# Antennas and Propagation Spring 2006

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```					D-ITET                                            Antennas and Propagation

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

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

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

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.

•     This is an open book exam.
•     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.
•     Possible further references of general interest will be written on the blackboard during
the examination.

Problem                        Points                 Initials

1

2

3

4

5

Total

— 1 / 15 —
D-ITET                            Antennas and Propagation                      March 23, 2006

Problem 1 (11 Points)
A pyramidal horn antenna ( a1 = 19.43 cm b1 = 14.35 cm, aperture efficiency eap = 0.5 ,
linearly polarized in ey -direction) should be measured in an anechoic chamber with the
length L = 8 m, the width W = 4 m and the height H = 4 m.

anechoic chamber

y
x
a1
z

b1
H

d

transmitting                                           receiving
antenna                                                antenna

L

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

A second antenna is mounted in a distance d = 5 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 Gt = 8 dBi and can be described
with the radiation resistance Rr = 220 Ω , the loss resistance RL = 5 Ω and it is connected
to a generator with the voltage Vg = 30 V and the inner resistance Rg = 75 Ω (assume the
reactances of antenna and generator X A = X g = 0 ). The relative electric field vector Et of
the transmitting antenna can be expressed as a superposition of two plane waves with
π                                                  π
E1,t = 9 cos(ωt + )ex + cos(ωt )ey          and     E 2,t = 5 cos(ωt − )ex + 2 cos(ωt )ey .
2                                                  2

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

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

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

— 2 / 15 —
D-ITET                            Antennas and Propagation                        March 23, 2006

Solution 1
2D 2
a) The far-field condition of the pyramidal horn is given by R >      , where D is the
λ
2     2
largest dimension of the antenna. Here: D = a1 + b1 = 0.2415 m, and thus R > 4.67 m.
Because the anechoic chamber has a length of L = 8 m, it is possible to perform a
measurement of the far-field radiation pattern.

b) The total electric field is Et = E1,t + E2,t . Together with
π
E1,t = 9 cos(ωt + )ex + cos(ωt )ey = −9 sin(ωt )ex + cos(ωt )ey
2
and
π
E2,t = 5 cos(ωt − )ex + 2 cos(ωt )ey = 5 sin(ωt )ex + 2 cos(ωt )ey
2
the total field can be written as
Et = −4 sin(ωt )ex + 3 cos(ωt )ey
π .
= 4 sin(ωt + π)ex + 3 sin(ωt + π − )ey
2
 −9 
Hence, the polarization vector of E1,t is ρt ,1 =   , and the polarization vector of E 2,t is
 
 
 j 
 
5                                −4 
ρt ,2 =   , and thus ρt = ρt,1 + ρt,2 =   .
                                 
 j3 
 
 j2 
                                 
 
At ωt = 0 the electric field is Et = 3ey , and at ωt = π / 2 the field is Et = −4ex .
Hence, the resulting wave propagating in +z -direction is elliptically right hand (or CW)
polarized.
OA      4
With the formula given on slide 2.22 the axial ratio is AR =       = .
OB       3

c) Using an equivalent circuit the power transmitted from the second antenna can be
expressed as
1 2
Pt = I 0 Rr
2
Uq              30
with I 0 =                 =       A = 0.1A . Therewith the transmitted power is
Rr + RL + Rq        300
1
Pt = (0.1)2 220 W = 1.1 W .
2

d) The received power can be determined by
Pr = Pt  ( )
λ 2
4πd
GrGt PLF ,
where the gain of the transmitting antenna is given Gt = 8 dBi , λ = 25 mm at
f = 12 GHz, d = 5 m is the distance between the two antennas, and the gain of the
receiving antenna can be calculated by
1 4π
Gr =        ab .
2 λ2 1 1

— 3 / 15 —
D-ITET                              Antennas and Propagation                      March 23, 2006

Here the factor eap = 1/ 2 is the aperture efficiency, and the polarization loss factor is
2                                 2
PLF = ρr ⋅ ρt
ˆ ˆ            = ey ⋅ ( 4ex − 3 jey ) / 25
ˆ       ˆ       ˆ               = 0.36 .
Pr = Pt  ( )
λ 2
4πd
GrGt PLF = 0.111 mW

— 4 / 15 —
D-ITET                               Antennas and Propagation                    March 23, 2006

Problem 2 (9 Points)
An airplane passing at an altitude H P is tracked by two independent (monostatic) radar
stations working at two different frequencies. Radar 1 works at f1 = 1 GHz and Radar 2 at
f 2 = 2 GHz, both have an output power P0 = 200 W and a gain G = 1 .
The radar cross section of the plane is σ = 100 m2. The geographical dimensions are given in
the following:
H P = H T = 100 m, H R = 0.5 m, H M = 90 m, DT = 1000 m, DP = 4000 m and DR = 8000 m.

Mountain

HR

DT                          DP                         DR

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

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.

ν

— 5 / 15 —
D-ITET                                 Antennas and Propagation            March 23, 2006

Solution 2
a.)
f1 = 1GHz , f 2 = 2GHz
c
⇒ λ1 =    = 0.3m, λ 2 = 0.15m
f
Free space loss for one way can be calculated as follows:
2
      λ1       
LLOS ,radar1 =                 = −106.4dB
 4π( DT + DP ) 
2
 λ2 
LLOS ,radar 2 =            = −116.5dB
 4π( DR ) 
4
      λ1         4πσ 
P = P0 G 
1
2
  2 
 4π( DT + DP )   λ1 
4
 λ   4πσ 
P2 = P0G  2   2  ⇒
2

 4πDR   λ 2 
P = 53.01dBm − 2 ⋅106.42dB + 41.45dB = −118.4dBm
1

P2 = 53.01dBm − 2 ⋅116.5dB + 47.47dB = −132.5dBm

b.)
The Fresnel zones are defined like:
DT DP
rn ≈ nλ1
DT + DP
When h < −0.6 ⋅ r1 the effect of the diffraction is very small.
DT DP
r1 ≈ λ1                ≈ 15.5m
DT + DP
⇒ 0.6 ⋅ 15.5m = −9.3m > 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.
4                            4
P2 '      
2  λ2    ⋅ 1 + Γ exp  jk 2H P H R   4πσ 
                                 
=G                      
               2  
      
                          
DR   λ2 
      
P0         4πDR              
4                  4
 ≈  2π 2H P H R 
 2H P H R                  
LLOS ,ground   = 1 + Γ exp  jk
          



λ



    DR       2    DR    
LLOS ,ground   = −11.2dB
P2' = −143.7dB

c.)
∆P = P2' − P1 = −25.3dB back and forward.
Read from the graph 0 ≤ v ≈ 0.8 ≤ 1

— 6 / 15 —
D-ITET                             Antennas and Propagation   March 23, 2006

Ld (dB) = 20 log (0.5e −0.95v ) =-12.65dB
⇒ e −0.95v = 0.466, v = 0.8
2 ( DT + DP )                      υ
υ =h                   ⇒h =
λ1DT DP                2 ( DT + DP )
λ1DT DP
h = 8.7 m
H M = H P + h = 108.7m

— 7 / 15 —
D-ITET                              Antennas and Propagation                       March 23, 2006

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 [ − j; − 1; j;1] corresponding to the element
λ
number [1; 2;3; 4] . The distance between the elements is a =        .
2
z

1
a
2
a/2
y
a/2
3
a
4

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 b = 2 ⋅ λ .
z

1          1                1         1           1
a
2          2                2         2           2
a/2
y
a/2
3          3                3         3           3
a
4          4                4         4           4
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.

— 8 / 15 —
D-ITET                                             Antennas and Propagation                                     March 23, 2006

Solution 5
Notation of the lecture notes:
dz = a
dy = b
a)
N   

sin  Ψ     
1         2 
AFn =
N sin  1 Ψ 
   
2 
   
Ψ = kdz cos θ + ξ
π
With ξ = + the linear phase progression.
2
sin (π(2 cos θ + 1))
AFn =
π
2 (
4 sin (2 cos θ + 1)                   )
b)
The nulls:
 λ            2n 
θn = cos−1          
−ξ ± N π   


 2πd                  
n = 1,2, 3,...



n ≠ N ,2N , 3N ,...


±n − 1 
⇒ θn = cos−1               ; θn = [0 , 60 , 90 ,180 ]
 2 
 λ                
θm = cos−1            (−ξ ± 2m π)
 2πdz             
                  
The maxima:              m = 0,1,2, 3,...
       1
θm = cos−1 ±2m −  ; θm = 120
      2 
There is no grating lobe in the visible region. The grating lobes depends on the phase
increment and the spacing.

c)
λ
For endfire arrays the condition is ξ = ±kd . So dz =                                     for a radiation towards θm = 180 .
4
d)
M
N
j (m −1)(kdz cos φ+ξz )  j (n −1)(kdy sin θ sin φ+ξy )
AF = ∑ I 1n  ∑ Im 1 e                         e
n =1
 m =1                            
M                                       N
j (m −1)(kdz cos θ +ξz )               j (n −1)(kdy sin θ sin φ +ξy )
=    ∑ I m 1e
m =1
⋅ ∑ I 1ne
n =1

AFz                                       AFy

— 9 / 15 —
D-ITET                              Antennas and Propagation             March 23, 2006

1 sin (M / 2) Ψ z  1 sin (N / 2) Ψy 
AF (θ, φ)n =                       ⋅                      =
M sin (1/ 2) Ψ z  N sin (1/ 2) Ψ y 
sin (π(2 cos θ + 1)) 1 sin (N / 2) Ψy 
π
2 (                )     
4 sin (2 cos θ + 1) N sin (1/ 2) Ψ y 


With Ψ y = 4π sin θ sin φ
sin (π(2 cos θ + 1)) 1 sin [10π sin θ sin φ ]
AF (θ, φ)n =
(π
)
4 sin (2 cos θ + 1) 5 sin [2π sin θ sin φ ]
2
e)
The only grating lobes come from the planar array in y directions because of the large
distance between the elements.

The maxima occurs at
ξz = −kdz cos θ0
ξy = −kdy sin θ0 sin φ0
         1
θ0 = cos−1 ±2m −  ; θ0 = 120
        2 
3
−4π      sin φ0 = 0, φ0 = 0
2
The grating lobes will occur:

sin θ sin φ − sin θ0 sin φ0 = ±nλ / dy
n
sin φ = ±   ,n > 0
3
φ = ±35.26 and θ = 120

— 10 / 15 —
D-ITET                                Antennas and Propagation                          March 23, 2006

Problem 4 (9 Points)

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

ρ1
b                                  b1

ψe

Points 2   a) Rewrite the expression for the E-plane horn’s directivity as a product of a function of
1
ρ1 and a function of x , i.e. DE = H ( ρ1 )⋅ F ( x) .
x
Note: ρ1 is the distance between the cylindrical center of the horn and its aperture. x is
given by: x = b1        2λρ1 , where b1 is the height of the horn.

Points 3   b) Using the graph below, determine the approximate value of ρ1 for which the E-plane
horn will have the maximum directivity of 14.77 dB.

Points 2   c) For the value of ρ1 found in b), find the horn height b1 and total flare angle 2ψ 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)?

2        2
[ C (x) + S (x) ]
F (x)
1
x
F ( x ) and

1 2          2
[ C (x) + S (x) ]
x

x

— 11 / 15 —
D-ITET                                           Antennas and Propagation                          March 23, 2006

Solution 4
a)
f = 11GHz ⇒ λ =2.7272 cm
Dimensions if the feeding waveguide are
a = 2.286 cm = 0.8382λ
b = 1.016 cm = 0.3725λ
This waveguide supports only TE10 mode because λ > a > 0.5λ and b < 0.5λ

The directivity of the E-plane horn is given by
64a ρ1  2  b1 


 + S 2  b1 
        


DE =             C                           
πλb1   2λρ1 

         
               
 2λρ1 
       

It can be rewritten as
64a ρ1 2λρ1  2  b1                           
DE =              ⋅             
          + S 2  b1 
             

C            
             
πλ 2λ         b1   2λρ1 
 
        2λρ1 
      

1
DE = H ( ρ1 )⋅ F ( x)
x
where
64a ρ1 53.6448 ρ1
H ( ρ1 ) =            =
πλ 2λ            π 2λ
1           1                                 b1
F ( x) = C 2 ( x) + S 2 ( x) , x =
x           x                                2λρ1
b)

2       2
[ C (x) + S (x) ]
F (x)
1
x
F ( x ) and

1 2          2
[ C (x) + S (x) ]
x

x
1
Function      F ( x) is graphically represented above. Its maximum value is approx. 0.8 for
x
x = 1 , i.e. b1 = 2λρ1 . In this case the directivity can be written as
53.6448 ρ1                        42.9158 ρ1
DE =                              ⋅ 0.8 =
π 2λ                              π 2λ
2
 D ⋅ π 2λ 
ρ1 =   E         
 42.9158 

— 12 / 15 —
D-ITET                              Antennas and Propagation                 March 23, 2006

The desired directivity is
DE = 14.77 dB = 101.477 = 30
Thus
2
 30 ⋅ π 2 
ρ1 =             ⋅λ
 42.9158 
ρ1 = 9.6458 ⋅ λ
ρ1 = 26.307 cm

c)

The horn height is given by:
b1 = 2λρ1 = 2λ ⋅ 9.6458 ⋅ λ
b1 = 4.3922 ⋅ λ
b1 = 11.9788cm

The flare angle is
    
 b1 
 2              4.3922 ⋅ λ 2 
              
2ψ E = 2 tan 
−1
     = 2 tan−1 
                           = 2 tan −1 (0.2277)

           
 9.6458 ⋅ λ  
 ρ1 
                             
                             

2ψ E = 25.65

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:
a = 0.5λTE10 ⇒ λTE10 = 2a = 4.572cm
fTE10 = 6.561 GHz
TE20:
a = λTE20 ⇒ λTE10 = a = 2.286cm
fTE10 = 13.12 GHz

The frequency range at which there are no higher order modes propagating is
6.561 GHz < f < 13.12 GHz

— 13 / 15 —
D-ITET                                      Antennas and Propagation                              March 23, 2006

Problem 5 (12 Points)
A rectangular microstrip patch antenna should be designed in order to be resonant at
f = 2.4 GHz. The patch in printed on a RT/duroid 6202 substrate with a relative
permittivity εr = 2.94 and a height h = 0.762 mm.

3 Points   a)   Determine the width W and the length L of the patch.
Assume that W             λ0 is valid.

W                                                             W

d                                                               d

L                                                              L

L1                                                           L1

d

x
line #1             line #2                                       line #1           line #2

z
line #0        50 W                                                   line #0     50 W

b)                                                        d)

Now two identical patch antennas as designed in a) are placed next to each other in a
distance d = 50 mm. The two patches should be matched to the 50 Ω 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 be
matched 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 L1 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 δ = 5 mm towards one patch, as shown in d)?
Assume a width of the microstrip lines #1 and #2 of w 0 = 0.53 mm.

— 14 / 15 —
D-ITET                            Antennas and Propagation                       March 23, 2006

Solution 5

a) Using the design rules from slide 8.22 yields the following results.
c       2
W = 0              = 44.499 mm,
2 f εr + 1
h − 2
1
ε + 1 εr − 1 
εreff = r       +          1 + 12       = 2.853
2         2           W
e + 0.3 W / h + 0.264
∆L = 0.412 ⋅ h reff              ⋅              = 0.378 mm
ereff − 0.258 W / h + 0.8
c0
L=              − 2∆L = 36.218 mm.
2 f εreff

b) Both, line #1 and line #2 are connected in parallel to line #0, and hence their impedance
must be 100 Ω .

c) The wavelength in air at f = 2.4 GHz is λ0 = 124.91 mm.
The conductance of a single slot is
1 W 
G1 =        = 0.00141S . This approximation is used because W / λ0 = 0.356        1 and
 
90  λ0 
h / λ0 = 0.0061      1.
Since mutual coupling between the slots is neglected, the resonance input impedance of the
microstrip antenna is determined by
1
Z1 =        = 354.6 Ω .
2G1
The patch is matched to the 100 Ω line by a recessed microstrip-line feed.
In order to match the antenna to line #1, the input impedance of the patch must be
Z in = 100 Ω , and thus the feed must be recessed by
L        Z in 

L1 =     cos−1 
 Z  = 11.655 mm.


π       
    1

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 ( ξ = 0 ), and
hence the main radiation occurs in broadside direction θ0 = 90 .
The effective permittivity of the microstrip line #1 and #2 is (with w 0 = 0.53 )
εreff,1 = 2.197 .
When the feeding line #0 is shifted by δ , this yields a patch differences in the feeding lines
between both elements of 2δ . This path difference translates into a phase shift of
2δ
ξ = 2π         = 0.746 with λeff,1 = 84.273 mm.
λeff,1
Hence, the main beam now looks in direction of
θ1 = cos−1 −(    ξ
kd)          (  ξλ
)
= cos−1 − 0 = 107.2 .
2πd

— 15 / 15 —

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