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5-1

Solutions for Chapter 5 Problems

1. General Wave Equations

P5.1: Starting with Maxwells equations for simple, charge-free media, derive theHelmholtz equation for H.

( )

2

2

t t

t t

= + =

EH E E + E

H H= -

Using a vector identity we also have: ( ) 2 = H H H

But 0 =H , leading to2

2

2t t

+

H H

H =

P5.2: Derive equation (5.10) by starting with the phasor point form of Maxwellsequations for simple, charge-free media.

For charge-free media the phasor form of Maxwells equations are:

( )

0

0

s

s

s s

s s

j

j

=

=

=

= +

D

B

E H

H E

Now we take the curl of both sides of Faradays Law,( ) ( ) ( )s s s sj j j j = = = +E H H E

Now since ( ) 2 = s s sE E E , and since 0s =E , we have

( )2 s sj j = +E E

P5.3: A wave with = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media.In the liquid, the wavelength is measured as 1.0 cm. What is the waves frequency (a) inair? (b) in the liquid? (c) What is the liquids relative permittivity?

(a)83 10

50.06

pu c x m sf GHz

m = = = =

(b) the frequency doesnt change with the media (the wavelength does) sof= 5 GHz(c)

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5-2

( )9 7

28

8

15 10 0.01 5 10

3 1036

0.5 10

p

r

r

m cu f x m x

s s

x

x

= = = =

= =

P5.4: Suppose Hs(z) =Hys(z) ay. Start with (5.14) and derive (5.29).

Since Hsis only a function ofz, (5.14) becomes2

2

20.s s

HH

z

=

(a)

If we let ,zsH Ae= then

22

2, and .z zs s

H HAe Ae

z z

= =

So (a) becomes ( ) ( )2 2 0, or 0. = + = This has two solutions:

(1) for 0, we have , , or .z zs s oH Ae H H e + + = = = =

(2) for 0, we have , , or .z zs s oH Ae H H e = = = =

The general solution is the linear superposition of the two, or

( ) .z zs o o yH e H e + = +H a

P5.5: Given = 1.0x10-5S/m , r= 2.0, r= 50., andf= 10. MHz, find , ,, and .

( )r o r oj j j = + = +

r o

r o

j

j

=

+

( )( ) ( )6 72 10 10 50 4 10 3948r oj j x x j = =

( )( )( )5 6 12 5 31 10 2 10 10 2 8.854 10 1 10 1.11 10r oj x j x x x j x + = + = + Inserting these into the expressions for and ,

3 3 2579.4 10 2.1 1 , 9.4 10 , 2.1 , 1880 jx j m x Np m rad m e = + = = =

These results are confirmed by ML0501.

P5.6: MATLAB: In some material, the constitutive parameters are constant over a largefrequency range and are given as = .10 S/m , r= 4.0, and r= 600. Write a MATLABroutine that will plot ,, and (magnitude and phase) versus the log of frequency from1 Hz up to 100 GHz.

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5-3

P5.7: Suppose E(x,y,t) = 5.0 cos(x106t 3.0x + 2.0y) az V/m. Find the direction ofpropagation, ap, and H(x,y,t).

3 25 j x j ys ze e=E a

We assume nonmagnetic material and therefore have3 2 3 210 15j x j y j x j ys s x yj j e e j e e

= = +E H a a

3 2 3 2 3 2 3 210 15 2.53 3.8j x j y j x j y j x j y j x j ys x y x yo

j je e e e e e e e

j j

= + =

H a a a a

( ) ( )6 6A

( , , ) 2.53cos 10 3 2 3.80cos 10 3 2m

x yx y t x t x y x t x y = + +H a a

To find the direction of propagation,s s

P

s s

=

E H

aE H

6 4 6 419 12.65j x j y j x j ys s x ye e e e = E H a a

And with the exponential terms canceling in the top and bottom of the equation for ap, wehave:

0.83 0.55 .P x y= a a a

Fig. P5.6

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5-4

P5.8: Suppose in free space, H(x,t) = 100.cos(2x107tx+ /4) azmA/m. Find E(x,t).

( )0.100 , , 4

120 0.100 12

j x j

s z P x

j x j j x j

s P s x z y

e e

e e e e

= = =

= = =

H a a a

E a H a a a

( )12 cos yt x = +E a

Since free space is stated,2 2

2 30 rad mc f

= = =

and then

7 212 cos 2 1030 4

y

Vx t x

m

= +

E a

2. Propagation in Lossless, Charge-Free MediaP5.9: Start with the Helmholtz equation (5.11), and using = j, derive (5.41), thetraveling wave equation.

22 2 2

20, let ( ) , and with =j we have 0.xss s s xs x xs

EE z E

z

= = + =

E E E a

Let2

2

2, so andz x xxs xsxs

E EE Ae A e A e

z z

= = =

Now we have2 2 2 20, or 0z zA e Ae + = + =

This can be factored: ( ) ( )2 2 0j j + = + = ,suggesting two solutions. The first solution uses j = and

.j z j zxs o

E Ae E e + = =

Likewise, the second solution uses j = + and

.j z j zxs oE Ae E e = =

The complete solution is a linear superposition of these two solutions, or

.j z j z

s o o xE e E e + += +E a

P5.10: A 100 MHz wave in free space propagates in the y direction with an amplitude of1 V/m. If the electric field vector for this wave has only an az component, find theinstantaneous expression for the electric and magnetic fields.

From the given information we have 62 200 10 rad

f xs

= = and2

,3p

rad

u m

= =

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5-5

or 62

( , ) 1cos 200 103

z

Vy t x t y

m

=

E a .

Now to find H.1 1 1

1 , 1120 120

j y j y j y

s z s P s y z xe e e

= = = =E a H a E a a a

So

( ) 61 2

, cos 200 10120 3

x

Ay t x t y

m

=

H a

or

( ) 62

, 2.7cos 200 10 .3

x

mAy t x t y

m

=

H a

P5.11: In a lossless, nonmagnetic material with r= 16, H= 100 cos(t 10y) azmA/m.Determine the propagation velocity, the angular frequency, and the instantaneousexpression for the electric field intensity.

883 10 0.75 10

16p

r

c x mu x

s

= = = =

( ) ( )8 80.75 10 10 7.5 10prad

u x xs

= = =

( )8( , ) 100cos 7.5 10 10 zmA

y t x t ym

= H a

0.100 ,

120 0.100 3

j y

s z

j y j y

s P s y z x

r

e

e e

=

= = =

H a

E a H a a a

( )8( , ) 9.4cos 7.5 10 10 xV

y t x t ym

= E a

P5.12: Given E = 120 cos(6x106t 0.080y) az V/m and H = 2.00cos(6x106t

0.080y) axA/m, find rand r.

0.080120 , 2j y j ys z s xe e

= =E a H a

1 1 120 120

120

j y j y j yrs P s y z x x

rr r

e e e

= = = =H a E a a a a

so we know

2r

r

=

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5-6

Now,6

61 6 10 75 100.080

p

r r

c xu x

= = = = =

64

75 10r r

c

x = =

And now(2)(4) 8r r r r

r

= = =

42

2r r

r

r r

= = =

3. Propagation in Dielectrics

P5.13: Work through the algebra to derive equation and equations (5.52) fromequations (5.50) and (5.51).

( )2 2 2 2 2 ;j j = + = +

Comparing the imaginary parts, we see 2 , or ,2

= =

and comparing the real parts, 2 2 2 0 + = .Rearranging and inserting our value for:

2 2 24 2 2 0

4

+ =

This is a quadratic expression (x2+ bx + c = 0), where here

22 2, ,2

x b c = = =

Solving the quadratic:2

2

2

4 1 1 1 44 1 1

2 2 2 2

b b c cx b c b b

b

= = =

Reinserting the a, b and c values:22 2 2

2 2 2

4 2 2

1 41 1 1 1

2 4

= + = +

2

1 1

= +

Now for :2

2 2 2 2 20, = , so 02 2

= =

Rearranging,

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5-7

2

4 2 2 02

=

Solving this quadratic we find

2

1 1

= + +

P5.14: MATLAB: Write a routine to prompt the user for a materials constitutiveparameters and an operating frequency, and calculate the andfrom (5.52). Verify theprogram by running Drill 5.6.

Now run the program for Drill 5.6:(a)

relative permeability: 1

real part of rel permittivity: 10complex part of rel permittivity: .01conductivity (S/m): 1e-12frequency (Hz): 100

alpha =3.3730e-009

beta =6.6268e-006

( b) relative permeability: 1real part of rel permittivity: 10complex part of rel permittivity: .01conductivity (S/m): 1e-12frequency (Hz): 1e6

alpha =3.3134e-005

beta =0.0663

These results agree with Drill 5.6.

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5-8

P5.15: Given a material with = 1.0x10-3 S/m, r = 1.0, and r = 3.0, r = 0.015,compare a plot of versus frequency from 1 Hz to 1 GHz using (5.52) to a similar plotusing (5.54). At what frequency does the % error exceed 2%?

P5.16: In a media with properties= 0.00964 S/m , r= 1.0, r= 100., and f= 100.MHz, a 1.0 mA/m amplitude magnetic field travels in the +x direction with its fieldvector in the z direction. Find the instantaneous form of the related electric field

intensity.

( )1 cos ;x x j xz s o zmA

e t x H e em

= =

H a H a

x j x x j x

s P s x o z o yH e e H e e = = =E a H a a a

( ) ( )( )( )( )

6 7

30

6 12

2 100 10 100 4 102664

0.00964 2 100 10 8.854 10

jj x xj

ej j x x

= = =

+ +

( )1

14.8 25.7j j jm

= + = +

Finally,

( )15 6( , ) 2.66 cos 200 10 26 30x yV

x t e x t xm

= +E a

Fig. P5.15

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5-9

P5.17: MATLAB: Make a pair of plots similar to Figure 5.4 for the 3 materials of Table5.1. Instead of loss tangent, one plot is to contain the magnitude of and the other is tohave the phase of .

4. Propagation in Conductors

P5.18: Starting with (5.13), show that =for a good conductor.

( ) for a good conductorj j j = +

( )1

, 1

2 2 22

2

j jj j j

+= = + = +

= =

(Note: we get the same result starting with (5.52) and assuming 1.

>>

Fig. P5.17

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5-10

P5.19: In seawater, a propagating electric field is given by E(z,t) = 20.e-zcos(2x106tz+ 0.5) ayV/m. Assuming =0, find (a) and, and (b) the instantaneous form of H.

For seawater we have r= 72,= 5, and r= 1.So: 7.896, 0.004o r oj j j j = =

44.981.257 jj

ej

= =

+

( ) 4.441 4.445 1 mj j j = + = +

14.4

m = =

0.5 28.620 20z z j radians z z js y yV V

e e e e e em m

= =E a a

28.6 28.61 1 2020 z z j z z js P s z y xA

e e e e e em

= = =H a E a a a

( )4.4 6( , ) 15.9 cos 2 10 4.4 28.6 45z xA

z t e x t zm

= + H a

or with appropriate significant digits:

( )4.4 6( , ) 16 cos 2 10 4.4 16z xA

z t e x t zm

= H a

P5.20: Calculate the skin depth at 1.00 GHz for (a) copper, (b) silver, (c) gold, and (d)nickel.

6

9 7 7

1 ;as an example, for copper at 1 GHz:

12.1 10 2.1

1 11 10 4 10 5.8 10

f

x m mH Vs A

x x xs m m HA V

=

= = =

Table P5.19

(S/m) r (m)Cu 5.8x107 1 2.1Ag 6.2x107 1 2.0Au 4.1x107 1 2.5

Ni 1.5x107 600 0.17

P5.21: For Nickel (= 1.45 x 107, r= 600), make a table of ,, , up, and for 1Hz,1kHz, 1MHz, and 1 GHz.

For Ni we have = 1.45x107S/m, r= 600

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5-11

( ) ( ) ( ) ( )7 7 3600 4 10 1.45 10 34.35 10 ( )f f Hz x x x f Hz = = = = = 1/

45 6 452 18.08 10 ( )j je x f Hz e

= =

612 10pr r

c mu xs

= =

Table P5.21

f(Hz)= 1 103 106 109

(Np/m) 185 5860 185x103 5.9x106

(rad/m) 185 5860 185x103 5.9x106 18ej45 570ej45 18ej45m 0.57ej45 5.4mm 170m 5.3m 170nm

up(m/s) 12x106 12x106 12x106 12x106

P5.22: A semi-infinite slab exists forz> 0 with = 300 S/m, r= 10.2, and r= 1.0. Atthe surface (z= 0),

E(0,t) = 1.0 cos(x 106t) axV/m.Find the instantaneous expressions for Eand Hanywhere in the slab.

The general expression for Eis: ( )6( , ) 1.0 cos 10z xV

z t e x t zm

= E a

( )( )6 710 4 10 3.948j j x x j = =

( )( )( )6 12 6

10 10.2 8.854 10 284 10j j x x j x

= = Here, >> (i.e. it is a good conductor), so

124.3f

m = = =

45 452 0.115j je e

= =

So now we have

( )24 6( , ) 1.0 cos 10 24z xV

z t e x t zm

= E a

To find Bwell work in phasors.

1 1 11 , 1z j z z j z z j zs x s P s z x ye e e e e e

= = = =E a H a E a a a

( )24 61

( , ) cos 10 24 450.115

z

y

Az t e x t z

m

= H a

( )24 6( , ) 8.7 cos 10 24 45z yA

z t e x t zm

= H a

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5-12

P5.23: In a nonmagnetic material, E(z,t) = 10.e-200zcos(2x 109t- 200z) axmV/m.Find H(z,t).

Since = , the media is a good metal. With r= 1 we have( )

( )( )

22

9 7

200, or 10.13

1 10 4 10o

o

Sf

f mx x

= = = =

45 452 28j je e

= =

1 1 1010 , 10z j z z j z z j zs x s P s z x ye e e e e e

= = = =E a H a E a a a

( )200 9( , ) 360 cos 2 10 200 45z ymA

z t e x t zm

= H a

P5.24: A 0.1 m layer of copper is deposited atop a very thick slab of nickel. For a fieldincident on the copper surface, (a) calculate Rsat 1.0 GHz. Compare this withRsat 1.0GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 m thickness of copper byitself.

Refer to Figure P5.24..In the copper portion the field is Cuzx xoE E e

=

In the nickel portion, ( ) ( )NiCu z ttx xoE E e e =

The current density in the copper is ,CuzxCu Cu xo

J E e

= and in the nickel is

( ) ( ).NiCu z ttxNi Ni xoJ E e e

= The current is

Fig. P5.24

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5-13

( )Cu Cu Niz t z tCu xo Ni xoI E e dydz E e e dydz

= + , or( )

,Cu Cu Nit

z t z t

Cu xo Ni xo

o t

I w E e dz w E e e dz

= + and upon evaluating

( )1 ,Cu Cut tCu Ni

xo

Cu NiI wE e e

= + and with V=ExoL,

we have ( )1

, where = 1 .Cu Cut tCu Ni

s s

Cu Ni

LR R R e e

w

= +

Now were ready to perform the calculations using the following data:

7 3Cu5.8 10 , 1, 479 10Cu r

S Npx x

m m = = =

7 6Cu1.5 10 , 600, 596 10Ni r

S Npx x

m m = = =

(a) 0.1m Cu over Ni:Rs= 176 m

(b) Semi-infinite Cu:Rs= 8.3 m(c) 0.1 m Cu:Rs= 177 m

P5.25: Calculate the DC resistance per meter length of a 4.0 mm diameter copper wire.Now find the resistance at 1.0 GHz.

( ) ( )22 71 1 1 1

DC: 1.375.8 10 0.002

R m

L a mx

= = =

1 GHz: 61

; 1 2.09 102 2

sRR f x mL a a

= = = =

( ) ( ) ( )7 61

0.665.8 10 2.09 10 2 0.002

R

L mx x

= =

5. The Poynting Theorem and Power Transmission

P5.26: In air, H(z,t) = 12.cos(x106t - z+ /6) ax A/m. Determine the power densitypassing through a 1.0 square meter surface that is normal to the direction of propagation.

( )

2

2

2

1 1

120 12 272 2avg xo z z z

A kW

H m m

= = =

P a a a

P5.27: A 600 MHz uniform plane wave incident in the z direction on a thick slab ofTeflon (r= 2.1, r= 1.0) imparts a 1.0 V/m amplitudey-polarized electric field intensityat the surface. Assuming = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c)Pav.

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5-14

( )( )6(0, ) 1cos 2 600 10 yV

t x t zm

= E a

( )( , ) 1 cosz yV

z t e t zm

= E a

Teflon: = 0 so = 0,

and( )6

8

2 600 102.1 18.2

3 10r

x rad

c x m

= = = =

(a) ( )9( , ) 1cos 1.2 10 18.2 yV

z t x t zm

= E a

(b)1 2.1

1 ,120

j z

s P s z y

Ve

m

= =

H a E a a

( )9( , ) 3.8cos 1.2 10 18.2 xmA

z t x t zm

= H a

(c) ( )2

2

1 2.11 1.92 120

avg z z

mW

m= =P a a

P5.28: Assume distilled water (= 10-4S/m, r= 81, r= 1.0) fills the regionz> 0. Atthe surface, we have E(0,t) = 8.0cos(2x108t) axV/m. Determine, for z > 0, (a) E(z,t),(b) H(z,t), and (c) Pavat z = 1.0 m. (d) Find the power passing through a 10 square metersurface located atz= 1.0 m.

(a) The general expression for Eis: ( )( , ) cos ,zo xV

z t E e t z

m

= +E a

and we can see from the given information that

8 88 , 2 10 , 10 , 0oV rad

E x f Hzm s

= = = = . Also

( )( ) ( )8 12 42 10 81 8.854 10 0.45, 10 , so 1 (low loss dielectric).x x

= = =

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5-15

0.0021 18.8

0.0021 18.8 0.0021 18.8

8 ,

1 8 191

41.9

z j z

s x

z j z z j z

s P s y y

Ve e

m

mAe e e e

m

=

= = =

E a

H a E a a

so ( )0.0021 8( , ) 191 cos 2 10 18.8z ymA

z t e x t z m= H a

(c)2

2 2(0.0021)(1)

2

10.764 0.761

2

zxoavg z z

E We e

m

= = =P a a

(d) 2(10 ) 7.6avgP P m W = =

P5.29: The density of solar radiation is approximately 150 W/m2at some locations on theearths surface. How much solar power is incident on a typical 100 Watt solar panel(.6 m x 1.6 m area) if the panel is normal to the radiation propagation direction? How

much power is incident if the panel is tilted 45to the radiation propagation direction?

144 , cos 45 102avg avgP P S W P P S W = = = =

P5.30: A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0mV/m amplitude at the surface. How much power passes through a square meter at thesurface? How much power passes through a square meter area 10. m beneath thesurface?

21

200 , 1 , 2

oo avg

EmV

f MHz E Pm = = =

Cu: 45 3 452 , 214 10 , so 5.22j jNp

e f x e mm

= = = =

( )23

3 2

10196 ; 96

2 5.22 10avg avg

WP P P S W

x m

= = = =

Now at 10 m beneath the surface, we have3(10 ) 3 (214 10 )(10 ) 6( 10 ) 10 118 10m x mo

VE z m E e e x

m

= = = =

( )

26

3 2118 101 1.3 ; 1.32 5.22 10avg

x WP P W

x m

= = =

6. Wave PolarizationP5.31: Suppose E(z,t) = 10.cos(t-z)ax + 5.0cos(t-z)ay V/m. What is the wavepolarization and tilt angle?

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5-16

Fig. P5.31

The figure indicates linear polarization.The tilt angle is:

1 5tan 2710

= =

P5.32: Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ayV/m, find the polarizationand handedness.

The field can be rewritten as E(z,t) = 10.cos(t-z)ax+ 20.cos(t-z-45-180)ayor E(z,t) = 10.cos(t-z)ax+ 20.cos(t-z+135)ay

Running ML0503:

Polarization Plot

enter x-amplitude: 10enter x-phase angle (degrees): 0enter y-amplitude: 20enter y-phase angle (degrees): 135

To determine direction of polarization,move from the o to + along the plot.

>>

From the figure, we have left-hand elliptical polarization.

Fig. P5.32

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5-17

P5.33: Given H(z,t) = 2.0cos(t-z)ax+ 6.0cos(t-z-120)ayA/m, find the polarizationand handedness.

Convert to E(z,t):

( )120 1202 6 2 6j z j z j j z j z js P s o z x y o y o xe e e e e e = = + = +E a H a a a a a

( ) ( )( )( , ) 6cos 120 2cos 180o x yE z t t z t z = + a a With this we can run ML0503:

Polarization Plot

enter x-amplitude: 6enter x-phase angle (degrees): -120enter y-amplitude: 2enter y-phase angle (degrees): -180

To determine direction of polarization,move from the o to + along the plot.

>> From the figure, we have right-hand elliptical polarization.

P5.34: Given

( ) ( )( , ) cos cos ,xo yoz t E t z E t z = + +x yE a a

Fig. P5.33

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5-18

we say that Ey leads Ex for 0 < < 180, and that Ey lags Exwhen 180 < < 0.Determine the handedness for each of these two cases.

For 0 < < 180, we have LHP

For 180 < < 360, we have RHPP5.35: MATLAB: For a general elliptical polarization represented by

( ) ( )( , ) cos cos ,xo yoz t E t z E t z = + +x yE a a the axial ratio and tilt angle can be

found from the following formulas (from K. R. Demarest, Engineering Electromagnetics,Prentice-Hall, 1998, pp. 451-453):

a=|Exo|, b=|Eyo|MAJ = length of majority-axisMIN = length of minority-axis

2 2 4 4 2 2

2 2 4 4 2 2

12 2 cos2

2

12 2 cos22

MAJ a b a b a b

MIN a b a b a b

= + + + +

= + + +

axial ratio=MAJ/MIN

1

2 2

1 2tan cos

2

ab

a b

= .

Compose a program that not only draws a polarization plot like MATLAB 5.3, but thatalso calculates the axial ratio and tilt angle. Run the program on Drill 5.11.

Now we run the program for Drill 5.11.

Polarization Plot

enter x-amplitude: 3enter x-phase angle (degrees): -30enter y-amplitude: 8enter y-phase angle (degrees): 90

To determine direction of polarization,move from the o to + along the plot.

AR = 3.1997

tiltangle = 11.7874

7. Reflection and Transmission at Normal IncidenceP5.36: Starting with (5.107) and (5.109), derive (5.110) and (5.111).

(1) i r to o oE E E+ =

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5-19

(2) 1

2

i r t

o o oE E E

=

Add (1) and (2): 1 1 2

2 2 1 2

22 1 , soi t t t t io o o o o oE E E E E E

= + = + = +

Now subtract (2) from (1):1 1 2 2 1

2 2 1 2 2 1

22 1 1 ,r t i r i

o o o o oE E E E E

= = = + +

P5.37: A UPW is normally incident from media 1 (z< 0, = 0, r= 1.0, r= 4.0) tomedia 2 (z > 0, = 0, r = 8.0, r = 2.0). Calculate the reflection and transmissioncoefficients seen by this wave.

2 1

1 22 1

120 8; 60 , 120 240

24

= = = = = = =

+

240 60 30.60

240 60 5

= = =

+

1 1.60= + =

P5.38: Suppose media 1 (z< 0) is air and media 2 (z> 0) has r= 16. The transmittedmagnetic field intensity is known to be Ht= 12 cos (t-2z)aymA/m. (a) Determine theinstantaneous value of the incident electric field. (b) Find the reflected average powerdensity.

2 2

2

12t

j z j zt os y y

EmA mAe e

m m

= =H a a

2t t2 o s

2

30 , so 12 , E 0.36 , and 1.13t

j zox

E mA V V e

m m m

= = = =E a

( ) 2 12 1

3 21 ; , 1

5 5t i i

o o oE E E

= = + = = = + =

+

12.83, so 2.83t

j zi ioo s x

EE e

= = =E a

( )1( , ) 2.83cos .x Vz t t zm

= E a

11.70, so 1.70 j zr i r

o o s xE E e += = = E a

( ) ( )1 11 1

1.70 4.5120

j z j zr r

s P s z x y

mAe e

m

+ += = =H a E a a a

( ) ( )( ) ( )3 21

1.70 4.5 10 3.82

r

avg z z

mWx

m

= =P a -a

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5-21

Et= 0

P5.42: A UPW given by E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m isincident from air (forz< 0) onto a perfect conductor (forz> 0). Find the instantaneous

expression for the reflected electric field intensity and the SWR.

As in the previous problem, = -1. We then haveE(z,t) = -10.cos(t+1z)ax- 20.cos(t+1z+/3)ayV/m

1

1SWR

+ = =

P5.43: The wave Ei= 10.cos(2 x 108t - 1z) axV/m is incident from air onto a copper

conductor. FindE

r

,E

t

and the time-averaged power density transmitted at the surface.

For copper we have

( )( ) ( )

4522

2

8 7 7 32 2 2 2

2

where 10 4 10 5.8 10 151 10

je

Npf x x x

m

=

= = = =

so 452 3.7 j

e m =

We find 6 452 2

1 2 1

2 21, and = 19.6 10 jx e

=+

So Er= -10.cos(2 x 108t + 1z) axV/m

( ) 2 2 45196 ,z j zt js xV

e e em

=E a

and ( )2 2196 cos 45zt

x

Ve t z

m

= +E a

( )( )

( )26

23

196 101cos 45 3.7 .

2 3.7 10

t

avg z z

x V m W

mx

= =

P a a

P5.44: Given a UPW incident from medium 1 (= 0, r= 1.0, r= 25.) to medium 2 (=0.0080, r= 1.0, r= 81.), calculate ,SWRand at 1 kHz, 1 MHz, and 1 GHz.

6

1 2 9

120 7.896 10 ( )24 ;

0.008 4.506 10 ( )25

j j x f Hz

j j x f Hz

= = = =

+ +

Table P5.44f 2() 2 1

2 1

=

+

1

1SWR

+ =

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5-22

Fig. P5.45

1kHz 0.994ej44.98 0.9815ej178.9 107.31MHz 29.3ej30.3 0.513ej155 3.111GHz 41.9ej0.05 0.286ej179.9 1.80

P5.45: MATLAB: Write a program that prompts the user for the constitutive parametersin medium 1 and medium 2 separated by a planar surface. You are to assume a wave isnormally incident from media 1 to media 2. The program is to plot and versus afrequency range supplied by the user. Use this program to plot and from 100 Hz to10 GHz for the pair of media specified in the previous problem.

Run the program:

P5.46: A wavespecified by Ei =

100.cos(x107

t-1z)axV/m is incident from air(at z < 0) to anonmagnetic media (z> 0, = 0.050 S/m, r= 9.0). Find E

r, Etand SWR. Also find theaverage power densities for the incident, reflected and transmitted waves.

7 6

1 1

1

2120 , 10 so 5 10 , 0.105

rad rad x f x Hz

s c m

= = = = = =

In this problem we find in medium 2 (z > 0) that = 0.0025 and = 0.05. These valuesare too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate:

43.6

2 2 20.969 , 1.019 , 28.1 jNp rad

em m = = =

.

Then,

174 40.82 1

2 1

10.898 , 18.6, 1 0.141

1j j

e SWR e

+ = = = = = + =

+

1100 j zis x

Ve

m

=E a

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5-23

1 1 174100 89.8j z j zr js x xV V

e e em m

+ += =E a a

,

so ( )7( , ) 89.8cos 10 0.105 174 .r xV

z t x t zm

= + +E a

2 2 40.8

100 14.1j z j zt j

s x x

V V

e e em m

= =E a a

,

so ( )7( , ) 14.1cos 10 1.02 40.8 .t xV

z t x t zm

= +E a

2 243.6 40.8 2.814.1

0.50228.1

j z j zt j j j

s y y

A Ae e e e e

m m

= =H a a

( ) ( ) ( ) 21

14.1 0.502 cos 40.8 2.8 2.62

t

z z

W

m= + =P a a

( )( )

2

2

10013.3

2 120i

z

W

m= =P a

( )( )

( )2

2

89.810.7

2 120

r

z

W

m

= =P -a

(check: 13.3 W/m2= 10.7 W/m2+ 2.6 W/m2)

P5.47: A wave specified by Ei = 12 cos(2x107t-1z+/4)ax V/m is incident from anonmagnetic, lossless, r= 9.0 media (atz< 0) to a media (z > 0) with = 0.020 S/m, r= 2.0, and r = 16.). Find H

i, Er, Hr, Et, Ht, and the average power densities for theincident, reflected and transmitted waves.

We use ML0501 in each media to find:

1 1 10; 0.628 ; 40rad

m = = =

332 2 21.01 ; 1.56 ; 84.9

jNp rade

m m = = =

We also will need reflection and transmission coefficients:

126 19.82 1

2 1

0.353 ; 1 0.84j j

e e

= = = + =

+

Incident:

1 412 j zi js x Ve em

=E a

1 14 4

1

120.300j z j zi j j

s y y

A Ae e e e

m m

= =H a a ,

7( , ) 0.300cos 2 10 0.628 .4

i

y

Az t x t z

m

= +

H a

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5-24

( )2

21

1215.655

2i

avg z z

W

m

= =P a a

Reflected:

( ) 1 1 14 45 126 17112 13.3 13.3j z j z j zr j j j js x x xV V V

e e e e e e e

m m m

+ + += = =E a a a

( )7( , ) 13.3cos 2 10 0.628 171 .r xV

z t x t zm

= + +E a

1 1171 17113.3

0.10640

j z j zr j j

s y y

A Ae e e e

m m

+ += =H a a

( )7( , ) 0.106cos 2 10 0.628 171 .r yA

z t x t zm

= + +H a

( )( )

2

2

13.310.704

2 40r

avg z z

W

m= =P a -a

Transmitted:

( ) 1 24 64.812 31.67j z j zt j js x xV Ve e e em m

= =E a a

,

( )7( , ) 31.7cos 2 10 1.56 64.8 .t xV

z t x t zm

= +E a

2 233 64.8 31.831.67

0.37384.9

j z j zt j j j

s y y

A Ae e e e e

m m

= =H a a

,

( )7( , ) 0.373cos 2 10 1.56 31.8 .t yA

z t x t zm

= +H a

( )( )( ) 2

31.67 0.373cos 64.8 31.8 4.954

2

t

avg z z

W

m= =P a a

(Check: 5.655W/m2= 0.704W/m2+ 4.954W/m2)

8. Reflection and Transmission at Oblique Incidence

P5.48: A 100 MHz TE polarized wave with amplitude 1.0 V/m is obliquely incident fromair (z< 0) onto a slab of lossless, nonmagnetic material with r= 25 (z > 0). The angle ofincidence is 40. Calculate (a) the angle of transmission, (b) the reflection andtransmission coefficients, and (c) the incident, reflected and transmitted fields.

(a) ( )6

1 282 100 10 2.09 , 10.45 .3 10rx rad rad

c x m c m = = = = =

1

2 2

1 1 1; sin sin 40 ; 7.4

5 5t t

r

= = = =

(b) Now we need to calculate the reflection and transmission coefficients.

1 2

120120 ; 24

25

= = =

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5-25

2 1

2 1

cos cos0.732; 1 0.268

cos cosi t

TE TE TE

i t

= = = + =

+

(c) The fields,Incident:

( )2.09 sin40 cos40 1.34 1.601 1

j x zi j x j z

s y y

V

e e e m

+

= =E a a

( )( , ) 1cos 1.34 1.60i yV

z t t x zm

= E a

( )1.34 1.601

cos 40 sin 40120

i j x j z

s x ze e

= +H a a

( ) 1.34 1.602.03 1.71i j x j zs x zmA

e em

= +H a a

( ) ( )( , ) 2.03 1.71 cos 1.34 1.60i x zmA

z t t x zm

= + H a a

Reflected:

0.732r io TE oE E= =

( ) 1.34 1.600.732r j x j zs yV

e em

+= E a

( )( , ) 0.732cos 1.34 1.60r yV

z t t x zm

= +E a

( )( )

( )

1.34 1.60

1.34 1.60

0.732cos 40 sin 40

120

1.49 1.25

r j x j z

s x z

r j x j z

s x z

Ae e

m

mAe e

m

+

+

= +

=

H a a

H a a

( ) ( )( , ) 1.49 1.25 cos 1.34 1.60r x z mAz t t x zm

= +H a a

transmitted:

0.268t io TE oE E= =

( )2 sin cos 1.35 10.40.268 0.268t tj x zt j x j z

s y y

Ve e e

m

+ = =E a a

( )1.35 10.40.268

cos 7.4 sin 7.424

t j x j z

s x z

Ae e

m

= +H a a

( ) 1.35 10.43.5 0.46t j x j zs x zmA

e em

= +H a a

( ) ( )( , ) 3.5 0.46 cos 1.35 10.4t x zmA

z t t x zm

= + H a a

P5.49: A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incidentfrom air (z< 0) onto a slab of lossless, nonmagnetic material with r= 25 (z> 0). The

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5-26

angle of incidence is 40. Calculate (a) the angle of transmission, (b) the reflection andtransmission coefficients, and (c) the incident, reflected and transmitted fields.

(a) The material parameters in this problem are the same as for P5.48. So, once again wehave t= 7.4. Also,1= 2.09 rad/m and2= 10.45 rad/m.

(b)2 1

2 1

cos cos0.589

cos cost i

TM

t i

= =

+

2

2 1

2 cos0.318

cos cos

iTM

t i

= =

+

(c)Incident:

( )1.34 1.601 cos 40 sin 40i j x j zs x ze e = E a a

( ) ( )( , ) 0.766 0.643 cos 1.34 1.60i x zV

z t t x zm

= E a a

1.34 1.601

120

i j x j z

s y

Ae e

m

=H a

( )( , ) 2.65cos 1.34 1.60i ymA

z t t x zm

= H a

Reflected:

( )1.34 1.600.589 cos 40 sin 40r j x j zs x ze e += +E a a

( ) ( )( , ) 0.452 0.379 cos 1.34 1.60r x zV

z t t x zm

= +E a a

1.34 1.600.589

120

r j x j z

s y

Ae e

m

+=H a

( )( , ) 1.56cos 1.34 1.60r ymA

z t t x zm

= +H a

transmitted:

( )1.35 10.40.318 cos 7.4 sin 7.4t j x j zs x ze e = E a a

( ) ( )( , ) 0.315 0.041 cos 1.35 10.4t x zV

z t t x zm

= E a a

( )( , ) 4.22cos 1.35 10.4t ymA

z t t x zm

= H a

P5.50: A randomly polarized UPW at 200 MHz is incident at the Brewsters angle fromair (z< 0) onto a thick slab of lossless, nonmagnetic material with r= 16 (z> 0). Thewave can be decomposed into equal TE and TM parts, each with an incident electric fieldamplitude of 10. V/m. Find expressions for the instantaneous value of the incident,reflected and transmitted electric fields.

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5-27

First we calculate the Brewsters angle:1

sin ; 761 1 16

BA BA = =+

Also, we calculate1= 4.19 rad/m,2= 16.8 rad/m, 1= 120 , and 2= 30 .

TE

( )1 sin cos 4.06 1.0110 10i ij x zi j x j z

s y y

Ve e e

m

+ = =E a a

At the Brewsters angle of incidence, we have from Snells Law:

1 1

2

sin sin 14t i

= =

2 1

2 1

cos cos0.883; 1 0.117

cos cosi t

TE TE TE

i t

= = = + =

+

V V=-8.83 ; =1.17

m mr i t i

o TE o o TE oE E E E= =

4.06 1.018.83r j x j zs y

Ve e

m

+= E a

( ) ( )2 sin cos 4.06 16.310 1.17t tj x zt j x j zs TE y yV

e e em

+ = =E a a

TM:

At the Brewsters angle, TM= 0 and 0.r

s =E

( )4.06 1.0110 cos sini j x j zs i x i ze e = E a a ,

( )4.06 1.01

2.42 9.70 .i j x j z

s x z

Ve e m

= E a a

( )4.06 16.310 cos sint j x j zs t x t ze e = E a a ,

( ) 4.06 16.39.70 2.40 .t j x j zs x zV

e em

= E a a

Combining the results we arrive at:

( ) ( )( , ) 2.4 10 9.7 cos 4.06 1.01i x y zV

z t t x zm

= + E a a a

( )( , ) 8.83cos 4.06 1.01r yV

z t t x z

m

= +E a

( ) ( )( , ) 9.7 1.2 2.4 cos 4.06 16.3t x y zV

z t t x zm

= + E a a a