Final Exam Antenna 20061

8/9/2019 Final Exam Antenna 20061

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D-ITET Antennas and Propagation

Student-No.: ...................................................................

Name: .....................................................................

Address: .....................................................................

.....................................................................

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Antennas and Propagation

Spring 2006

March 23, 2006, 09:00 am 12:00 noon

Dr. Ch. Fumeaux, Prof. Dr. R. Vahldieck

This exam consists of 6 problems. The total number of pages is 15, including

the cover page. You have 3 hours to solve the problems. The maximum

possible number of points is 53.

Please note:

This is an open bookexam. Attach this page as the front page of your solution booklet. All the calculations should be shown in the solution booklet to justify the solutions. Please, do not use pens with red ink. Do not forget to write your name on each solution sheet. Please, put your student card (LEGI) on the table. Possible further references of general interest will be written on the blackboard during

the examination.

Problem Points Initials

1

2

3

4

5

Total


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D-ITET Antennas and Propagation March 23, 2006

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Problem 1 (11 Points)

A pyramidal horn antenna ( 1 19.43a = cm 1 14.35b = cm, aperture efficiency 0.5ape = ,

linearly polarized in ye

-direction) should be measured in an anechoic chamber with the

length 8L = m, the width 4W = m and the height 4H = m.

a1

b1

transmittingantenna

receivingantenna

anechoic chamber

d

x

y

z

H

L

2 Points a) Is it possible to measure the radiation pattern of the pyramidal horn at 12f = GHz in

the anechoic chamber? Calculate!

A second antenna is mounted in a distance 5d = m from the pyramidal horn. In a

measurement this second antenna is used as the transmitting antenna and the horn as the

receiving antenna. The transmitting antenna has a gain of 8dBitG = and can be described

with the radiation resistance r 220R = , the loss resistance 5LR = and it is connected

to a generator with the voltage 30gV = V and the inner resistance 75gR = (assume the

reactances of antenna and generator 0A gX X= = ). The relative electric field vector tE

of

the transmitting antenna can be expressed as a superposition of two plane waves with

1, 9 cos( ) cos( )2t x yE t e t e

= + +

and 2, 5 cos( ) 2 cos( )2t x y

E t e t e

= +

.

3 Points b) Determine the polarization and the axial ratio AR of the transmitted wave tE

!

2 Points c) Calculate the power tP radiated from the transmitting antenna.

4 Points d) Determine the power rP received by the pyramidal horn antenna.


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Solution 1

a) The far-field condition of the pyramidal horn is given by22D

R

> , where D is the

largest dimension of the antenna. Here:2 21 1 0.2415D a b= + = m, and thus 4.67R > m.

Because the anechoic chamber has a length of 8L = m, it is possible to perform ameasurement of the far-field radiation pattern.

b) The total electric field is 1, 2,t t tE E E= +

. Together with

1, 9 cos( ) cos( ) 9 sin( ) cos( )2t x y x y E t e t e t e t e

= + + = +

and

2, 5 cos( ) 2 cos( ) 5 sin( ) 2 cos( )2t x y x y E t e t e t e t e

= + = +

the total field can be written as4 sin( ) 3 cos( )

4 sin( ) 3 sin( )2

t x y

x y

E t e t e

t e t e

= +

= + + +

.

Hence, the polarization vector of 1,tE

is ,1

9t

j

=

, and the polarization vector of 2,tE

is

,2

5

2t j

=

, and thus ,1 ,2

4

3t t t j

= + =

.

At 0t = the electric field is 3t yE e=

, and at / 2t = the field is 4t xE e=

.Hence, the resulting wave propagating in z+ -direction is elliptically right hand (orCW)

polarized.

With the formula given on slide 2.22 the axial ratio is4

AR3

OA

OB= = .

c) Using an equivalent circuit the power transmitted from the second antenna can be

expressed as

20

1

2t rP I R=

with 030

A 0.1A300q

r L q

U

I R R R= = =+ + . Therewith the transmitted power is

21 (0.1) 220 W 1.1W2t

P = = .

d) The received power can be determined by

( )2

PLF4r t r t

P P G G d

= ,

where the gain of the transmitting antenna is given 8dBitG = , 25 = mm at

12f = GHz, 5d = m is the distance between the two antennas, and the gain of the

receiving antenna can be calculated by

1 12

1 4

2rG a b

= .


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Here the factor 1/ 2ape = is the aperture efficiency, and the polarization loss factor is

( )22 PLF 4 3 / 25 0.36r t y x y e e je = = = .

The received power is therewith

( )

2

PLF 0.1114r t r t P P G G d

= = mW


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An airplane passing at an altitudeP

H is tracked by two independent (monostatic) radar

stations working at two different frequencies. Radar1 works at1

1f = GHz andRadar2 at

2 2f = GHz, both have an output power 0 200P = W and a gain 1G = .The radar cross section of the plane is 100 = m2. The geographical dimensions are given in

the following:

100P T

H H= = m, 0.5R

H = m, 90M

H = m, 1000T

D = m, 4000P

D = m and 8000R

D = m.

HR

HT HP

Radar 1

Mountain

DPDT DR

Radar 2

HM

Points 4 a) Neglect the mountain and the earth surface influence and compute the free space loss

(one way) and the received power at the two radar stations.

Points 3 b) Calculate the power received by both radars if all geographical entities are taken into

account.

Points 2 c) Estimate a virtual mountain height so that both radar stations receive the same power.


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Solution 2

a.)

1 2

1 2

1 , 2

0.3 , 0.15

f GHz f GHz

c m mf

= =

= = =

Free space loss for one way can be calculated as follows:2

1, 1 106.4

( )LOS radar

T P

L dBD D

= =

4 +

2

2, 2

116.5( )

LOS radar

R

L dBD

= =

4

The Power received at the two radar station is:4

2 11 0 2

1

4

2 22 0 2

2

1

2

( )

53.01 2 106.42 41.45 118.4

53.01 2 116.5 47.47 132.5

T P

R

P P GD D

P P GD

P dBm dB dB dBm

P dBm dB dB dBm

4= 4 +

4=

4

= + =

= + =

b.)

The Fresnel zones are defined like:

1T P

n

T P

D Dr nD D

+

When 10.6h r< the effect of the diffraction is very small.

1 1 15.5

0.6 15.5 9.3

T P

T P

D Dr m

D D

m m h

+

= >

The first Fresnel zone is not covered up to 40% so the diffraction can be neglected.

We take into account the same propagation for outgoing and incoming wave. The free space

losses can NOT be neglected.44

' 22 22

0 2

4 4

,

21 exp4

2 221 exp

P R

R R

P R P RLOS ground

R R

H HP G jkP D D

H H H H L jk

D D

2

4 = +

= +

,

'2

11.2

143.7

LOS groundL dB

P dB

=

=

c.)'

2 1 25.3P P P dB = = back and forward.Read from the graph 0 0.8 1v


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( )dB =-12.65dB0.95

0.95

( ) 20 log 0.5

0.466, 0.8

v

d

v

L e

e v

=

= =

( )

( )11

2

2

8.7m

108.7

T P

T P T P

T P

M P

D Dh h

D D D DD D

h

H H h m

+= =

+

=

= + =


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Problem 3 (Points12)

Four isotropic sources are placed along the z-axis as shown in the following figure. The

amplitudes of the four isotropic elements are [ ]; 1; ;1j j corresponding to the element

number[ ]1;2;3;4 . The distance between the elements is2

a = .

z

a/2

a/2

4

2

3

1

y

a

a

Points 2 a) Find the array factor of the linear array.

Points 3 b) Find the nulls and maxima of the array factor. Are there any grating lobes, how can one

influence the grating lobes?

Points 2 c) Determine a new distance 'a between the elements so that the array becomes an endfire

array. In what direction is the main beam pointing now?

The linear array from above is used for a planar array in zy plane as shown in the figure

below. The distance between the single linear arrays is 2b = .

4

2

3

1

4

2

3

1

z

a/2

a/2

4

2

3

1

y

a

a

4

2

3

1

4

2

3

1

4

2

3

1

4

2

3

1

4

2

3

1

4

2

3

1

4

2

3

1

b b b b

Points 3 d) Determine the array factor of the planar array.

Points 2 e) Give the position of the main beam and the number of grating lobes of this array.


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Solution 5

Notation of the lecture notes:

z

y

d a

d b

=

=

a)

sin1 2

1sin

2

n

N

AFN

=

coszkd = +

With2

= + the linear phase progression.

( )

( )sin (2 cos 1)

4 sin (2 cos 1)2

nAF

+

= +

b)

The nulls:

1

1

2cos

2

1,2,3,...

, 2 , 3 ,...

1cos ; [0 , 60 , 90 ,180 ]2

n

n n

n

d N

n

n N N N

n

=

=

= =

The maxima:

( )1

1

cos 22

,1,2,3,...

1cos 2 ; 1 0

2

0

2

m

z

m m

md

m

m

=

=

= =

There is no grating lobe in the visible region. The grating lobes depends on the phase

increment and the spacing.

c)

For endfire arrays the condition is kd = . So zd

=4

for a radiation towards 180m = .

d)

( )( ) ( )( )1 1

1 1

1 sin sin1 cosI y yz z

N M

n m

n m

j n kd j m kd AF I e e

= =

+ + =

( )( ) ( )( )cos1 1

1 1

1 sin sin1 y yz zM N

m n

m n

z y

j n kd j m kd I e I e

AF AF

= =

+ +=


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( )

( )

( )

( )

( )

( )

( )

( )

sin / 2sin / 21 1(

sin 1/2 sin 1/2

sin / 2sin (2 cos 1) 1

sin 1/24 sin (2 cos 1)2

yz

n

z y

y

y

NMAF

M N

N

N

, ) = =

+

+

With 4 sin siny =

( )

( )[ ]

[ ]

sin (2 cos 1) sin 10 sin sin1(

5 sin 2 sin sin4 sin (2 cos 1)2

nAF

+, ) =

+

e)

The only grating lobes come from the planar array in y directions because of the large

distance between the elements.

The maxima occurs at0

0 0

cos

sin sin

z z

y y

kd

kd

=

=

10 0

1cos 2 ; 120

2m

= =

0 0

34 sin 0, 0

2 = =

The grating lobes will occur:

0 0sin sin sin sin /sin , 0

3

35.26 120

yn d

nn

and

=

= >

= =


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An E-plane sectoral horn operating at 11 GHz and having directivity of 14.77 dB should be

designed.

The horn is fed by an X-band WR 90 rectangular waveguide with inner dimensions a =2.286 cm and b = 1.016cm. The E-plane view of the antenna is given in the figure below

Points 2a) Rewrite the expression for the E-plane horns directivity as a product of a function of

1 and a function ofx , i.e. ( ) ( )1

1ED H F x

x= .

Note:1

is the distance between the cylindrical center of the horn and its aperture. x is

given by:1 12x b = , where b1 is the height of the horn.

Points 3 b) Using the graph below, determine the approximate value of 1 for which the E-plane

horn will have the maximum directivity of 14.77 dB.

Points 2 c) For the value of 1 found in b), find the horn height b1 and total flare angle 2 e .

Points 2 d) What is the frequency bandwidth at which there are no higher order modes propagating

in the horn (assume no fabrication imperfections)?

1

x(

)

(

)

an

d

1

F

x

F

x

x

[ C ( ) + S ]2

x2( )x

[ C ( ) + S ]2

x2( )x

x

1

e

1bb


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Solution 4

a)11GHz =2.7272 cmf =

Dimensions if the feeding waveguide are2.286cm 0.8382

1.016cm 0.3725

a

b

= =

= =

This waveguide supports only TE10 mode because 0.5 and 0.5a b > > <

The directivity of the E-plane horn is given by

2 21 1 1

1

64

2 2E

a b bD C S

b

1 1

= +

It can be rewritten as

( ) ( )

1 2 21 1

1

1

64 22 2 2

1

E

E

a b bD C S b

D H F xx

1

1 1

= +

=

where

( )

( ) ( ) ( )

1 1

1

2 2 1

64 53.6448

2 2

1 1,

2

aH

bF x C x S x x

x x

1

= =

= + =

b)

1

x(

)

(

)

an

d

1

F

x

F

x

x

[ C ( ) + S ]2

x2( )x

[ C ( ) + S ]2

x2( )x

x

Function ( )1

F xx

is graphically represented above. Its maximum value is approx. 0.8 for

1x = , i.e.1

2b 1= . In this case the directivity can be written as

1 1

2

1

53.6448 42.91580.8

2 2

2

42.9158

E

E

D

D

= =

=


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The desired directivity is1.47714.77dB 10 30ED = = =

Thus2

1

1

1

30 2

42.9158

9.6458

26.307 cm

=

=

=

c)

The horn height is given by:

1

1

1

2 2 9.6458

4.3922

11.9788cm

b

b

b

1= =

=

=

The flare angle is

( )1

1 1 1

1

4.39222 22 2 tan 2 tan 2 tan 0.2277

9.6458

2 25.65

E

E

b

= = =

=

d)

The bandwidth of the horn is limited by the bandwidth of the feeding waveguide. For the X-

band WR 90 waveguide the cutoff frequencies of the fundamental and higher order modes

are

TE10:

10 10

10

0.5 2 4.572cm

6.561GHz

TE TE

TE

a a

f

= = =

=

TE20:

20 10

10

2.286cm

13.12 GHz

TE TE

TE

a a

f

= = =

=

The frequency range at which there are no higher order modes propagating is

6.561 GHz 13.12 GHzf< <


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A rectangular microstrip patch antenna should be designed in order to be resonant at

2.4f = GHz. The patch in printed on a RT/duroid 6202 substrate with a relative

permittivity r 2.94 = and a height 0.762h = mm.

3 Points a) Determine the width W and the length L of the patch.

Assume that 0W is valid.

50 W

W

L

d

L1

d

d

W

L

L1

line #0

line #1 line #2

50 Wline #0

line #1 line #2

z

x

b)

Now two identical patch antennas as designed in a) are placed next to each other in a

distance 50d = mm. The two patches should be matched to the 50 feeding line #0 with

the help of the matching network shown in Fig. b).

Note: Mutual coupling can be neglected.

1 Point b) What is the impedance of the two (identical) feeding lines #1 and #2 in order to bematched to line #0?

3 Points c) The patch antenna should be matched to the line #1 by means of a recessed feeding line.

Design the inset 1L in such a way that the patch is matched to the feeding line #1!

5 Points d) In which direction does the main radiation of this arrangement occur? What happens if

the line #0 is moved by 5 = mm towards one patch, as shown in d)?

Assume a width of the microstrip lines #1 and #2 of 0 0.53w = mm.


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Solution 5

a) Using the design rules from slide 8.22 yields the following results.

0 244.4992 1r

cW f = =+ mm,

12

eff1 1

1 12 2.8532 2

r rr

h

W

+ = + + =

eff

eff

0.3 / 0.2640.412 0.378

0.258 / 0.8r

r

e W hL h

e W h

+ + = =

+mm

0

eff

2 36.2182 r

cL L

f = = mm.

b) Both, line #1 and line #2 are connected in parallel to line #0, and hence their impedance

must be 100 .

c) The wavelength in air at 2.4f = GHz is 0 124.91 = mm.

The conductance of a single slot is

10

10.00141S

90

WG

= = . This approximation is used because 0/ 0.356 1W = and

0/ 0.0061 1h = .

Since mutual coupling between the slots is neglected, the resonance input impedance of the

microstrip antenna is determined by

1 1

1

354.62Z G= = .The patch is matched to the 100 line by a recessed microstrip-line feed.

In order to match the antenna to line #1, the input impedance of the patch must be

in 100Z = , and thus the feed must be recessed by

in11

1

cos 11.655L Z

LZ

= = mm.

d) The two microstrip patches form an antenna array with two elements. Because of the

symmetric feeding of the array, both antennas radiate with the same phase ( 0 = ), and

hence the main radiation occurs in broadside direction 0 90 = .The effective permittivity of the microstrip line #1 and #2 is (with 0 0.53w = )

eff,1=2.197r .

When the feeding line #0 is shifted by , this yields a patch differences in the feeding linesbetween both elements of 2. This path difference translates into a phase shift of

eff,1

22 0.746

= = with eff,1 84.273 = mm.

Hence, the main beam now looks in direction of

( ) ( )

01 11 cos cos 107.2

2kd d

= = = .

Final Exam Antenna 20061

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