Antenna Basics Ppt

Fundamental Antenna Parameters

1. Radiation Pattern An antenna radiation pattern is defined as “a graphical representation of the radiation properties of the antenna as a function of space coordinates. In most cases, the radiation pattern is determined in the far-field region. Radiation properties include radiation intensity, field strength, phase or polarization.

Coordinate System

Types of Radiation Patterns Idealized Point Radiator

Isotropic

Vertical Dipole

Omnidirectional

Radar Dish

Directional

Radiation Pattern Lobes

Main lobe

Full Null Beamwidth Between 1st NULLS Side lobes Back lobes

Radiation Pattern Lobes

Field Regions Reactive near-field region

Far-field (Fraunhofer) region

3 R1  0.62 D



D

R1 R2

Radiating near-field (Fresnel) region R2  2

D2



Radiation Intensity

Aside on Solid Angles surface area  r 2

   1.0 rad

arc length  

total circumfrance  2 radians

r   1.0 sr

total surface area  S o  4 r 2   r 2 S   2o sr r infinitesimal area ds  r 2 sin( ) d d of surface of sphere ds d  2  sin( ) d d r

Radiation Intensity

tot dPrad W U sr d tot dPrad W 2 Prad  m ds

tot  Prad   U d 4

tot  Prad   Prad ds

U  r 2 Prad since Prad ( ,  , r ) decays as 1/r2 in the far field

U ( ,  ) will be independent of r

Radiation Intensity

1 ~ ~* 1 ~ 2 1 2 2 Prad ( ,  , r )  E  H  E  E  E 2 2 2 r2 2 2 U ( ,  )  E  E 2 U ( ,  ) U ( ,  )  U max

Radiation Intensity Examples 1. Isotropic radiator tot Prad Prad ( ,  , r )  4 r 2 tot Prad U ( ,  )  r Prad ( ,  , r )   const 4 U ( ,  ) U ( ,  )   1.0 U max 2

2. Hertzian Dipole

 l I 0 e  jr E ( ,  , r )  j sin( ) 4 r E ( ,  , r )  0 2

 l I 0 e  jr    l I 0 2 2 2 1 2 1 U ( ,  )  r E  E  r  sin( )   2 2 4 r 2  4 U ( ,  ) U ( ,  )   sin 2 ( ) U m ax

2

  sin 2 ( ) 

Directive Gain

U ( ,  ) U ( ,  ) U ( ,  )  tot  4 tot Prad U ave Prad 4 U Dm ax  Do  4 mtotax  1 (directivity ) Prad D( ,  ) 

Directivity Examples

1. Isotropic radiator

U ( ,  )  U o 

tot Prad

4 U ( ,  ) D ( ,  )  4  1 .0 tot Prad Do  1.0 2. Hertzian Dipole

 l e  jr E ( ,  , r )  j sin( ), E ( ,  , r )  0 4 r 1    l I 0 2 2 U ( ,  )  r 2 E  E   2 2  4

   lI 0 P   U ( ,  )d   2  4 4 tot rad

D( ,  )  4 Do 

3 2

  

U ( ,  ) 3 2  sin ( ) tot Prad 2

2

  sin 2 ( ) 

2 2 

2

   l I 0  8   sin (  )  sin(  ) d  d   0 0 2  4  3 2

Antenna Gain

U ( ,  ) G ( ,  )  4 Pinput DIRECTIVITY

POWER DENSITY IN A CERTAIN DIRECTION DIVIDED BY THE TOTAL POWER RADIATED

GAIN

POWER DENSITY IN A CERTAIN DIRECTION DIVIDED BY THE TOTAL INPUT POWER TO THE ANTENNA TERMINALS (FEED POINTS)

IF ANTENNA HAS OHMIC LOSS… THEN, GAIN < DIRECTIVITY

Antenna Gain Sources of Antenna System Loss 1.

losses due to impedance mismatches

2.

losses due to the transmission line

3.

conductive and dielectric losses in the antenna

4.

losses due to polarization mismatches

According to IEEE standards the antenna gain does not include losses due to impedance or polarization mismatches. Therefore the antenna gain only accounts for dielectric and conductive losses found in the antenna itself. However Balanis and others have included impedance mismatch as part of the antenna gain.

The antenna gain relates to the directivity through a coefficient called the radiation efficiency (et) impedance mismatch conduction losses dielectric losses

G ( ,  )  et  D( ,  )  er ec ed  D( ,  ) et  1

Overall Antenna Efficiency The overall antenna efficiency is a coefficient that accounts for all the different losses present in an antenna system. et   e  e p er ec ed  e p  er ecd

e p  polarization mismatches er  reflection efficiency(impedance mismatch) ec  conduction losses ed  dielectric losses ecd  conductor & dielectric losses

Reflection Efficiency The reflection efficiency through a reflection coefficient (G) at the input (or feed) to the antenna.

er  1  G G

2

Rinput  Rgenerator Rinput  Rgenerator

Rinput  antenna input impedance () Routput  generator output impedance ()

Radiation Resistance The radiation resistance is one of the few parameters that is relatively straight forward to calculate.

Rrad 

total rad 2 o

2P I



2  U ( ,  )d 4

Io

2

Example: Hertzian Dipole

   l I o  tot  Prad   U ( ,  )d   2  4  4 Rrad 

   l I o 2  2  4 2

Io

2 2 

   l I o 2  sin (  )  sin(  ) d  d   0 0 2  4

2

 8   3     l  4 

2

 8 2  l      3 3    

2

2

 8   3

Radiation Resistance Example: Hertzian Dipole (continued)

Rrad  let Rrad

2

   l I o 2  2  4

l

 8   3     l  4 

2

Io

1 and   377  100 2 1  377  0.079  3 10000 

 50  7.9  er  1     0.0063  50  7.9  2

2

 8 2  l      3     3

2

Antenna Radiation Efficiency Conduction and dielectric losses of an antenna are very difficult to separate and are usually lumped together to form the ecd efficiency. Let Rcd represent the actual losses due to conduction and dielectric heating. Then the efficiency is given as

ecd 

Rrad Rcd  Rrad

For wire antennas (without insulation) there is no dielectric losses only conductor losses from the metal antenna. For those cases we can approximate Rcd by:

Rcd 

 o 2b 2 l

where b is the radius of the wire,  is the angular frequency,  is the conductivity of the metal and l is the antenna length

Example Problem: A half-wavelength dipole antenna, with an input impedance of 73 is to be connected to a generator and transmission line with an output impedance of 50. Assume the antenna is made of copper wire 2.0 mm in diameter and the operating frequency is 10.0 GHz. Assume the radiation pattern of the antenna is

U ( ,  )  Bo sin 3 ( ) Find the overall gain of this antenna SOLUTION First determine the directivity of the antenna. D ( ,  )  4

U ( ,  ) tot Prad

Bo sin 3 ( ) 16 3 D( ,  )  4  sin ( ) 2  3  3  B0  4   D0  Dm ax 

16  1.697 3

Example Problem: Continued SOLUTION Next step is to determine the efficiencies

et  er ecd 73  50 er  (1  G )  (1  )  0.965 73  50 2

2

ecd 

Rrad Rcd  Rrad

 o 0.015 2 10 109  4 107 Rcd    0.0628 7 2b 2 2 (0.001) 2  5.7 10 l

73  0.9991 73  0.0628 et  er ecd  0.965  0.9991  0.964 ecd 

Example Problem: Continued SOLUTION Next step is to determine the gain

G ( ,  )  er ecd D( ,  ) 16 3 sin ( ) 3 16 G0  Gmax  0.964  1.636 3 G0 (dB)  10 log 10 (1.636)  2.14 dB G ( ,  )  0.964

Antenna Type Gain (dBi)

Gain over Isotropic

Half Wavelength Dipole

1.76

1.5x

Cell Phone Antenna (PIFA)

3.0

2.0x

Standard Gain 15 Horn

31x

Cell phone tower antenna

6

4x

Large Reflecting Dish

50

100,000x

Small Reflecting Dish

40

10,000x

Power Levels

0.6 Watts

Effective Aperture

Aphysical

Pload

Question:

plane wave incident

?

Pload  AphysicalWinc

Answer: Usually NOT

Pload  Aeff Winc  Aeff 

Pload Winc

Directivity and Maximum Effective Aperture (no losses) Antenna #1

transmit

Atm, Dt

Antenna #2 Direction of wave propagation

R

2 Aem  Do 4

receiver

Arm, Dr

Directivity and Maximum Effective Aperture (include losses) Antenna #1

Antenna #2 Direction of wave propagation

transmit

Atm, Dt

receiver

Arm, Dr

R

2 * 2 Aem  ecd (1  G ) Do ˆ w  ˆ a 4 2

conductor and dielectric losses

reflection losses (impedance mismatch)

polarization mismatch

Friis Transmission Equation (no loss) Antenna #1 Antenna #2 (r,r)

(t,t)

R The transmitted power density supplied by Antenna #1 at a distance R and direction (r,r) is given by:

Wt 

Pt Dgt ( t , t ) 4 R 2

The power collected (received) by Antenna #2 is given by:

Pr  Wt Ar 

Pt Dgt ( t ,  t ) 4 R 2 2

Ar 

Pt Dgt ( t ,  t ) Dgr ( r ,  r )2 4 R 2

Pr     Dgt ( t ,  t ) Dgr ( r ,  r )   Pt  4 R 

4

Friis Transmission Equation (no loss) Antenna #1 Antenna #2 (r,r)

(t,t)

R 2

Pr     Dgt ( t ,  t ) Dgr ( r ,  r )   Pt  4 R  If both antennas are pointing in the direction of their maximum radiation pattern: 2

Pr     Dto Dro   Pt  4 R 

Friis Transmission Equation ( loss) Antenna #1 Antenna #2 (r,r)

(t,t)

R conductor and dielectric losses receiving antenna

reflection losses in receiving (impedance mismatch)

free space loss factor

2

Pr 2 2    * 2  Dgt ( t , t ) Dgr ( r , r ) ˆ w  ˆ a  ecdtecdr (1  Gr )(1  Gt ) Pt  4 R  conductor and dielectric losses transmitting antenna

reflection losses in transmitter (impedance mismatch)

polarization mismatch

Friis Transmission Equation: Example #1 A typical analog cell phone antenna has a directivity of 3 dBi at its operating frequency of 800.0 MHz. The cell tower is 1 mile away and has an antenna with a directivity of 6 dBi. Assuming that the power at the input terminals of the transmitting antenna is 0.6 W, and the antennas are aligned for maximum radiation between them and the polarizations are matched, find the power delivered to the receiver. Assume the two antennas are well matched with a negligible amount of loss. 2

Pr 2 2    * 2 max max ˆ ˆ  Dt Dr  w   a  ecdtecdr (1  Gr )(1  Gt ) Pt  4 R  =1



=1

c 3e8   0.375 m f 800 e6

Dtm ax  10 3 /10  2.0 Drm ax  10 6 /10  4.0

=0

=0

=1

2

0.375   Pr  0.6 watts     2  4  1.65 nW  4 1 609.344 

Friis Transmission Equation: Example #2 A half wavelength dipole antenna (max gain = 2.14 dBi) is used to communicate from an old satellite phone to a low orbiting Iridium communication satellite in the L band (~ 1.6 GHz). Assume the communication satellite has antenna that has a maximum directivity of 24 dBi and is orbiting at a distance of 781 km above the earth. Assuming that the power at the input terminals of the transmitting antenna is 1.0 W, and the antennas are aligned for maximum radiation between them and the polarizations are matched, find the power delivered to the receiver. Assume the two antennas are well matched with a negligible amount of loss. 2

Pr 2 2    * 2 max max ˆ ˆ  Dt Dr  w   a  ecdtecdr (1  Gr )(1  Gt ) Pt  4 R  =1



=1

c 3e8   0.1875 m f 800 e6

Dtm ax  10 2.14 /10  1.64 Drm ax  10 24 /10  251 .0

=0

=0

=1 2

 0.1875  Pr  1.0 watts    1.64  251  0.15 pW  4  781,000 

Friis Transmission Equation: Example #2 A roof-top dish antenna (max gain = 40.0 dBi) is used to communicate from an old satellite phone to a low orbiting Iridium communication satellite in the Ku band (~ 12 GHz). Assume the communication satellite has antenna that has a maximum directivity of 30 dBi and is orbiting at a distance of 36,000 km above the earth. How much transmitter power is required to receive 100 pW of power at your home. Assume the antennas are aligned for maximum radiation between them and the polarizations are matched, find the power delivered to the receiver. Assume the two antennas are well matched with a negligible amount of loss. 2

Pr 2 2    * 2 max max ˆ ˆ  Dt Dr  w   a  ecdtecdr (1  Gr )(1  Gt ) Pt  4 R  =1

=1

c 3e8    0.025 m f 800 e6 Drm ax  10 40 /10  10 ,000 Dtm ax  10 30 /10  1000 .0

=0

=0

Pt 

=1

100 1012 watts 2

0.025     10,000 1000  4  36,000,000 

 82 W

Radar Range Equation

Definition: Radar cross section or echo area of a target is defined as the area when intercepting the same amount of power which, when scattered isotropically, produces at the receiver the same power density as the actual target.

   Winc  2 Ws  Ws  lim     lim 4 R m2   2 R  4 R R  Winc    

 (radar cross section) m2 R (distance from target) m Ws (scattered power density) W/m2 Winc (incident power density) W/m2

Radar Range Equation (no losses) Power density incident on the target is a function of the transmitting antenna and the distance between the target and transmitter:

Winc 

Pt Dgt ( t , t ) 4 Rt

2

The amount of power density scattered by the target at the location of the receiver is then given by: The amount of power delivered by the receiver is then given by:

Ws  Winc

Pt Dgt ( t , t )   2 (4 Rt Rr ) 2 4 Rr

Pt Dgt ( t , t )

2 Pr  Ws Ar   Dgr ( r ,  r ) 2 (4 Rt Rr ) 4 Dgt ( t , t ) Dgr ( r ,  r ) Pr   Pt (4 Rt Rr ) 2 4 2

Note that in general:

   ( t ,  t , r ,  r )

Radar Range Equation (losses)

Dgt ( t , t ) Dgr ( r , r )    Pr 2 2 * 2   ˆ w  ˆ a  ecdtecdr (1  Gr )(1  Gt ) Pt 4  4 Rt Rr  2

Radar Cross Section (RCS) Definition: Radar cross section or echo area of a target is defined as the area when intercepting the same amount of power which, when scattered isotropically, produces at the receiver the same power density as the actual target.

   Winc  2 Ws  Ws  lim     lim 4  R m2    2 R  4 R R  Winc        lim 4 R 2 R   

2  E scat 

 E inc

2

scat 2   E  m 2  lim 4 R 2  m2 2 R   inc   E   

Transmitter and receiver not in

   ( t ,  t , r ,  r )

 t   r ,  t   r the same location (bistatic RCS)  t   r ,  t   r Transmitter and receiver in the

same location (usually the same antenna) called mono-static RCS

Radar Cross Section (RCS) RCS Customary Notation: Typical RCS values can span 10-5m2 (insect) to 106 m2 (ship). Due to the large dynamic range a logarithmic power scale is most often used.

 dBsm   dBm

2

  m2  10 log10    ref

    10 log 10  m 2  1   

100 m2

20 dBsm

10 m2

10 dBsm

1 m2

0 dBsm

0.1 m2

-10 dBsm

0.01 m2

-20 dBsm

  