5-1

Solutions for Chapter 5 Problems

1. General Wave Equations

P5.1: Starting with Maxwell’s equations for simple, charge-free media, derive the

Helmholtz 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 to 2

2

2t t

H HH =

P5.2: Derive equation (5.10) by starting with the phasor point form of Maxwell’s

equations for simple, charge-free media.

For charge-free media the phasor form of Maxwell’s 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 Faraday’s Law,

s s s sj j j j E H H E

Now since 2

s s s

E 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 wave’s frequency (a) in

air? (b) in the liquid? (c) What is the liquid’s relative permittivity?

(a) 8

3 1 05

0 .0 6

pu c x m s

f G H zm

(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz

(c)

5-2

9 7

28

8

15 1 0 0 .0 1 5 1 0

3 1 03 6

0 .5 1 0

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 Hs is only a function of z, (5.14) becomes 2

2

20 .

s

s

HH

z

(a)

If we let ,z

sH A e

then

2

2

2, a n d .

z zs sH H

A e A ez z

So (a) becomes 2 2

0, o r 0 . This has two solutions:

(1) for 0 , w e h a v e , , o r .z z

s s oH A e H H e

(2) for 0 , w e h a v e , , o r .z z

s s oH A e H H e

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

.z z

s o o yH e H e

H a

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

r o r oj j j

r o

r o

j

j

6 7

2 1 0 1 0 5 0 4 1 0 3 9 4 8r o

j j x x j

5 6 1 2 5 3

1 1 0 2 1 0 1 0 2 8 .8 5 4 1 0 1 1 0 1 .1 1 1 0r o

j x j x x x j x

Inserting these into the expressions for and , 3 3 257

9 .4 10 2 .1 1 , 9 .4 10 , 2 .1 , 1880j

x j m x N p m rad m e

These results are confirmed by ML0501.

P5.6: MATLAB: In some material, the constitutive parameters are constant over a large

frequency range and are given as = .10 S/m , r = 4.0, and r = 600. Write a MATLAB

routine that will plot , , and (magnitude and phase) versus the log of frequency from

1 Hz up to 100 GHz.

% M-File: MLP0506

%

5-3

% This program is a modification of ML0501.

% For a given material, it will plot the attenuation,

% phase constant and intrinsic impedance vs f.

%

% Wentworth, 1/23/03

%

clc %clears the command window

clear %clears variables

% Initialize Variables

uo=pi*4e-7;

eo=8.854e-12;

sig=0.10;

er=4;

ur=600;

% Perform Calculation

for i=1:10

for j=1:10

m=(i-1)*10+j;

f(m)=j*10^(i-1);

w(m)=2*pi*f(m);

A(m)=i*(w(m)*ur*uo);

B(m)=complex(sig,w(m)*er*eo);

gamma(m)=sqrt(A(m)*B(m));

alpha(m)=real(gamma(m));

beta(m)=imag(gamma(m));

eta(m)=sqrt(A(m)/B(m));

meta(m)=abs(eta(m));

aeta(m)=180*angle(eta(m))/pi;

end

end

subplot(3,1,1)

plot(f,alpha,'-o',f,beta,'-*')

ylabel('1/m')

xlabel('frequency (Hz)')a

legend('alpha','beta')

subplot(3,1,2)

semilogx(f,meta)

ylabel('magnitude of eta (ohms)')

subplot(3,1,3)

semilogx(f,aeta)

ylabel('phase of eta (degrees)')

xlabel('frequency (Hz)')

5-4

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

propagation, ap, and H(x,y,t).

3 2

5j x j y

s ze e

E a

We assume nonmagnetic material and therefore have 3 2 3 2

10 15j x j y j x j y

s s x yj j e e j e e

E H a a

3 2 3 2 3 2 3 21 0 1 52 .5 3 3 .8

j x j y j x j y j x j y j x j y

s x y x y

o

j je e e e e e e e

j j

H a a a a

6 6 A

( , , ) 2 .5 3 co s 1 0 3 2 3 .8 0 co s 1 0 3 2 m

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 Ha

E H

6 4 6 419 12 .65

j x j y j x j y

s 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, we

have:

0 .8 3 0 .5 5 .P x y a a a

Fig. P5.6

5-5

P5.8: Suppose in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).

0 .1 0 0 , , 4

1 2 0 0 .1 0 0 1 2

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 cosy

t x E a

Since free space is stated,

2 22 3 0 ra d m

c f

and then

7 21 2 c o s 2 1 0

3 0 4y

Vx t x

m

E a

2. Propagation in Lossless, Charge-Free Media

P5.9: Start with the Helmholtz equation (5.11), and using = j, derive (5.41), the

traveling wave equation.

2

2 2 2

20 , le t ( ) , a n d w ith = j w e h a v e 0 .

x s

s s s x s x x s

EE z E

z

E E E a

Let 2

2

2, s o a n d

z x xx s x s

x s

E EE A e A e A e

z z

Now we have 2 2 2 2

0 , o r 0z z

A e A e

This can be factored: 2 2

0j j ,

suggesting two solutions. The first solution uses j and

.j z j z

x s oE A e E e

Likewise, the second solution uses j and

.j z j z

x s oE A e 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 of

1 V/m. If the electric field vector for this wave has only an az component, find the

instantaneous expression for the electric and magnetic fields.

From the given information we have 62 2 0 0 1 0

ra df x

s and

2,

3p

r a d

u m

5-6

or 6 2( , ) 1 c o s 2 0 0 1 0

3z

Vy t x t y

m

E a .

Now to find H.

1 1 11 , 1

1 2 0 1 2 0

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

, c o s 2 0 0 1 01 2 0 3

x

Ay t x t y

m

H a

or

6 2

, 2 .7 c o s 2 0 0 1 0 .3

x

m Ay t x t y

m

H a

P5.11: In a lossless, nonmagnetic material with r = 16, H = 100 cos(t – 10y) az mA/m.

Determine the propagation velocity, the angular frequency, and the instantaneous

expression for the electric field intensity.

8

83 1 00 .7 5 1 0

1 6p

r

c x mu x

s

8 8

0 .7 5 1 0 1 0 7 .5 1 0p

ra du x x

s

8

( , ) 1 0 0 co s 7 .5 1 0 1 0z

m Ay t x t y

m H a

0 .1 0 0 ,

1 2 00 .1 0 0 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 .4 co s 7 .5 1 0 1 0x

Vy t x t y

m E a

P5.12: Given E = 120 cos(xt – y) az V/m and H = 2.00cos(xt –

y) ax A/m, find r and r.

0 .0 8 0

1 2 0 , 2j y j y

s z s xe e

E a H a

1 1 1 2 0 1 2 0

1 2 0

j y j y j yr

s P s y z x x

rr r

e e e

H a E a a a a

so we know

2r

r

5-7

Now, 6

61 6 1 07 5 1 0

0 .0 8 0p

r r

c xu x

64

7 5 1 0r r

c

x

And now

( 2 )( 4 ) 8r

r r r

r

42

2

r r

r

r r

3. Propagation in Dielectrics

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

equations (5.50) and (5.51).

2 2 2 2

2 ;j j

Comparing the imaginary parts, we see 2 , o r ,2

and comparing the real parts, 2 2 20 .

Rearranging and inserting our value for : 2 2 2

4 2 20

4

This is a quadratic expression (x2 + bx + c = 0), where here 2

2 2, ,

2x 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 0

2 2

Rearranging,

5-8

2

4 2 20

2

Solving this quadratic we find

2

1 1

P5.14: MATLAB: Write a routine to prompt the user for a material’s constitutive

parameters and an operating frequency, and calculate the and from (5.52). Verify the

program by running Drill 5.6.

% MLP0514

%

% Prompts user for material's constitutive

% parameters and an operating frequency, then

% calculates alpha(Np/m) and beta(rad/m).

%

% Wentworth, 1/24/03

%

clc

clear

ur=input('relative permeability: ');

erp=input('real part of rel permittivity: ');

erdp=input('complex part of rel permittivity: ');

s=input('conductivity (S/m): ');

f=input('frequency (Hz): ');

w=2*pi*f;

uo=pi*4e-7;

eo=8.854e-12;

seff=s+w*erdp*eo;

A=sqrt(1+(seff/(w*erp*eo))^2);

B=ur*uo*erp*eo/2;

alpha=w*sqrt(B*(A-1))

beta=w*sqrt(B*(A+1))

Now run the program for Drill 5.6: (a)

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

5-9

alpha = 3.3730e-009 beta = 6.6268e-006

(b)

relative permeability: 1 real part of rel permittivity: 10 complex part of rel permittivity: .01 conductivity (S/m): 1e-12 frequency (Hz): 1e6 alpha = 3.3134e-005 beta = 0.0663

These results agree with Drill 5.6.

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 plot

using (5.54). At what frequency does the % error exceed 2%?

% MLP0515

%

% Compares alpha calculated using (5.52) to

% that calculated using (5.54).

%

% Wentworth, 1/25/03

%

clc

clear

% Initialize variables

ur=1;

erp=3;

erdp=.015;

s=1e-3;

uo=pi*4e-7;

eo=8.854e-12;

B=ur*uo*erp*eo/2;

5-10

% Perform calculations

for i=1:10

for j=1:10

m=(i-1)*10+j;

f(m)=j*10^(i-1);

w(m)=2*pi*f(m);

seff(m)=s+w(m)*erdp*eo;

A(m)=sqrt(1+(seff(m)/(w(m)*erp*eo))^2);

alpha1(m)=w(m)*sqrt(B*(A(m)-1));

alpha2(m)=(seff(m)/2)*sqrt(ur*uo/(erp*eo));

diff(m)=abs(100*(alpha1(m)-alpha2(m))/alpha1(m));

C(m)=diff(m)<2;

if diff(m)<2

if diff(m-1)>2

fdiff=f(m);

Fstr=num2str(fdiff);

end

end

end

end

% generate plot

loglog(f,alpha1,'-o',f,alpha2,'-*')

legend('(5.52)','(5.54)')

xlabel('frequency (Hz)')

ylabel('alpha(Np/m)')

S=strcat('Error drops below 2% when frequency > ',Fstr);

title(S)

grid on

Fig. P5.15

5-11

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 field

vector in the z direction. Find the instantaneous form of the related electric field

intensity.

1 c o s ; x x j x

z s o z

m Ae t x H e e

m

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

3 0

6 1 2

2 1 0 0 1 0 1 0 0 4 1 02 6 6 4

0 .0 0 9 6 4 2 1 0 0 1 0 8 .8 5 4 1 0

jj x xj

ej j x x

1

1 4 .8 2 5 .7 j j jm

Finally,

1 5 6

( , ) 2 .6 6 co s 2 0 0 1 0 2 6 3 0x

y

Vx t e x t x

m

E a

P5.17: MATLAB: Make a pair of plots similar to Figure 5.4 for the 3 materials of Table

5.1. Instead of loss tangent, one plot is to contain the magnitude of and the other is to

have the phase of .

%ML P5.17

clc;clear

%want to plot intrinsic impedance vs frequency for

%the data listed in table 5.1

%Here, we'll plot the magnitude and phase of the

%intrinsic impedance.

%enter data from Table 5.1

sigC=5.8e7; %conductivity of copper in S/m

sigS=4; % conductivity of seawater

sigG=1e-12; % conductivity of glass

er1C=1; %real part of rel perm for Copper

er1S=72; %real part of rel perm for seawater

er1G=10; %real part of rel perm for glass

er2C=0; %imag part of rel perm for Copper

er2S=12; %imag part of rel perm for seawater

er2G=0.010; %imag part of rel perm for glass

%enter constant values

eo=8.854e-12; %free space permittivity, F/m

uo=pi*4e-7; %free space permeability, H/m

5-12

%calculations

n=2:.2:14;

f=10.^n;w=2*pi*f;

seffC=sigC+w*er2C*eo;

seffS=sigS+w*er2S*eo;

seffG=sigG+w*er2G*eo;

etaC=sqrt(i*w*uo./(seffC+i*er1C*eo))

etaS=sqrt(i*w*uo./(seffS*er1S*eo))

etaG=sqrt(i*w*uo./(seffG+i*er1G*eo))

magC=abs(etaC);

angC=180*angle(etaC)/pi;

subplot(3,2,1)

semilogx(f,magC)

ylabel('mag, ohms')

title('copper')

subplot(3,2,2)

semilogx(f,angC)

ylabel('phase, deg')

magS=abs(etaS);

angS=180*angle(etaS)/pi;

subplot(3,2,3)

semilogx(f,magS)

ylabel('mag, ohms')

title('seawater')

subplot(3,2,4)

semilogx(f,angS)

ylabel('phase, deg')

magG=abs(etaG);

angG=180*angle(etaG)/pi;

subplot(3,2,5)

semilogx(f,magG)

ylabel('mag, ohms')

xlabel('freq (Hz)')

title('glass')

subplot(3,2,6)

semilogx(f,angG)

xlabel('freq (Hz)')

ylabel('phase, deg')

5-13

4. Propagation in Conductors

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

fo r a good c on d u c to rj j j

1

, 12 2 22

2

j jj j j

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

P5.19: In seawater, a propagating electric field is given by E(z,t) = 20.e-z cos(xt –

z + 0.5) ay V/m. Assuming ’’=0, find (a) and , and (b) the instantaneous form of H.

For seawater we have r = 72, = 5, and r = 1.

Fig. P5.17

5-14

So: 7 .8 9 6 , 0 .0 0 4o r o

j j j j

4 4 .9 81 .2 5 7

jje

j

4 .4 4 1 4 .4 4 5 1 mj j j

14 .4

m

0 .5 2 8 .62 0 2 0

z z j ra d ia n s z z j

s y y

V Ve e e e e e

m m

E a a

2 8 .6 2 8 .61 1 2 02 0

z z j z z j

s P s z y x

Ae e e e e e

m

H a E a a a

4 .4 6

( , ) 1 5 .9 c o s 2 1 0 4 .4 2 8 .6 4 5z

x

Az t e x t z

m

H a

or with appropriate significant digits:

4 .4 6

( , ) 1 6 c o s 2 1 0 4 .4 1 6z

x

Az t e x t z

m

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;a s a n e x a m p le , fo r c o p p e r a t 1 G H z :

12 .1 1 0 2 .1

1 11 1 0 4 1 0 5 .8 1 0

f

x m m

H V s Ax x x

s m m H A V

Table P5.19

(S/m) r (m)

Cu 5.8x107 1 2.1

Ag 6.2x107 1 2.0

Au 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

7 7 3

6 0 0 4 1 0 1 .4 5 1 0 3 4 .3 5 1 0 ( )f f H z x x x f H z

= 1/

4 5 6 4 52 1 8 .0 8 1 0 ( )

j je x f H z e

5-15

61 2 1 0

p

r r

c mu x

s

Table P5.21

f(Hz)= 1 103 106 109

(Np/m) 185 5860 185x103 5.9x106

(rad/m) 185 5860 185x103 5.9x106

18ej45º 570ej45º 18ej45ºm 0.57ej45º

5.4mm 170m 5.3m 170nm

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

P5.22: A semi-infinite slab exists for z > 0 with = 300 S/m, r = 10.2, and r = 1.0. At

the surface (z = 0),

E(0,t) = 1.0 cos( x 106t) ax V/m.

Find the instantaneous expressions for E and H anywhere in the slab.

The general expression for E is: 6

( , ) 1 .0 co s 1 0z

x

Vz t e x t z

m

E a

6 7

1 0 4 1 0 3 .9 4 8j j x x j

6 1 2 6

1 0 1 0 .2 8 .8 5 4 1 0 2 8 4 1 0j j x x j x

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

12 4 .3f

m

4 5 4 52 0 .1 1 5

j je e

So now we have

2 4 6

( , ) 1 .0 co s 1 0 2 4z

x

Vz t e x t z

m

E a

To find B we’ll work in phasors.

1 1 11 , 1

z j z z j z z j z

s x s P s z x ye e e e e e

E a H a E a a a

2 4 61

( , ) co s 1 0 2 4 4 50 .1 1 5

z

y

Az t e x t z

m

H a

2 4 6

( , ) 8 .7 co s 1 0 2 4 4 5z

y

Az t e x t z

m

H a

P5.23: In a nonmagnetic material, E(z,t) = 10.e-200z cos(2 x 109t - 200z) ax mV/m.

Find H(z,t).

5-16

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

22

9 7

2 0 0, o r 1 0 .1 3

1 1 0 4 1 0o

o

Sf

f mx x

4 5 4 52 2 8

j je e

1 1 1 01 0 , 1 0

z j z z j z z j z

s x s P s z x ye e e e e e

E a H a E a a a

2 0 0 9

( , ) 3 6 0 co s 2 1 0 2 0 0 4 5z

y

m Az t e x t z

m

H a

P5.24: A 0.1 m layer of copper is deposited atop a very thick slab of nickel. For a field

incident on the copper surface, (a) calculate Rs at 1.0 GHz. Compare this with Rs at 1.0

GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 m thickness of copper by

itself.

Refer to Figure P5.24..

In the copper portion the field is C uz

x x oE E e

In the nickel portion, N iC u

z tt

x xoE E e e

The current density in the copper is ,C uz

x C u C u x oJ E e

and in the nickel is

.N iC u

z tt

xN i N i xoJ E e e

The current is

( )C u C u N i

z t z t

C u xo N i xoI E e d yd z E e e d yd z

, or

( ),C u C u N i

t

z t z t

C u xo N i xo

o t

I w E e d z w E e e d z

and upon evaluating

Fig. P5.24

5-17

1 ,C u C ut tC u N i

x o

C u N i

I w E e e

and with V=ExoL,

we have

1

, w h e re = 1 .C u C ut tC u N i

s s

C u N i

LR R R e e

w

Now we’re ready to perform the calculations using the following data:

7 3

C u5 .8 1 0 , 1, 4 7 9 1 0

C u r

S N px x

m m

7 6

C u1 .5 1 0 , 6 0 0 , 5 9 6 1 0

N i r

S N px 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 7

1 1 1 1D C : 1 .3 7

5 .8 1 0 0 .0 0 2

R m

L a mx

1 GHz: 61; 1 2 .0 9 1 0

2 2

sRR

f x mL a a

7 6

10 .6 6

5 .8 1 0 2 .0 9 1 0 2 0 .0 0 2

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 density

passing through a 1.0 square meter surface that is normal to the direction of propagation.

2

2

2

1 11 2 0 1 2 2 7

2 2a vg xo z z z

A kWH

m m

P a a a

P5.27: A 600 MHz uniform plane wave incident in the z direction on a thick slab of

Teflon (r = 2.1, r = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field intensity

at the surface. Assuming = 0 for Teflon, find in the Teflon (a) E(z,t), (b) H(z,t) and (c)

Pav.

6

(0 , ) 1 co s 2 6 0 0 1 0y

Vt x t z

m E a

5-18

( , ) 1 co sz

y

Vz t e t z

m

E a

Teflon: = 0 so = 0,

and

6

8

2 6 0 0 1 02 .1 1 8 .2

3 1 0r

x ra d

c x m

(a) 9

( , ) 1 co s 1 .2 1 0 1 8 .2y

Vz t x t z

m E a

(b) 1 2 .1

1 ,1 2 0

j z

s P s z y

Ve

m

H a E a a

9

( , ) 3 .8 co s 1 .2 1 0 1 8 .2x

m Az t x t z

m H a

(c)

2

2

1 2 .111 .9

2 1 2 0a vg z z

m W

m P a a

P5.28: Assume distilled water ( = 10-4 S/m, r = 81, r = 1.0) fills the region z > 0. At

the surface, we have E(0,t) = 8.0cos(2x108t) ax V/m. Determine, for z > 0, (a) E(z,t),

(b) H(z,t), and (c) Pav at z = 1.0 m. (d) Find the power passing through a 10 square meter

surface located at z = 1.0 m.

(a) The general expression for E is: ( , ) co s ,z

o x

Vz t E e t z

m

E a

and we can see from the given information that

8 88 , 2 1 0 , 1 0 , 0

o

V ra dE x f H z

m s . Also

8 1 2 4

2 1 0 8 1 8 .8 5 4 1 0 0 .4 5, 1 0 , so 1 (lo w lo ss d ie lec tric ).x x

41 0 1

1 2 0 0 .0 0 2 12 2 8 1

N p

m

1 8 .8r

ra d

c m

11 2 0 4 1 .9

8 1

so 0 .0 0 2 1 8

( , ) 8 co s 2 1 0 1 8 .8z

x

Vz t e x t z

m

E a

(b)

0 .0 0 2 1 1 8 .8

0 .0 0 2 1 1 8 .8 0 .0 0 2 1 1 8 .8

8 ,

1 8 1 9 1

4 1 .9

z j z

s x

z j z z j z

s P s y y

Ve e

m

m Ae e e e

m

E a

H a E a a

5-19

so 0 .0 0 2 1 8

( , ) 1 9 1 co s 2 1 0 1 8 .8z

y

m Az t e x t z

m

H a

(c) 2

2 2 ( 0 .0 0 2 1 )(1 )

2

10 .7 6 4 0 .7 6 1

2

zx o

a v g z z

E We e

m

P a a

(d) 2(10 ) 7 .6

avgP P m W

P5.29: The density of solar radiation is approximately 150 W/m2 at some locations on the

earth’s 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 45 to the radiation propagation direction?

144 , cos 45 102avg avg

P P S W P P S W

P5.30: A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0

mV/m amplitude at the surface. How much power passes through a square meter at the

surface? How much power passes through a square meter area 10. m beneath the

surface?

2

12 0 0 , 1 ,

2

o

o a vg

Em Vf M H z E P

m

Cu: 4 5 3 4 52 , 2 1 4 1 0 , so 5 .2 2

j jN pe f x e m

m

2

3

3 2

1 019 6 ; 9 6

2 5 .2 2 1 0a vg a vg

WP P P S W

x m

Now at 10 m beneath the surface, we have 3

(1 0 ) 3 ( 2 1 4 1 0 )(1 0 ) 6( 1 0 ) 1 0 1 1 8 1 0

m x m

o

VE z m E e e x

m

2

6

3 2

1 1 8 1 011 .3 ; 1 .3

2 5 .2 2 1 0a v g

x WP P W

x m

6. Wave Polarization

P5.31: Suppose E(z,t) = 10.cos(t-z)ax + 5.0cos(t-z)ay V/m. What is the wave

polarization and tilt angle?

The figure indicates linear polarization.

The tilt angle is:

1 5ta n 2 7

1 0

5-20

Fig. P5.31

P5.32: Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m, find the polarization

and handedness.

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

or E(z,t) = 10.cos(t-z)ax + 20.cos(t-z+135°)ay

Running ML0503: Polarization Plot enter x-amplitude: 10 enter x-phase angle (degrees): 0 enter y-amplitude: 20 enter 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.

P5.33: Given H(z,t) = 2.0cos(t-z)ax + 6.0cos(t-z-120)ay A/m, find the polarization

and handedness.

Convert to E(z,t):

Fig. P5.32

5-21

1 2 0 1 2 0

2 6 2 6j z j z j j z j z j

s P s o z x y o y o xe e e e e e

E a H a a a a a

( , ) 6 co s 1 2 0 2 co s 1 8 0o x y

E z t t z t z a a

With this we can run ML0503:

Polarization Plot enter x-amplitude: 6 enter x-phase angle (degrees): -120 enter y-amplitude: 2 enter 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 yo

z t E t z E t z x y

E a a

we say that Ey leads Ex for 0 < < 180, and that Ey lags Ex when –180 < < 0.

Determine the handedness for each of these two cases.

For 0 < < 180°, we have LHP

For 180° < < 360°, we have RHP

Fig. P5.33

5-22

P5.35: MATLAB: For a general elliptical polarization represented by

( , ) cos cos ,xo yo

z t E t z E t z x y

E 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-axis

MIN = length of minority-axis

2 2 4 4 2 2

2 2 4 4 2 2

12 2 c o s 2

2

12 2 c o s 2

2

M A J a b a b a b

M IN a b a b a b

axial ratio=MAJ/MIN

1

2 2

1 2ta n c o s

2

a b

a b

.

Compose a program that not only draws a polarization plot like MATLAB 5.3, but that

also calculates the axial ratio and tilt angle. Run the program on Drill 5.11.

% M-File: MLP0535

%

% This program modifies ML0503. As before, it will

% trace polarization ellipses, given the amplitude

% and phase of a pair of linearly polarized waves.

% Now it will also calculate axial ratio and tilt %angle.

%

% Wentworth 1/28/03

% Variables:

% Exo,Eyo amplitudes for the pair of waves

% fxd,fyd phase angle for each wave

% fx,fy phase (radians) for each wave

% wtd ang freq * time, in degrees

% wtr ang freq * time, in radians

% x,y superposed position

% x0,y0 position at wtd=0 degrees

% x45,y45 position at wtd=45 degrees

% a,b shorthand for Exo,Eyo

% MAJ,MIN majority,minority axis length

% AR,tiltangle axial ration, tilt angle

%

clc %clears the command window

clear %clears variables

% Prompt for input values

disp('Polarization Plot')

disp(' ')

Exo=input('enter x-amplitude: ');

5-23

Fig. P5.35

fxd=input('enter x-phase angle (degrees): ');

fx=fxd*pi/180;

Eyo=input('enter y-amplitude: ');

fyd=input('enter y-phase angle (degrees): ');

fy=fyd*pi/180;

disp(' ')

disp('To determine direction of polarization,')

disp('move from the o to + along the plot.')

disp(' ')

%Perform calculations

wtd=0:360; %wt in degrees

wtr=wtd*pi/180;

x=Exo*cos(wtr+fx);

y=Eyo*cos(wtr+fy);

x0=Exo*cos(fx);

y0=Eyo*cos(fy);

x45=Exo*cos(fx+pi/4);

y45=Eyo*cos(fy+pi/4);

fdiff=fy-fx;

a=abs(Exo);b=abs(Eyo);

temp=sqrt(a^4+b^4+2*a^2*b^2*cos(2*fdiff));

MAJ=2*sqrt(0.5*(a^2+b^2+temp));

MIN=2*sqrt(0.5*(a^2+b^2-temp));

AR=MAJ/MIN

temp2=(2*a*b/(a^2-b^2))*cos(fdiff);

tiltangle=(0.5*atan(temp2)*180/pi)

%Make the plot

plot(x,y,x0,y0,'ok',x45,y45,'+k')

xlabel('x')

ylabel('y')

title('Polarization Plot')

axis('equal')

Now we run the program for Drill 5.11.

Polarization Plot enter x-amplitude: 3 enter x-phase angle (degrees): -30 enter y-amplitude: 8 enter y-phase angle (degrees): 90 To determine direction of polarization,

5-24

move from the o to + along the plot. AR = 3.1997 tiltangle = 11.7874

7. Reflection and Transmission at Normal Incidence P5.36: Starting with (5.107) and (5.109), derive (5.110) and (5.111).

(1)i r t

o o oE E E

(2) 1

2

i r t

o o oE E E

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

2 2 1 2

22 1 , so

i t t t t i

o 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) to

media 2 (z > 0, = 0, r = 8.0, r = 2.0). Calculate the reflection and transmission

coefficients seen by this wave.

2 1

1 2

2 1

1 2 0 8; 6 0 , 1 2 0 2 4 0

24

2 4 0 6 0 30 .6 0

2 4 0 6 0 5

1 1 .60

P5.38: Suppose media 1 (z < 0) is air and media 2 (z > 0) has r = 16. The transmitted

magnetic field intensity is known to be Ht = 12 cos (t-2z)ay mA/m. (a) Determine the

instantaneous value of the incident electric field. (b) Find the reflected average power

density.

2 2

2

1 2

t

j z j zt o

s y y

Em A m Ae e

m m

H a a

2t t

2 o s

2

3 0 , so 1 2 , E 0 .3 6 , an d 1 .1 3

t

j zo

x

E m A V Ve

m m m

E a

2 1

2 1

3 21 ; , 1

5 5

t i i

o o oE E E

5-25

12 .8 3, so 2 .8 3

t

j zi io

o s x

EE e

E a

1( , ) 2 .8 3 co s .

x

Vz t t z

m E a

11 .7 0 , s o 1 .7 0j zr i r

o o s xE E e

E a

1 11 1

1 .7 0 4 .51 2 0

j z j zr r

s P s z x y

m Ae e

m

H a E a a a

3

2

11 .7 0 4 .5 1 0 3 .8

2

r

a vg z z

m Wx

m

P a -a

P5.39: Suppose a UPW in air carrying an average power density of 100 mW/m2 is

normally incident on a nonmagnetic material with r = 11. What is the time-averaged

power density of the reflected and transmitted waves?

1 2

1 11 2 0 1 1

1 2 0 ; ; 0 .5 3 7

1 1 1 11 1

o

1 0.463

2

2P P 2 8 .8

r i

a vg a vg

m W

m

2

2

2

2

1P P 1 1 7 1 .2

2

x ot i

a v g a v g

E m W

m

P5.40: A UPW in a lossless nonmagnetic r = 16 media (for z < 0) is given by

E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m.

This is incident on a lossless media characterized by r = 12, r = 6.0 (for z > 0). Find the

instantaneous expressions for the reflected and transmitted electric field intensities.

1 1

310 20

j z j zi j

s x ye e e

E a a

1 13

10 20j z j zr j

s x ye e e

E a a

1 2

1 2 0 1 23 0 ; 1 2 0 1 2 0 2

61 6

2 1

2 1

0 .7 0 0 ; 1 1 .7 0

1 13

7 14j z j zr j

s x ye e e

E a a

( , ) 1 17 c o s 1 4 c o s

3

r

z t x y

Vt z t z

m

E a a

5-26

2 23

10 20 ,j z j zt j

s x ye e e

E a a or 2 2

317 34

j z j zt j

s x ye e e

E a a , so

( , ) 2 21 7 c o s 3 4 c o s

3

t

z t x y

Vt z t z

m

E a a .

P5.41: The wave Ei = 100 cos(x 106t - 1z + /4) ax V/m is incident from air onto a

perfect conductor. Find Er and Et.

For the perfect conductor, 2 = 0. So = -1 and

Er = -100 cos(x 106t + 1z + /4) ax V/m

Et = 0

P5.42: A UPW given by E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m is

incident from air (for z < 0) onto a perfect conductor (for z > 0). Find the instantaneous

expression for the reflected electric field intensity and the SWR.

As in the previous problem, = -1. We then have

E(z,t) = -10.cos(t+1z)ax - 20.cos(t+1z+/3)ay V/m

1

1S W R

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

conductor. Find Er, Et and the time-averaged power density transmitted at the surface.

For copper we have

4 52

2

2

8 7 7 3

2 2 2 2

2

w h e re 1 0 4 1 0 5 .8 1 0 1 5 1 1 0

je

N pf x x x

m

so 4 5

23 .7

je m

We find 6 4 52 2

1 2 1

2 21, a n d = 1 9 .6 1 0

jx e

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

2 2 4 51 9 6 ,

z j zt j

s x

Ve e e

m

E a and 2

21 9 6 co s 4 5

zt

x

Ve t z

m

E a

26

23

1 9 6 1 01co s 4 5 3 .7 .

2 3 .7 1 0

t

a vg z z

x V m W

mx

P a a

5-27

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 SWR and at 1 kHz, 1 MHz, and 1 GHz.

6

1 2 9

1 2 0 7 .8 9 6 1 0 ( )2 4 ;

0 .0 0 8 4 .5 0 6 1 0 ( )2 5

j j x f H z

j j x f H z

Table P5.44

f 2() 2 1

2 1

1

1S W R

1kHz 0.994ej44.98° 0.9815ej178.9° 107.3

1MHz 29.3ej30.3° 0.513ej155° 3.11

1GHz 41.9ej0.05° 0.286ej179.9° 1.80

P5.45: MATLAB: Write a program that prompts the user for the constitutive parameters

in medium 1 and medium 2 separated by a planar surface. You are to assume a wave is

normally incident from media 1 to media 2. The program is to plot and versus a

frequency range supplied by the user. Use this program to plot and from 100 Hz to

10 GHz for the pair of media specified in the previous problem.

%ML P0545

clear

clc

%prompt user for constit parameters of media 1 & 2

%then plot ref & trans coeff over a freq range.

%We'll plot mag and angle of each.

%enter constant values

eo=8.854e-12; %free space permittivity, F/m

uo=pi*4e-7; %free space permeability, H/m

%enter media 1 values

er1=input('enter er1: ');

ur1=input('enter ur1: ');

s1=input('enter s1: ');

%enter media 2 values

er2=input('enter er2: ');

ur2=input('enter ur2: ');

s2=input('enter s2: ');

%calculations

n=2:.5:10;

f=10.^n;

5-28

Fig. P5.45

w=2*pi.*f;

eta1=sqrt(i*w*ur1*uo./(s1+i*w*er1*eo));

eta2=sqrt(i*w*ur2*uo./(s2+i*w*er2*eo));

Gamma=(eta2-eta1)./(eta2+eta1);

Gmag=abs(Gamma);

Gang=180*angle(Gamma)/pi;

Tau=1+Gamma;

Tmag=abs(Tau);

Tang=180*angle(Tau)/pi;

subplot(2,1,1)

semilogx(f,Gmag,'-o',f,Tmag,'-*')

xlabel('frequency (Hz)')

ylabel('magnitude')

legend('reflection','transmission')

subplot(2,1,2)

semilogx(f,Gang,'-o',f,Tang,'-*')

xlabel('frequency (Hz)')

ylabel('phase angle (degrees)')

legend('reflection','transmission')

Run the program:

enter er1: 25

enter ur1: 1

enter s1: 0

enter er2: 81

enter ur2: 1

enter s2: .008

>>

P5.46: A wave specified by Ei = 100.cos(x107t-1z)ax V/m is incident from air (at z < 0)

to a nonmagnetic media (z > 0, = 0.050 S/m, r = 9.0). Find Er, Et and SWR. Also find

the average power densities for the incident, reflected and transmitted waves.

5-29

7 6

1 1

1

21 2 0 , 1 0 s o 5 1 0 , 0 .1 0 5

r a d r a dx f x H z

s c m

In this problem we find in medium 2 (z > 0) that = 0.0025 and = 0.05. These values

are too close to allow for simplifying assumptions. Using (5.13) and (5.31), we calculate:

4 3 .6

2 2 20 .9 6 9 , 1 .0 1 9 , 2 8 .1

jN p ra de

m m .

Then,

1 7 4 4 0 .82 1

2 1

10 .8 9 8 , 1 8 .6 , 1 0 .1 4 1

1

j je S W R e

11 0 0j zi

s x

Ve

m

E a

1 1 1 7 41 0 0 8 9 .8

j z j zr j

s x x

V Ve e e

m m

E a a ,

so 7

( , ) 8 9 .8 co s 1 0 0 .1 0 5 1 7 4 .r

x

Vz t x t z

m E a

2 2 4 0 .81 0 0 1 4 .1

j z j zt j

s x x

V Ve e e

m m

E a a ,

so 7

( , ) 1 4 .1 co s 1 0 1 .0 2 4 0 .8 .t

x

Vz t x t z

m E a

2 24 3 .6 4 0 .8 2 .81 4 .10 .5 0 2

2 8 .1

j z j zt j j j

s y y

A Ae e e e e

m m

H a a

2

11 4 .1 0 .5 0 2 co s 4 0 .8 2 .8 2 .6

2

t

z z

W

m P a a

2

2

1 0 01 3 .3

2 1 2 0

i

z

W

m P a

2

2

8 9 .81 0 .7

2 1 2 0

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 a

nonmagnetic, lossless, r = 9.0 media (at z < 0) to a media (z > 0) with = 0.020 S/m, r

= 2.0, and r = 16.). Find Hi, Er, Hr, Et, Ht, and the average power densities for the

incident, reflected and transmitted waves.

We use ML0501 in each media to find:

1 1 10; 0 .6 2 8 ; 4 0

ra d

m

3 3

2 2 21 .0 1 ; 1 .5 6 ; 8 4 .9

jN p ra de

m m

We also will need reflection and transmission coefficients:

5-30

1 2 6 1 9 .82 1

2 1

0 .3 5 3 ; 1 0 .8 4j j

e e

Incident:

1 41 2

j zi j

s x

Ve e

m

E a

1 14 4

1

1 20 .3 0 0

j z j zi j j

s y y

A Ae e e e

m m

H a a ,

7( , ) 0 .3 0 0 c o s 2 1 0 0 .6 2 8 .

4

i

y

Az t x t z

m

H a

2

2

1

1 215 .6 5 5

2

i

a vg z z

W

m

P a a

Reflected:

1 1 14 4 5 1 2 6 1 7 11 2 1 3 .3 1 3 .3

j z j z j zr j j j j

s x x x

V V Ve e e e e e e

m m m

E a a a

7

( , ) 1 3 .3 co s 2 1 0 0 .6 2 8 1 7 1 .r

x

Vz t x t z

m E a

1 11 7 1 1 7 11 3 .30 .1 0 6

4 0

j z j zr j j

s y y

A Ae e e e

m m

H a a

7

( , ) 0 .1 0 6 co s 2 1 0 0 .6 2 8 1 7 1 .r

y

Az t x t z

m H a

2

2

1 3 .310 .7 0 4

2 4 0

r

a vg z z

W

m P a -a

Transmitted:

1 24 6 4 .81 2 3 1 .6 7

j z j zt j j

s x x

V Ve e e e

m m

E a a ,

7

( , ) 3 1 .7 co s 2 1 0 1 .5 6 6 4 .8 .t

x

Vz t x t z

m E a

2 23 3 6 4 .8 3 1 .83 1 .6 70 .3 7 3

8 4 .9

j z j zt j j j

s y y

A Ae e e e e

m m

H a a ,

7

( , ) 0 .3 7 3 co s 2 1 0 1 .5 6 3 1 .8 .t

y

Az t x t z

m H a

2

3 1 .6 7 0 .3 7 3c o s 6 4 .8 3 1 .8 4 .9 5 4

2

t

a v g 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 from

air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The angle of

5-31

incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and

transmission coefficients, and (c) the incident, reflected and transmitted fields.

(a)

6

1 28

2 1 0 0 1 02 .0 9 , 1 0 .4 5 .

3 1 0

rx ra d ra d

c x m c m

1

2 2

1 1 1; s in s in 4 0 ; 7 .4

5 5t t

r

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

1 2

1 2 01 2 0 ; 2 4

2 5

2 1

2 1

c o s c o s0 .7 3 2 ; 1 0 .2 6 8

c o s c o s

i t

T E T E T E

i t

(c) The fields,

Incident:

2 .0 9 s in 4 0 co s 4 01 .3 4 1 .6 0

1 1j x z

i j x j z

s y y

Ve e e

m

E a a

( , ) 1 c o s 1 .3 4 1 .6 0i

y

Vz t t x z

m E a

1 .3 4 1 .6 01

co s 4 0 s in 4 01 2 0

i j x j z

s x ze e

H a a

1 .3 4 1 .6 0

2 .0 3 1 .7 1i j x j z

s x z

m Ae e

m

H a a

( , ) 2 .0 3 1 .7 1 co s 1 .3 4 1 .6 0i

x z

m Az t t x z

m H a a

Reflected:

0 .7 3 2r i

o T E oE E

1 .3 4 1 .6 0

0 .7 3 2r j x j z

s y

Ve e

m

E a

( , ) 0 .7 3 2 co s 1 .3 4 1 .6 0r

y

Vz t t x z

m E a

1 .3 4 1 .6 0

1 .3 4 1 .6 0

0 .7 3 2c o s 4 0 s in 4 0

1 2 0

1 .4 9 1 .2 5

r j x j z

s x z

r j x j z

s x z

Ae e

m

m Ae e

m

H a a

H a a

( , ) 1 .4 9 1 .2 5 co s 1 .3 4 1 .6 0r

x z

m Az t t x z

m H a a

transmitted:

0 .2 6 8t i

o T E oE E

2s in co s 1 .3 5 1 0 .4

0 .2 6 8 0 .2 6 8t t

j x zt j x j z

s y y

Ve e e

m

E a a

5-32

1 .3 5 1 0 .40 .2 6 8

co s 7 .4 s in 7 .42 4

t j x j z

s x z

Ae e

m

H a a

1 .3 5 1 0 .4

3 .5 0 .4 6t j x j z

s x z

m Ae e

m

H a a

( , ) 3 .5 0 .4 6 co s 1 .3 5 1 0 .4t

x z

m Az t t x z

m H a a

P5.49: A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident

from air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The

angle of incidence is 40. Calculate (a) the angle of transmission, (b) the reflection and

transmission 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 we

have t = 7.4°. Also, 1 = 2.09 rad/m and 2 = 10.45 rad/m.

(b)

2 1

2 1

c o s c o s0 .5 8 9

c o s c o s

t i

T M

t i

2

2 1

2 c o s0 .3 1 8

c o s c o s

i

T M

t i

(c)

Incident:

1 .3 4 1 .6 0

1 co s 4 0 s in 4 0i j x j z

s x ze e

E a a

( , ) 0 .7 6 6 0 .6 4 3 co s 1 .3 4 1 .6 0i

x z

Vz t t x z

m E a a

1 .3 4 1 .6 01

1 2 0

i j x j z

s y

Ae e

m

H a

( , ) 2 .6 5 co s 1 .3 4 1 .6 0i

y

m Az t t x z

m H a

Reflected:

1 .3 4 1 .6 0

0 .5 8 9 co s 4 0 sin 4 0r j x j z

s x ze e

E a a

( , ) 0 .4 5 2 0 .3 7 9 co s 1 .3 4 1 .6 0r

x z

Vz t t x z

m E a a

1 .3 4 1 .6 00 .5 8 9

1 2 0

r j x j z

s y

Ae e

m

H a

( , ) 1 .5 6 co s 1 .3 4 1 .6 0r

y

m Az t t x z

m H a

transmitted:

1 .3 5 1 0 .4

0 .3 1 8 co s 7 .4 s in 7 .4t j x j z

s x ze e

E a a

( , ) 0 .3 1 5 0 .0 4 1 co s 1 .3 5 1 0 .4t

x z

Vz t t x z

m E a a

5-33

( , ) 4 .2 2 co s 1 .3 5 1 0 .4t

y

m Az t t x z

m H a

P5.50: A randomly polarized UPW at 200 MHz is incident at the Brewster’s angle from

air (z < 0) onto a thick slab of lossless, nonmagnetic material with r = 16 (z > 0). The

wave can be decomposed into equal TE and TM parts, each with an incident electric field

amplitude of 10. V/m. Find expressions for the instantaneous value of the incident,

reflected and transmitted electric fields.

First we calculate the Brewster’s angle: 1

s in ; 7 61 1 1 6

B A B A

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

TE

1s in co s 4 .0 6 1 .0 1

1 0 1 0i i

j x zi j x j z

s y y

Ve e e

m

E a a

At the Brewster’s angle of incidence, we have from Snell’s Law:

1 1

2

s in s in 1 4t i

2 1

2 1

c o s c o s0 .8 8 3; 1 0 .1 1 7

c o s c o s

i t

T E T E T E

i t

V V= -8 .8 3 ; = 1 .1 7

m m

r i t i

o T E o o T E oE E E E

4 .0 6 1 .0 18 .8 3

r j x j z

s y

Ve e

m

E a

2

s in co s 4 .0 6 1 6 .31 0 1 .1 7

t tj x zt j x j z

s T E y y

Ve e e

m

E a a

TM:

At the Brewster’s angle, TM = 0 and 0 .r

sE

4 .06 1 .01

10 cos sini j x j z

s i x i ze e

E a a ,

4 .0 6 1 .0 1

2 .4 2 9 .7 0 .i j x j z

s x z

Ve e

m

E a a

4 .06 16 .3

10 cos sint j x j z

s t x t ze e

E a a ,

4 .0 6 1 6 .3

9 .7 0 2 .4 0 .t j x j z

s x z

Ve e

m

E a a

Combining the results we arrive at:

5-34

( , ) 2 .4 1 0 9 .7 co s 4 .0 6 1 .0 1i

x y z

Vz t t x z

m E a a a

( , ) 8 .8 3 co s 4 .0 6 1 .0 1r

y

Vz t t x z

m E a

( , ) 9 .7 1 .2 2 .4 co s 4 .0 6 1 6 .3t

x y z

Vz t t x z

m E a a a