of a dielectric medium ( z 0) with r = 9.Solution:

(a) Hi= y 2 102e j(8 x+ 6 z) .

Since H i is along y, which is perpendicular to the plane of incidence, the wave isTM polarized, or equivalently, its electric eld vector is parallel polarized (parallel tothe plane of incidence).

(b) From Eq. (8.65b), the argument of the exponential is

jk 1( xsin i + zcos i) =

j(8 x+ 6 z) .

Hence,k 1 sin i = 8, k 1 cos i = 6,

from which we determine

i = tan186

= 53 . 13,

k 1 = 62 + 82 = 10 (rad/m) .Also,

= upk = ck = 3108 10 = 3 109 (rad/s) .(c)

1 = 0 = 377 ,

2 = 0 r2 =

03

= 125 . 67 ,

t = sin1 sin i r2 = sin1 sin53 . 13 9 = 15 . 47,

= 2 cos t 1 cos i 2 cos t + 1 cos i

= 0 . 30 , = ( 1 + )

cos i

cos t= 0. 44 .

In accordance with Eqs. (8.65a) to (8.65d), E i0 = 2 102 1 andE i = ( x cos i z sin i) 2 102 1 e j(8 x+ 6 z) = ( x 4 . 52 z 6 . 03) e j(8 x+ 6 z) .

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E r is similar to E i except for reversal of z-components and multiplication of amplitudeby . Hence, with

= 0 . 30,

E r = Re [Ere j t ] = (x 1 . 36 + z 1 . 81) cos (3 109t 8 x+ 6 z) V/m ,

H r = y 2 102 cos (3 109t 8 x+ 6 z)= y 0 . 6 102 cos (3 109t 8 x+ 6 z) A/m .

(d) In medium 2,

k 2 = k 1 2 1 = 10 9 = 30 rad/m , t = sin1

2

1sin i = sin1

1

3 sin53 . 13 = 15 . 47,

and the exponent of E t and H t is

jk 2( xsin t + zcos t) = j30( xsin15 . 47 + zcos15 . 47) = j(8 x+ 28 . 91 z) .Hence,

E t = ( x cos t z sin t) E i0 e j(8 x+ 28 . 91 z)= ( x 0 . 96 z 0 . 27) 2 102 377 0 . 44 e j(8 x+ 28 . 91 z)= ( x 3 . 18 z 0 . 90) e j(8 x+ 28 . 91 z) ,

H t = y E i0 2

e j(8 x+ 28 . 91 z)

= y 2 . 64 102 e j(8 x+ 28 . 91 z) ,E t = Re {E te j t }

= ( x 3 . 18 z 0 . 90) cos (3 109t 8 x28 . 91 z) V/m ,H t = y 2 . 64 102 cos (3 109t 8 x28 . 91 z) A/m .

(e)

S tav = | E t0|2

2 2= | H t0|2

2 2 =

(2 . 64 102)22 125 . 67 = 44 mW/m 2 .

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Hence,

k 1 = 22 + 32 = 3.6 (rad/m)

i = tan1(2/ 3) = 33.7.

Also,

k 2 = k 1 r2 = 3.6 2.25 = 5.4 (rad/m) 2 = sin1 sin i 12.25 = 21.7.

(b)

k 1 = 2 f

c

f = k 1c2

= 3.6 3 108

2 = 172 MHz .

(c) In order to determine the electric eld of the reected wave, we rst have todetermine the polarization of the wave. The vector argument in the given expression

for Ei

indicates that the incident wave is a mixture of parallel and perpendicularpolarization components. Perpendicular polarization has a y-component only (see8.46a), whereas parallel polarization has only x and z components (see 8.65a). Hence,we shall decompose the incident wave accordingly:

Ei

= Ei

+ Ei

with

Ei

= y 4e j(2 x+ 3 z) (V/m)E

i = ( x 9 z 6)e j(2 x+ 3 z) (V/m)

From the above expressions, we deduce:

E i 0 = 4 V/m E i 0 =

92 + 62 = 10.82 V/m .

Next, we calculate and for each of the two polarizations:

=cos i ( 2/ 1) sin 2 icos i + ( 2/ 1) sin 2 i

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Using i = 33.7 and 2/ 1 = 2.25 / 1 = 2.25 leads to:

= 0.25 = 1 + = 0.75 .

Similarly,

= ( 2/ 1) cos i + ( 2/ 1) sin 2 i( 2/ 1) cos i + ( 2/ 1) sin 2 i = 0.15 ,

= ( 1 + )cos icos t

= ( 1 0.15)cos33 .7cos21 .7

= 0.76 .

The electric elds of the reected and transmitted waves for the two polarizations aregiven by (8.49a), (8.49c), (8.65c), and (8.65e):

Er

= y E r 0e jk 1( xsin r zcos r )

Et

= y E t 0e jk 2( xsin t+ zcos t)

Er = ( x cos r + z sin r) E r0e jk 1 ( xsin r zcos r)

Et = ( x cos t z sin t) E t 0e jk 2 ( xsin t+ zcos t)

Based on our earlier calculations:

r = i = 33.7 t = 21.7k 1 = 3.6 rad/m , k 2 = 5.4 rad/m ,

E r 0 = E i

0 = ( 0.25) (4) = 1 V/m . E t 0 = E

i0 = 0.75 (4) = 3 V/m .

E r0 = E i0 = ( 0.15) 10 .82 = 1.62 V/m .

E t 0 = E i0 = 0.76 10 .82 = 8.22 V/m .

Using the above values, we have:

E

r

= Er

+ Er

= ( x E r0 cos r + y E r

0 + z E r0 sin r)e j(2 x3 z)

= ( x 1.35 + y z 0.90)e j(2 x3 z) (V/m) .

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(d)

E t = Et + E t

= ( x 7.65 y3 z 3.05)e j(2 x+ 5 z) (V/m) .(e)

S t = | E t0|22 2

| E t0|2 = ( 7.65)2 + 32 + ( 3.05)2 = 76.83 2 =

0 r2 =

3771.5

= 251 .3

S t = 76.832 251 .3

= 152 .86 (mW/m 2).

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Problem 9.14 For a short dipole with length l such that l , instead of treatingthe current

I

( z

) as constant along the dipole, as was done in Section 9-1, a more

realistic approximation that ensures the current goes to zero at the dipole ends is todescribe I ( z) by the triangular function

I ( z) = I 0(1 2 z/ l), for 0 z l/ 2 I 0(1 + 2 z/ l), for l/ 2 z 0

as shown in Fig. P9.14. Use this current distribution to determine the following:

(a) The far-eld E ( R, , ).(b) The power density S ( R, , ).(c) The directivity D.

(d) The radiation resistance Rrad .

l I 0

I ( z)~

Figure P9.14: Triangular current distribution on a shortdipole (Problem 9.14).

Solution:(a) When the current along the dipole was assumed to be constant and equal to I 0 ,

the vector potential was given by Eq. (9.3) as:

A( R) = z 04

e jkR

R l/ 2 l/ 2 I 0 d z.If the triangular current function is assumed instead, then I 0 in the above expressionshould be replaced with the given expression. Hence,

A = z 04

e jkR

R I 0 l/ 20 1 2 zl dz + 0 l/ 2 1 + 2 zl dz = z 0 I 0l8 e

jkR

R,

which is half that obtained for the constant-current case given by Eq. (9.3). Hence,the expression given by (9.9a) need only be modied by the factor of 1 / 2:

E = E = jI 0lk 0

8 e jkR

Rsin .

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(b) The corresponding power density is

S ( R, ) = | E |2

20=

0k 2 I 20 l2

128 2 R2sin 2 .

(c) The power density is 4 times smaller than that for the constant current case, butthe reduction is true for all directions. Hence, D remains unchanged at 1.5.

(d) Since S ( R, ) is 4 times smaller, the total radiated power Prad is 4-timessmaller. Consequently, Rrad = 2Prad / I 20 is 4 times smaller than the expression givenby Eq. (9.35); that is,

Rrad = 20 2(l / )2 ( ).

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Problem 9.18 A car antenna is a vertical monopole over a conducting surface.Repeat Problem 9.5 for a 1-mlong car antenna operating at 1 MHz. The antennawire is made of aluminum with c = 0 and c = 3.5 107 S/m, and its diameter is1 cm.

Solution:(a) Following Example 9-3, = c/ f = ( 3 108 m/s )/ (106 Hz ) = 300 m. As

l / = 2 (1 m)/ (300 m ) = 0.0067, this antenna is a short (Hertzian) monopole.From Section 9-3.3, the radiation resistance of a monopole is half that for acorresponding dipole. Thus,

Rrad = 12 80 2(

l

)2

= 40 2(0.0067 )2 = 17.7 (m ),

Rloss = l2 a f c

c= 1 m

(10 2 m) (106 Hz )(4 10

7 H/m )3.5 107 S/m

= 10.7 m ,

= Rrad

Rrad + Rloss=

17.7 m17 .7 m + 10 .7 m

= 62% .

(b) From Example 9-2, a Hertzian dipole has a directivity of 1.5. The gain, fromEq. (9.29), is G = D = 0.62 1.5 = 0.93 = 0.3 dB.

(c) From Eq. (9.30a),

I 0 = 2Prad Rrad = 2(80 W )17 .7 m = 95 A,and from Eq. (9.31),

Pt = Prad

=

80 W0.62

= 129 .2 W .

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Problem 9.21 A half-wave dipole TV broadcast antenna transmits 1 kW at 50 MHz.What is the power received by a home television antenna with 3-dB gain if located ata distance of 30 km?

Solution: At f = 50 MHz, = c/ f = ( 3 108 m/s )/ (50 106 Hz ) = 6 m, for whicha half wave dipole, or larger antenna, is very reasonable to construct. Assuming theTV transmitter to have a vertical half wave dipole, its gain in the direction of thehome would be G t = 1.64. The home antenna has a gain of G r = 3 dB = 2. From theFriis transmission formula (Eq. (9.75)):

Prec = Pt 2G rG t(4 )2 R2

= 103 (6 m )2 1.64 2(4 )2(30 103 m)2

= 8.3 10 7 W .

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Problem 9.22 A 150-MHz communication link consists of two vertical half-wavedipole antennas separated by 2 km. The antennas are lossless, the signal occupies abandwidth of 3 MHz, the system noise temperature of the receiver is 600 K, and thedesired signal-to-noise ratio is 17 dB. What transmitter power is required?

Solution: From Eq. (9.77), the receiver noise power is

Pn = KT sys B = 1.38 10 23 600 3 106 = 2.48 10 14 W .

For a signal to noise ratio S n = 17 dB = 50, the received power must be at least

Prec = S nPn = 50(2.48 10 14 W) = 1.24 10 12 W .

Since the two antennas are half-wave dipoles, Eq. (9.47) states D t = Dr = 1.64, and

since the antennas are both lossless, Gt = Dt and Gr = D r . Since the operatingfrequency is f = 150 MHz, = c/ f = ( 3 108 m/s )/ (150 106 Hz ) = 2 m. Solvingthe Friis transmission formula (Eq. (9.75)) for the transmitted power:

Pt = Prec(4 )2 R2

2G rG t= 1.24 10 12

(4 )2(2 103 m)2

(2 m )2(1.64)(1.64)= 75 ( W) .

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