E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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E&CE 476 Mid-term Examination Date: February 20, 2006 Time: 1 ½ hours Instructor: S. Safavi-Naeini P1. An antenna system is composed of three half-wave (resonant) dipoles, located along x-axis, y-axis and z-axis respectively. All are centered at the origin and fed by three identical 0.1 tje ω [Amp] current sources at a frequency of 1GHz. The input impedances are 73 ohm.

a) Find the total electric field and the radiation pattern in xy-plane ( °= 90θ ). Draw an approximate plot of the pattern as a function of φ in this plane.

b) What are the maximum radiation directions in xy-plane. c) Estimate the gain in xy-plane. (Hint: Find the input power from the knowledge of input impedances)

Solution: a) Electric field due to z-directed dipole in xy-plane:

)()(

: thereforeplane,-on xyˆˆ,2

)2/sin()2/(sin

)2/cos(2

cos

2)(

0000

200

rErE

zr

eIjZr

eIjZrE

z

rjkrjk

rr

r

−=

−==

≈−−

θ

θ θππ

π

ππ

π

Electric fields due to x-directed dipole:

φ

φφ

φπ

πφ

ˆ is plane-in xy dipole

ofdirection therespect to with conical dipole directed-for x becausesinsin

cos2

cos

2)(

200

0

≈−

reIjZ

rErjkr

Electric fields due to y-directed dipole:

.̂- isr unit vecto its and

)2/( is plane-in xy dipole ofdirection therespect to with angle conical dipole, directed-yfor because

coscos

sin2

cos

2)2/sin(

)2/(sin

)2/cos(2

cos

2)( 2

002

0000

φ

φπ

φφ

φπ

πφπ

φπ

φππ

πφ

−

−=−

−

−

−≈−−

reIjZ

reIjZ

rErjkrjkr

z

y

x

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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Therefore the total field is:

±

+

=

≤≤+≤≤−

±

+−=−

2

200

2

2200

cos

sin2

cos

sin

cos2

cos1

2

2for )( ,0for )(,ˆcoscos

sin2

cossin

sin

cos2

cosˆ

2

0

φ

φπ

φ

φπ

π

πφππφφφφ

φπ

φφ

φπ

π

rIZ

E

zreIjZ

Erjk

r

r

b) To draw a rough sketch of the pattern in xy-plane we note that θEEz = is constant. The ϕE part, given

below:

πφππφφ

φπ

φ

φπ

πφ2for )( ,0for )(,,

cos

sin2

cos

sin

cos2

cos

200 ≤≤+≤≤−

±

=

rIZ

E , is simply the sum or

difference two dipole patterns (red dotted line), one rotated by o90 . This is plotted blow in arbitrary unit.

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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The plot of total field,2

Er

, ignoring the constant terms, is simply obtained by adding 1 (a circle of radius one in

polar coordinate system) to the previous polar plot squared, which is shown below (arbitrary unit).

The above plots show that the radiation maxima are along oo 315,135=φ . c) Gains along the above directions is:

[ ] [ ]

( ) )5.1(707.0219

12029.121929.1

]733/[29.1

)]4/(733/[58.222

)]4/(/[58.222)4/(

)2/(

:have we/2or 0for

)]4/(/[cos

sin2

cos

sin

cos2

cos1

22)4/(

)2/(

020

20

0

220

2002

200

2

0

2

2

2

200

2

0

2

dBZ

IIZ

rIr

IZrP

rIZ

rP

ZEG

rPr

IZrP

ZEG

inin

inin

−=×

≈×

=××=

=×

=

==

=

−

+

==

ππ

ππ

πππ

πφ

πφ

φπ

φ

φπ

ππ

r

r

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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P2. Shown below are two half-wave resonant dipoles at a distance of 0.4 λ , radiating at a frequency of 2.4 GHz. Dipole #1 is driven by a current source of amplitude 0.05 [A]. Dipole # 2 is terminated by a variable capacitor, C.

a) Assuming that the self-impedances ( 2211 ZZ = ) are approximately the same as the input impedance of one resonant dipole in free space, namely 73 ohm, write down the impedance matrix relationship between the antennas’ terminal currents and voltages.

b) Find the input impedance at the antenna #1, and the input power in terms of C respectively. c) Find the input (excitation) current at antenna # 2 as a function of C. d) Determine the array factor and antenna system radiation pattern in terms of C. e) Estimate C such that the antenna system has an end-fire pattern (maximum radiation along z). Find

directivity of the antenna array for this case. Solution: a)

][344 where,7373

1221122

21211

Ω−≈+=+=

jZIIZVIZIV

b) , c) From above equation and the V-I relationship for a capacitor:

jXjXZ

IV

ZjX

IZICXjXICjIV

−+≈

−−==⇒

−−=⇒××===−=

731156

7373

7373

104.22,/1,)/(2

12

1

11

1122

9222 πωωω

d)

φθαθπα

απ

θπθ

cossincos,)(/)cos8.0exp(73

3441

sin

cos2

cos

)cos8.0exp(73

344105.0)cosexp(

73105.0

2max

2

2

2

2

12

=−

−−

=

−

−−=

−

−=

CFjjX

jF

jjX

jjkd

jXZ

AF

e) (+10%) For end-fire radiation ( o0max =θ ), C, should maximize the following expression:

( )

dB16.461.279206

}Re{

31.1120}97for {

}Re{

)8.0exp(73

3441

}Re{

)8.0exp(73

3441

42/)}(Re{

)0(42

1

)4/(

)0(}excitationcurrent unity with dipole one frompower radiated{

)4/(

)2/(

pF][7.0]F[107.09773

)30(56)8.0exp(

73344

1

1

2

1

2

0

211

)0(

22

10

2

211

2

22

20

02

2

2

0

2

1222

222

2

===≈=−

−−

=−

−−

=

×=

×==

=×≈⇒≈⇒+

++≈

−−

− −

ZX

Z

jjX

jZ

IZ

jjX

jI

Z

rIZ

AFr

ZZ

rP

AF

rP

ZED

CXX

Xj

jXj

AF

radrad

ππ

ππ

π

πππ

π

44444 844444 76

r

=1I 0.05tje ω C

2I z

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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P3. A pyramidal horn has a square aperture of dimensions 3 [cm] by 3 [cm]. The horn is connected, through a TE10 rectangular waveguide, to a perfectly impedance-matched 10GHz source with 1 [W] output power.

a) Find the amplitude of electric field at: 1) the center of the horn aperture, and 2) a distance of 1 [km], along the axis (z-axis) from the horn.

b) Determine the far-field pattern in yz-plane ( o90=φ ) Solution: a)

1) 23

3/2003/100

)/(,3

100/,10

32

00

0010222

0102

0 ZZ

ZkYakk w ====−=×= ππ

βπ

πβπ

aperture) ofcenter at the field(]V/m[7.323

4109

[W] 14 0

200

420 =⇒××

×=== EEZEY

abP

-

wrad

2) Along the z-axis ( o0=θ ):

( )

( )( )

( ) ]V/m[71]V/m[101.77.31091000

11032

)0],m[1000(

1)0],m[1000(

1 cos

1 sin

542

2

100002

022

200

200

0

0

µπ

π

θ

πθ

πϕ

πϕ

θθ

ϕ

θ

=×=×××

===

==+===⇒

=

=

−−

=−

−

rE

abEr

kEErE

abEr

ejkE

abEr

ejkE

rrjk

rjk

r

r

b)

( )

2

0

0

2

20

0

00

sin2

sin2

sin)90,(

1

sin2

sin2

sin

0

==⇒

=−

θ

θφθ

πθ

θ

θ bk

bk

Fbk

bk

abEr

ejkE

rjko

P4. A high-gain narrow beam reflector antenna is required for a 30 GHz radio link. The Half-Power Beam-width (HPB) should be less than o5.1 .

a) Assuming a second degree (n = 2) “Parabolic taper on a pedestal” aperture illumination, find an estimate of the edge illumination that results in the minimum reflector diameter, 2a, which is required for such a narrow beam and for a SLL better than -30 dB.

b) Assuming that ae = se = 90%, and e = 96%, find the taper efficiency, te , the total aperture efficiency, ape , and the gain of your design.

Solution: a) The -30 dB SLL falls between the lines corresponding to SSL=-29.5 dB and -31.7 dB in Table 7.1. A simple linear estimate for HP is )2/(207.1 aλ . The corresponding edge illumination becomes: C=-12.45 dB (0.238). The taper efficiency is estimated as 824.0=tε .

b) The total aperture efficiency becomes: 64.096.09.09.0824.0 =×××=ape . The minimum diameter required is obtained from the HP = 026.05.1 =o [rad] and 464.0026.0/01.0207.1026.0/207.12 =×== λa [m] or 46.4 [cm]. The gain is therefore 13600)()/4(64.0 22 =××= aG πλπ or 41.3 [dB].

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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Appendix Half-wave dipole antenna:

Far-field:

≈−

θ

θπ

πµ

20

00

sin

cos2

cos

2)(

0

rkeI

rArjk

zr , dB)15.2(64.1,/,sin)( 0 === DZEHrAjE z θϕθ θω

r,

Mutual impedance plots: Horn Antenna (Rectangular Aperture)

( )( )

( )( )

−

=

−

=

−

−

20

2

0

0

0

00

20

2

0

0

0

00

cossin

cossin2

cos

sinsin2

sinsin2

sincoscos

cossin

cossin2

cos

sinsin2

sinsin2

sinsin

0

0

ϕθπ

ϕθ

ϕθ

ϕθϕθ

ϕθπ

ϕθ

ϕθ

ϕθϕ

ϕ

θ

ak

ak

bk

bk

abEr

ejkE

ak

ak

bk

bk

abEr

ejkE

rjk

rjk

204

EYab

P wrad = , ab

64

)4/(

)2/(20010

2

0

2

max

λλβπ==

rP

ZED

rad

r

, )/(,/ 00222

010 ZkYak w βπβ =−=

E&CE 476, University of Waterloo S. Safavi-Naeini, 2006

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