Chapter 7 Antennascourses.egr.uh.edu/ECE/ECE3317-02053/docs/3317Chap7Lect0914student.pdfWhat is an...

Chapter 7 Antennas

ECE 3317Dr. Stuart Long

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7-1

What is an “antenna”?

An antenna is a structure that provides electrical current that thenproduces an electromagnetic wave. They can also be thought ofas transducers that transfer electromagnetic energy between atransmission line and free space. They are the usual sources ofEM waves.

In this chapter we will only focus on three types of antennas:

1. Infinitesimal Current Sources.

2. Linear Wire Antennas (Dipoles).

3. Uniform Linear Arrays.

7-2

Examples of antennas

http://www.cfas.org/NASA_Space_Place/

www-nutev.phyast.pitt.edu

http://www.ece.utah.edu/~cfurse/APS.html

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7-3

Vector Potential

The vector potential is defined somewhat indirectly as:

vector potenti

N

al

:ot

e

= ∇×

∇ ⋅ =

A

A

B A

B ( )

( )is uniquely defined. If we let

then

and give same which means that w

not

e need to specify

0

0

ψψ

′

∇ ⋅ ∇× ≡

′ = +′× = × +

′× = × +

B

A A B

A

A AA A

A A

∇

∇ ∇ ∇ ∇

∇

also.

⋅ A ∇

7.1

7-4

Scalar Potential

( )

From Ma

xwell's equations

reca l l:

j

j

ω

ω

= ∇ ×

∇ × = −

∇ × = − ∇ ×

B A

E B

E A

( )

( )but:

So let

0

Φ r

0

o

ψ

j

j

ω

ω

∇ × ∇ =

∇ × + =

+ = −

E A

E A ∇

Φwhere is the scalar poten ial

Φt

jω= − −E A ∇

7.2

7.3

7-5

Lorentz Condition

Thus far we have defined

vector potential

calar potential

To uniquely define we also have to specify

Φ Φ s

jω

∇ ⋅

= ∇×

= − − ∇

A

A

A

B A

E A

Φ Lorentz condition (gauge) 0 jω µ ε∇ ⋅ + =A 7.4

7.1

7.3

7-6

2 2

2 2 v

Using vector identities and the Lorentz condition we obtain:

Differential equation for

Di eΦ ffΦ

k

k

µ

ρε

∇ + =

∇ + =

A

A A - J

- rential equation for Φ

Vector and Scalar Potential

7.6

7.5

7-7

The solution to the previous equations are given by:

Solution(

)( ) 4

jk

V

e dVµπ

′− −

′

′′=

′−∫∫∫

JAr rrr

r r

v

for

Solution for

Φ

1 ( )

( )

4

jk

V

e dVρπε

′− −

′

′′=

′− Φ∫∫∫

A

r rrrr r

field point source point k ω µε

′→

→

= rr

7.7

7.8

Vector and Scalar Potential

7-8

Spherical Coordinates

θr

z

x

φ

θ̂

r̂

φ̂

Note: unit vectors depend on the location of the vector.

y

(fig.7.2)

( , , )r θ φ

7-9

y

′−r r

′r

r

field pointsource point→′

→

r

r

z

x

Spherical Coordinates( , , )r θ φ

7-10

Infinitesimal Hertzian Dipole

I

z∆

2a → ←

y

′r

r

z

x

Electrically short

Physically th in a

z

z

λ∆ <<

∆<<

7-11

Fields Due to an Infinitesimal Hertzian Dipole

I

z∆

2a → ←

x

y

z

′r

r

dV’

dV’r’

r’

r’r’

7-12

To find :

use to obtain ˆˆˆ cos sin

we first in spherical coordinates

I ˆ4

jkrz er

θ θ

µ

µπ

−

= −

= = ∇ ×

∆=

B H A

A

A

z r

z

θ

we then evaluate with vector in spherical coordinates

as seen i (eqn. 2n 7.1 )

I ˆˆ cos sin

4

jkrzer

µ θ θπ

−∆ = −

∇×

A

A A

r θ

I

z∆

2a → ←

x

y

z

′r

r

Fields Due to an Infinitesimal Hertzian Dipole

7-13

( ) 1sinr

Ar φθ θ

∂∇ × =

∂A ( )sin Aθθ

φ∂

−∂

( ) 1 1sin

rArθ θ φ

∂∇ × =

∂A rA

r φ∂

−∂ ( )

( ) ( )1

1 I 1ˆ 1 sin4

r

jkr

ArAr r

jk zer jkr

θφ θ

θµ π

−

∂ ∂ ∇ × = − ∂ ∂

∆= ∇ × = +

A

H A φ

7.12

with vector in spherical

co

ordi

nat

s

e∇ × A A

7.13

I

z∆

2a → ←

x

y

z

′r

r

Fields Due to an Infinitesimal Hertzian Dipole

7-14

( ) ( )

2

2 2

,1 1

r r

The electric field outside the dipole is given by :

: has only a component with two terNote ms ˆ

1

I 1 1 1 1 ˆˆ 2cos 1 sin4

jkr

j

jk zer jkr jkrjkr jkr

ωε

µθ θ

ε π

−

= ∇ ×

∆= + + + +

H

E H

E r

φ

θ

2 3 2 3

1 1 1 1 1

r r,

r,

r,

rhas an component and ˆˆ a com ponent

E r θ

7.14

I

z∆

2a → ←

x

y

z

′r

r

Fields Due to an Infinitesimal Hertzian Dipole

7-15

7.17a

7.17b

I

z∆

2a → ←

x

y

z

′r

r

Radiation Field (Far Field)

7-16

: field , field , and the direction of propagation are all perpendicular to each other.

and

field , and fiel

d are

sin proportional t

Note

o

vk

r

µ ωηε

θ

= = =

E H

EH

E H

The wave propagates in the radial direction.

The surfaces of constant phase are spherical.

I

z∆

2a → ←

x

y

z

′r

r

Radiation Field (Far Field)

7-17

Hertzian Dipole (Ideal Current Source)

I(z)

zz∆

I0

7-18

Real Electrically Short Antenna (I→0 at ends)

I(z)

z∆ z

I0

7-19

Practical Approximation to Hertzian Dipole

I(z)

z∆

z∆

z∆

Capacitive Plate Antenna

7-20

Radiation Patterns

|E| vs. angle θ at a constant r

|E|

θ0

(rectangular plot)

ππ−

0

IE sin4

|r

|k zθ

µ θε π

∆=

7-21

x

Field Patterns

z

θ

IE sin4

|r

|k zθ

µ θε π

∆=

(fig.7.4)

θE

(polar plot)

7-22

Power Pattern

r

2r

S versus angle plot

Not : S ne si

θ

θ

0 dB

θ -30°30°

60°

120°

150° -150°

-120°

-60°

(polar plot)

7-23

3 dB Beamwidth

Angle in degrees between points 3 dB down from

maximum.

HPBW

7-24

max

max

Electric Field

Power

E20log 20log 0.707 3 dBE

P 1 10log 10lo

g 3 dBP 2

= = −

= = −

max0.707← Emax0.5 →P

HPBW

3 dB Beamwidth

7-25

2

Surface Element sindS r d dθ θ φ=

Coordinate System

rdθ

sinr θ

sinr dθ φ

dr

dφ

7-26

Directive Gain

( )2

rad

22

rad0 0

Directive Gain

Total Radiated P

ower

4,

P

P

sin

r

r

S rD

S r d dπ π

φ θ

πθ φ

θ θ φ= =

=

= ∫ ∫

7.19

7.19

rdθ

sinr θ

sinr dθ φ

dr

dφ

7-27

Directivity of Hertzian Dipole

22

2

rad

2

0Directivity (maxi1.5 mun directive gain s i in

I z sin2 4

4 I z P3 4

3 ( ,

) si

n2

r

D

kSr

k

D

η θπ

π ηπ

θ φ θ

= =

∆ =

∆ =

=

= 90 direction)θ

7.19

7-28

Linear Antenna

z

z∆

0z =

2z h=

1z h= −

7-29

2

1

1 2

cos

If total length is comparable to a wavele ngth

( ) I I( )

ˆ sin U( )4

ˆ

U( ) Ι ( z)

jkr

h jkzh

h h z

ejkr

e dzθ

λ

µ θ θε π

η

θ

−

−

+ ≈ ⇒ =

=

=

= ∫

E

EH

θ

φ

z

z∆

0z =

2z h=

1z h= −

7.23a

7.23b

7.24

Linear Antenna

7-30

z

z∆

0z =

2z h=

1z h= −

Same field as Hertzian dipole with ½ current

Electrically short dipole

Example

( )01 2

ˆ sin4 2

jkr Ijke h hr

µθ θε π

− = + E

I(z)

7-31

Example

Half-Wave Dipole

1 2

0

0 2

h h h4

I( ) cos

cos cos2U( ) 2

sin

z I kz

Ik

λ

π θθ

θ

= = =

=

=

7.25

7.26

0II(z)

z∆

4z λ

=

4z λ

= −

7-32

Example

-9 -3-6

0 dB

infin. dipole 2λ

θ

Note: This is a normalized plot of the radiated fields. Equal

driving currents result in much higher fields for dipoles 2λ

-30°30°

60°

120°

150° -150°

-120°

-60°

7-33

Example 7-34

Monopole Over a Ground Plane

Ground plane

7-35

Monopole Over a Ground Plane

z

hz∆

2h

z

monopole dipole

same radiation pattern in upper half plane

same current at driving point

the voltage1 1Z Z2 2

•

•

⇒ =

7-36

Total of NElements

z

y

x d← →

rγθ

φ

Uniform Linear Array

7-37

1) Identical radiators oriented in direction and centers along the -axis.

2) All elements are equally spaced a distance .

3) All elements are driven by currents with equal magnitude and with

ˆˆ

d

p

z y-

20 1 2

rogressive phase shifts .

I I( ) I I( ) ; ; I ( I )j j

ψ

z z e z eψ ψ= = =

z

y

x d← →

rγθ

φ

Uniform Linear Array

7-38

( ) ( )

( )

The total field of the antenna array can be found by adding the fields from each antenna (principle of superposition).

Total field of an arrayF ,

.

t e

e

θ θ φ

θ

=

⇒

E

E E E

E

( )

Element factor field due to a single element located at origin

Array factor.

F ,

θ φ ⇒

7.35

z

y

x d← →

rγθ

φ

Uniform Linear Array

7-39

total number of elements

phase shift in driving current between adjacent elements

spacing between elements

cos sin

N

d

ψ

γ θ

=

=

=

= sinφ

( )( cos )

( cos ) Array Factor 1F , 1

jN kd

j kd

ee

ψ γ

ψ γθ φ+

+

−=

−7.36

z

y

x d← →

rγθ

φ

Uniform Linear Array

7-40

7.37

z

y

x d← →

rγθ

φ

Uniform Linear Array

( )

( )

( )

Array Factor

not

1ing that

Arr

( cos )1F , ( cos

=2 sin 2

)1

cossin2F , ay Factor M

cossiagnitu de

2

n

j

jN kdej kde

kdN

kd

x xe

ψ γθ φ ψ γ

γ ψ

θ φγ ψ

+−=

+−

+ =

+

−

total number of elements

phase shift in driving current between adjacent elements

spacing between elements

cos sin

N

d

ψ

γ θ

=

=

=

= sinφ

7-41

2-Element Array

( ) ( )

with

in pl

2

x-yane ( plane)

sin sin sinF , 2cos

2sin

2

sinF , 2cos

2

cos

co

s2

2

N

kd

kd

kdkd

θ

θ φ ψθ φ

π

φ ψφ

γ ψγ ψ

π

=

=

+= =

+=

+ +

( )

cossin

2F ,cos

sin2

kdN

kd

γ ψ

θ φγ ψ

+

=+

z

y

x d← →

rγθ

φ

7-42

2-Element Array

special case in plane ( plane) with phase shift x-y 0

sinF , 2cos

2

no

2

2

kd

θ ψ

π φφ

π= =

=

maxima always to line of array for Broadsid =0 eψ ⇒⊥

z

y

x d← →

rγθ

φ

sinF , 2cos

22kd φ ψ

φπ +

=

7-43

F 2cos sin2

20

2

2 =2

d

kd

πφ

πθ

ψλ

π λ πλ

=

=

=

=

=

x

maxima always to line of array for Broadsid =0 eψ ⇒⊥

z

y

x d← →

rγθ

φ

x

y

2-Element Array || Example

φ

7-44

maxima always to line of array for Broadsid =0 eψ ⇒⊥

z

y

x d← →

rγθ

φ

x

y

F 2cos sin4

20

4

2 =4 2

d

kd

πφ

πθ

ψλ

π λ πλ

=

=

=

=

=

φ

2-Element Array || Example

7-45

maxima always to line of array for Broadsid =0 eψ ⇒⊥

z

y

x d← →

rγθ

φ

x

y

( )F 2cos sin

20

2 =2

d

kd

π φ

πθ

ψλ

π λ πλ

=

=

=

=

=

2-Element Array || Example

φ

7-46

sin2 2F 2cos

2

2

2

4

2 =4 2

d

kd

π πφ

πθ

πψ

λ

π λ πλ

=

=

=

=

=

+

z

y

x d← →

rγθ

φ

x

y

2-Element Array || Example

Endfire

φ

(with phase shift)

7-47

z

y

x d← →

rγθ

φ

x

y

sinF 2cos

2

2

2

2 =2

d

kd

π π φ

πθ

ψ πλ

π λ πλ

=

=

=

=

=

+

2-Element Array || Example

φ

(with phase shift)

7-48

sinF 2cos

2

2

2

2 =2

22

d

kd

φ

πθ

πψ

λ

π λ πλ

π π

=

=

=

=

=

+

z

y

x d← →

rγθ

φ

x

y

2-Element Array || Example

φ

(with phase shift)

7-49

Phased Arrays

-1

= sin = 02

in plane max field at where

By adjusting can change direction of mai n

b

= sin

e am

m m

m

kd

kd

πθ φ φ ψ

ψφ

ψ

+

−

electronically

z

y

x d← →

rγθ

φ

sinF , 2cos

22kd φ ψ

φπ +

=

7-50

Uniform Linear Array (Handy Facts)

( )

1) | | is symmetric about line of array ( - axis)

2) Principal maximun occurs when (magnitude of this max is )

3) Secondary maxima

ˆF

cos 0

cos = 2m +1 m = 1,2,3...ne2

ar2

w

kd N

kd

N

γ ψ

γ ψ π

+ =

+±

y

-1

ith magnitude at of

4) Nulls when

3 3 sin 2 2

cos = m = 1,2,3... 2

N

kd N m

π π

γ ψ π

+±

z

y

x d← →

rγθ

φ( )

cossin

2F ,cos

sin2

kdN

kd

γ ψ

θ φγ ψ

+

=+

7-51

0.75

0.75

20

2 3=2

3cos = sin2

d

kd

kd

πθ

ψλ

π λ πλ

γ ψ π φ

=

=

=

=

+x

y

4-Element Array || Example

z

y

x d← →

rγθ

φ

φ

7-52

( ) -1

o o

o o

1) Symmetric about - axis

2) Principal max at of

3) Secondary max of a

ˆ

0 4

3 sin3 2 sin = 1.08 4 = 2m +1 = 3t ,

4) Nulls at

0 56.4 2 2 2

3 sin24 = = 19.5 44.

2, , 8

m

m

n

n

N

n

φ

π φπ π φ

π φπ φ

⇒

⇒

=

±

y

o90

4-Element Array || Example

x

y

φ

7-53

=at 0

down from max.

F =4.0

3db F =(0.707)(4.0)=2.83

sinFsin

2 . 83

4

φ

αα

=

⇒

=

max0.707← Emax0.5 →P

HPBW

3 dB Beamwidth

( )sinF

sin

3 sin3 sin4

π φ

π φ=

7-54

sinF

sin2.83

4

αα

=

=

max0.707← Emax0.5 →P

3 dB Beamwidth

HPBW

HPBW

=1.43

3 sin =1.43

=8.73

for rad

3db

ians

HPBW=17.5

α

π φ

φ

HPBW

HPBW=2

φ

HPBWφ

7-55