This document contains solutions to multiple physics problems involving electromagnetic waves. Problem 1 involves calculating the conductivity and penetration depth of graphite at different frequencies. Problem 2 involves propagating an electromagnetic wave in seawater and calculating various parameters like attenuation constant and phase velocity. It provides the solutions and steps for parts a, b, and c of this problem. Problem 3 involves analyzing the behavior of electromagnetic waves on a finite transmission line terminated by a load impedance and derives relevant equations.

1. ©Haris H.
Q1. Given that the skin depth for graphite at 100MHz is 0.16mm, determine (a) the conductivity of graphite, and
(b) the distance that a 1GHz wave travels in graphite such that its field intensity is reduced by 30(dB).
(Sol.) (a) mS
f
/1099.01016.0
1 53
 



(b) At f=109
Hz, mNpf /1098.1 4
 
m
e
zedB z 4
10
10 1075.1
log
5.1
log20)(30 



Q2. )10cos(100ˆ),( 7
txztE 

V/m at z=0 in seawater: εr=72, μr=1, σ=4S/m. (a) Determine α, β, vp, and ηc. (b) Find the
distance at which the amplitude of E is 1% of its value at z=0. (c) Write E(z,t) and H(z,t) at z=0.8m, suppose it propagates
in the +z direction.
(Sol.)  7
10 , f=5×106
Hz, σ/ωε0εr=200>>1, ∴ Seawater is a good conductor in this case.
(a)   mNpf /89.8 ,



f
jc )1( 
smvp /1053.3 6



, m707.0
2



 , m112.0
1



(b) mze z
518.0)100ln(
1
01.0 


(c) )cos(100ˆ])(Re[),( ztexezEtzE ztj

 
)11.710cos(082.0ˆ)8.0cos(100ˆ),8.0(8.0 78.0
 
txtextEmz 
),8.0(ˆ
1
),8.0( tEatH n



, )61.110cos(026.0ˆ]
)8.0(
Re[ˆ),8.0( 7
 tye
E
ytH tj
c
x



Q2 a). Outline salient properties of the r-circles:
Several salient properties of the r-circles:
1. The centers of all r-circles lie on the Γr-axis.
2. The r=0 circle, having a unity radius and centered at the origin, is the largest.
3. The r-circles become progressively smaller as r increases from 0 toward ∞, ending at the (Γr=1, Γi=0) point
for open-circuit.
4. All r-circles pass through the (Γr=1, Γi=0) point.
Salient properties of the x-circles:
1. The centers of all x-circles lie on the Γr=1 line, those for x>0 (inductive reactance) lie above the Γr–axis, and
those for x<0 (capacitive reactance) lie below the Γr–axis.
2. The x=0 circle becomes the Γr–axis.
3. The x-circle becomes progressively smaller as |x| increases from 0 toward ∞, ending at the (Γr=1, Γi=0) point
for open-circuit.
4. All x-circles pass through the (Γr=1, Γi=0) point.
Summary
1. All |Γ|–circles are centered at the origin, and their radii vary uniformly from 0 to 1.
2. The angle, measured from the positive real axis, of the line drawn from the origin through the point
representing zL equals θΓ.
3. The value of the r-circle passing through the intersection of the |Γ|–circle and the positive-real axis equals the
standing-wave radio S.
b) Illustrate circle of constant reactance on smith chart.

2. ©Haris H.
Q3 a) draw a finite transmission line terminated with load impedance
(Proof)








)2.....()(
)1...()(
00
00
zz
zz
eIeIzI
eVeVzV


, 




0
0
0
0
0
I
V
I
V
Z
Let z=l, V(l)=VL, I(l)=IL































eZIVV
eZIVV
e
Z
V
e
Z
V
I
eVeVV
LL
LL
L
L
)(
2
1
)(
2
1
00
00
0
0
0
0
00











])()[(
2
)(
])()[(
2
)(
)(
0
)(
0
0
)(
0
)(
0
z
L
z
L
L
z
L
z
L
L
eZZeZZ
Z
I
zI
eZZeZZ
I
zV























)'cosh'sinh()'(
)'sinh'cosh()'(
])()[(
2
)'(
])()[(
2
)(
0
0
0
'
0
'
0
0
'
0
'
0
'
zZzZ
Z
I
zI
zZzZIzV
eZZeZZ
Z
I
zI
eZZeZZ
I
zV
L
L
LL
z
L
z
L
L
z
L
z
L
L




'tanh
'tanh
)'(
0
0
0
zZZ
zZZ
ZzZ
L
L




 , Zi=


 

tanh
tanh
)(
0
0
0
'
0
L
L
z
z
ZZ
ZZ
ZZ





Lossless case (α=0, γ=jβ, Z0=R0, tanh(γl)=jtanβl): Zi=
ljZR
ljRZ
R
L
L


tan
tan
0
0
0



Q3 b)
Illustrate circle of constant resistance on smith chart.
Q3 a) Q4 b)
What should be the size of a hollow cubic cavity made of copper in order for it to have a dominant resonant frequency
of 10GHz? (b) Find the Q at that frequency.
(Sol.) (a) For a cubic cavity, a=b=d, TM110, TE011, and TE101 are degenerate dominant modes. )(10
2
103 10
8
101 Hz
a
f 

 ,
ma 2
10
8
1012.2
102
103 



 .
(b) 

0101
0101
101
33
f
a
R
af
Q
s

For copper, )/(1080.5 7
mS , .10700)1080.5)(104(10)10
3
12.2
( 77102
101  
Q
Q3. b)

3. ©Haris H.
A standard rectangular waveguide WG-16 is to be designed for the Xband (8-12.4 GHz) radar application. The
dimensions are a= 2.29 cm and b= 1.02 cm. If only the lowest mode TE10mode is to propagate inside the waveguide
and that the operating frequency be at least 25% above the cutoff frequency of the TE10mode but no higher than
95% of the next higher cutoff frequency, what is the allowable operating frequency range of this waveguide?
Q4 a) Classify the propagating waves in a uniform waveguide according to whether their component is zero or non-
zero β–ω curve for waveguide TE and TM modes
Q4 a)
• Half Power Beam Width (HPBW) is defined as the angular difference between the points where the radiation
intensity reaches half of its maximal value (3 dB difference in decibels).
• First Null Beam Width (FNBW) is defined as the angular difference between the two nulls enclosing the main
beam.
• The Side Lobe Level (SLL) is a parameter used to describe the level of side lobe suppression. As previously
mentioned, high side lobes are often not desired, since they represent radiation outside the main beam sector. Side
lobe level is defined as the difference in decibels between the main beam peak value & the side lobe peak value.
• Half-power beam width: Angular width of main beam between the half-power (-3dB) points
• Sidelobe level: (|Emax| in one sidelobe)/( |Emax| in main beam)
• Null positions: Directions which have no radiations in the far-field zone.

4. ©Haris H.
b)
i. Define Friis Formula
Define σbs= radar cross section of target, 2
2
21
22
21
2
1
2
2
12
2
)4(
4
4
)
4
(
r
GG
r
AAA
r
A
G
r
A
P
P DDeeee
D
e
t
L






ii. Ratio of the directive gain and the effective area of antenna is a universal constant what it is?
Q5 a)
Derive the equations Electric and magnetic Field intensities for Elemental Electric Dipole


Qdzp ˆ


j
I
QQj
dt
dQ
I  ,
R
eId
aa
R
eId
zA
Rj
R
Rj 






 

4
)sinˆcosˆ(
4
ˆ 00 
 AaAaAa RR
ˆˆˆ 

5. ©Haris H.















0
sin)(
4
sin
cos)(
4
cos
0
0












A
R
eId
AA
R
eId
AA
Rj
z
Rj
zR



Rj
R
e
RjRj
Id
a
A
RA
RR
aAH















]
)(
11
[sin
4
ˆ
])([
1
ˆ
1
2
2
00


)](
1
ˆ)sin(
sin
1
ˆ[
11
00
 

RH
RR
aH
R
a
j
H
j
E R






















)(120/,0
]
)(
1
)(
11
[sin
4
]
)(
1
)(
1
[cos2
4
000
32
2
0
32
2
0











whereE
e
RjRjRj
Id
E
e
RjRj
Id
E
Rj
Rj
R


Far field of a Hertzian dipole: if βR=2πR/λ>>1



 sin)(
4 R
eId
jH
Rj


, 


 sin)(
4
0
R
eId
jE
Rj


b)
i. Width of main beam (or simply beamwidth). The main-beam beamwidth describes the sharpness of the main
radiation region. It is generally taken to be the angular width of a pattern between the half-power, or - 3 (dB), points. In
electric-intensity plots it is the angular width between points that are 0.707 times the maximum intensity. Of course, the
main beam must point in the direction where the antenna is designed to have its maximum radiation.
Sidelobe levels. Sidelobes of a directive (nonisotropic) pattern represent regions of unwanted radiation; they should
have levels as low as possible. Generally, the levels of distant sidelobes are lower than the levels of those near the main
beam. Hence, when one talks about the sidelobe level of an antenna pattern, one usually refers to the first (the nearest
and highest) sidelobe.
Directivity. The beamwidth of an antenna pattern specifies the sharpness of the main beam, but it does not provide us
with any information about the rest of the pattern. For example, the sidelobes may be very high-an undesirable feature.
A commonly used parameter to measure the overall ability of an antenna to direct radiated power in a given direction is
directive gain, which may be defined in terms of radiation intensity. Radiation intensity is the time-average power per
unit solid angle. The SI unit for radiation intensity is watt per steradian (W/sr).
ii. is referred to as Friis Transmission Formula
iii. An anechoic chamber ("an-echoic" meaning non-reflective, non-echoing or echo-free) is a room designed to
completely absorb reflections of either sound or electromagnetic waves.
b) For a uniform linear array, give the general expression of normalized array factor. Sketch the normalized array factor
for a five-elemental array. For Broadside

6. ©Haris H.
Normalized array factor in the xy-plane (θ=π/2):

 )1(2
...1
1
)( Njjj
eee
N
A




 j
jN
e
e
N 1
11
=
)2/sin(
)2/sin(1

N
N
, where Ψ=βdsin(θ)cosφ+ξ=βdcosφ+ξ if θ=π/2
Mainbeam direction, φ0: ∵ Max at Ψ=0, ∴ βdcosφ0+ξ=0
d



0cos
Null locations: 

k
N

2
, k=1,2,3,…
Sidelobe locations:
2
)12(
2



m
N
, m=1, 2, 3, …
The first sidelobe level:
2
3
2


N
, )(212.0
)3/2sin(
11
)(  Nas
NN
A

Broadside array )0,
2
( 0  

 : |Emax| occurs at a direction ⊥ the line of arrays.
Endfire array ),0( 0 d  : |Emax| occurs at a direction // the line of arrays.
Beamwidth between two first nulls:
N
NN 

4
2
,
2
21
21




2 
N
ddd


4
)cos(cos)cos()cos( 2121 
Let   0201 ,
)(sin)
2
( 1
0
Nd



 
 for a broadside array.
Nd


2
)0( 0  for an endfire array.
Q5 a)
Plot the H-plane radiation patterns of two parallel dipoles for the following two cases: (a) 0,2/  d , (b)
2/,4/  d .
(Sol.) Let the dipole is z-directed
In the H-plane )2/(   : )cos(
2
1
cos
2
cos)( 

  dA
(a) )cos
2
cos()( 

 A , (b) )1(cos
4
cos)(  

A
Others:
A 100MHz uniform plane wave xExE ˆ

propagates in the +z direction. Suppose εr=4, μr=1, σ=0, and it has a maximum
value of 10-4
V/m at t =0 and z=0.125m. (a) Write the instantaneous expressions for E

and H

. (b) Determine the
location where E

is a positive maximum when t=10-8
sec.
(Sol.)
3
4
00

  rrk , zan ˆˆ  , 


 60
0
0

r
r

7. ©Haris H.
(a) )102cos(10ˆˆ),( 84
  
kztxExtzE x

has the maximum in case of
0102 8
  kzt
6

  )
63
4
102cos(10ˆ),( 84 
  
ztxtzE

,
)
63
4
102cos(
60
10
ˆ),(ˆ
1
),( 8
4





ztytzEatzH n

(b) 1)2cos( n ,
2
3
8
13
2
63
4
)10(102 maxmax
88 n
znz 



A 3GHz, y-polarized uniform plane wave propagates in the +x direction in a nonmagnetic medium having a dielectric
constant 2.5 and a loss tangent 10-2
. (a) Determine the distance over which the amplitude of the propagating wave will
be cut in half. (b) Determine the intrinsic impedance, the wavelength, the phase velocity, and the group velocity of the
wave in the medium. (c) Assuming )
3
106sin(50ˆ 9 
  tyE

V/m at x=0, write the instantaneous expression for H

for all t and x.
(Sol.) 39922
10166.45.210
36
1
103210110 





It is a low–loss dielectric material: mrad /34.99])(
8
1
1[ 2




)
2
1(




 jc  =  
29.0238
(a)



2
 =0.497, mde d
395.1
2
1497.0

(b) smvp /108973.1 8



, sm
ddd
d
vg /108975.1
)/(
1 8



(c)
)0016.0
3
(
497.03497.0
21.0ˆˆ
1
50ˆ






j
x
n
t
j
x
eezEaHeeyE

)332.06.31106sin(21.0ˆ),( 9497.0
  
xteztxH x

A/m
A TE10 wave at 10GHz propagates in a brass σc=1.57×107
(S/m) rectangular waveguide with inner dimensions a=1.5cm
and b=0.6cm, which is filled with εr=2.25, μr=1, loss tangent=4×10-4
. Determine (a) the phase constant, (b) the guide
wavelength, (c) the phase velocity, (d) the wave impedance, (e) the attenuation constant due to loss in the dielectric,
and (f) the attenuation constant due to loss in the guide walls.
(Sol.) f=1010
Hz, m
f
v
02.0
10
102
1025.2
103
10
8
10
8






For TE10 mode, Hz
a
v
fc
10
2
8
10667.0
)105.1(2
102
2



 
mrad
f
f
v
c
/234)(1 2


 ,. m
ffc
g 0268.0
)(1 2





sm
ff
v
v
c
p /1068.2
)(1
8
2


 , )(4.337
)(1 210



ff
Z
c
TE

mS /105104 44 
  , mdBmNpZTEd /73.0/084.0
2 10



)(05101.0 
c
c
s
f
R


,
mdBmNp
ffb
ffabR
c
cs
c /457.0/0526.0
)(1
]))(2(1[
2
2





 .
An air-filled a×b (b<a<2b) rectangular waveguide is to be constructed to operate at 3GHz in the dominant mode. We
desire the operating frequency to be at least 20% higher than the cutoff frequency of the dominant mode and also at
least 20% below the cutoff frequency of the next higher-order mode. (a) Give a typical design for the dimensions a and
b. (b) Calculate for your design β, vp, λg and the wave impedance at the operating frequency.

8. ©Haris H.
(Sol.) (a) 22
)()(
2
1
b
n
a
m
fc 

. b<a<2b, the dominant mode: TE10, the next mode: TE01
a
f TEc
2
1
)( 10
 ,
b
f TEc
2
1
)( 01
 , %20
)21(
)21(103 9




a
a
,
%20
)21(
103)21( 9




b
b
ma 06.0 , mb 04.0 , and a<2b
(b) Choose a=0.065m, b=0.035m, )(103.2)( 9
10
Hzf TEc  , 679.0)(1 2

f
fc
,
mrad
f
fc
/15.40)(1 2
  , sm
ff
v
c
p /107.4
)(1
11 8
2




,
m
f
vp
g 157.0 ,  590639.0/120)(1 2
010
 ffZ cTE
The magnetic field intensity of a linearly polarized uniform plane wave propagating in the +y direction in seawater εr=80,
μr=1, σ=4S/m is )
3
10sin(1.0ˆ 10 
  txH

A/m. (a) Determine the attenuation constant, the phase constant, the
intrinsic impedance, the phase velocity, the wavelength, and the skin depth. (b) Find the location at which the amplitude
of H is 0.01 A/m. (c) Write the expressions for E(y,t) and H(y,t) at y=0.5m as function of t.
(Sol.) (a) σ/ωε=0.18<<1, ∴ Seawater is a low-loss dielectric in this case.



2
 mNp/96.83 )
2
1(




 jc  0283.0
8.41 j
e
])(
8
1
1[( 2


  300 , smvp /1033.3 7



, m2
1019.1
1 


 , m3
1067.6
2 




(b) mye y 2
1074.210ln
1
1.0
01.0 



(c) )
3
10sin(1.0ˆ),( 10 

 
ytextyH y
,  300,5.0 y
)
3
10sin(1075.5ˆ),5.0( 1020 
  
txtH

)0283.0
3
10sin(1041.2ˆ),5.0(ˆ),5.0(ˆˆ 1018


  
tztHatEya ncn


9. ©Haris H.
What are guided waves? Give examples
The electromagnetic waves that are guided along or over conducting or dielectric surface are called guided waves.
Examples: Parallel wire, transmission lines

10. ©Haris H.
What is TE wave or H wave?
Transverse electric (TE) wave is a wave in which the electric field strength E is entirely transverse. It has a magnetic field
strength Hz in the direction of propagation and no component of electric field Ez in the same direction
What is TH wave or E wave?
Transverse magnetic (TM) wave is a wave in which the magnetic field strength H is entirely transverse. It has a electric
field strength Ez in the direction of propagation and no component of magnetic field Hz in the same direction What is a
TEM wave or principal wave?
TEM wave is a special type of TM wave in which an electric field E along the direction of propagation is also zero. The
TEM waves are waves in which both electric and magnetic fields are transverse entirely but have no components of Ez
and Hz .it is also referred to as the principal wave.
What is a dominant mode?
The modes that have the lowest cut off frequency is called the dominant mode.
What is cut-off wavelength?
It is the wavelength below which there is wave propagation and above which there is no wave propagation.
What is an evanescent mode?
When the operating frequency is lower than the cut-off frequency, the propagation constant becomes real i.e., γ = α. The
wave cannot be propagated. This non- propagating mode is known as evanescent mode.
What is the dominant mode for the TE waves in the rectangular waveguide?
The lowest mode for TE wave is 𝑇𝐸10 (m=1 , n=0)
What is the dominant mode for the TM waves in the rectangular waveguide?
The lowest mode for TM wave is 𝑇𝑀11(m=1 , n=1)
What is the dominant mode for the rectangular waveguide?
The lowest mode for TE wave is TE10 (m=1 , n=0) whereas the lowest mode for TM wave is TM11(m=1 , n=1).
The TE10 wave have the lowest cut off frequency compared to the TM11 mode. Hence the TE10 (m=1 , n=0) is the
dominant mode of a rectangular waveguide. Because the TE10 mode has the lowest attenuation of all modes in a
rectangular waveguide and its electric field is definitely polarized in one direction everywhere.
Why TEM mode is not possible in a rectangular waveguide?
Since TEM wave do not have axial component of either E or H, it cannot propagate within a single conductor waveguide
Explain why TM01 and TM10 modes in a rectangular waveguide do not exist.
For TM modes in rectangular waveguides, neither m or n can be zero because all the field equations vanish (i.e., Hx, Hy
,Ey. and Ez.=0). If m=0, n=1 or m=1, n=0 no fields are present. Hence TM01 and TM10 modes in a rectangular waveguide
do not exist.
What are degenerate modes in a rectangular waveguide?
Some of the higher order modes, having the same cut off frequency, are called degenerate modes. In a rectangular
waveguide, 𝑇𝐸 𝑚𝑛 𝑎𝑛𝑑 𝑇𝑀 𝑚𝑛 modes (both m ≠ 0 and n ≠ 0) are always degenerate.
What is Radiation Pattern and Power Pattern?
A radiation pattern (or field pattern)is a graph that describes the relative far field value, E or H, with direction at a fixed
distance from the antenna. A field pattern includes a magnitude pattern |E| or |H| and a phase pattern ∠E or ∠H. A power
pattern is a graph that describes the relative (average) radiated power density |Pav| of the far-field with direction at a fixed
distance from the antenna.
Directivity: D=

 


0
22
0
2
maxmax
sin),(
44
ddE
E
P
U
r


, where U=R2
Pav
2
2
ER


and Pr=   UddSPav 
 
ddER sin
22
0 0
2
 

is the time-average radiated power
Directivity gain: GD(θ, )=

 


0
22
0
2
sin),(
),(4),(4
ddE
E
P
U
r


, ∴ D=(GD)max
Power gain: GP =
iP
Umax4
, where Pi= Pr+Pl, Pi: total input power, Pl: loss
Radiation efficiency: ηr= GP/D=Pr/Pi
Find the directive gain and the directivity of a Hertzian dipole.
(Sol.)  HEHEPav
2
1
*Re
2
1
 , 

22
02
2
sin
32
)( Id
U  .



 
2
0
2
2
0
2
sin
2
3
sin)(sin
sin4
),( 
 dd
GD , ),
2
( 

DGD  =1.5=1.76 (dB).

11. ©Haris H.
Eg. Find the radiation resistance of a Hertzian dipole.
(Sol.) 




0
2*
2
0
sin
2
1
ddRHEPr
= rR
IdIdI
dd
dI
2
])(80[
212
)(
sin
32
)( 2
22
2
2
0
22
3
0
2
0
2
02
22
 




 
∴ 22
)(80


d
Rr 
Find the radiation efficiency of an isolated Hertzian dipole made of a metal wire of radius a, length d, & conductivity σ.
(Sol.) The ohmic power loss is  RIP 2
2
1
 . The radiated power is rr RIP 2
2
1

)/(1
1
rr
r
r
RRPP
P
 


 , )
2
(
a
d
RR s


  ,
where

 0f
Rs  
))((
160
1
1
3
da
Rs
r





Assume that a=1.8mm, md 2 , MHzf 5.1 , and  = )/(1080.5 7
mS
)(200 m
f
c
 , )(1020.3
1080.5
)104()1050.1( 4
7
76



 


sR ,
)(057.0)
108.12
2
(1020.3 3
4


 


R , )(079.0)
200
2
(80 22
 rR and %58
057.0079.0
079.0


r
Eg. A 1MHz uniform current flows in a vertical antenna of the length 15m. The antenna is a center-fed copper rod having
a radius of 2cm. Find (a) the radiation resistance, (b) the radiation efficiency, (c) the maximum electric field intensity at
a distance of 20km, the radiated power of the antenna is 1.6kW.
(Sol.) dmm 

 15300
10
103
6
8
 , a=0.02m, σcopper=5.8×107
,
c
c
s
f
R


 =2.6×10-4
(a)  97.1)300/15(80 22
rR , (b) %98)
)/)(/(160
1/(1 3

da
Rs
r


(c) 2
0
22
12
)(


dI
Pr  =1600 mV
R
Id
E /109.1)
4
( 20
max





Consider a five-element broadside binomial array. (a) Determine the relative excitation amplitudes in the array
elements. (b) Plot the array factor for d=λ/2. (c) Determine the half-power beamwidth and compare it with that of a
five-element uniform array having the same element spacings.
(Sol.) 1:4:6:4:1, broadside 0 
(a) 
 432
4641
16
1
)( jjjj
eeeeA  2cos2cos86
16
1
, where   cosd
(b)
2

d ,  d , and 0 2
)]coscos(1[
4
1
)(  A
(c)
2
1
)]coscos(1[
4
1 2
  ,  86.74 , ∴  28.30)86.7490(22 
Eg. Draw the far-field pattern of a phased array of dipoles with N=5, d=λ/2.
(Sol.) The effective scan range is about from 
600  to 
1200  as follows.
2

  
3
0

  0 
2
0

 
2

  
3
2
0

 

12. ©Haris H.
Communication is to be established between two stations 1.5km apart that operate at 300MHz. Each is equipped with
a half-wave dipole. (a) If 100W is transmitted from one station, how much power is received by a matched load at the
other station? (b) Repeat (a) assuming that both antennas are Hertzian dipoles.
(Sol.) (a)
22
2
21
16 r
GG
P
P DD
t
L


 . Half-wave dipole: GD=1.64, f=300×106
λ=1m
Pt=100W, WWPP tL 

76.0106.7
)1500(16
164.1 7
22
22



 
(b) GD=1.5 WWPL 633.01033.6 7
 