©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) mSf

/1099.01016.01 53

(b) At f=109Hz, mNpf /1098.1 4

me

zedB z 4

10

10 1075.1log

5.1log20)(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.) 710 , f=5×106Hz, σ/ωε0εr=200>>1, ∴ Seawater is a good conductor in this case.

(a) mNpf /89.8 ,

fjc )1(

smv p /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(ˆ1

),8.0( tEatH n

, )61.110cos(026.0ˆ]

)8.0(Re[ˆ),8.0( 7 tye

EytH 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.

©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

00

I

V

I

VZ

Let z=l, V(l)=VL, I(l)=IL

eZIVV

eZIVV

eZ

Ve

Z

VI

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

eZZeZZZ

IzI

eZZeZZI

zV

)'cosh'sinh()'(

)'sinh'cosh()'(

])()[(2

)'(

])()[(2

)(

0

0

0

'

0

'

0

0

'

0

'

0

'

zZzZZ

IzI

zZzZIzV

eZZeZZZ

IzI

eZZeZZI

zV

LL

LL

z

L

z

LL

z

L

z

LL

'tanh

'tanh)'(

0

00

zZZ

zZZZzZ

L

L

, Zi=

tanh

tanh)(

0

00

'0

L

L

zz

ZZ

ZZZZ

Lossless case (α=0, γ=jβ, Z0=R0, tanh(γl)=jtanβl): Zi=ljZR

ljRZR

L

L

tan

tan

0

00

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. )(102

103 108

101 Hza

f

,

ma 2

10

8

1012.2102

103

.

(b)

01010101

10133

fa

R

afQ

s

For copper, )/(1080.5 7 mS , .10700)1080.5)(104(10)103

12.2( 77102

101 Q

Q3. b)

©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.

©Haris H.

b)

i. Define Friis Formula

Define σbs= radar cross section of target, 2

2

21

22

21

2

1

2

212

2

)4(

4

4)

4(

r

GG

r

AAA

r

AG

r

A

P

P DDeeeeD

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

IQQj

dt

dQI ,

R

eIdaa

R

eIdzA

Rj

R

Rj

4

)sinˆcosˆ(4

ˆ 00 AaAaAa RR

ˆˆˆ

©Haris H.

0

sin)(4

sin

cos)(4

cos

0

0

A

R

eIdAA

R

eIdAA

Rj

z

Rj

zR

Rj

R

eRjRj

Ida

ARA

RRaAH

])(

11[sin

4ˆ

])([1

ˆ1

2

2

00

)](1

ˆ)sin(sin

1ˆ[

11

00

RHRR

aHR

aj

Hj

E R

)(120/,0

])(

1

)(

11[sin

4

])(

1

)(

1[cos2

4

000

32

2

0

32

2

0

whereE

eRjRjRj

IdE

eRjRj

IdE

Rj

Rj

R

Far field of a Hertzian dipole: if βR=2πR/λ>>1

sin)(4 R

eIdjH

Rj

,

sin)(4

0R

eIdjE

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

©Haris H.

Normalized array factor in the xy-plane (θ=π/2):

)1(2 ...11

)( Njjj eeeN

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=0d

0cos

Null locations:

kN

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

NNA

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,

221

21

2 N

ddd

4

)cos(cos)cos()cos( 2121

Let 0201 ,

)(sin)2

( 1

0Nd

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

1cos

2cos)(

dA

(a) )cos2

cos()(

A , (b) )1(cos4

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 10-4V/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=10-8sec.

(Sol.) 3

400

rrk , zan

ˆˆ ,

60

0

0 r

r

©Haris H.

(a) )102cos(10ˆˆ),( 84 kztxExtzE x

has the maximum in case of

0102 8 kzt6

)

63

4102cos(10ˆ),( 84

ztxtzE

,

)63

4102cos(

60

10ˆ),(ˆ

1),( 8

4

ztytzEatzH n

(b) 1)2cos( n , 2

3

8

132

63

4)10(102 maxmax

88 nznz

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

1103210110

It is a low–loss dielectric material: mrad /34.99])(8

11[ 2

)2

1(

jc = 29.0238

(a)

2 =0.497, mde d 395.1

2

1497.0

(b) smv p /108973.1 8

, sm

ddd

dvg /108975.1

)/(

1 8

(c) )0016.0

3(

497.03497.0 21.0ˆˆ1

50ˆ

jx

n

t

jx 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=1010Hz, mf

v02.0

10

102

1025.2

10310

8

10

8

For TE10 mode, Hza

vf c

10

2

8

10667.0)105.1(2

102

2

mradf

f

v

c /234)(1 2

,. mff c

g 0268.0)(1 2

smff

vv

c

p /1068.2)(1

8

2

, )(4.337

)(1 210

ffZ

c

TE

mS /105104 44 , mdBmNpZTEd /73.0/084.02 10

)(05101.0 c

cs

fR

,

mdBmNpffb

ffabR

c

csc /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.

©Haris H.

(Sol.) (a) 22 )()(2

1

b

n

a

mf c

. b<a<2b, the dominant mode: TE10, the next mode: TE01

af TEc

2

1)(

10 ,

bf TEc

2

1)(

01 , %20

)21(

)21(103 9

a

a,

%20)21(

103)21( 9

b

bma 06.0 , mb 04.0 , and a<2b

(b) Choose a=0.065m, b=0.035m, )(103.2)( 9

10Hzf TEc , 679.0)(1 2

f

fc ,

mradf

fc /15.40)(1 2 , smff

v

c

p /107.4)(1

11 8

2

,

mf

v p

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 )

21(

jc 0283.08.41 je

])(8

11[( 2

300 , smvp /1033.3 7

, m21019.1

1

, m31067.62

(b) mye y 21074.210ln1

1.0

01.0

(c) )3

10sin(1.0ˆ),( 10 ytextyH y , 300,5.0 y

)3

10sin(1075.5ˆ),5.0( 1020 txtH

)0283.03

10sin(1041.2ˆ),5.0(ˆ),5.0(ˆˆ 1018

tztHatEya ncn

©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

©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, 𝑇𝐸𝑚𝑛 𝑎𝑛𝑑 𝑇𝑀𝑚𝑛 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=R2Pav

22 ER

and Pr= UddSPav

ddER sin22

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.) HEHEPav2

1*Re

2

1 ,

22

02

2

sin32

)( IdU .

2

0

22

0

2

sin2

3

sin)(sin

sin4),(

ddGD , ),

2(

DGD =1.5=1.76 (dB).

©Haris H.

Eg. Find the radiation resistance of a Hertzian dipole.

(Sol.)

0

2*2

0sin

2

1ddRHEPr

=rR

IdIdIdd

dI

2])(80[

212

)(sin

32

)( 222

22

0

223

0

2

0

2

02

22

∴ 22 )(80

dRr

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

rr

RRPP

P

, )

2(

a

dRR s

,

where

0fRs

))((160

1

1

3 da

Rs

r

Assume that a=1.8mm, md 2 , MHzf 5.1 , and = )/(1080.5 7 mS

)(200 mf

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

1530010

1036

8

, a=0.02m, σcopper=5.8×107,

c

cs

fR

=2.6×10-4

(a) 97.1)300/15(80 22rR , (b) %98))/)(/(160

1/(13

da

Rsr

(c) 2

0

22

12

)(

dIPr =1600 mV

R

IdE /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 464116

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

30

0

20

2

3

20

©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=300×106λ=1m

Pt=100W, WWPP tL

76.0106.7)1500(16

164.1 7

22

22

(b) GD=1.5 WWPL 633.01033.6 7