antennas and microwaves engineering (650427) · antenna (hertzian dipole) if the center of this...

Philadelphia University

Faculty of Engineering Communication and Electronics Engineering

Part 6 Dr. Omar R Daoud 1

ANTENNAS and MICROWAVES

ENGINEERING

(650427)

4/15/2018 Dr. Omar R Daoud 2

Antenna Types

Short Antennas If the current distribution of a radiating element is known, it’s possible to

calculate the radiated fields by a direct integration but, the integrals can be

very complex.

For time harmonic fields, integration is performed to find a phasor called

retarded vector magnetic potential, which then followed by simple

differentiation to find the H field.

Electric fields, and analogous term for H field is the vector magnetic

potential A, often used for antenna calculations (we now seek a relation

between vector A and a current source).

The derivations lead to

VE

ddo

dsS dv

R

JA

4

00

The vector magnetic potential

SS 00 AB

The magnetic flux density

SSr 00Re2

1,, *

HEP

The time averaged power radiated

000 /SS BH

SrS 000 HaE


Antenna Types

Short Antennas

Hertzian Dipole

Suppose that a short line of current,

tIti cos)( 0

It’s placed along the z axis as shown.

z

jkRs

SR

eIaA

4

00

aH sin4

0R

keIj

jkRs

S

aE sin4

00R

keIj

jkRs

S

The time averaged power density at observation point is:

rR

kIr aP

2

22

22200 sin

32,

The term in brackets is the max power

density. The sin2θ term is the normalized

radiation intensity Pn(θ) plotted as :

38sinsinsin 22 dddp

The pattern solid angle

The directivity is 5.14

max

p

D

The total power radiated

2

0

2

2

22

22

0

2

02

max

2 4032

IR

IkrPrP pprad

2280

radR

For Hertzian dipole, where l<<λ, Rrad will be very small and the

antenna will not efficiently radiate power. Larger dipole antennas,

have much higher Rrad and thus more efficient.


Antenna Types

Short Antennas

Small Loop

The time averaged power density at observation point is:

38sinsinsin 22 dddp

The pattern solid angle

The directivity is 5.14

max

p

D

The total power radiated

aA sin1

4 2

00

jkRsS ejkR

R

SI

2aS

aH

jkRsS e

R

SkI sin

4 0

00

aE

jkRsS e

R

SkI sin4

00

rR

kSIr aP

2

220

2220

20

2

sin32

,

220

2220

20

2

max32 R

kSIP

Since the normalized power function is the same as Hertzian dipole, then

2

2

20

30

3

4

SIPrad

2

2

4320

SRrad

The equations also valid for

multi turn loop, as long as the

loop small compared to

wavelength. For N circular

loop, S=Nπa2 and for square

coil N loops, with each side

length b, S=Nb2

Radiation patterns of loops with various circumferences


Antenna Types

Short Antennas

Hertzian Dipole

Example

Suppose a Hertzian dipole antenna is 1 cm long and is excited by a 10 mA

amplitude current source at 100 MHz. What is the maximum power density

radiated by this antenna at a 1 km distance? What is the antenna’s radiation

resistance?

Solution:

The max. power radiated is:

2 2 22 2 2

max 2 2 2 2 2 2

2 0.010 0.0101200.052

32 32 3 1000

o oI pWP

r m

l

The antenna radiation resistance : 2 2

2 2 0.0180 80 8.8

3radR m

l

8

6

3 10, 3 .

100 10 1

c x m sc f m

f x s


Antenna Types

Dipole Antennas A drawback to Hertzian dipole as a practical antenna is its small

radiation resistance. A longer will have higher radiation resistance,

becomes more efficient. It as an L long conductor conveniently

placed along the z axis with current distribution i (z,t).

The time averaged power radiated is:

where, the pattern function is given by:

Therefore, the normalized power function is:

And the max time averaged power density is then :

Assume sinusoidal current distribution on

each arm

For simplicity, assume phase term =0,

and make use of current distribution

term with magnetic field equation for a

Hertzian dipole.

rFr

Ir a P

2

2015

,

2

sin

2coscos

2cos

kLkL

F

max

F

FPn

max2

20

max

15

F

r

IP


Antenna Types

Dipole Antennas

HPBW means half plane BW or 3-dB BW

As length of antenna increases:

beam becomes narrower,

and directivity increases

3D and 2D amplitude patterns for a thin dipole

of l=1.25λ


Antenna Types

Dipole Antennas

Half wave Dipole Antennas

It is a smallest resonant dipole antenna.

It has a convenient radiation resistance.

kL/2 = π/2

With the F(θ) is 1, the maximum power density

is:

Therefore, the normalized power density is:

rr

Ir a P

2

2

2

20

sin

cos2

cos15

,

2

2015

,r

Ir

P

2

2

cos

cos2

cos

nP

The current distribution and

normalized radiation pattern for

a half wave dipole antenna.


Antenna Types

Dipole Antennas Half wave Dipole Antennas

The pattern solid angle is

with L = λ/2, then the directivity is:

The radiation resistance is given by:

Its radiation resistance much higher than of Hertzian

dipole

Radiates more efficiently.

Easier to construct an impedance matching network for this

antenna impedance.

It contains a reactive components, Xant, where for a λ/2 dipole

antenna it is equal to 42.5Ω . Therefore, total impedance by

neglecting Rdiss.

The current distribution and

normalized radiation pattern for

a half wave dipole antenna. 658.7 p

640.14

max

p

D

Its directivity is slightly higher than

the directivity of Hertzian dipole.

pradrad PrRIP max22

02

1 2.73

30pradR

5.422.73 jZant

For impedance matching, need to make reactance zero (in resonant

condition). So, it can be achieved by making the antenna slightly

shorter (reduced in length until reactance vanishes).


Antenna Types


Example

Find the efficiency and maximum power gain of a λ/2 dipole antenna constructed with AWG#20

(0.406 mm radius) copper wire operating at 1.0 GHz. Compare your result with a 3 mm length dipole

antenna (Hertzian Dipole) if the center of this antenna is driven with a 1.0 GHz sinusoidal current.

Solution

first find the skin depth of copper at 1.0 GHz,

The wire area over which current is conducted by:

m

fcu

6

7790

1009.2108.5104101

11

It is much smaller than the

wire radius

291033.52 maS cu

At 1 GHz, the wavelength is

0.3m and the λ/2 is 0.15m long


Antenna Types


Example

Find the efficiency and maximum power gain of a λ/2 dipole antenna constructed with AWG#20

(0.406 mm radius) copper wire operating at 1.0 GHz. Compare your result with a 3 mm length dipole

antenna (Hertzian Dipole) if the center of this antenna is driven with a 1.0 GHz sinusoidal current.

Solution

The ohmic resistance is then:

for Hertzian Dipole, the ohmic resistance of the small dipole is :

485.01

SRdiss

the radiation resistance for

half wave dipole is 73.2Ω,

99.0485.02.73

2.73

e 63.1640.199.0maxmax eDG

mS

Rdiss 7.91

mm

mRrad 79

3.0

10380

23

2 89.07.979

79

mm

me

34.15.189.0maxmax eDG

Thus, the half wave dipole is clearly more efficient

with a higher gain than the short dipole.


Antenna Types

Dipole Antennas Dipole Antennas

Example

A dipole of the length 2l = 3 cm and

diameter d = 2 mm is made of copper wire

(s = 5.7 107 S/m) for mobile

communications. If the operational

frequency is 1 GHz, Then

calculate its input impedance,

radiation resistance and radiation

efficiency;

if this antenna is also used as a

field probe at 100 MHz for EMC

applications, find its radiation

efficiency again, and express it in

dB.

Solution

Since the frequency is 1 GHz, the wavelength λ = 30 cm, 2l /λ =

0.1 and βl = 0.1π.

The Skin depth

It is much smaller than the radius of the wire. Thus, we need to

compute the loss resistance of the dipole:

Since Za has not taken the loss resistance into account, the more

accurate input impedance is

And then, the radiation efficiency is

𝑍𝑎 ≈ 20 𝛽𝑙 2 −𝑗120(ln(

2𝑙𝑑− 1)

(𝛽𝑙)= 1.93 − 𝑗652Ω

𝛿 ≈1

𝜋𝑓𝜇𝜎 =

1

𝜋×109×4𝜋×10−7×5.7×107= 2.1 × 10−6m

𝑅𝐿 ≈2𝑙 𝑓𝜇

𝑑 𝜋𝜎= 0.04Ω

𝑍𝑎 = 1.97 − 𝑗652Ω

𝑒 =1.93

1.93 + 0.04= 97.97%


Antenna Types

Dipole Antennas Dipole Antennas

Example

A dipole of the length 2l = 3 cm and

diameter d = 2 mm is made of copper wire

(s = 5.7 107 S/m) for mobile

communications. If the operational

frequency is 1 GHz, Then

calculate its input impedance,

radiation resistance and radiation

efficiency;

if this antenna is also used as a

field probe at 100 MHz for EMC

applications, find its radiation

efficiency again, and express it in

dB.

Solution

The frequency is now 100 MHz and the wavelength λ = 300 cm,

thus 2l/λ = 0.01 and βl = 0.01π. This is an electrically small

antenna; we can use the same approach as in previous to

obtain:

𝑍𝑎 ≈ 20 𝛽𝑙 2 −𝑗120(ln

2𝑙𝑑− 1

(𝛽𝑙)= 0.0197 − 𝑗6524.3Ω

𝛿 ≈1

𝜋𝑓𝜇𝜎 = 6.67 × 10−6m

𝑅𝐿 ≈2𝑙 𝑓𝜇

𝑑 𝜋𝜎= 0.0126Ω


Antenna Types

Monopole Antennas A monopole antenna is excited by a current source at

its base.

By image theory, the current in the image will be

the same with the current in actual monopole. The

pair of monopole resembles a dipole antenna.

A monopole antenna placed over a conductive plane

and half the length of a corresponding dipole

antenna will have identical field patterns in the

upper half plane.

Consider the construction of half wave

dipole for an AM radio station

broadcasting at 1 MHz. At this f, the

wavelength is 300m long and the half

wave dipole antenna must be 150m tall.

We can cut this in half, by employing

image theory to build a quarter wave monopole antenna that is only 75m

tall!!

For the upper half plane (00<

θ<900), the time averaged power,

max power density and normalized

power density for the quarter wave

monopole is the same with half

wave dipole. But the pattern solid

angle is different.

dPnp 28.34

829.3 max

p

p D

6.3630

pradR

25.216.36 jZant Normalized power density is

zero for (900< θ<1800)


Antenna Types-Summary 13


Antenna Types

Antenna Arrays The motivation is to

achieve desired high gain or radiation pattern,

the ability to provide an electrically scanned beam.

It consists of more than one antenna element

They are strategically placed in space to form an array with desired characteristics which are achieved by varying the feed (amplitude and phase) and relative position of each radiating element;

The main drawbacks are

the complexity of the feeding network required and

the bandwidth limitation (mainly due to the feeding

network)

where An is the amplitude, n is the relative

phase, Ee is the radiated field of the antenna

element, and AF is called array factor.

Thus the radiation pattern of an array is the product of the pattern of individual element antenna with the (isotropic source) array pattern.


Antenna Types

Antenna Arrays A properly spaced collection of antennas, can have

significant variation in φ leading to dramatic

improvements in directivity. It can be designed to give a particular shape of

radiating pattern.

Control of the phase and current driving each array

element along with spacing of array elements can

provide beam steering capability.

For simplification:

All antenna elements are identical

The current amplitude is the same feeding each element.

The radiation pattern lies only in xy plane, θ=π/2

The radiation pattern then can be controlled by: controlling the spacing between elements or

controlling the phase of current driving for each element

consider a pair of dipole antennas driven

in phase current source and separated by

λ/2 on the x axis.

Assume each antenna radiates

independently, at far field point P, the

fields from 2 antennas will be 180 out-of-

phase, owing to extra λ/2 distance travel

by the wave from the farthest antenna

fields cancel in this direction.

At point Q, the fields in phase and adds.

The E field is then twice from single

dipole, fourfold increase in power

broadside array max radiation is

directed broadside to axis of elements.


Antenna Types

Antenna Arrays A properly spaced collection of antennas, can have

significant variation in φ leading to dramatic

improvements in directivity. It can be designed to give a particular shape of

radiating pattern.

Control of the phase and current driving each array

element along with spacing of array elements can

provide beam steering capability.

For simplification:

All antenna elements are identical

The current amplitude is the same feeding each element.

The radiation pattern lies only in xy plane, θ=π/2

The radiation pattern then can be controlled by: controlling the spacing between elements or

controlling the phase of current driving for each element

Modify with driving the pair of dipoles

with current sources 180 out of phase.

Then along x axis will be in phase and

along y axis will be out of phase, as

shown by the resulting beam pattern

endfire array max radiation is

directed at the ends of axis containing

array elements.


Antenna Types

Antenna Arrays Pair of Hertzian Dipoles

The find radiated power

Consider a pair of z oriented Hertzian

dipole, with distance d, where the total

field is the vector sum of the fields for

both dipoles and the magnitude of

currents the same but a phase shift

between them.

aaEEE2

20

1

1020100

44

21

R

keIj

R

keIj

jkRs

jkRs

SStotS

jss eIIII 0201

Assumption,

2121 RrR

cos2

cos2

21d

rRd

rR

r

rtotSSS

dk

R

kI

Er

a

aHEP

2cos

2cos4

32

2

1Re

2

1,

2,

2

22

22200

200

*

rarrayunit FFr a P

,

2,

Unit factor, Funit is the max time

averaged power density for an

individual antenna element at

θ=π/2

22

22200

32 R

kIFunit

The array factor, Farray is

2cos4 2

arrayF coskd


Antenna Types


Example

The λ/2 long antennas are driven in phase and are λ/2 apart. Find:

The far field radiation pattern for a pair of half wave dipole shown.

The maximum power density 1 km away from the array if each antenna is driven by a 1mA

amplitude current source at 100 MHz.

Solution:

At 100 MHz, λ = 3m, so that 1 km away is definitely in far field. For a half wave dipole.

At θ = π/2

2

sin

2coscos

2cos

kLkL

F

rFr

Ir a P

2

2015

,

rrr

IrPr a aP

2

2

2

20

sin

cos2

cos15

,,2

2015

r

IFunit

The array factor, with d = λ/2 and =0 (due to

antennas are driven at the same phase) :

cos

2cos4 2

arrayF

cos

2cos

60,

2, 2

2

20

r

IFFr rarrayunit a P


Antenna Types


Example

The λ/2 long antennas are driven in phase and are λ/2 apart. Find:

The far field radiation pattern for a pair of half wave dipole shown.

The maximum power density 1 km away from the array if each antenna is driven by a 1mA

amplitude current source at 100 MHz.

Solution: The normalized power function is:

This can be plotted as :

2

sin

2coscos

2cos

kLkL

F

rFr

Ir a P

2

2015

,

cos2

cos

,2

,

,2

,

,2

2

max

rP

rP

nP

The maximum radiated power density at 1000m is :

22

23

2

20

max 191000

106060

m

pW

r

IP


Antenna Types

Antenna Arrays N-Element Linear Arrays

The procedure of two-element array can be

extended for an arbitrary number of array

elements, by simplifying assumptions: The array is linear (antenna elements are evenly

spaced, d along a line).

The array is uniform (each antenna element driven

by same magnitude current source, constant phase

difference, between adjacent elements).

10

2030201 ,....,, Nj

sNj

sj

ss eIIeIIeIIII

rNjjj

jkR

totS eeeR

keIj aE

12000 ...1

4

coskd

2sin

2sin

2

2 N

Farray

2

maxNFarray

The normalized power pattern for these elements is :

2sin

2sin

1

2

2

2max

N

NF

FP

array

arrayn


Antenna Types


Example

Five antenna elements spaced λ/4 apart with progressive phase steps 300. The antennas

are assumed to be linear array of z oriented dipoles on the x axis. Find:

The normalized radiation pattern in xy plane

The plot of the radiation pattern.

Solution:

To find the array factor, first need to find psi, Ψ:

Inserting this ratio to array factor,

6cos

218030cos

4

2

cos

kd

12cos

4sin

12

5cos

4

5sin

2sin

2sin

2

2

2

2

N

Farray 252

max NFarray


Antenna Types


Example

Five antenna elements spaced λ/4 apart with progressive phase steps 300. The antennas

are assumed to be linear array of z oriented dipoles on the x axis. Find:

The normalized radiation pattern in xy plane

The plot of the radiation pattern.

Solution:

The normalized radiation pattern is :

12cos

4sin

12

5cos

4

5sin

25

1

2

2

nP


Antenna Types

Helical Antennas Modes of Operation

Normal (Broadside) (the maximum radiation of the

array is directed normal to its axis ( = 00))

Axial (End-fire) – Most practical (if the maximum

radiation is directed along the axis of the array ( =

900)) Circular polarization can be achieved over a wider

bandwidth (usually 2:1)

More efficient

Important Parameters

C

S

D

S 11 tantan

α = 0o (flat loop)

α = 90o (linear wire)

220 CSL = single turn

220 CSNNLLn

D = diameter of helix

S = spacing between turns

C = circumference of helix

L = length of one turn

AR = axial ratio

= pitch angle


Antenna Types


Normal (Broadside) (NL0 << λ)

Dipole:

Loop:

Δφ = j = 90o

sin4

0

r

SeIkjE

rjko

o

sin

4

2 0

22

r

eIDkE

rjko

o

20

2

24

D

S

Dk

S

E

EAR

o

For this special case,

1

22

0 D

SAR

SCD 02

SC 02 000222

tan

DS

S

S

D

S

The radiated field is circularly polarized in all

directions other than θ = 00





AR = axial ratio

= pitch angle


Antenna Types


Axial (End-fire) – Most practical

oo 1412 00

3

4

4

3 C

3N

(C ≈ λ0 near optimum)

0

140

CR

Accuracy (± 20%) NSC

HPBW2

3

052

NSCFNBW

23

0115(deg)

3

0

2

15

SCNDo

N

NAR

2

12





AR = axial ratio

= pitch angle

HPBW = half power beamwidth

FNBW = first null beamwidth

Do = Directivity

R = Radiation Resistance


Antenna Types


Axial (End-fire) – Most practical

Example:

Design a circularly polarised helix antenna of an end-fire radiation pattern with a

directivity of 13 dBi. Find out its radiation resistance, HPBW, AR and radiation pattern.


Antenna Types

Helical Antennas


Antenna Types

Helical Antennas Feed Design

The nominal impedances of ordinary helices is 100-200 Ω. However, for many

practical TLs, it is desired to make it 50 Ω, and can be accomplished in many

ways.

One way is to properly design the first ¼ turn of the helix next to the feed.

This is done by flattening the wire in the form of a strip width, w, and nearly

touching the ground plane which is covered by a dielectric slab of height (h):

The helix transitions from the strip to the regular wire gradually during the ¼ to

½ turns.

2377

00

Z

wh

where

w – width of the strip starting at feed

εr – dielectric constant of the dielectric slab

Z0 – characteristic impedance of the input TL


Antenna Types

Yagi Uda Array Antennas The driven element (feeder) is the very heart of the

antenna. It determines the polarisation and centre

frequency. For a dipole, the recommended length is

about 0.47 to ensuring a good input impedance to a

50 Ω feed line.

The reflector is longer than the feeder to force the

radiated energy towards the front. The optimum

spacing between the reflector and the feeder is

between 0.15 to 0.25 wavelengths.

The directors are usually 10 to 20% shorter than the

feeder and appear to direct the radiation towards the

front. The director to director spacing is typically 0.25

to 0.35 wavelengths,

The number of directors determines the maximum

achievable directivity and gain.

Typical Sizes

Director lengths: (0.4 – 0.45) λ

Feeder length: (0.47 – 0.49) λ

Reflector length: (0.5 – 0.525) λ

Reflector – feeder spacing:

(0.2 – 0.25) λ

Director spacing: (0.3 – 0.4) λ


Antenna Types

Yagi Uda Array Antennas Applications

Amateur radio

TV antenna (usually single or a few channels)

Frequency ranges HF (3-30 MHz)

VHF (30-300 MHz)

UHF (300-3000 MHz)

Design sizes and Directivities

D0 (λ/2) = 1.67 = 2.15 dB 0.001 ≤ d/λ0 ≤ 0.04

0.002 ≤ D/λ0 ≤ 0.04

Typical Sizes





(0.2 – 0.25) λ



Antenna Types

Yagi Uda Array Antennas Design Procedure

Specify: Center frequency (fo)

Directivity

do/λ (diameter of parasitic elements)

D/ λ (diameter of boom)

Find: Lengths of directors and reflector(s)

Spacing of directors and reflector(s)

Typical Sizes





(0.2 – 0.25) λ



Antenna Types

Yagi Uda Array Antennas Example

Given: Center frequency (fo) = 50.1 MHz,

Directivity (relative to λ/2) = 9.2 dB, d =

2.54 cm, D = 5.1 cm. Using these

parameters, design a Yagi Uda Array

antenna by finding the element spacing,

lengths and total array length.

Solution

λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ

= 8.52 x 10-3

Step 1: Find L and N from Table 10.6 L = 0.8 λ (from the given

directivity of 9.2 dB)

N = 5 (3 directors, 1 reflector, 1 feeder)


Antenna Types








Solution

λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ

= 8.52 x 10-3

Step 1:

For d/λ = 0.0085:

424.0

428.0

0482.0

''4

''5

''3

''1

l

ll

l

Marked by dot ()

Step 2: From Fig., draw vertical line

through d/λ = 0.00424

442.0

0485.0

''5

''3

''1

ll

lMarked in the Fig. by

(x)


Antenna Types








Solution

λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ

= 8.52 x 10-3

Step 3:

Measure Δl between l3’’, l5’’ and l4’’. Transpose

to l3’, l5’ to find l4’. Read l4’.

438.0'4 l


Antenna Types








Solution

λ = 5.988m = 598.8 cm; d/λ = 0.00424; D/λ

= 8.52 x 10-3

Step 4:

From the Fig., for D/λ = 0.00852

Δl = 0.005 λ

443.0005.0438.0005.0

447.0005.0442.0005.0

490.0005.0485.0005.0

'44

'353

'11

ll

lll

ll

Therefore,

Total array length: 0.8 λ

The spacing between directors: 0.2 λ

The reflector spacing: 0.2 λ

The actual elements length:

L3 = L5 : 0.447λ

L4 : 0.443λ

L1 : 0.490λ


Antenna Types

Log-Periodic Antennas The antenna is divided into the so called active region

and inactive regions.

The role of a specific dipole element is linked to the

operating frequency: if its length, L, is around half of the

wavelength, it is an active dipole and within the active

region; Otherwise it is in an inactive region and acts as a

director or reflector as in Yagi-Uda antenna

The driven element shifts with the frequency – this is why

this antenna can offer a much wider bandwidth than the

Yagi-Uda. A travelling wave can also be formed in the

antenna.

The highest frequency is basically determined by the

shortest dipole length while the lowest frequency is

determined by the longest dipole length (L1).

Other parameters such as the directivity or the length

of the antenna) are required to produce an optimised

design.

In practice, the most likely scenario is that the

frequency range is given from fmin to fmax, the

following equations may be employed for

design


Antenna Types

Log-Periodic Antennas Example Design a log-periodic dipole antenna to cover all UHF

TV channels, which is from 470 MHz for channel 14 to

890 MHz for channel 83. Each channel has a

bandwidth of 6 MHz. The desired directivity is 8 dBi

Solution The given three parameters are: fmin = 470MHz, fmax = 890MHz, and

D = 8 dBi.

From the table, we can see that, for the optimum design,

the scaling factor τ = 0.865,

the spacing factor σ = 0.157, and

the apex angle α = 12.13.

That means at least six elements are required. To be on the safe side we

should use seven or even eight elements to be sure the desired directivity

will be achieved.

If N = 8, we can afford to start from a lower

frequency, say 400 MHz, thus

Ln = the length of element n, and n = 1, 2, . . , N;

sn = the spacing between elements n and (n + 1);

dn = the diameter of element n;

gn = the gap between the poles of element n.

Zo= Characteristic Impedance


Antenna Types

Log-Periodic Antennas Example Design a log-periodic dipole antenna to cover all UHF

TV channels, which is from 470 MHz for channel 14 to

890 MHz for channel 83. Each channel has a

bandwidth of 6 MHz. The desired directivity is 8 dBi

Solution The given three parameters are: fmin = 470MHz, fmax = 890MHz, and

D = 8 dBi.

The spacing between the elements is as follows

The total length of the antennas is

Ln = the length of element n, and n = 1, 2, . . , N;

sn = the spacing between elements n and (n + 1);

dn = the diameter of element n;

gn = the gap between the poles of element n.

Zo= Characteristic Impedance

Aperture-Type Antennas They are often used for higher frequency applications (> 1GHz)

than wire-type antennas.

The relation between the aperture E field to the radiated field is

as follows:

Then , the directivity will be


Antenna Types


Antenna Types

Aperture-Type Antennas Example An open waveguide aperture of dimensions a long x

and b along y located in the z = 0 plane. The field in

the aperture is TE10 mode and given by

Find

the radiated far field

the directivity.

Solution The far field can be obtained using

In the φ = 0 plane (H-plane), i.e. the xz plane, we

have kx = β sin θ, ky = 0, and

In the φ = π/2 plane (the E-plane), i.e. the yz plane, we have kx = 0, ky = β sin θ, and


Antenna Types

Aperture-Type Antennas Example An open waveguide aperture of dimensions a long x

and b along y located in the z = 0 plane. The field in

the aperture is TE10 mode and given by

Find

the radiated far field

the directivity.

Solution

Typical radiation patterns of an open waveguide in

(a) the H-plane; (b) the E-plane

Aperture-Type Antennas Horn Antennas

Horn antennas are the simplest and one of the

most widely used microwave antennas – the

antenna is nicely integrated with the feed line

(waveguide) and the performance can be easily

controlled. They are mainly used for standard antenna gain

and field measurements, feed element for reflector

antennas, and microwave communications.

Design In the H-Plane, the dimensions are linked by

In the E-Plane, the dimensions are linked by


Antenna Types

To make this horn, we must have

The directivity:

The design equation

For the optimum design, use a first guess approximation


Example:

Design a standard gain horn with a directivity of 20 dBi at

10 GHz. WR-90 waveguide will be used to feed the horn.

Solution:

The directivity is D = 20 dBi = 100, wavelength λ = 30 mm and the

dimensions of the waveguide are a = 22.86 mm and b = 10.16 mm.

Step 1: Compute the dimension A from the design Equation

• The parameters for the optimum horns are:

• The design equation

• Plot the function


Antenna Types

Function Y(A) vs. A

around A = 135 (mm). As seen in the Figure, the

actual solution is readily found as


Example:

Design a standard gain horn with a directivity of 20 dBi at

10 GHz. WR-90 waveguide will be used to feed the horn.

Solution: Step 2: Find B

B = 104.82 mm

Step 3: Find the remaining dimensions

R1 = 199.41mm;

R2 =183.14 mm;

lH = 210.36 mm and RH = 165.38 mm;

lE = 190.49 mm and RE = 165.38 mm.

The half-power beamwidths in the two principal planes for this

antenna are


Antenna Types

Step 4: Check if RH and RE are the same. If not,

it means that the solution of A in Step 1 is not accurate enough.

From the results in Step 3, we can see that RH and RE are identical, thus the design is very good.

However, if we used the guessing value A = 0.45λ √D = 135 (mm) as the solution, it would give RH =

168.21 mm and RE = 162.73 mm. They are

obviously different, which means that the design

needs to be revised.

Aperture-Type Antennas Reflector Antennas

Reflector antennas can offer much higher gains

than horn antennas and are easy to design and

construct.

The most widely used antennas for high frequency

and high gain applications in radio astronomy,

radar, microwave and millimetre wave

communications, and satellite tracking and

communications.

The most popular shape is the paraboloid –

because of its excellent ability to produce a pencil

beam (high gain) with low sidelobes and good

cross-polarisation characteristics

The reflector design problem consists primarily of matching the feed antenna pattern to the reflector. The usual goal is to have the feed pattern about 10 dB down in the direction of the rim, that is the edge taper = (the field at the edge)/(the field at the

centre) ≈10 dB.


Antenna Types Paraboloidal and Cassegrain reflector antennas

Directivity:

Half-power beamwidth

Let us assume that the diameter of the reflector is 2a and the focal length is F. Any point on this paraboloid

must satisfy the following condition:

where the subtended/angular aperture angle θ0 (also

known as the edge angle) is determined by the

reflector diameter and the focal length:

If g(θ) is the power radiation pattern of the feed located at

the focus – it is circularly symmetrical (not a function of φ).

The aperture efficiency is


The aperture efficiency is generally viewed as

the product of the

Spillover efficiency: the fraction of the total

power intercepted and collimated by the reflector;

it reduces the gain and increases the side-lobe

levels.

Taper efficiency: the uniformity of the amplitude

distribution of the feed pattern over the surface of

the reflector.

Phase efficiency: the phase uniformity of the field

over the aperture plane; it affects the gain and side

lobes.

Polarization efficiency: the polarization uniformity

of the field over the aperture plane.

Blockage efficiency: by the feed; it reduces gain

and increases side-lobe levels. The support

structure can also contribute to the blockage.

Random error efficiency: over the reflector

surface.


Antenna Types Paraboloidal and Cassegrain reflector antennas

Directivity:

Half-power beamwidth

Let us assume that the diameter of the reflector is 2a and the focal length is F. Any point on this paraboloid

must satisfy the following condition:

where the subtended/angular aperture angle θ0 (also

known as the edge angle) is determined by the

reflector diameter and the focal length:

If g(θ) is the power radiation pattern of the feed located at

the focus – it is circularly symmetrical (not a function of φ).

The aperture efficiency is


Example:

A circular parabolic reflector has F/2a = 0.5. The field

pattern of the feed antenna is E(θ) = cos θ, θ < π/2. Find the

edge taper, spillover efficiency and aperture efficiency.

Solution:

The edge angle is

The edge taper is

, but

The Spillover efficiency is


Antenna Types

the spillover loss is about −1.08 dB.

The aperture efficiency

Antenna Equivalent circuit

Slots Antennas They are very low-profile and can be

conformed to basically any configuration,

thus they have found many applications, for

example, on aircraft and missiles.

The radiated field by the slot is the same as

the field radiated by its equivalent surface

electric current and magnetic current which

were given by

where E and H are the electric and magnetic fields within

the slot and is the unit vector normal to the slot surface

S.

For a half-wavelength slot, its equivalent

electric surface current JS = 0, the remaining

source at the slot is its equivalent magnetic

current MS (it would be 2MS if the conducting

ground plane were removed using the imaging

theory).


Antenna Types

Slot waveguide antenna array: widely used for radar

n

Antenna Equivalent circuit

Slots Antennas From Babinet’s Principle, the input

impedance of the slot antenna is given as


Antenna Types

Slot waveguide antenna array: widely used for radar

Microstrip/Patch Antennas Ease of construction and integration, relatively low

cost, compact low profile configuration and good

flexibility

Typical applications for 1 - 20 GHz

To be a resonant antenna, the length L should be

around half of the wavelength (the antenna can be

considered as a l/2 transmission line resonant cavity

with two open ends where the fringing fields from the

patch to the ground are exposed to the upper half

space (z > 0) and are responsible for the radiation.)

The radiation mechanism is the same as the slot

line (there are two radiating slots on a patch antenna).

As a resonant cavity, there are many possible

modes (as waveguides), thus a patch antenna is

multi-mode and may have many resonant frequencies


Antenna Types

Microstrip/Patch Antennas


Antenna Types

Directivity

Input impedance

Bandwidth

Optimised width:

Resonant freq.:

Length:

Microstrip/Patch Antennas Example

RT/Duroid 5880 substrate ( and d = 1.588 mm) is to be used to make a

resonant rectangular patch antenna of linear polarisation;

Design such an antenna to work at 2.45 GHz for Bluetooth

applications;

Estimate its directivity;

If it is to be connected to a 50 ohms microstrip using the same

PCB board, design the feed to this antenna;

Find the fractional bandwidth for VSWR < 2.


Antenna Types




Design such an antenna to work at 2.45 GHz for Bluetooth

applications;

Solution:

Step #1: Find W Step #2: Calculate the effective permittivity

Step #3: Compute the extension of the length Step #4: Determine the length L


Antenna Types




Estimate its directivity;

Since the wavelength at 2.45 GHz is 122.45 mm > W, from

It is about 6.6 or 8.2 dBi.


Antenna Types





PCB board, design the feed to this antenna;

Step #1: Input Impedance

Step #2: Characteristics Impedance

Step #3: Transition line width

Step #4: Transition line length


Antenna Types

Does not match well

with a 50 standard

microstrip and therefore

a quarter-wavelength

transformer is used to

connect them





PCB board, design the feed to this antenna; Step #5: The microstrip width

Find the fractional bandwidth for VSWR < 2.


Antenna Types

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