Sixth Semester B.E. Degree Examination Antennas and ... · PDF fileDirectivity is also de ned...

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June/July 2013Sixth Semester B.E. Degree Examination

Antennas and PropagationTime: 3 hrs. Max. Marks: 100

Note: Answer any FIVE full questions, selecting at least two questions from each part.

PART – A1. a. De� ne the following terms with respect to antenna: i) Directivity ii) Beam solid angle iii) Radiation resistance (09 Marks)Ans: i) Directivity: The directivity of an antenna is de� ned as the ratio of maximum radiation intensity to the

average radiation intensity of the antenna. It is given by

max

av

P( , )D

P( , )

θ φ=

θ φ

Where P(�, �)max is the maximum power density expressed in watts/m2.

P(�, �)av is the average value of the power density over a sphere at the far � eld of an antenna.

It is a dimensionless quantity.

Directivity is also de� ned as the ratio of the maximum radiation intensity of an antenna under test to the radiation intensity of an isotropic antenna.

max

iso

P( , )D

P( , )

θ φ=

θ φ

Directivity also can be shown to be equal to the ratio of the area of a sphere to the beam area.

Mathematically A

4D

π=Ω

The Directivity can also be expressed in dBi as iA

410log dB

πΩ

Where dBi is stands for decibels over isotropic. For example, if an antenna radiates over half a sphere then

�A = 2� and 4

D 22

π= =π

. This is equal to 3.01 dBi. For a short dipole the value of directivity is 1.5 and that

of a �/2 antenna is 1.64.

Solid angle d� is the angle subtended by the surface area dA and is given by

d� = sin� d�d�

Surface area dA = r2 sin� d�d�

= r2 d�

Beam area or the beam solid angle is given by,

A nP ( , )dΩ = θ φ Ω∫∫

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June/July 2013 Antennas and Propagation

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ii) Beam solid angle:

r sin �

2�r sin � rd�

d�

r

�

r sin �d�

Polar angle� = 0

rd�

dA = r2 sin � d� d� = r2 d�

y

�

��

d�

rd�

Latitude

z

Longitude

x Azimuth angle

Half power beam width �

HP

Actual pattern of beam area

Equivalent solid angle �A

Fig: 1.a II(i) Fig: 1.a II(ii)

This is the integral of the normalized power pattern.2

A n

0 0

P ( , )sin d d (sr)π π

Ω = θ φ θ θ φ∫ ∫The beam solid angle �A is defined as the solid angle through which all the power radiated by the antenna

would flow if P(�, �) is zero else where.

Therefore the power radiated = P(�, �) �A watts. The beam solid angle can also be approximately expressed as

Beam solid angle �HP �HP (sr)

where �HP and �HP are the half power beam widths in the two planes, vertical and horizontal respectively.

Beam solid angle is also called as beam area

The equivalent solid angle �A can be thought of as a cap of the cone formed by extending the lines passing

through the half power points as shown in the fig 1a II(ii).

The unit of beam solid angle is steradian

1 steradian = 3283 square degrees

Solid angle in a sphere = 41253 square degrees

iii) Radiation resistance:

Consider a transmitter and an antenna connected by a transmission line as shown in the fig. 1a III. For the transmission line it appears as a resistance Rr. This resistance is not the one of the antenna itself but is the resistance coupled to the antenna terminals from the space. In the case of a transmitter the radiated power is absorbed by, trees, buildings, sky, ground and other antennas. In the case of a receiver the apparent temperature of Rr raises due to passive radiation from distant objects and active radiation from other

2-Antennas and Propagation June July 2013.indd 282-Antennas and Propagation June July 2013.indd 28 26/01/2014 07:49:18 PM26/01/2014 07

Antennas and Propagation June/July 2013

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antennas.

Tx/Rx

Rr

T

Antenna pattern

response in the space

Virtual transmission line

Virtual resistance

AntennaTransmission line

Transmission line terminals

Rr

Fig. 1a III Virtual transmission line

The radiation resistance of an antenna can be regarded as that virtual resistance which couples the antenna to the distant regions in the space through a virtual transmission line as shown in the � gure.

Radiation resistance can also be defined as that virtual resistance which consumes same amount of power

as the power actually radiated by the antenna. This can be given by

2rms

WRr

I=

where W = Power radiated

Irms = RMS value of current that flows in the antenna

The radiation resistance of a short dipole is given by 2

2

80Rr ohm

π=λ

2l

where l = length of the conductor used for antenna and � the wave length

The radiation resistance of a halfwave dipole is found to be equal to 73 �.

1. b. State and prove Frii's transmission formula. (05 Marks)Ans: Statement of Friis transmission formula. The ratio of received power to the transmitted power is given by

( )2

R T R2

T

P G .G

P 4 r

λ=

π

where PR = received power PT = transmitted power GT = gain of the transmitting antenna GR = gain of the receiving antenna r = distance between transmitting and receiving antenna � = wavelength

T R

rRec antennaTrans

antenna

Aet Aer

Fig. 1 (c)



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Consider a radio communication system as shown in fig 1(c).

Let Pt be the power fed by the transmitter to the transmitting antenna of effective aperture A

et.

At a distance ÂrÊ a receiving antenna of effective aperture Aer intercepts some of the power radiated by the

transmitting antenna. This antenna delivers it to the receiver „R‰. Assuming that the transmitting antenna is an isotropic antenna, the power per unit area at the receiving antenna is

tr 2

PS .....(1)

4 r=π

Let Gt be the gain of the transmitting antenna. The power per unit area is increased in proportion.

t tr 2

P GS .....(2)

4 r=π

Now the power collected by the receiving antenna of effective aperture Aer is given by

t tr er2

P GP A .....(3)

4 r=π

Gain of the transmitting antenna can be expressed in terms of its aperture Aet as

ett 2

4 AG

π=

λ

Substituting this in (3) We get

t etr er2 2

t et err 2 2

P 4 AP A

4 r

P A .AP .....(4)

r

π= ×λ π

=λ

This is one form of FRIIS Formula

Now we know that in general, the relationship between gain and aperture is given by

e2

4 AG

π=

λ

Using this we can write2 2

t ret er

G GA and A

4 4

λ λ= =

π π

Substituting these expressions in (4) we get2 2

t rr t 2 2

2t t r t t r

r 22

G G 1P P

4 4 r

P G G P G GP .....(5)

(4 r) 4 r

λ λ= ×

π π λλ

= =π π⎛ ⎞

⎜ ⎟⎝ ⎠λ

This is another form of FRIIS formula.

1. c. Show that maximum effective aperture of short dipole is 0.119 �2 (06 Marks)Ans: Consider a short dipole of length 'l' placed along y axis with center at '0'.

The field distribution w.r.t origin is given by

e = E cos wt



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2

E cos yπ=λ ..........(1)

y

z

E

-l/2 l/2

x

dy

y

e = E cos cot

Consider a small length dy of a dipole. As the length is too small the magnitude of the field can be assumed to be constant then p.d across dy can be written as

dv = e.dy ..........(2)

The total p.d aross the terminals is

2

2

V dv edy

2e cos y dy

Esin

−

= =

π⎛ ⎞= ⎜ ⎟⎝ ⎠λ

λ π⎛ ⎞= ⎜ ⎟⎝ ⎠π λ

∫ ∫

∫l

l

l

............(3)

For a short dipole when l << � v can he approximately written as

Ev .

E volts

λ π=π λ

=

l

l ............(4)

Power density is given by2E

S120

=π ............(5)

We know that for a short dipole 2 2

r 2

80R

2

π=λ

l ............(6)

( )22

em 2 2 2r

2

2 2

EVA

E 804SR4.

1203

0.119 square units8

= =π×

π λ

= λ = λπ

l

l

2. a. State and prove power theorem and its application. (05 Marks)Ans: If the poynting vector is known at all points on the sphere of radius r from a point source in a lossless

medium, the total power radiated by the source is the integral over the surface of the sphere of the radial component Sr of the average poynthing vecotor .



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P = s.ds = Sr ds ......(1)

Where P = Power radiated, w

Sr = radial component of average poynting vector w/m2

ds = infinitesimal element of area of sphere

= r2 sin�d�d� m2

For an isotropic source, Sr is independent of � and ��and therefore,

P = Sr ds = S

r x 4�r«2 .......(2)

and

2r 2

PS w / m

4 r=π

.......(3)

Equation (3) indicates that the magnitude of the poynting vector varies inversely as the square of the distance from the point source radiator.

2. b. Show that the directivity for unidirectional operation is 2(n+1) for an intensity variation of U = Um cosnn� (05 Marks)

Ans: U = Um cosn� .......(1)

2

2

T

2n

T m

0 0

nm

0

P Ud

P U cos sin d d

2 U cos sin d

π

π

θ φ

π

θ= φ=

= Ω

= θ θ θ φ

= π θ θ θ

∫ ∫

∫ ∫

∫

( )0 1

n nT m m

x 1 x 0

m

Let cos x

sin d dx sin d dx

When 0, x 1

When , x 02then,

P 2 U x dx 2 U x dx

2 U

n 1

= =

θ =− θ θ = ∴ θ θ = −

θ = =πθ = =

= π − = π

π=

+

∫ ∫

The power given by the isotropic antenna is given by 4«U0

( )

( )

mO

m

O

2 U4 U

n 14 n 1U

DU 2

2 n 1

π∴ = π

+π +

= =π

= +

2. c. Derive an expression and draw the � eld pattern for isotropic point sources of the same amplitude and same phase. (10 Marks)



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Ans:

/2

/2E

E0 e-j /2

E0 e+j/2

Fig. (b)

0 2 x

�d/2

1

d/2

r

To distant point

Fig. (a)Let the two sources 1 and 2 be separated by a distance ÂdÊ and located symmetrically with respect to the origin of the co-ordinates as shown in fig. (a). The angle � is measured anticlockwise direction from the positive x axis. The origin ÂoÊ is taken as reference for phase. Then at a distant point in the direction � the

field from source 1 is retarded by dr

2 cos �, while the field from source 2 is advanced by

dr

2 cos �, where

dr is the distance between the sources expressed in radians.

2 ddr .....(1)

π=λ

The total field at a distant point P placed at a large distance ÂrÊ is given by

E = EOe-j/2 + E

Oe+j/2 .....(2)

Where = dr cos �.

Amplitude of the field components at a distance r is EO.

The first term in (2) is the component of the field due to source 1 and the second term is the component due to source 2.

Equation (2) may be re-written as +600

-600 -1200

+1200

� = 0

Fig. (c) Filed pattern

j j2 2

O

O

O

e eE 2E

2

2E cos .....(3)2dr

2E cos cos .....(4)2

ψ − ψ+⎛ ⎞= ⎜ ⎟⎝ ⎠ψ=

⎛ ⎞= φ⎜ ⎟⎝ ⎠

To make a maximum value unity set 2 EO = 1

Then

drE cos cos

2

2 dcos cos cos d cos

2

When d 2

E cos cos ......(5)2

⎛ ⎞= φ⎜ ⎟⎝ ⎠

π π⎛ ⎞ ⎛ ⎞= φ = φ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠λ λλ=

π⎛ ⎞= φ⎜ ⎟⎝ ⎠

We know that = dr cos � + �



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� = 0 Since phase difference is zero.For direction of maxima put E = 1

cos cos 12

π⎛ ⎞∴ φ =⎜ ⎟⎝ ⎠

This is satisfied when � = � 90�For direction of minima set E = 0

cos cos 02

π⎛ ⎞∴ φ =⎜ ⎟⎝ ⎠

� = 0 or 180�

Direction of Half Power points

1E

2

1cos cos

2 2

=

π⎛ ⎞φ =⎜ ⎟⎝ ⎠

This is satisfied when � = � 60� or � 120�

The approximate field pattern based on the values calculated above can be obtained as in the figure (c) above.

3. a. Starting from electric and magnetic potentials, obtain the far � eld components for a short dipole. (12 Marks)Ans: Electric and magnetic � elds can be expressed in terms of vector and scalar potentials. Usually retarded

potentials are considered as the distance to the point at which the � elds have to be determined is much larger than a wave length.Fig (a) represents a dipole and its relation to coordinates.

Fig (b) gives a geometry of a short dipole. With reference to this figure, the retarded vector potential of the electric current has only one component, A

z.

yL

E«

E�

z

�

Dipole

E�

�

r

x

Fig (a)

y

P

rS

�

d

z

Ls

2

Fig (b)

s1

Its value can be given by

[ ]L2

L2

0z

IA dz

4 S−

μ=π ∫ ......(1)



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where [I] is the retarded current given by

[ ] jwO

sI I e t

c

⎡ ⎤⎛ ⎞= − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ ......(2)

In equations (1) and (2)

Z = Distance to a point on the conductor

IO = Peak value in time of current (uniform along dipole)

ø0 = Permeability of free space

= 4« × 10–7 H/m

If the distance from the dipole is large compared to its length (« >> L) and the wavelength is large compared to the length (� >> L), we can take s = r and neglect the phase differences of the field contributions from different parts of the wire. In this case the integrand in (1) can be taken as a constant.

Therefore

( )jw r0 o c

z

LI e tA

4 r

μ −=

π ......(3)

The retarded scalar potential is given by

0 V

1 [ ]V dt

4 s

ρ=π ∈ ∫ ......(4)

Where [�] is the retarded charge density given by

[�] = �0ejw (t–s/c) ......(5)

and dt = infinitesimal volume element.

�0 = Permittivity of free space

= 8.854 × 10–12 F/m

It is evident the region of charge in the case of a dipole is confined to the points at the two ends.

[ ] [ ]

[ ]

0 1 2

q q1V .....(6)

4 s s

Now

[q] I dt .....(7)

⎡ ⎤∴ = −⎢ ⎥π∈ ⎣ ⎦

= ∫Using equation (2), this can be written as

( )( )

[ ]

sc

sc

tjwtjw O

O

I e[q] I e dt

jw

1......(8)

jw

−−= =

=

∫

Substituting (8) into (6)

( ) ( )s1 s2C Ct tjw jw

O

0 1 2

I e eV ......(9)

4 jw s s

− −⎡ ⎤⎢ ⎥= −⎢ ⎥π ∈⎢ ⎥⎣ ⎦

Referring to fig. (c), when r >> L, the lines connecting the ends of the dipole and the point p may be considered as parallel so that



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L/2 cos �

� r

S1

S2

L

DipoleL/2 cos �

To point P

Fig (c)

1

2

LS r cos ......(10)

2L

S r cos ......(11)2

= − θ

= + θ

Substituting (10) and (11) into (9) we get

( )( ) ( )L L

2C 2C

ttjw cw cos w cosj jO

02

I e L Le r cos e r cos

4 jw 2 2V .......(12)

r

−θ θ⎧ ⎫⎡ ⎤ ⎡ ⎤× + θ − − θ⎨ ⎬⎢ ⎥ ⎢ ⎥π ∈ ⎣ ⎦ ⎣ ⎦⎩ ⎭=

The denominator is actually equal to 2 2 2 2

2 L cos L cosr .sin ceas r L, can beneglected.

4 4

θ θ− >> Hence r2

only is retained in the denominator.

By de MoivreÊs theorem equation (12) becomes

( )rcjw t

O2

0

I e wLcos wLcos LV cos jsin r cos

4 jwr 2c 2c 2

wLcos wLcos Lcos jsin r cos .......(13)

2c 2c 2

− ⎡ θ θ ⎤⎛ ⎞ ⎛ ⎞= + + θ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠π ∈ ⎣ ⎦θ θ⎛ ⎞ ⎛ ⎞− − − θ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

If the wavelength is much greater than the length of the dipole (� >>L), then

wLcos Lcoscos cos 1 .....(14)

2cand

wLcos wLcossin ......(15)

2c 2c

θ π θ= ≈λ

θ θ≈

Introducing (14) and (15) into (13) we obtain.

( )rctjw

O2

0

I Lcos e 1 c 1V ......(16)

4 C r jw r

Weknow that E jwA V

Where E the electric field A Vector potential

V Scalar potential

1H A ......(18)

−θ ⎛ ⎞= +⎜ ⎟π∈ ⎝ ⎠

= − − ∇− −−

− ∇ ×μ



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Where H – The magnetic field ø – PermeabilityIt is desirable to obtain E and H in polar co–ordinates.For vector potential the polar coordinates are given byA = a

r A

r + a� A� + a� A� .......(19)

Since vector potential for dipole has only Az components A� = 0 A

r and A� are given by

Az �

(a)

A�

Az

�

(b)

From fig (a)

Ar = A

z cos � ......(20)

From fig (b)

A� = – Az sin� .....(21)

Here Az is as given by equation (4)

( )jw rct

0 Oz

LI eA

4 r

−μ=

π

In polar coordinates gradient of V

’

V is given by

r

r r

v 1 v 1 vV a a . a .....(22)

r r r sin

Now

E a E a E a E .....(23)

θ φ

θ θ φ φ

∂ ∂ ∂∇ = + +∂ ∂θ θ ∂φ

= + +

From equation (17)

E = – jwA –

’

V

� arE

r + a�E� + a�E� = –jw a

rA

r–jwa�A�–jw��

r

v 1 v 1 va a a

r r r r sinθ φ∂ ∂ ∂− − −∂ ∂ θ ∂φ

.....(24)

Equating the respective components on either side we can write

r r

vE jwA ....(25)

r1 v

E jwA ....(26)r

1 vE jwA ....(27)

r sin

θ θ

φ φ

∂= −∂∂= −∂θ

∂= −θ ∂φ

In equation (27) A� = 0

Since V given by equation (16) is independent of v

0∂φ =∂φ

� E� = 0

substituting (20) into (25)

r z

vE jwA cos

r

∂= − θ −∂ ....(28)



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Substituting (21) into (26)

z

1 vE = jw A sin

r

∂θ θ −∂θ

.....(29)

To find out Er let us first find out

v

r

∂∂

when V is given by equation (16)

( )

( ) ( )( )

( )

rc

r rc c

rc

jw tO

2o

t jw tjwO O

2 3 2o o

jw tO

2 3 2o

I Lcos e 1 cV

4 c r jwr

I L cos e I L cos ev 1 2c 1 1 jwcr 4 c r jwr r jwr 4 c

I L cos e 1 2c jw 1

4 c r jwr cr r

−

− −

−

θ ⎛ ⎞= +⎜ ⎟π∈ ⎝ ⎠

θ ⎛ ⎞ ⎛ ⎞ θ∂ = − − + + × −⎜ ⎟ ⎜ ⎟∂ π∈ ⎝ ⎠ ⎝ ⎠ π∈

θ ⎡ ⎤= − − − −⎢ ⎥π∈ ⎣ ⎦

( )

( ) ( )

( ) ( )

( )

rc

r rc c

r rc c

rc

jw to

2 30

jw t jw t0 O O

z 2 3o

jw t jw t0 O O

2o

jw tO O

3o

I Lcos 2 2c jw

4 c r jwr cr

jw LI e cos I L cos ev 2 2c jwEr jwA cos

r 4 r 4 c r jwr cr

jw LI e cos I L cos e

4 r 2 cr

I L cos e I L cos e

2 jwr

−

− −

− −

−

− θ ⎡ ⎤= + +⎢ ⎥π ∈ ⎣ ⎦

μ θ θ ⎡ ⎤∂∴ = − θ − = − + + +⎢ ⎥∂ π π∈ ⎣ ⎦μ θ θ

= − +π π∈

θ θ+ +

π∈

( )rcjw t

2o

0 2

.jw

4 c r

1We know that =

c

−

π∈

μ∈

Substituting this value in the above equation the first term becomes ( )r

cjw tO

2o

I L jw cos e

4 c r

−θ= −

π∈

This is equal to the fourth term in magnitude but opposite in sign. They cancel each other.rc

rc

jw(t )O

r 2 3O

jw (t )O

2O

I Lcos e 1 1E ....(30)

2 cr jwr

vFor finding out E let us find out

Weknow that

I L cos e 1 cV

4 c r jwr

−

θ

−

θ ⎡ ⎤∴ = +⎢ ⎥π ∈ ⎣ ⎦

∂∂θ

θ ⎛ ⎞= +⎜ ⎟π ∈ ⎝ ⎠

( )

( ) ( )

rj t c0

20

z

r rj t j tc c0 0 0

2 20

I Lsin eV 1 c

4 c r jwr

1 vE j A

r1 v

j A sinr

j I Lsin e I Lsin e 1 c

4 r 4 r r j r

ω −

θ

ω − ω −

− θ ⎛ ⎞∂ = +⎜ ⎟∂θ π ∈ ⎝ ⎠∂= − ω θ −∂θ∂= ω θ −∂θ

⎛ ⎞ωμ θ θ ⎛ ⎞⎜ ⎟= + +⎜ ⎟π π ∈ ⎝ ω ⎠⎜ ⎟⎝ ⎠



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Substituting 0 20

1

cμ =

∈ we can write

( ) ( )

( )

r rj t j tc c0 0

20 0

rj t c0

2 2 30

j I Lsin e I Lsin e 1 cE

4 r 4 r c r j r

I Lsin e j 1 1

4 c r cr j r

ω − ω −

θ

ω −

ω ω θ ⎛ ⎞= + +⎜ ⎟π ∈ π ∈ ⎝ ω ⎠

θ ⎛ ⎞ω= + +⎜ ⎟π ∈ ⎝ ω ⎠ ........(31)

Magnetic field :- Magnetic field can be calculated by using

1H A= ∇ ×

μ

��

..........(1)

’ × A is gives by

z

ar ra r sin a

1A

r sin r

Ar A r sin A

θ φθ

∂ ∂ ∂∇ × =θ ∂ ∂θ ∂φ

θ θ φ

We know that as has only two components namely Ar and A� and they are given by

Ar = A

z cos �

A� = _ Az sin �

Since Az has no components is A� direction:

0∂ =∂φ

and r sin � A� = 0

2

ar ra r sin a

1A 0

r sin rAr A 0

θ φθ

∂ ∂∇ × =θ ∂ ∂θ

θ

∵

( )

2

r.sin .aA Ar

r sin r

1a rA .Ar

r rφ

θ φ ∂ ∂⎡ ⎤= θ −⎢ ⎥θ ∂ ∂θ⎣ ⎦∂ ∂⎡ ⎤= θ −⎢ ⎥∂ ∂θ⎣ ⎦

( )( )

( )

( )

z

rj t c0 0

rj t sinc0 0

rj t sin jc0 0

r.A r. A sinr r

r LI esin

r 4

LI e j

4 c

LI e

4 c

θ

ω −

ω − θ

ω − θ ω

∂ ∂= ⎡ − θ⎤⎣ ⎦∂ ∂⎡ ⎤− μ∂ ⎢ ⎥= θ

∂ π⎢ ⎥⎣ ⎦

μ − ω⎛ ⎞= − × ⎜ ⎟⎝ ⎠π

μ=

π ..........(3)



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( )

( )

rj t c0 0

r z

rj t c0 0

I LeA A cos co

4 r

I Le sin

4 r

ω −

ω −

μ∂ ∂ ∂= θ =∂θ ∂θ ∂θ π

−μ θ=

π

( ) ( )

( )

r rj t jw tc c0 0 0 0

rjw t c0 0

02

0

LI e sin j I Le sinaA

r 4 c 4 r

1I Le

1 jw 1H A sin

4 cr r

ω − −

−

⎡ ⎤μ θ ω μ θφ ⎢ ⎥∴∇ × = +π π⎢ ⎥

⎣ ⎦

μμ ⎡ ⎤= ∇ × = θ +⎢ ⎥μ π ⎣ ⎦

( )rjw t c0 2

This has only component

jw 1H I Lsin e

cr r

Hr 0 and H 0

−

φ

⎡ ⎤∴ φ = θ +⎢ ⎥⎣ ⎦= θ =

when r is very large, the terms 2

1

r and

3

1

r can be neglected.

Hence we effectively have only two components.

( ) ( )

( ) ( )

rjw t c rjw t0 0 c2

0 0

rjw t c rjw t0 0 c

jwI Lsin e jI LE sin e

4 c r 4 cr

jwI Lsin e jI LH sin e

4 cr 4 r

−−

−−

θ βθ = = θ

π ∈ π ∈

θ βφ = = = θ

π π Far field case.

3. b. Derive an expression for radiation resistance of a short electric dipole. (08 Marks)Ans: The average poynting vector is given by

( )e1S R E H *2= × ......(1)

The far field components are E� and H� that the radial component of the poynting vector is

r

1 *S ReE .H2 θ φ= .......(2)

The far field components are related by the intrinsic impedance of the medium, Hence

OE H Z H ......(3)

120 H

θ φ φ

φ

μ= =∈

= π

Therefore (2) can be written as

2*r O e O

2

1 1S Re Z H H H R Z

2 21

H 1202

φ φ φ

φ

= =

= π

Total power PT



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2 2 2 2 2 2 22 30 0

T 2 2 20 0 0

2 2 2 2 2 20 0

2 2

15 I L sin 30 I LP r sin d d sin d

r

30 I L 40 I L4.....(4)

3

π π ππ θ π= θ θ φ = θ θ

λ λ

π π⎛ ⎞= =⎜ ⎟⎝ ⎠λ λ

∫ ∫ ∫

If Rr is the radiation resistance

2

OT r

IP R .....(5)

2

⎛ ⎞= ⎜ ⎟⎝ ⎠

Equating (4) and (5) we get

O

2IO

2 240 IRr

2

π=

2

2

2 2

r 2

L

80 LR

λπ∴ =λ

4. a. Derive an expression for far � eld components of a loop antenna. (10 Marks)Ans:

2a d

(b)(a)

z

x

Dipole4

d

�

z

y

To distant point

(d)

Dipole 2

The field pattern of a small circular loop of radius ÂaÊ can be determined by considering a square loop of the same area, that is

«a2 = d2 .......(1)

Where d = side of the square loop, (fig b).

It is assumed here that the loop dimensions are small when compared to the wave length. It can be shown that the far field patterns of a circular and square loops of the same area are the same when the loops are small butt can differ when they are large in terms of wavelength.

Consider a square loop whose area is same as the circular loop which is oriented as in the fig. (c). In the case of the orientation shown far electric field has only E� component. To find the far–field pattern in the yz plane, it is enough if we consider only two of the small linear dipoles (2 and 4). The cross section through the loops in y–z plane is illustrated in fig. (d). The field pattern of the loop in this plane is the same as that



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of two isotropic point sources.j /2 j /2

0 0E E e E eψ − ψφ = − φ + φ ......(2)

Where E�O = electric field from individual dipole and

2 dsin dr sin

πψ = θ = θλ ......(3)

Therefore,

�/2�/2

E

E�0 ej�/2 (Due to dipole 4)

–E�0 e-j�/2 (Due to dipole 2)

( )j j2 2

O

O

O

O

e eE 2jE

2j

2j E sin 2

2 d2j E sin sin 2

dr2jE sin sin .....(4)

2

ψ − ψ+φ = φ

ψ= − φ

π ψ= − φλ

⎛ ⎞= − φ θ⎜ ⎟⎝ ⎠

The j factor in equation (4) indicates that the total field E� in phase quadrature with the field E�O of the

individual dipole. Now if d << �, equation (4) can be written as

E� = – j E�O dr sin � .........(5)

The far field of individual dipole can be given by

O

[I]j60 LE

r

πφ =λ

.........(6)

where [I] is the retarded current on the dipole and r is the distance from the dipole.

Substituting (6) in (5), we get

60 LE [I] dr sin

r

πφ = θλ

.........(7)

For a short dipole

L = d and substituting 2 d

drπ=λ in equation (7) we can write for a small loop

2

2

2

2

2

60 [I] 2 d sinE

r

120 [I]sin . A........(8)

rE [I]sin A

H .........(9)120 r

π π θφ =λ

π θ=λ

φ π θφ = =π λ

4. b. The radius of a circular loop antenna is 0.02�. How many turns of the antenna will give a radiation resistance of 35 �. (05 Marks)

Ans: r = 0.02� �A = �r2 = �(0.02)2�2

We know that 2

2

nARr 31200

⎛ ⎞= ⎜ ⎟⎝ ⎠λ



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Substituting the value of A and Rr we get22

32

6 2

35 31200 n 1.2566 10

31200 1.579 10 n

−

−

⎛ ⎞λ= × ×⎜ ⎟λ⎝ ⎠

= × ×

2 6

6

35n 10

31200 1.579

35n 10

49266

∴ = ××

= ×

30.02665 10

26.65

say n 27 turns

= ×==

4. c. Write a note on slot antenna. (05 Marks)Ans:

� �4 4

w

Slot antenna

Metal sheet

Fig. 5c – 1The figure shows the arrangements in a slot antenna. A �/2 slot is cut in a large sheet of metal. It is observed that the width of the slot is small (w << �) when compared to the wavelength.

However the currents are not confined to the edges of the slot they spread out over the sheet. The slot can radiate to both sides, of the sheet or one side of the sheet only. This depends on the feed configuration. It can be fed by a coaxial transmission line. Several ways of feed are available. One of them is as shown in the figure 5c–2.

�/2

Fig. 5c.2

cable bonded to sheet

If the slot is horizontal as in the figure 5C–1 the radiation normal to sheet is vertically polarized.

Slot antennas are used in applications in which law–profile or flush mountings are required. (high speed air craft).

Any slot has its complementary form in wires are strips. A complementary to a thin rectangular slot is its dual electric dipole. This would simply fill the slot. Their electrical and magnetic quantities are exchanged.

Hence the data on pattern and impedance of these complementary antennas can be used to predict the patterns and impedances of corresponding slots.



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PART – B

5. a. Explain the features of an helical antenna and the practical design considerations of the helical antenna. (10 Marks)

Ans:

L

HelixaxisD

d

Surface of imaginary Helix cylinder

S

A

An extension of a conductor in the form of helix acts like an antenna. Its radiation and directivity depends on the dimensions.

Helix has uniform resistive input impedance over a wide bandwidth. It acts as a super gain end fire array over the same bandwidth. It has negligible mutual impedances. Because of its simplicity, high gain circular polarization it can be used as a feed antenna for parabolic reflectors. Array of helical antenna is used for satellite communication. Helical antennas operate in two modes, namely axial mode and the normal mode.

The helix basically has a three dimensional form. The end view of the helix is a circle. It is considered to be wound on a cylinder. Helix is a combination of a geometry of a circle, a straight line and a cylinder. Some of the physical features of a helix are shown in the figure above.

D = diameter of the helixC = circumference = «DS = spacing between turns as shownL = Length of 1 turnn = Number of turnsA = axial length = nsd = diameter of the helix conductor

The geometry of the helix is shown in the figure.* When � = 0, the helix becomes a loop of

diameter D

��

L

S

C = «D

When � � = 90� helix becomes a linear straight conductor

As the pitch angle � � increases circumference gets reduced for a given length L

Helix modes: Transmission mode describes the way in which the electromagnetic waves propagate along the helix. It behaves as if it were an infinite transmission line or a waveguide. Usual modes are T

0, T

1, T

2

etc, To being the lowest in the order. Transmission mode is known as ÂTÊ mode.

Radiation (R) mode describes the far field pattern in general. Among many radiation modes, Normal and axial modes are more significant.

Fig. (a) above shows a normal (RO) mode or ominidirectional mode. Fig. (b) represents the axial (R

1) or a

beam mode.



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Direction of maximum reradiation

R1 mode

� g.(b)

Direction of maximum reradiation

Helix

R0 mode

� g.(a)

Practical design considerations of the helical antennaThe following are the important parameters for designing a helical antenna to be considered.

1. Beam width

2. Gain

3. Impedance

4. Axial ratio

The first two are interdependent and the other parameters are the functions of the number of turns, turn spacing and frequency. The parameters are also the functions of

i. Size

ii. Shape

iii. Helical conductor diameter

iv. Helical support structure

v. Feed arrangement

The nominal centre frequency corresponds to circumference of around 1�. All the four design parameters must be satisfactory over the centre frequency.Helical antennas are generally used in conjunction with parabolic corner or flat-sheet reflector.The conductor size is not critical and it usually ranges from 0.005� to 0.05� where � is the wavelength of radiation.The following figures show a few ground plane aspects of helical antennas.John D Kraus developed the following quasi empirical formula in 1949 after conducting several measurements.

( ) ( )

( ) ( )

52HPBW

c n s

115

c n s

=λ λ

λ λ

degrees

BWFN = degrees



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3/4�

Flat ground plane (circular or

require)

Flat ground plane Shallowcupped

ground plane

taper for mach

50 ohm coaxial wire general purpose flush

mounted two turn antenna

� = 140

S� = 0.25

0.75�

ground planesPlastic cover

0.5�

d 2.5nSλ= − λ

5. b. Write note on: i) Ultra wide band antenna, ii) Lens antenna (10 Marks)Ans: i) Ultra - wide band antennas: Ultra - wide band antennas are mainly used in digital system. It can be

E - Plane

Feed

Side view

H - planeFeed

Top view

Shown that a short pulse is made up of a sine wave of fundamental frequency and its infinite harmonics, by using Fourier analysis. It follows that if a short pulse is to be transmitted through a system, we require an in finite bandwidth.

A top view of a rhombic 'v' antenna and its side view is shown in the figure. This is very much suitable for pulse applications.

The rhombic and vee shapes help in achieving impedance matching over wide band of frequencies range.

The gradual floring out and the gradual tapering may be compared with the shape of the horn antennas which produces good impedance matching over a large bandwidth.

ii) Lens antenna: When the feed antenna is kept at focal point of the lens antenna, the diverging rays (spherical wave fronts) are collimated (parallel rays) forming a plane wave front, after they incident on the lens and passing through it. Collimation occurs because of refraction. For the incoming mode the arrows are reversed. Here incoming parallel rays converge at the focal point after passing through the lens due to refraction.



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Feed antenna

Focal Point

LensPlane wave

front

Lens axis

Focal length

Lens antennas are used in association with a point source, ideally. But in practice, It is used with horn like antennas. There are two types of lens antennas, namely, dielectric and metal plate lens.

Some salient features of dielectric lens are

i) Made of polythylene

ii) They are bulky

iii) Uniform illumination for lens antenna is better if focal length is long.

iv) At f < 10GHz lens becomes thick. Main disadvantages of lens antenna is its frequency sensitivity.

6. a. Explain: i) Yagi Uda antenna, ii) Parabolic re� ectors (10 Marks)Ans: i) Yagi Uda antenna:

Director Main lobe

Supporting rod

Driven elementMinor lobes

Reflector

Fig 6a I(i) Yagi-Uda antenna

A Yagi-Uda antenna array consists of i) Driven elementii) Directoriii) reflectorThe driven element usually is a folded dipole and it is an active element. The signal is fed to the driven element. The reflector and the director are induced are the parasitic elements in which the currents are induced due to the current in the driven element. The length of the driven element is �/2. The length of the reflector is more than that of the driven element. This acts as an inductive element. The director has a shorter length than the driven element and it is capacitive in nature. The purpose of reflector and the director is to produce a maximum radiation in the direction required. In Practice there can be many directors, but only one reflector. Maximum number of elements is usually less than 16. A three element Yagi-Uda antenna is illustrated in the figure 6a I(i).



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0.31 � 0.31 � 0.31 � 0.31 �

0.01 �0.01 �300 �

line

Driver elementDirectors

Metal support0.475 �

0.25 �

0.46 � 0.44 � 0.44 � 0.43 � 0.40 �

1.5 �

Reflector

Fig 6.a I(ii) Six element Yagi-Uda arrayA typical six element Yagi-Uda antenna has been shown in the fig. 6a I (ii). Typical spacings between the elements and length of each element is also shown in the figure. The maximum directivity of this array is about 12 dBi. The Yagi-Uda antenna are mostly for TV applications.ii) Parabolic reflectors: These use the properties of a parabola.

D

F

O�

B

d

x Plane wave

F

C

O

A

y

-y

-x

Tangent

E

Fig. 6aII Geometry of a parabolaOF = Focal length = f

F = Focus

O = Vertex

OO� = Parabola axis

A parabola is the locus of a point moving in such a way that its distance fixed point F (focus) plus its distance from a straight line is constant. This straight line is called a directrix. This is illustrated in the fig. 6aII.

From the figure it can be written as

FA + AB = FC + CD = K

where K is a constant

The equation of a parabola is given by

Y2 = 4fx

The open mouth d is known as the aperture. The ratio f/d is important parameter in designing a parabolic reflector and typically in between 0.25 and 0.50.



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When a source is kept at a focus point the rays stricking the parabola will be parallel to each other and to the parabola axis. All the rays being in phase they produce a strong concentrated beam of radiation along the parabolic axis.Now if a beam of parallel rays strikes the parabolic surface, they will be meeting at focus. In this case the parabolic reflector acts as a receiving antenna reflector. The rays which are perpendicular to the directrix only will be meeting at the focus. Others will not meeting due to path differences.The parabolic reflectors are used at microwave range of frequencies.If the primary antenna is a non-directional then,

BWFN 140d

HPBW 70d

λ=

λ=

The different types of reflectors are cut or truncated paraboloid, parabolic cylinder, pill box and cheese antenna, offset paraboloidAntenna feed systems arei) Half wave dipoleii) Array of collinear dipolesiii) Yagi-Uda antenna arrayiv) Hornv) Cassegrain feedMost popularly used feed is the horn antenna. For low noise receiver applications cassegrain feed is used.

6. b. Write short notes on: i) Turnstile antenna ii) Antennas for ground penetrating radar (10 Marks)Ans: i) Turnstile antenna: The dipole antennas placed perpendicular to each other as shown in � g. 6b(i).

Feed line

Dipole 1

Dipole 2

Fig. 6b(i)

The dipoles intersect at their mid points. They carry equal currents but differ 900 in phase. The antenna is exited by different non-resonant lines of unequal length. The radiation is omni directional.Electrical field is given by

( )cos sin cos cos2 2E( ) sin t cos tcos sin

π⎛ ⎞θ λ⎜ ⎟ θ⎝ ⎠θ = ω + ω

θ θ



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Directivity can be improved by arranging an array of turnstile antennas.Produces horizontal polarisation and best suited for matching 70� dual coaxial line. It is used as a must mounted television transmission of antenna for frequencies above 50 MHz. The improvement in bandwidth is obtained by arranging an array of super-turnstile antennas with a wavelength of spacing between them. This array provides more horizontal gain.ii) Radars can be used to detect underground anomalies both natural and man-made. The anomalies include buried

metallic or non-metallic objects, earth abnormalities etc. Pulse and the pulse echo are used in the system. The logic difference between GPR with that of above earth radar system arei) Total distance traveled by the wave is not large.

Feeding Bow-tie dipole terminals

Resistance loadii) Ground is a lossy medium.iii) Large mismatch at air ground interface.iv) Pulse width should be very small as the distance or total time is very small.

Usually two dipole antennas are used, placed very close to earth surface of which, one is for transmission and the other is for reception. To reduce bouncing and ringing, the elements of the dipoles are constructed with resistance loaded in the form of bow-tie as shown in the figure.

7. a. Discuss the propagation characteristics of radio waves for different frequencies. (10 Marks)Ans: The radio waves can be divided into several ranges as, ELF, VLF, LF, MF, HF, VHF, UHF, SHF and EHF.

Their characteristics and applications are brie� y accounted below:Extremly low frequency (ELF): This frequency band ranges from 3 to 30 Hz. The range of transmission in this band is about 8000 kms. The high power waves produced can penetrate deep into sea. Huge antennas are required to handle high powers in 100 MW range. The range of distances covered is much greater than other frequency ranges. Very less affected by atmospheric disturbances. The data rate is very low compared to other communication systems. Mainly used in underwater communications.Very low frequency (VLF), 3 to 30 kHz : Range of transmission is 8000 kms. Can penetrate vegitation and water. These waves can be used for navigation. They are used for low speed secure teletypewriters. The VLF transmitters are usually at shores. The data rates are higher than that of ELF. Interrupted continuous wave is the common mode of operation.Low frequency (LF), 30-300 kHz: The transmission range of these waves is 1600 to 8000 kms. commonly uses FSK modulation technique. Even though it can pass through vegitation and water, less effective than ELF and VLF.0Medium frequency (MF), 300-3000 kHz: These waves can be propagated through ground wave, shy wave and space wave communication system. The transmission range is from 160 kms to 1600 kms. When used in ground wave propagation. The range is about 1600 kms to about 5000 kms in the sky wave propagation. This depends on the output power and atmosphere conditions. The principal applications are in Medium distance communication, and radio navigation. The medium frequency waves may support low capacity multinational circuits for both voice and data. This range mostly adopts amplitude modulation.High frequency (HF), 3 to 30 MHz: This mode of wave is mostly employed for long distance communication. These can transmit by ground wave or sky wave propagation. Transmission range when ground wave propagation is used is about 50 to 500 kms. Transmitter power ranges from 2 kW to about 100 kW, depending on the particular application. The HF when used with sky wave propagation it is highly verlberable to interference.Very high frequency: VHF, 30 to 300 MHz: This band is mainly used for line of sight (LOS) communication. Range of communication distance can be increased by using repeaters. It is to be noted that at higher antenna heights greater distances can be covered. Higher the frequency range, lower is the power requirement. 40 to



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50 km range of transmission is possible through ground wave propagation. For hand held portable systems power required is about 0.25 W. For a 12/24 multichannel LOS systems power is about 120 Watts.Ultrahigh frequency (UHF), 300 to 3000 MHz: The principal methods of propagation are tropospheric scattering, LOS systems used mainly in satellite communication systems. Transmitter power range from 10 to 100 watts when LOS systems are used and troposcatter systems use 2500 to 10,000 watts. UHF systems provide high quality, reliable and high capacity transmissions using data rates of 2.4 kbps and above. UHF and VHF are mainly used in TV.Super high frequency (SHF), 3 to 30 GHz: These ranges handle high data rate microwaves troposcatter and satellite systems. Microwave systems highly reliable long distance communication. SHF carrier signals can provide large band widths. Thus assuring data rates of 2.4 kbps or higher.Extremely high frequency (EHF), 30 to 300 GHz: Very good potential for research in this range is available. EHF band can provide secure voice and high speed data rates upto 100 mbps. The systems can choose single of multichannel for working. 600 channels per link can be derived. The EHF can provide high capacity, low power and narrow beam widths. It can provide an excellent mobility.

7. b. Explain the principle of surface wave propagation. Obtain in equation for tilt angle � of the wave. (10 Marks)Ans: i) Surface waves: Consider a vertically polarized dipole placed above the ground. Let us assume that both

the dipole and the � eld points are on the surface of the earth and separated by a distance ‘d’. If the ground has � nite conductivity, typical values being 10–3 S/m to 30 × 10–3 S/m, electric � elds due to the dipole and its image cancell each other. The total electric � eld due to the direct and the ground re� ected waves is zero on the surface of the ground. In such a case waves can propagate as surface waves. They constitute the primary mode of propagation for frequencies ranging from a few KHz to several MHz. For instance in the AM broadcast application, the vertical monopole above ground is used to radiate power in the MW frequency band. The receivers are usually placed very close to the surface of earth and therefore they receive the broadcast signals via surface waves. Surface wave propagation can be used to cover a distance of several hundred kms for the communication.

Earth

The receivers are placed very close to the earth surface

T R

Fig. 7a.1

103

100

10–1

10–2

10–3

10–4

10–2 10–1 100 101 102

Numerical distance

Gro

und

wav

e at

tenu

atio

n fa

ctor

Fig. 7a.2



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The attenuation factor of the wave is a function of the numerical distance between the transmitter and the receiver. It also depends on the frequency and the electrical properties of the ground (relative permittivity or and conductivity �) over which the wave is propagating. At the surface of the earth the attenuation factor is known as ground wave attenuation factor and it is denoted by Asu. A behavioral curve of attenuation factor as a functions of numerical distance is shown in figure above. This curve has a power factor angle b = 0°. The numerical distance p and the power factor angle are expressed as

1 r

RP cos B

x1

b tanx

−

π=λ

∈ +⎛ ⎞= ⎜ ⎟⎝ ⎠

Where R = the distance between the transmit and receive antennas and x is given by

O

xw

σ=∈

For x >> �r, the power factor angle is nearly zero and the ground is mostly resistive.

For large numerical distances, the attenuation factor decreases by a factor of 10 for every decade. (20 dB/decade).

Therefore the attenuation factor is inversely proportional to P. As the numerical distance is proportional to R, the attenuation factor also is proportional to R.

The electric field intensity due to the surface wave is proportional to the product of Asu

and ( )jkRRe

−

.

Thus the electric field intensity due to the surface wave at large distances from a vertically polarized antenna is inversely proportional to the square of the distance or the power is inversely proportional to R4.

Tilt: The electric field of a vertically polarized wave near the surface of the earth will have a forward tilt. Its magnitude depends on the conductivity and the permittivity of the earth. Usually the horizontal component is much smaller than the vertical component and they are not in phase. Hence the electric field is elliptically polarized very close to the surface of the earth.

Electric vector due to wave tilt.

ii) Diffractions: Electromagnetic waves bend around sharp edges or corners of obstacles that appear in their path. This is known as “diffraction”. Diffraction allows the signals to propagate in regions lying behind the obstructions. If the receiver is placed behind such obstructions in such a way that the line of sight path is completely obstructed, the diffracted � eld still have suf� cient strength at the receiver location. This enables to establish a communication link. The strength of the received signal in this obstructed or shadow region depends on the position of the receiver and the geometrical shape of the obstruction.

Eh = Js Zs

vertical component of the � eld Ev = H�0

where Zs = surface impedance of earth Eh

Ev

r 12 2 22 2 2

1

1

11 tan2 37710 Etan DF

1tilt angle tan2 w

−

−

−

ω σ ω∈=σ + ω ∈ω∈σ +

σ=∈



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Diffracted wave

obstructedLOS

RT

Hill knife edge

Fig. 7a.3 Knife edge diffraction model

Bui

ldin

g

Obstructed LOS

Diffracted wave

T

Fig. 7a.4 Rounded – surface Diffraction model

Fig. (a) above shows a knife edge diffraction model. Here shadowing is caused by a single hill or a mountain. This model is used to compute the diffracted field.

Fig. (b) shows a rounded surface model. Diffraction here is due to a building.

Consider a transmitter and a receiver placed as shown in figure 7a.5.

RT

�

�

h

D

d1

d2

obstacle

knife edge

Fig. 7a.5There is an obstacle between the two and it is at a distances of d

1 from the transmitter and d

2 from the

receiver. This obstacle blocks the line of sight.

For this analysis it is assumed that the knife edge obstacle extends to infinity into and out of the plane of the paper. The tip of the obstacle is at a height h from LOS path.

Let us assume that d1 >> h and d

2 >> h.

It can then be shown that the field strength at the receiver is given by

22j t 2

O

Vd

1 jE E e dt

2

π∞

−+⎛ ⎞= ⎜ ⎟⎝ ⎠ ∫ ......(1)

Where EO is the free–space field strength in the absence of the knife edge and the ground. The parameter V

d

is known as Fresnel–kirchhoff diffraction parameter and is given by

( )1 2d

1 2

2 d dV h

d d

+=

λ

Vd is directly proportional to height; h. The integral is equation (1) is known as the Fresnel integral. It is

usually evaluated by using tables, graphs or numerical methods for a given value of Vd.

The knife edge diffraction gain is defined by



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d 10O

EF 20log dB

E

⎛ ⎞= ⎜ ⎟⎝ ⎠

Fresnel–kirchnoff diffraction parameters.Diffraction gain of the knife edge obstacle.

The diffraction, gain varies as a function of Vd as shown in the graph, figure 7a.6. Three cases are considered

here.

i) Vd = 0. This corresponds to h = 0. The tip of the knife edge touches the direct line of sight path between

the transmitter and the receiver and the diffracted signal is 6dB below that of the free space signal.

ii) Vd > 0, In this case h > 0. The knife edge blocks the LOS path. Diffraction gain decreases

monotonically.

iii) For Vd < 0, the signal at the receiver can be larger than the free space signal for certain values of V

d.

For example, the knife–edge diffraction gain is 1.37dB for Vd = –1.22. When V

d further decreases the

diffraction gain oscillates about 0 dB. The knife edge diffraction gain is 0 dB.

2

–4–5 –3 –2 –1.2–1 0 1 2–20

–18

–16

–14

–12

–10

–8

–6

–4

–2

01.37

Km

ife

edge

dif

frac

tion

gain

Fd(

dB)

Fig. 7a.6 Frernel – kirchhoff diffraction parameter Vd

8. a. Draw and explain different ionized layers in an ionospheric propagation. (10 Marks)Ans: Ionosphere is an ionized region above earth. It consists of several layers as shown in � gure 8b.

The radio waves radiated from the transmitting antennas bend and get reflected to the ground due to different refractive index in the ionized region. The layers get affected during day, night and there exists seasonal variations also.D-layer is a phenomena of day time. Largely absent during night. Ionization in the D layer is low because of less ultra-violet rays penetrating this layer. Suitable for VLF.E layer: Exists between 90 to 140 km from the earth surface. It is closely controlled by the amount of ultra-violet light from the sun and uniformly decays with time at night. Medium distance communication in LF and HF bands.F1 layer: Exists between 150 to 250 kms above earth surface in summer and 150 to 300 km in winter. This layer is almost constant. Littre diurnal and seasonal variations exist. F2 layer: Between 250 to 400 km. At night F1 & F2 layers combine and form a single F layer. More variable in nature. Responsible for long distance HF communication.There is yet another type of ionospheric layer which is called sporadic ‘E’ layer. Most unpredictable layer. The existence of this layer is due to the anomalous phenomena. It appears in and around E layer at some



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discrete locations and then disappears. It has some negative values. It may prevent the use of more favorable layers. It may also result in additional absorption and multi path problems. There may be an additional delay in return of path at some frequencies.

F2 layer exists during

Day and night

Combined E layer 150 - 250 Km in

summer and 150 - 300km in winter

250 - 400 Km F2

250 - 300 Km F1

90 - 140 Km E

50 - 90 Km D

Earth Earth(i) Day (ii) night

Fig. 8.b Ionospheric Layers

8. b. A distance of 1500 km one is to be covered along earth surface using a communication link of the re� ection region of ionosphere has fc 6 MHz and fMUF 7.5 MHz, calculate the height of the region.

(05 Marks)Ans: d = 1500 km, fc = 6 MHz, MUF = 7.5 MHz

h = ?We know that

2

MUF c

df f 1

2h⎛ ⎞= + ⎜ ⎟⎝ ⎠

Squaring both sides,

MUF c

22 2

2

2 2MUF

2c

df f 1

4h

f d1

f 4h

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

2 2MUF

2c

22

2

MUF

c

f d1

f 4h

d4h

f1

f

⎛ ⎞− =⎜ ⎟⎝ ⎠

=⎛ ⎞

−⎜ ⎟⎝ ⎠

2

MUF

c

d2h

f1

f

=⎛ ⎞

−⎜ ⎟⎝ ⎠



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2

MUF

c

dh

f2 1

f

=⎛ ⎞

−⎜ ⎟⎝ ⎠

332 2

1500 750 75010 kms

750 107.5 7.52 1 1

6 6

−= = = =×⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

8. c. Write a note on skip distance. (05 Marks)Ans: Skip distance:

The distance from transmitter at which the ray returns to ground reduces as the angle of incidence �1

reduces. The fig. (8c) shows the paths of the signal at different angles of incidence.

The minimum distance that is measured from the transmitter at which it is possible to get the signal at the surface of the earth is called the skip distance. With in this distance it is not possible to get the wave at an intermediate point. The distance 'd' is the "skip distance"as shown in the figure.Expression for skip distance: With reference to fig. 8c(i) above distance between T and R is the distant d, which is the skip distance. For a communication upto a certain distance the earth surface is considered to be flat. For calculating skip distance this assumption is made.From equation (3) above

2

muf

c

fd 2h 1

f

⎛ ⎞= ′ −⎜ ⎟⎝ ⎠

..........(4)

dT R

1 2

3

4

�1�r

A d/z d/2Flat earth

B

0C

h

Fig. 8c Fig. 8c(i)

1, 2, 3 are the rays at different angles


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