Final Exam Antenna 2008

8/9/2019 Final Exam Antenna 2008

1/18

Solution Manual

Antenna and Propagation - Final examSpring 2008

22nd August 2008

Dr. M. Gimersky, Dr. Chr. Fumeaux, Prof. Dr. R. Vahldieck

This exam contains 4 problems on 7 pages including this cover page. You have 3 hours to solve theproblems. A total of 72 points can be collected. You are free to use any auxiliary material, e.g.

the script, a calculator, . . . , other than communication devices.

Please follow these guidelines to earn maximum credit:

Submit your work in orderof problem number, and attach this page as the front page of yoursolution booklet.

Place your student identification (Legi) on your desk. Please do not use pens with red ink. Write your name on each solution sheet.

State all equations that you use in general form before you substitute numbers or parametersinto them. Specify units with all numerical answers. Distinguish vectors from scalar quantities, using arrows or other obvious identifying marks. Show all of your work and draw a box around your final answers . Possible further references of general interest will be written on the blackboard during the

examination.

Good Luck!

Problem Points Signature

1

2

3

4

Total:


2/18

D-ITET-IFH Antenna and Propagation - Final exam 22nd August 2008

Problem 1

The aim of this problem is to investigate the radiation properties of a planar antenna array arranged

in a non-rectangular grid. An example of such an array is depicted in Figure (1).

2 points a) Start by finding the non-normalized and normalized array factors for the lineararray shown in Figure (2). Suppose M isotropic point sources are arranged in

the array and fed with identical amplitudes and phase difference 1.

4 points b) Use the result of (a) to find the normalized array factor of the planar arrayshown in Figure (1). Again, assume isotropic point sources with identical am-plitudes and phase differences 1 and 2 along the x and y directions, respec-tively.

2 points c) Suppose M = N, d1 = d2 = d and = 0. Find relations for 1 and 2 in sucha way that the main beam is directed towards 0 and 0.

4 points d) Assuming 1 and 2 remain the same as above, find the direction of the mainbeam (, ) in terms of

= 0, 0 and 0. Again suppose M = N, d1 = d2 = d.

3 points e) Draw curves of and versus 0 < < 60 for 0 = 30 and 0 = 90

.

2 points f) Explain how one can obtain the direction of the main beam when = 70.

3 points g) Find the direction of the radiation-pattern nulls assuming M = N = 5, d1 =d2 = /8, 1 = 0

, 2 = 0 and = 30.

Total: 20 points

-2-


3/18


Solution for Problem 1

a) The array factor can be written as the superposition of all the contributions in the far field:

AF = 1 + ej(kd1 cos()sin +) + e2j(kd1 cos()sin +) + . . . + ej(M1)(kd1 cos()sin +)

=ejM 1

ej 1 = ej(M1)

2

sinM

2

sin

2

where is defined as = kd1 cos( )sin + 1. The normalized array factor is obtained asfollows:

AFn =1

M

sinM

2

sin

2

b) The planar array behaves in the far-field region as an array of identical point sources alongthe y-axis with radiation pattern equivalent to AFn found in part (a). The overall array factor

becomes:

AFn =1

M

sinM1

2

sin

1

2

1N

sinN2

2

sin

2

2

where

1 = kd1 cos( )sin + 1 and 2 = kd2 sin sin + 2

c) The main beam is directed along 0 and 0, if 1 and 2 vanish along this direction. AssumingM = N, d1 = d2 = d and = 0, one can write

1 = 0 kd cos 0 sin 0 + 1 = 0 1 = kd cos 0 sin 02 = 0

kd sin 0 sin 0 + 2 = 0

2 =

kd sin 0 sin 0.

d) The direction of the beam is obtained by following the same approach as in the previous part:

1 = 0 kd cos( )sin kd cos 0 sin 0 = 02 = 0 kd sin sin kd sin 0 sin 0 = 0

Therefore, and are obtained by solving the following system of equations:cos( )sin = cos 0 sin 0sin sin = sin 0 sin 0

cos(

)

sin = cot 0

= arctan

cos

cot 0 sin

and = arcsin

sin 0 sin 0

sin

e) First, the values for 0 and 0 should be inserted in the obtained equations for and .

0 = 90 cot 0 = 0 = arctan( cot ) = 90 +

= arcsin

1 12sin(90 + )

= arcsin

1

2cos

The curves of and in terms of are depicted in the following figure.

-3-


4/18


f) When = 70, the system of equations has no real solution. This means that there exists noangle in which 1 = 0 and 2 = 0. To obtain the direction of the main beam, one should obtainthe angles and along which the array factor has its global maximum. This can be carried outby setting the gradient of the array factor equal to zero.

g)The nulls are obtained by assuming

51/2 = m m = 0, 1, 2,... or52/2 = n n = 0, 1, 2,...

m and n should not be equal to zero at the same point

where

1 =2

8cos( 30)sin =

4cos( 30)sin

2 =2

8sin sin =

4sin sin .

By inserting the above equations for 1 and 2 in the first relations, one obtains

cos( 30)sin = 8m5

sin sin = 8n5

The above equations have solutions only when m = 0 or n = 0. The first equation leads to = 120, = 300 or = 0. The second equation leads to = 0, = 180 or = 0. The solution = 0

makes both m and n equal to zero, which is not acceptable. Therefore there exist the followingplanes on which the array factor is equal to zero, except for the point = 0:

= 120, = 300, = 0, and = 180

-4-


5/18


Problem 2

An aperture is located in an infinite PMC plane with field distribution

Ha = H0ax b2 x b2a2 y a2

b

a

x

y

z

Ha

PMCS

The plane S corresponds to the infinite xy-plane, including the center region where the aperture islocated.

1 point a) Does the field distribution in the aperture correspond to a waveguide mode? Ifso, fill in the appropriate values X, m, and n into TXmn.

2 points b) Use the field equivalence principle to define equivalent sources Js and Ms overS.

5 points c) Compute the vector potentials A and F using the far-field approximation. Hint:the following integral may help you.

c2z= c

2

ejz = csin(2 c)

2 c

2 points d) Provide the far-field expressions for E and H in terms of A and F.

3 points e) Provide the complete far-field expressions for E and H.

3 points f) Compute the far-field radiation intensity Wrad for this aperture.

Total: 16 points

-5-


6/18


Solution to Problem 2

a) This distribution does not correspond to any waveguide mode. No!

b) For an aperture on a PMC surface, the equivalence principle and image theory provide thefollowing equivalent sources:

in aperture: JS = 2az Ha = 2H0ayMS = 0

outside aperture: JS = 0

MS = 0

c) We compute the vector potentials using the far field approximation. This means that for am-plitude variations, R r, and for phase variations R r r cos . For this geometry, r cos expands to x sin cos + y sin sin in cartesian coordinates:

A =

4

S

JSejkR

Rds

f.f.a.

ejkr

4r

S

Jsejkr cosds

NN =

S

Jejkr cosds

=

S

Jy cos sin ejkr cosds

= 2H0 cos sin b

2

y= b2

a2

x=a2

ejk(x sin cos+y sin sin)dxdy

using the exponential integral

c

2

z= c2

ejz = c

sin(

2c)

2c

, this reduces to

= 2abH0 cos sin sin

ka2 sin cos

ka2 sin cos

sinkb2 sin sin

kb2 sin sin

.

-6-


7/18


Similarly, for N,

N =S

Jejkr cosds

= 2H0 cos S

ejkr cosds.

same as above

Combining these, we find

A =H0ejkr

2r(ab cos sin )

sin XX

sin YY

,

where X = ka2 sin cos and Y =kb2 sin sin ,

A =H0ejkr

2r(ab cos )

sin X

X sin Y

Y, and

F =

4

S

>

0MS

ejkR

Rds = 0.

d) The radiated fields are derived from the vector potentials as follows:

E =j A +:0

jar F = j A

H =j

ar A>

0j F = j ar A

e) We can write out the far fields using the results from the previous two parts:

E = j A =

C jH0kabe

jkr

2r

sin X

X

sin Y

Y

[cos cos a + cos a]

H = j

ar A =

jH0kabe

jkr

2r

sin X

X

sin Y

Y

C/

[cos cos a + cos a]

-7-


8/18


f) The far-field radiation intensity can be found from the far-fields as Wrad =12 Re[

E H]:

Wrad =1

2Re[ E H] = 1

2

C2

cos2 (cos2 1)ar

=

2

H0kabe

jkr

2r

2sin X

X

2 sin YY

2cos2 (cos2 1)ar

-8-


9/18


Problem 3

A rectangular patch antenna is intended for a GPS application (f0 = 1570 MHz). The antenna is

fed by a microstrip line residing on the same plane as the patch (Figure 1).

Figure1

L1 L2

L

Figure1Figure1

Wm

1 3 5 7 9 11 13

30

50

70

90

110

130

Width Wm

[mm]

Figure2

er=2.2

er=10.2

[Oh

m]

Impedance

2 points a) There is a choice of two different substrates, on which the patch and the feedlinecan be fabricated: one with thickness t1 = 4 mm and dielectric constant r1 =2.2 and one with thickness t2 = 4 mm and dielectric constant r2 = 10.2. Which

of the two substrates would you choose, in order to obtain as high efficiencyand as large bandwidth as possible? Justify your answer.

2 points b) For the chosen substrate, determine the width W and the length L of the patchantenna for operation with the T M010 mode.

2 points c) Figure 2 illustrates the characteristic impedance of the feed (microstrip) lineversus its width Wm for the two different types of substrate (r1 = 2.2, r2 =10.2, thickness t = 4 mm). For the chosen substrate and a recessed microstripline of depth L1 = 1.9 cm, find the necessary width Wm of the microstrip lineso that the resonant patch antenna is perfectly matched. Neglect the mutualcoupling between the slots and assume that W


10/18


2 points f) We wish to place an H-plane sectoral horn antenna in the far field of the patchantenna and orient it for maximum transmission. Using the same coordinatesystem as of the patch antenna, explain how exactly you will need to place the

horn antenna. You can add a representative drawing to your explanation.

L

L/2

Figure3

z

xy

PEC

a) b)

3 points g) We reduce the patchs length to one half (L = 12 L) and add a perfectly conduct-ing ground plane parallel to the xz-plane, attached to the patch, as shown inFigure 3b. The patch is positioned on the yz-plane. We assume that the groundplane in the xz-plane is infinitely large. Will the presence of the ground planeaffect the fields of the T M010 mode? Will this affect the resonant frequency,and how?

Total: 16 points

-10-


11/18


Solution 3

a) We choose the substrate with the lowest dielectric permittivity r1 = 2.2, so that the antenna

has a better efficiency, a larger bandwidth and a larger gain.

b)

W =c0

2 f0

2

r + 1= 7.55cm

reff =r + 1

2+

r 12

1 + 12

t

W

1/2= 2.07

L = t 0.412(reff + 0.3)(W/t + 0.264)

(reff 0.258)(W/t + 0.8)= 2.1 mm

L =c0

2 f0reff 2L = 6.22cm

c)

Since there is perfect matching,

=Zin ZmZin + Zm

= 0 Zin = Zm

where Zin is the input impedance of the patch and Zm the characteristic impedance of the microstripline. To estimate Zin, we have the following:

Zin = Rin = Rin(0) cos

2 L L1 =1

2G1 cos

2 L L1Zin =

90

2

0W

2cos2

L

L1

= 94.9

where 0 = 19.11cm

From Figure 2, the width of the microstrip line should be Wm 3.1 mm.d)

It is the T M030 mode. The current distribution Jb is shown in Figure 4:

Figure4

-11-


12/18


e)

It is not possible, because the microstrip line cannot support this mode. In the T M011 mode, the

electric field is minimum at the middle of the patchs width W.f)

The H-plane horn antenna should look in the x direction and be aligned so that its height b (thesmaller dimension) is parallel to the length L of the patch. In this way, the electric fields of thetwo antennas are aligned. Figure 5 shows the configuration.

x

y

z

Figure5

b

L

g)

The groundplane will not affect the near fields, since the electric field at the position L/2 is zero.The resonant frequency will also remain the same.Such a technique is used to make the patch electrically smaller.

-12-


13/18


Problem 4

3km

4km 500km

200km

D-layer

E-layer

F-layer( ,f )rp pe

Tx Rxs 8

Figure 4.1: Sketch of the communication link

Two amateur radio operators in Switzerland want to establish a radio link between their homes (see

sketch) during daytime with two similar lossless 2 dipole antennas which are placed 11.5 m aboveground. Their communication is established in the 28.5 m-band. The line of sight between theoperators is disturbed by a 3 km-high mountain. The F-layer of the ionosphere is situated 200 kmabove ground and contains roughly one million electrons per cubic centimeter. The D and E layersof the ionosphere contain much less electrons at the given time of the day and are approximatedby a damping factor of Li = 1 dB per path.

4 points a) Sketch at least four different communication paths between the operators.

2 points b) Assuming that the polarisation vector of the transmitting antenna is unknown,which polarisation vector would be most favorable for the receiving antenna?Explain your answer!

If both antennas are oriented vertically to the ground and one is fed by 100 W of input power,

14 points c) which of all possible communication paths results in the strongest signal? Whatis the received power?

Total: 20 points

See hints on the other page.

-13-


14/18


- 5 . 0 - 2 . 5 0 . 0 2 . 5 5 . 0 7 . 5 1 0 . 0 1 2 . 5 1 5 . 0

K

n

f

e

-

e

d

g

e

d

f

f

r

a

c

t

o

n

g

a

n

(

d

B

)

P a r a m e t e r

- 2 . 5 - 2 . 0 - 1 . 5 - 1 . 0 - 0 . 5 0 . 0

- 7 . 5

- 5 . 0

- 2 . 5

Figure 4.2: knife-edge diffraction gain

Hints:In order to solve this problem efficiently, round and approximate reasonably. For the sake ofsimplicity, neglect the curvature of earth and ionosphere and consider them to be parallel to eachother. Also consider the mountain as an infinitely long absorbing half plane (knife-edge diffraction).The F-layer of the ionosphere above them can be approximated by a dielectric slab with an effectiverelative permittivity of

r = 1

fpf

2,

where fp denotes the plasma frequency fp. Furthermore consider the ionosphere as a neutral plasma,consisting of a gas of positively charged ions and negatively charged electrons. If the electrons arecold, it is possible to show that the charge density oscillates at the plasma frequency defined as

fp =1

2

q2Ne0me

,

where Ne is the number of electrons per cubic meter, q is the electric charge of one electron, me isthe mass of one electron, and 0 is the permittivity of free space.The directivity for a vertical 2 dipole antenna at the height hant above ground is defined as:

D () = 2sin2 cos2 (khant cos )

13 cos(2khant)(2khant)2 +

sin(2khant)

(2khant)3

.

-14-


15/18


Solution for Problem 4

3km

4km 500km

200km

D-layer

E-layer

F-layer( ,f )rp pe

Tx Rxs 8

path1

path2path3path4

Figure 4.3: Sketch of the four possible propagation paths

a) There is an infinite number of propagation paths between these two antennas, since the waves

are reflected at the ground ( = 1) and at the interface between air and ionosphere. Thus theproblem can be approximated by a transmission-line structure containing two parallel dielectricslabs (air and ionosphere) with perfect electric conductor at one side and an absorbing obstacle(height: 3 km) on the ground (see sketch). The wave is propagating in the air layer as well as leakingenergy into the ionosphere, depending on the incident angle. Thus the following propagation pathsare possible:

path 1:Direct link between Tx and Rx diffracted around the obstacle.

path 2:Link with one reflection from the ionosphere, also diffracted at the obstacle.

path 3:Link with two reflection from the ionosphere and one from the ground.

path n:Link with n 1 reflection from the ionosphere and n 2 from the ground.

Due to the geometric symmetry of the configuration, the elevation angles for the transmitted andthe received waves are the same.

-15-


16/18


b) Since the path 1 is very unlikely to contribute energy to the receiver due to the large obstacle,the received wave has to be reflected from the ionosphere. Its plasma frequency for the F-layeryields:

fp = 12

q2Ne0me

= 8.966 MHz,

where q = 1.6 1019 C, me = 9.11 1031 kg, 0 = 8.854 1012 Fm and Ne = 1 1012 1m3 .Consequently the effective relative permittivy for the F-layer ionosphere results in:

io = 1

fpf

2= 0.2744.

Since the ionosphere is electrically thinner than the air slab, an incident angle to the surface normalvector for total reflection can be found as followed:

c = arcsinioio

airair = 0.5514 rad = 31.59.

Since total reflection only occurs for parallel reflection, i.e. the electric field vector is coplanar tothe plane of incidence, the polarisation vector of the receiving antenna shall be as well coplanar tothe plane of incidence. For example, the dipole antenna could be placed vertical to the ground.

c) In combination with the results of a) and b), a power link budget is determined for each pathas a product of the input power, the gain values of transmitting and receiving antennas and theindividual path losses LPn, yielding:

PRn = PT GT (n) LPn GR (n) .Since the radiation patterns of both antennas only depend on the elevation angle n (see figure 4.4)

and both antennas are losses, their gain equals their directivity and is defined as:

GT (n) = GR (n) = ecdDT (n) = 2sin2 n cos

2 (khant cos n)13 cos(2khant)(2khant)2 +

sin(2khant)

(2khant)3

,where hant = 11.5 m is the antenna height and k =

2 = 0.22

1m is the wavenumber. LPn denotes

the path losses for each path defined as

LPn = io (n)2(n1) gr (n)2(n2) 4sn

2L

2(n1)io Ldn,

where each term is denoted as follows:

io (n)2(n1) reflection losses at the ionosphere gr (n)2(n2) reflection losses at the ground

4sn

2path losses due to the distance sn = 2 (n 1) hiocos(n) for n > 1

(s1 = 504 103 m)

L2(n1)io ionosphere D- and E- layer losses only, depending on the number of passes

Ldn diffraction losses due to the obstacle (mountain) for each path

-16-


17/18


Tx

path4 path2path3

q4

x path1

h3

h1

d11 d21

d12

d22

e

h2

d13

q3q2

q1

Figure 4.4: The figure depicts the geometrical sketch of the problem whereas the height of the

antenna (11.5 m) is disregarded in relation to the mountain height.

As depicted in figure 4.4 the elevation angle is determined for each path as:

n = arctan

s1

2 (n 1) 1

hio

,

where hio = 2 105 m denotes the height of the F layer ionosphere. Consequently the distance eequals 5000 m and the angle yield to = arctan

3000 m4000 m

= 0.6435 rad = 36.9. Furthermore the

path is diffracted at the obstacle, thus a diffraction height according to the Fresnel zones for eachindividual path is found as:

hn = e sin2 n ,

as well as the corresponding direct path fractions d1 and d2

d1n = e cos

2 n

,

d2n =hio

cos(n) d1n.

Note that d1 and d2 are only denoted to a fraction of the whole propagation path, i.e. the distancebetween the transmitter and the first ionospheric reflection. In the result, the diffraction parameter

-17-


18/18


n can be found as

n = hn

2 (d1n + d2n)

d1nd2n.

Finally, the diffraction loss is found with the help of the figure 4.2. Furthermore, the ionosphereD- and E-layer loss Lio is given as Lio = 1 dB.Since the ground is approximated by a perfect electric conductor layer, its reflection coefficient gr (n) = 1. The reflection coefficient for the F layer is defined as

io (n) =io sin

2 n

io cos2 2 nio sin

2 n

+

io cos22 n

,but it only important to consider for path 4 and higher, since total reflection occurs in the remainingpaths (see b)).In the result the following values are computed:

parameter path 1 path 2 path 3 path 4

PR [fW] 26214.4 0.00641 65941.0 20.76

GT [1] 6.4 0.00011 0.5366 0.4619

[rad] 2 0.8999 0.5622 0.3976

[] 90 51.6 32.2 22.8

LP [1015] 6.4 5299.2 2290.1 0.973

Ld [1] 0.0003162 0.676 1.0 1.0

io [1] -1 -1 -1 -0.3294

s [km] 504 643.4 945.5 1301.5

h [m] -3000 137.0 1785.2 2526.4

d1 [m] 4000 4998.1 4670.4 4314.8

d2 [m] 500000 316706.5 231712.6 212606.6

[m] -12.56 0.5172 6.9893 10.2913

Lio [1] 0.0 0.79 0.79 0.79

It is not required to compute all values, a reasonable guess is also accepted. In the result, thestrongest signal is transmitted via path 3.

-18-

Documents

Final Exam Antenna 2008