spring 2013 hw 9 solns

8/14/2019 Spring 2013 HW 9 Solns

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8/15

Problem 7.4 The electric field of a plane wave propagating in a nonmagnetic

material is given by

E= [y 3sin(107t0.2x)+ z 4cos(107t0.2x)] (V/m)

Determine

(a) The wavelength.

(b) r.

(c) H.

Solution:

(a)Since k=0.2,

= 2k

= 20.2

=10 m.

(b)

up=

k=107

0.2 =5107 m/s.

But

up= cr

.

Hence,

r=

c

up

2=

31085

107

2=36.

(c)

H= 1

kE= 1

x

y3sin(107t0.2x) + z4cos(107t0.2x)

=z3

sin(107t0.2x) y4

cos(107t0.2x) (A/m),

with

= 0r 120

6 =20= 62.83 ().


9/15

Problem 7.6 The electric field of a plane wave propagating in a lossless,

nonmagnetic, dielectric material withr=

2.56 is given by

E= y 20cos(6109t kz) (V/m)

Determine:

(a) f,up,,k, and .

(b) The magnetic fieldH.

Solution:

(a)

= 2f=6109 rad/s,

f=3109

Hz=3 GHz,

up= cr

=3108

2.56=1.875108 m/s,

=up

f=

1.8751086109 =3.12 cm,

k=2

=

2

3.12102 =201.4 rad/m,

= 0r

= 377

2.56=

377

1.6=235.62 .

(b)

H=x20

cos(6109t kz)

=x 20235.62

cos(6109t201.4z)

=x 8.49102 cos(6109t201.4z) (A/m).


10/15

Problem 7.12 The electric field of an elliptically polarized plane wave is given by

E(z, t) = [x 10sin(t kz60)

+y 30cos(tkz)] (V/m)

Determine the following:

(a) The polarization angles(,).

(b) The direction of rotation.

Solution:

(a)

E(z, t) = [x10sin(t kz60) +y30cos(t kz)]

= [x10cos(t kz+ 30

) +y30cos(t kz)] (V/m).

Phasor form: E= (x10ej30 +y30)ejkz.Sinceis defined as the phase ofEy relative to that ofEx,

= 30,

0=tan1

30

10

=71.56,

tan2= (tan20) cos= 0.65 or = 73.5,

sin2= (sin20) sin=

0.40 or =

8.73

.

(b)Since


11/15

Problem 7.3 The electric field phasor of a uniform plane wave is given by

E= y 10ej0.2z (V/m). If the phase velocity of the wave is 1.5 108 m/s and therelative permeability of the medium is r=2.4, find the following:(a) The wavelength.

(b) The frequency fof the wave.

(c) The relative permittivity of the medium.

(d) The magnetic fieldH(z, t).

Solution:

(a)FromE= y10ej0.2z (V/m), we deduce thatk=0.2 rad/m. Hence,=

2

k=

2

0.2= 10= 31.42 m.

(b)

f=up

=

1.510831.42

=4.77106 Hz=4.77 MHz.

(c)From

up= crr

, r= 1

r

c

up

2=

1

2.4

3

1.5

2=1.67.

(d)

=

120

r

r=120

2.4

1.67=451.94 (),

H= 1

(z) E= 1

(z) y10ej0.2z = x 22.13ej0.2z (mA/m),

H(z, t) = x 22.13cos(t+ 0.2z) (mA/m),

with= 2f=9.54106 rad/s.


12/15

Problem 7.5 A wave radiated by a source in air is incident upon a soil surface,

whereupon a part of the wave is transmitted into the soil medium. If the wavelength

of the wave is 60 cm in air and 20 cm in the soil medium, what is the soils relative

permittivity? Assume the soil to be a very low-loss medium.

Solution:From = 0/

r,

r =

0

2=

60

20

2= 9.


13/15

Problem 7.10 The electric field of a uniform plane wave propagating in free space

is given by E= (x +j y)30ejz/6 (V/m)Specify the modulus and direction of the electric field intensity at the z =0 plane att= 0, 5, and 10 ns.

Solution:

E(z,t) = Re[Eejt]=Re[(x +j y)20ejz/6ejt]

=Re[(x +yej/2)20ejz/6ejt]

=x20cos(tz/6) +y20cos(tz/6 +/2)

=x20cos(tz/6)y20sin(tz/6) (V/m),

|E|=E2x+E

2y

1/2=20 (V/m),

= tan1Ey

Ex

=(tz/6).

From

f= c

=

kc

2 =

/63108

2 =2.5107 Hz,

= 2f=5107 rad/s.

Atz=0,

= t=5107t=

0 at t= 0,0.25=45 at t= 5 ns,0.5= 90 at t= 10 ns.

Therefore, the wave is LHC polarized.


14/15

Problem 7.11 A linearly polarized plane wave of the formE= xaxejkz can be

expressed as the sum of an RHC polarized wave with magnitudeaR

, and an LHC

polarized wave with magnitudeaL. Prove this statement by finding expressions for

aRand aLin terms ofax.

Solution:

E= xaxejkz,

RHC wave: ER=aR(x+yej/2)ejkz =aR(x jy)e

jkz,

LHC wave: EL=aL(x+yej/2)ejkz =aL(x+j y)e

jkz,

E=ER+ EL,

xax=aR(x jy)+aL(x+j y).

By equating real and imaginary parts,ax= aR+ aL, 0= aR+ aL, oraL= ax/2,aR= ax/2.


15/15

Problem 7.13 Compare the polarization states of each of the following pairs of

plane waves:

(a) Wave 1: E1= x 2cos(t kz)+y 2sin(t kz).Wave 2: E2= x 2cos(t+ kz)+y 2sin(t+ kz).

(b) Wave 1: E1= x 2cos(t kz)y 2sin(t kz).Wave 2: E2= x 2cos(t+ kz)y 2sin(t+ kz).

Solution:

(a)

E1= x 2cos(tkz)+y 2sin(tkz)

= x 2cos(tkz)+y 2cos(t kz/2),

E1= x 2ejkz

+y 2ejkzej/2

,0=tan

1ayax

=tan1 1=45,

= /2.

Hence, wave 1 is RHC.

Similarly, E2= x 2ejkz +y 2ejkzej/2.Wave 2 has the same magnitude and phases as wave 1 except that its direction is

along z instead of+z. Hence, the locus of rotation ofE will match the left handinstead of the right hand. Thus, wave 2 is LHC.

(b)

E1= x 2cos(tkz)y 2sin(tkz),

E1= x 2ejkz +y 2ejkzej/2.Wave 1 is LHC. E2= x 2ejkz +y 2ejkzej/2.Reversal of direction of propagation (relative to wave 1) makes wave 2 RHC.

spring 2013 hw 9 solns

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