Uniform Plane Wave
1. 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?
3x108 m s
(a) f 
 
 5GHz
 
0.06m
(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz
(c)
1
m
c

u p   f   5 x109   0.01m   5 x107 
s
s
r

up
c
2
 3x108 
 r  
 36
8 
 0.5 x10 
_____________________________________________________________________
2. Given  = 1.0x10-5 S/m , r = 2.0, r = 50., and f = 10. MHz, find , , , and .
  jr o   j r o     j

jr o
  j r  o
jr o  j 2 10 x106   50   4 x107   j 3948
  j r  o  1x105  j 2 10 x106   2   8.854 x1012   1x105  j1.11x103
Inserting these into the expressions for  and ,
  9.4x103  j 2.1 1 m ,   9.4 x103 Np m ,   2.1 rad m ,   1880e j 257 
_____________________________________________________________________
3. Suppose in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).
H s  0.100e j  x e j a z , a P  a x ,
   4 
E s   a P  H s  120 a x  0.100e j  x e j a z  12 e  j  x e j a y
E  12 cos t   x    a y
Since free space is stated,
2 2


 2 30 rad m
 c f
and then
2
 V

E  12 cos  2 x107 t 
x  ay
30
4 m

_____________________________________________________________________
4. 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


up
the

given
information
we
have
  2 f  200 x106
2 rad
,
3 m
rad
s
and
2  V

y  az .
or E( y, t )  1cos  200 x106 t 
3  m

Now to find H.
1
1
1  j y
Es  1e j y a z , H s  a P  Es 
a y 1e j y a z 
e ax

120
120
So
1
2  A

H  y, t  
cos  200 x106 t 
y  ax
120
3  m

or
2  mA

H  y, t   2.7 cos  200 x106 t 
y  ax
.
3 
m

_____________________________________________________________________
5. 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.
up 

c
3x108
m


 0.75 x108

s
r
16
  u p    0.75 x108  10   7.5 x108
rad
s
mA
H ( y, t )  100 cos  7.5 x108 t  10 y  a z
m
 j y
H s  0.100e a z ,
Es  a P  H s 
120
r
a y  0.100e j y a z  3 e j y a x
E( y, t )  9.4 cos  7.5 x108 t  10 y  a x
V
m
_____________________________________________________________________
6. 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.
 mA   x
 x  j  x
H  1
 e cos t   x  a z ; H s  H o e e a z
 m 
Es  aP  Hs  a x  Hoe x e j xa z   Hoe x e j xa y
j 2 100 x106  100   4 x107 
j


 2664e j 30 
6
12
  j
0.00964  j 2 100 x10  8.854 x10 

j   j   14.8  j 25.7
1
m
Finally,
E( x, t )  2.66e 15 x cos  200 x106 t  26 x  30  a y
V
m
_____________________________________________________________________
7. In seawater, a propagating electric field is given by E(z,t) = 20.e-z cos(xt –
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.
So: jo  j 7.896, j r o  j 0.004

j
 1.257e j 44.98 
  j
 
j   j   4.441  j 4.445 1 m
    4.4
1
m
V
V
 20e  z e   z e j 28.6 a y
m
m
1
1

20
A
H s  a P  Es  a z  20e  z e   z e j 28.6 a y 
e  z e   z e j 28.6 a x



m
A
H( z, t )  15.9e 4.4 z cos  2 x106 t  4.4 z  28.6  45  a x
m
or with appropriate significant digits:
A
H( z, t )  16e 4.4 z cos  2 x106 t  4.4 z  16  a x
m
_____________________________________________________________________
8. For Nickel ( = 1.45 x 107, r = 600), make a table of , , , up, and  for 1Hz,
1kHz, 1MHz, and 1 GHz.
Es  20e  z e   z e j 0.5radians a y
For Ni we have = 1.45x107S/m, r = 600
     f    f  Hz  600  4 x107 1.45x107   34.35x103 f ( Hz )
= 1/
 j 45
e   18.08 x106 f ( Hz )e j 45 

c
m
up 
 12 x106
s
r r
Table
f(Hz)=
1
103
185
5860
(Np/m)
185
5860
(rad/m)
j45º
18e 
570ej45º

5.4mm
170m

up(m/s)
12x106
12x106
 2
106
185x103
185x103
18ej45ºm
5.3m
12x106
109
5.9x106
5.9x106
0.57ej45º
170nm
12x106
9. 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: E( z, t )  1.0e  z cos  x106 t   z  a x
j  j  x106  4 x107   j 3.948
V
m
j  j  x106  10.2  8.854 x1012   j 284 x10 6
Here,  (i.e. it is a good conductor), so
1
   f   24.3  
m
 j 45
  2 e  0.115e j 45 

So now we have
V
E( z, t )  1.0e 24 z cos  x106 t  24 z  a x
m
To find B we’ll work in phasors.
Es  1e z e j z a x , H s 
1

a P  Es 
1

a z 1e z e j z a x 
1

e z e j z a y
1
A
e 24 z cos  x106 t  24 z  45  a y
0.115
m
A
H( z, t )  8.7e 24 z cos  x106 t  24 z  45  a y
m
_____________________________________________________________________
10. 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.
H( z, t ) 


E(0, t )  1cos 2  600 x106  t   z a y
E( z, t )  1e  z cos t   z  a y
Teflon:  = 0 so  = 0,
and     

c
r 
V
m
V
m
2  600 x106 
8
3x10
(a) E( z, t )  1cos 1.2 x109 t  18.2 z  a y
2.1  18.2
V
m
1
2.1
V
(b) H s  a P  Es 
a z 1e j z a y ,

120
m
mA
H ( z, t )  3.8cos 1.2 x109 t  18.2 z  a x
m
rad
m
1 1 2.1
mW
(c) Pavg 
a z  1.9 2 a z
2 120
m
_____________________________________________________________________
11. 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
mV
1 Eo 2
f  200MHz , Eo  1
, Pavg 
m
2 
Cu:   2
 j 45
Np
e ,    f   214 x103
, so   5.22e j 45 m

m
3
1 10 
W
Pavg 
 96 2 ; P  Pavg S  96W
3
2 5.22 x10
m
Now at 10 m beneath the surface, we have
2
E ( z  10 m)  Eo e  (10  m )  103 e  (214 x10 )(10  m )  118 x106
3
V
m
1 118 x10 
W
Pavg 
 1.3 2 ; P  1.3W
3
2 5.22 x10
m
_____________________________________________________________________
12. Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m, find the polarization
and handedness.
6 2
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
_____________________________________________________________________
13. Given
E( z, t )  Exo cos t   z  ax  E yo cos t   z    ay ,
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
_____________________________________________________________________
14. 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
 1


120
11
  0.537
1  o  120; 2 
; 


11
 1
 1
11 

  1   0.463
r
i
Pavg
  Pavg
 28.8
2
mW
m2
2
1 Exo
mW
i
P 
  2 Pavg
11  71.2 2
2 2
m
_____________________________________________________________________
15. 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.
t
avg
Eis  10e j1z a x  20e j1z e j 3a y
Ers  10e j1za x  20e j1z e j 3a y
120
12
 30; 2  120
 120 2
6
16
 
  2 1  0.700;   1    1.70
2  1
1 
Ers  7e j1z a x  14e j1z e j 3a y
 V

E(rz ,t )  7 cos t  1 z  a x  14 cos  t  1 z   a y
3
m

 j 2 z
 j 2 z j 3
 j 2 z
t
t
Es  10 e
a x  20 e
e a y , or Es  17e
a x  34e j2 z e j 3a y , so
 V

Et( z ,t )  17 cos t   2 z  a x  34 cos  t   2 z   a y
.
3
m

_____________________________________________________________________
16. The wave Ei = 10.cos(2x 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
2  2
 2 j 45
e 
2
where  2   f 2 2   108  4 x107  5.8 x107   151x103
Np
 2
m
so 2  3.7e j 45 m
22
2
 2  19.6 x106 e j 45
1  2 1
r
8
So E = -10.cos(2x 10 t + 1z) ax V/m
V
V
Ets  196  e  2 z e  j 2 z e j 45 a x
, and Et  196e  2 z cos t   2 z  45  a x
m
m
We find   1, and  =
6
1 196 x10 V m 
W

cos  45  a z  3.7 2 a z .
3
2  3.7 x10  
m
2
t
avg
P
17. 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.
rad
2 
rad
so f  5 x106 Hz, 1 
  0.105
s
1 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:
Np
rad
 2  0.969
,  2  1.019
, 2  28.1e j 43.6  .
m
m
Then,
1 
 
  2 1  0.898e j174 , SWR 
 18.6,   1    0.141e j 40.8
2  1
1 
1  120,    x107
V
m
V
V
Ers  100e j 1z a x  89.8e  j 1z e j174 a x ,
m
m
Eis  100e  j 1z a x
so Er ( z, t )  89.8cos  x107 t  0.105 z  174  a x
Ets  100 e j2 z a x
V
.
m
V
V
 14.1e j 2 z e j 40.8 a x ,
m
m
so Et ( z , t )  14.1cos  x107 t  1.02 z  40.8  a x
V
.
m
14.1  j 43.6  j 2 z j 40.8
A
A
e
e
e
a y  0.502e  j 2 z e  j 2.8 a y
28.1
m
m
1
W
P t  14.1 0.502  cos  40.8  2.8  a z  2.6 2 a z
2
m
H ts 
100   13.3 W a
P 
z
2 120 
m2
2
  89.8 
W
r
P 
 10.7 2  -a z 
2 120 
m
2
i
(check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2)
_____________________________________________________________________
18. 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)
 cos t  1 cos i
TM  2
 0.589
2 cos t  1 cos i
 TM 
22 cos i
 0.318
2 cos t  1 cos i
(c)
Incident:
Eis  1e j1.34 x e j1.60 z  cos 40 a x  sin 40 a z 
Ei ( z, t )   0.766a x  0.643a z  cos t  1.34 x  1.60 z 
H is 
V
m
1  j1.34 x  j1.60 z A
e
e
ay
120
m
H i ( z, t )  2.65cos t  1.34 x  1.60 z  a y
mA
m
Reflected:
Ers  0.589e j1.34 x e j1.60 z  cos 40 a x  sin 40 a z 
Er ( z , t )   0.452a x  0.379a z  cos t  1.34 x  1.60 z 
H rs 
0.589  j1.34 x  j1.60 z A
e
e
ay
120
m
H r ( z, t )  1.56 cos t  1.34 x  1.60 z  a y
V
m
mA
m
transmitted:
Ets  0.318e j1.35 x e j10.4 z  cos 7.4 a x  sin 7.4 a z 
Et ( z, t )   0.315a x  0.041a z  cos t  1.35 x  10.4 z 
V
m
mA
m
_____________________________________________________________________
H t ( z, t )  4.22 cos t  1.35 x  10.4 z  a y