ap2011 solutions 11 - Universiti Malaysia Pahangee.ump.edu.my/hazlina/teaching_ANT/teaching_ANT_exam_tutorial_A.pdf · Solutions 1 – Fundamentals ... The intensity exists only in

Solutions 1 – Fundamentals Antennas and Propagation, Frühjahrssemester 2011

Problem 1.1 :

If a transmitter produces 50 W of power, express the transmit power in units of

a) dBm,

b) dBW.

Note: For the unit dBmW, the abbreviation dBm is commonly used.

Transmitter power is 50WtP = .

a)[ ]

[ ]3,dBm

mW10 log 10 log 50 10 47.0dBm

1mW

t

t

PP

¹ Ì¼ Í= Õ = Õ Õ =¼ Í¼ Í½ Î

b)[ ]

[ ],dBW

W10 log 10 log 50 17.0dBW

1W

t

t

PP

¹ Ì¼ Í= Õ = Õ =¼ Í¼ Í½ Î


Problem 1.2 :

If 50 watts are transmitted by a unity gain antenna at 900 MHz carrier frequency, find the

received power (assuming unity gain receiver antenna) in [dBm] and [W] at a free space

distance of:

a) 100 m from antenna,

b) 10 km from antenna.

Power density as a function of distance:24

tPpowerS

area Rπ= =

Unity-gain / isotropic receive antenna described by its effective area: 2

4isotropicA

λ

π=

Received power:

2

4rec tP S A P

R

λ

π

Ã Ô= ⋅ = Ä Õ

Å Ö

a)

received power = 3.5 x 10–6

W ≅ –24.5 dBm

b)

received power = 3.5 x 10–10

W ≅ –64.5 dBm


Problem 1.3 :

Assume an air-filled metallic rectangular waveguide of cross-section 22.86 mm x 10.16 mm.

Calculate

a) the monomode frequency range,

b) the guided wavelength at 10 GHz,

c) group velocity and phase velocity at 10 GHz.

a)

From [FuK II, 4.39], with a = 22.86 mm : , 10

16557

2c TEf MHz

a µε= =

Similarly, fcTE20 = 13114 MHz and fcTE01 = 14753 MHz. Therefore, the monomode

frequency range is limited by the cutoff frequencies of modes TE10 and TE20, respectively.

b)

From [FuK II, 4.26], considering TE10 and β=2π/λg :

2 22

g c a

π ω π

λ

Ã Ô Ã Ô= −Ä Õ Ä Õ

Å Ö Å Ö , λg = 39.7 mm

c)

From [FuK II, 4.7, 4.9] : 8

03.97 10 1.32mphase g s

v f x cλ= = = ⋅

28

02.27 10 0.76mgroup s

phase

cv x c

v= = = ⋅


Problem 1.4 :

A hypothetical isotropic antenna is radiating in free space. At a distance of 100 m from the

antenna, the total electric field is measured to be 5 V/m.

a) Find the power density at this location.

b) Determine the total power radiated by the antenna.

a) power density (vectorial!) 2 2

* 2rad

1 50.03315 W/m

2 2 2(120 )r r r

EW E H a a a

h p¹ Ì= fl = = =¼ Í½ Î

f f f f f f

b) radiating power2

2rad rad

0 0

2

2 2rad

0 0

rad

0.03315 sin

0.03315 (100) sin 2 0.03315 100 2

4166.67 W

S

P W dS r d d

P d d

P

p p

f q

p p

q q f

q q f p

= =

= Õ = Õ Õ Õ

= Õ Õ Õ = Õ Õ Õ

=

ÄÄ Ä Ä

Ä Ä

•


Problem 1.5 :

A dipole of length 3λ/2 is resonant at f = 150 MHz. Calculate its mechanical length

a) in air,

b) in water (εrel = 81).

a)

0

r r

cfλ

ε µ= with λ=2m, the dipole length in air is 3 m.

b)

with λ=2/9 m, the dipole length in air is 1/3 m = 0.33 m.


Problem 2.4 :

The power radiated by a lossless antenna is 10 Watts. The directional characteristics of the

antenna are represented by the radiation intensity of

30 cos (W/sr) 0 /2, 0 2U B q q p f p= ‘ ‘ ‘ ‘

a) Find the maximum power density (in watts per square meter) at a distance of 1000 m

(assume far field distance). Specify the angle where this occurs.

b) Find the directivity of the antenna (dimensionless and in dB).

c) Calculate the half-power beamwidth (HPBW).

d) Find the first-null beamwidth (FNBW).

a)

Prad

= U sinθ ⋅ dθ ⋅ dφθ =0

π2

Ðφ =0

2π

Ð = B0

⋅ cos3θ ⋅ sinθ ⋅ dθ ⋅ dφθ =0

π2

Ðφ =0

2π

Ð

Prad

= 2π B0

cos4 θ

4

Ã

ÅÄÔ

ÖÕ0

π

2

=π

2⋅ B

0= 10 µ B

0=

20

π= 6.3662

å 3 3

0( ) cos 6.3662cosU Bθ θ θ= =

W =U

r 2=

6.3662

r 2cos3θ =

6.3662

10002cos3θ = 6.3662 × 10−6 ⋅ cos3θ

Wmax

= Wcosθ =1

= 6.3662 × 10−6 Watts/m2 @ θ = 0

b)

max0

4 4 6.36628 9.03

10rad

UD dB

P

π π ⋅= = = =

c)

The radiation half-power occurs when :

1

3 13

1

1 1cos cos 37.5 0.21

2 2θ θ

θ θ π−

=

Ã Ô= µ = = ° =Ä Õ

Å Ö

The half power beamwidth is given as : 1 12 0.42 74.9d θ πΘ = = = ° . The pattern is

independent on φ, thus HPBW in a plane at right angle to the first one is Θ

2d= Θ

1d. The

directivity can be estimated using an approximate formula (see lecture slides) as

0

1 2

41,253 41,2537.35 8.66

74.9 74.9d d

D dB= = = =Θ Θ ⋅

. This compares quite well with the result of (b).

d)

! cos3θ = 0

θ =θ2

µ θ2

=π

2= 90o µ FNBW = 180°


Problem 2.5 :

The normalized radiation intensity of a given antenna is given by:

1. 1 sin sinU q f= Õ



The intensity exists only in 0 180 , 0 180q f‘ ‘ ‘ ‘c c c c region, and is zero elsewhere.

a) Find the exact directivity (dimensionless and in dB).

b) Find the Azimuthal and elevation plane half-power beamwidths (in degrees).

c) Find the directivity by using approximate formulas.

a)

max1,2,3

,1,2,3

4

rad

UD

P

π= where Umax = 1, and it occurs when θ = φ = π / 2 (in all 3 cases 1,2,3)

2

,1

0 0 0 0

sin sin sin 22

radP U d d d d

π π π π

φ θ φ θ

πθ θ φ φ φ θ θ π

= = = =

= = ⋅ = ⋅ =Ð Ð Ð Ð , 1

4 (1)4 6.02D dB

π

π= = =

2 2

,2

0 0 0 0

sin sin sin2 2

radP U d d d d

π π π π

φ θ φ θ

π πθ θ φ φ φ θ θ

= = = =

= = ⋅ = ⋅Ð Ð Ð Ð , 2

165.09 7.07D dB

π= = =

3

,3

0 0 0 0

4sin sin sin 2

3radP U d d d d

π π π π

φ θ φ θ

θ θ φ φ φ θ θ= = = =

= = ⋅ = ⋅Ð Ð Ð Ð , 3

34.71 6.73

2D dB

π= = =

using

b)

The half-power beamwidths are equal to

( )0 1 0

1, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0

1, 2 90 sin (1/ 2) 120elevationHPBW −= ⋅ − =

( )0 1 0

2, 2 90 sin (1/ 2) 90azimuthHPBW −= ⋅ − = ( )0 1 0


( )0 1 0

3, 2 90 sin (1/ 2) 120azimuthHPBW −= ⋅ − = ( )0 1 0


c)

Directivity using approx. formulas ( 1,2,3

41253

azimuth elevation

D ≈Θ ⋅Θ

, 1,2,3 2 2

72815

azimuth elevation

D ≈Θ + Θ

) :

This gives for case 1 (HPBWs of 120º,120º) : D ≈ 2.86 = 4.6 dB and D ≈ 2.53 = 4.0 dB

It gives for cases 2,3 (HPBWs of 90º,120º) : D ≈ 3.82 = 5.8 dB and D ≈ 3.24 = 5.1 dB

Smith Chart Problems

! The % ! length line shown has a characteristic impedance of 5%6 and isterminated with a load impedance of ZL 7 5 8 j956!

:a; Locate zL 7ZL

Z

7 % 8 j% 5 on the Smith chart!

See the point plotted on the Smith chart!

:b; What is the impedance at % 7 % !@

Since we want to move away from the load :i!e!C toward thegenerator;C read read % %DE! on the wavelengths towardgenerator scale and add % 7 % ! to obtain % DE! on thewavelengths toward generator scale! A radial linefrom the center of the chart intersects the constant reGecHtion coeIcient magnitude circle at z 7 % KL 8 j LL HenceZ 7 zZ 7 5%:% KL 8 j LL; 7 N 8 NE6!

:c; What is the VSWR on the line@

Find VSWR 7 zmax 7 K on the horizontal line to the rightof the chartTs center! Or use the SWR scale on the chart!

:d; What is VL@

From the reflection coefficient scale below the chartCWnd jVLj 7 % L55! From the angle of reflection coef3ficient scale on the perimeter of the chartC Wnd the angle ofVL 7 9X 5 ! Hence VL 7 % L55ej!"#"$

!

:e; What is V at % 7 % ! from the load@

Note that jVj 7 jVLj 7 % L55! Read the angle of the reGectioncoeIcient from the angle of reflection coefficientscale as 55 % ! Hence V 7 % L55ej$$"

!

Lz 0.1 0.5j= +

0.1λ=!

0 1

0.074 . . .T Gλ0.174 . . .T Gλ

z 0.38 1.88j= +

maxVSWR 13z= =

0

L 0.855126.5Γ =

0

LAng 126.5Γ =

0Ang 55.0Γ =

00.855 55.0Γ =

L 0.855Γ = Γ =

VSWR 13=

LZLZ 5 25 [ ]j= + Ω0Z 50= Ω

0.1λ=!

0 1

Problem 1

! A transmission line has Z . /!01 ZL . zL . 0! j0! 2!

3a4 What is z at $ .%

6. 0! 7%8

From the chart1 read 0!6<=% from the wavelengths to+ward generator scale! Add 0! 7% to obtain 0!=/=% onthe wavelengths toward generator scale! This is noton the chart1 but since it repeats every half wavelength1 itis the same as 0!=/=% 0!700% . 0! /=%! Drawing a raHdial line from the center of the chart1 we Ind an intersecHtion with the constant reJection coeKcient magnitude circleat z . Z . !7 L j !7!

3b4 What is the VSWR on the line8

From the intersection of the constant reJection coeKcient cirHcle with the right hand side of the horizontal axis1 read VSWR. zmax . 7!R!

3c4 How far from the load is the Irst voltage minimum8 The Irst currentminimum8

The voltage minimum occurs at zmin which is at a distance of0!700% 0!6<=% . 0!0RR% from the load! Or read this distancedirectly on the wavelengths toward load scale!The current minimum occurs at zmax which is a quarter of awavelength farther down the line or at 0!0RR%L0! 7% . 0! VR%from the load!

Lz 0.2 0.2j= −

0.25λ=!

0.467 . . .T Gλ

0.217 . . .T Gλ

maxVSWR 5.3z= =

VSWR 5.3=

LZL LZ z 0.2 0.2 [ / ]j= = − Ω Ω0Z 1.0= Ω

0.25λ=!

Problem 2

z Z 2.5 2.5j= = +

0.033λ

minz

! The air()lled two(wire line has a characteristic impedance of 567 and isoperated at f 8 GHz! The load is ZL 8 <66 = j>67!

?a@ For the line aboveD )nd zL on the chart!

The normalized load is zL 8ZLZ

8<66 = j>6

568 E$6 = j6$F!

See the Smith chart for location of point!

?b@ What is the line impedance E$5 cm from the loadI

Note that % 8c

f8

<6! cmKsec

<6" Hz8 <6 cm$ Since we are

going to move toward the generator ?away from the load@D atthe normalized load positionD )rst read 6$E<N% on the wave$lengths toward generator scale! Then add E!5 cmK<6cm 8 6$E5% to this value to obtain 6$>PN% on the wave$lengths toward generator scale! A radial line fromthe center at this point intersects the constant reRection co(eScient magnitude circle at z 8 6$> ! j6$<ND so Z 8 zZ 856?6$> = j6$<N@ 8 E<$5! jF$57!

?c@ What is the VSWR on the lineI

From the intersection of the reRection coeScient circle andthe horizontal axis on the right hand side of the chartD readVSWR 8 E!>! Or use the SWR scale below the chart!

Solutions 9 – microstrip patch antennas Antennas and Propagation, Frühjahrssemester 2011

Microstrip transmission line

Problem 9.1 :

A microstrip line is used as a feed line to a microstrip patch. The substrate of the line is

alumina (εrel = 10) while the dimensions of the line are w0 / h = 1.2.

a) Determine the effective dielectric constant εreff and

b) the characteristic impedance Z0 of the line.

c) Calculate the reflection coefficient if a line of the computed characteristic

impedance is connected to a 50 Ω line.

d) Calculate the respective lengths of half-wavelength long lines of impedance Z0

on alumina, of a line in free-space, and of a line fully embedded in alumina halfspace,

at f = 1.5 GHz.

a)

For 0 / 1.2w h = and 10re = , the effective dielectric constant can be computed as 1

2

0

1 11 12 6.86

2 2r r

reff

h

w

e ee

-+ - ¹ Ì= + + =¼ Í

¼ Í½ Î.

b)

For a microstrip line with 0 / 1w h > the characteristic impedance is given by:

( )0 0

120

1.393 0.667 ln 1.444

44.414

c

reff

Z w w

h h

p

e=

¹ Ì+ + +¼ Í½ Î= W

c)

The reflection coefficient from a 50W line is

50 44.4140.059 24.6dB

50 44.414

-G = = -

+5 .

d)

The free-space wavelength is λ0 = 200 mm. Using 0eff

reff

λλ

ε= ,

for the 44.4 Ω line on alumina, / 2 38.2 mmeffλ = ;

for a line in free-space, / 2 100mmeffλ = ;

for a line embedded in alumina, / 2 31.6mmeffλ = .


Microstrip patch antenna

Problem 9.4 :

Design a patch antenna working at the frequency 1.575 GHz and etched on Duroid 6010

substrate ( 10re = ) of thickness 0.508h = mm.

a) Determine W and L ,

b) Determine the inset length 1L needed for matching of antenna to the microstrip

feed line (50W )

a)

For an efficient radiator, a practical width that leads to good radiation efficiency is:

0

0

240.58mm

2 1r

cW

f e= =

+

The effective dielectric constant is

1/21 11 12 9.696

2 2r r

reff

h

W

e ee

-+ - ¹ Ì= + + =¼ Í½ Î

Because of the fringing effects, one has to take into account that the antenna looks

greater than its physical dimensions of 2 LD :

( )( )( )( )

0.3 0.2640.412 0.2202mm

0.258 0.8

reff

reff

W

hL h

W

h

e

e

+ +D = =

- +

So the real physical length will be

0

0

2 30.124mm2 reff

cL L

f e= - D = .

L1


b)

For the matching:

( )21 1

1

1( ) cos

2inR L LG L

p‚

where 1( )inR L has to match the microstrip line impedance 50W .

11 2 2

sin2 cos ( )

490.326S120 120

i

XX XS XI XG

p p

- + + += = =

0 1.3396X kW= = and 0 0.19ml = .

1

11019.73

2inRG

= = W

11

50cos 0.4289

1019.73

LL L

p- ² ÉÊ³= =Ê³ Ê³µ Ë

or if the approximate expression is used (with 0/ 0.2132W l = , 0/ 0.0027h l = ):

26

10

1505.05 10

90

WG

l-² ÉÊ³= = ÕÊ³ Ê³µ Ë

1

1990

2inRG

‚ = W

11

50cos 0.4278

990

LL L

p- ² ÉÊ³= =Ê³ Ê³µ Ë

D-ITET-IFH Antennas and Propagation - Final exam 13th August 2009

-7-

Solution to Problem 4

a) Distance divided by speed of light; i.e., 41000 seconds, or, 11 hours 23 minutes.

b) Friis transmission equation,

2

4r t t rP PG G

R

λ

π

Ã Ô= Ä Õ

Å Ö

Gain of a dish, , 2

4r tG efficiency area

π

λ= ⋅ ⋅

Transmit antenna gain = 2769 = 34.4 dBi

Receive antenna gain = 1’807’309 = 62.6 dBi

Received power = 3.14 x 10-20

W = -165 dBm

c) calculate first null beamwidth αfnbw from feed aperture directivity,

22

19.5 88a

D dBiπ

λ

Ã Ô= = = Ä Õ

Å Ö , a = 205 mm , a / λ = 1.5

the plot in (6.37) reads αfnbw = 48…50º (alternatively, the asymptotic formula

on (6.36) states αfnbw = 69.9º / (a / λ) = 46.6º, but this formula holds only if a>>λ).

From geometry it follows that distance = 340…360 cm for αfnbw = 47…50º.

dish

aperture

first null beamwidth

46cm

274cm

distance d

D-ITET-IFH Antennas and Propagation - Final exam 13th August 2009

-8-

d) this requires much larger feed aperture, leading to shadowing and therefore,

gain reduction (an uniform aperture with 29 dBi gain would have 20 % of the

area of the main dish).

Note: ray-optics assumption, such that the feed is in the focal point and the

waves leaving the aperture are forming (at the beginning) a parallel, that is,

non-divergent, beam.

e) corrugated horn

f) Half-power beamwidth of the large dish: the aperture distribution is as shown

in (6.86), that is, circular symmetric. The best approximation for the HPBW is

the H-plane HPBW of the TE11-aperture distribution in groundplane.

The E-plane HPBW is too narrow, as it equals those of a uniform distribution,

which cannot be realized on a dish aperture.

0370.145

/

o

HPBWa λ

= =

If 360º rotation occur in 24 hours, then 0.072º rotation takes 17.4 seconds.

Note: observation point is very far away.

g) non-constant field distribution (main factor), blocking by feed and mechanical

support structures, spillover at edge, roughness and shape imperfections,

conductive loss.

Solutions 11 – wave propagation Antennas and Propagation, Frühjahrssemester 2011

Propagation close to earth

Problem 11.1 :

A mobile phone is located 5 km away from a base station. It uses a vertical /4 monopole

antenna with a gain of 2.55 dB to receive cellular radio signals. The free space E-field at

1 km from the transmitter is measured to be 310 V/m. The carrier frequency used for this

system is 900f MHz.

a) Find the length and effective aperture of the receiving antenna.

b) Find the received power at the mobile using the 2-ray ground reflection model

assuming the height of the transmitting antenna is 50 m, and the receiving antenna is

1.5 m above ground.

a)

At frequency of 900 MHz, the wavelength is 0.333 m. The length of antenna

is /4 8.33L cm. The antenna gain can be expressed in terms of the effective

aperture as follows

2

4 eAG

where gain is given as 2.55dB 1.8G . Thus,

22

20.333 1.80.0159m

4 4e

GA .

b)

Transmitter-receiver separation distance is 5d km. The field at the receiver end is

a sum of the direct and the reflected rays. Thus, we have

' '' ' ''tot

jkd jkd jkd jkdd r d dE E e E e E e E e

where dE and rE are the magnitudes of the direct and reflected field components,

respectively. From here, we can find the magnitude of the total field as

' ''tot

'' '1

1

jkd jkdd d

jk d dd

jkd

E E e E e

E e

E e

If d is large compared to rh and th we have

2

'' ' r th hd dd


Thus, with 1

2

tot 1 1 2 sin ,2

r th hjkjk d

d d dE E e E e E

where k . Since free-space E-field is reversely proportional to the distance

d we can write

0 0d

E dE

d,

where 0E is the field at distance 0 1d km. Thus,

0 0tot 2 sin

2

E dE

d

For r th h d one can write

0 0 0 0tot

3

6

2 22 sin

2

2 10 1 1000 2 (50) (1.5)

5000 0.333 5000

113.1 10 V/m

r dE d E d h hE

d d d

The received power is

262tot 13113.1 10

0.0159 2.7 10 W2 120 240r e

EP A

In dB: ,dB 125.68 dBW 95.68 dBmrP


Edge diffraction and Fresnel zones

Problem 11.2 :

Compute the diffraction loss for the scenario shown in the figure below. Assume 1/3m,

1 1d km, 2 1d km, and

a) h = 25 m,

b) h = 0,

c) h = –25 m.

T R

a

h

d1 d2

Compare your answers using values from the graph showing the knife-edge diffraction in

function of the diffraction parameter , as well as the approximate solution given by the

equation on slide 8.68. For each of these cases identify the Fresnel zone within which the tip

of the obstruction lies.

a)

h = 25 m. The Fresnel diffraction parameter is obtained as

1 2

1 2

22.74

d dh

d d

From the figure (p.8.64) the diffraction loss is obtained as 22 dB. Using the numerical

approximation

0.225

(dB) 20 log 21.7 dBdL

The path length difference is given by

21 2

1 2

( )0.625m

2

h d d

d d, and we get

23.75n .

Therefore the tip of the knife edge completely blocks the first three Fresnel zones.

b)

h = 0. From the figure the diffraction loss is obtained as 6 dB. Using the numerical

approximation


0.95(dB) 20 log 0.5 6dBdL e

With 0h m and 0 the tip of the knife edge lies in the middle of the first

Fresnel zone.

c)

h = –25 m. The knife edge lies below LOS and 2.74 , 3.75n . From the figure

a diffraction loss of 1 dB is obtained. Using the numerical approximation

(dB) 0dBdL . The tip of the edge is in the 4th

Fresnel zone, leaving the first three

Fresnel zones unobstructed.


Path loss / Propagation close to earth

Problem 11.3 :

A base station transmits a power of 10 W into a feeder cable with a loss of 10 dB. The

transmit antenna has a gain of 12 dBd (dBd refers to a /2 dipole) in the direction of a

mobile receiver, with antenna gain 0 dBd and feeder loss 2 dB. The mobile receiver has a

sensitivity of –104 dBm.

a) Determine the maximum acceptable path loss.

b) Calculate the maximum range of the communication system, assuming 1 1.5h m,

2 30h m, 900f MHz and that propagation takes place over a plane earth.

c) How does this range change if the base station height is doubled?

Quantity Value in

original units

Value in

consistent units

TP : transmitted power 10 W 10 dBW

TG : gain of transmitting antenna 12 dBd 14.15 dBi

RG : gain of receiving antenna 0 dBd 2.15 dBi

RP : received power –104 dBm –134 dBW

TL : transmitter feeder loss 10 dB 10 dB

RL : receiver feeder loss 2 dB 2 dB

a)

The acceptable path loss is

,dB ,dB ,dB ,dB ,dB ,dB 148.3dBT T R R T RL P G G P L L

b)

The plane earth loss is given by 4

PEL 2 21 2

148.3 dBr

Lh h

. Therefore, 34 kmr .

c)

Doubling one heigt reduces the plain earth loss by a factor of four. The distance

inceases by the fourth root of this ratio, thus, by a factor of 1.41. Therefore,

48 kmr .

Documents

ap2011 solutions 11 - Universiti Malaysia Pahangee.ump.edu.my/hazlina/teaching_ANT/teaching_ANT_exam_tutorial_A.pdf · Solutions 1 – Fundamentals ... The intensity exists only in