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Wave propagation in wireless channels Principles of Antenna Systems:

• Antenna pattern

w(φl, φz) = power transmitted per unit of solid angle in the direction (φl, φz)

d

gw

wl zl z

i l z

( , )( , )

( , )φ φ φ φ

φ φ=

beam pattern

antenna

Transmitter

φl

d dφl

dCosφldφz

φz

2

wi(φl, φz) = w(φl, φz) for an “ isotropic” radiator solid angle = A

d 2 (steradians)

• P w dT l z= z ( , )φ φ Ω unit sphere

d d dl l zΩ = cosφ φ φ

• g

wPl z

l z

t

( , )( , )φ φ φ φ

π=

4

w

P

d

d

Pi l z

t t( , )φ φπ π

= =4 4

2

2

• Question: Very often people give you an antenna gain. Then,

how does one relate g(φl, φz) to antenna gain?

d A

3

g g l z∆ max ,φ φb gm r • Antenna Solid angle.

This is the solid angle into which most of the antenna power is sent.

Ω ∆fvt

l z

t

t

P

w

PgPmax( , )φ φ

πb g

=4

Ω f v g= 4π

• Often, one defines an antenna “beamwidth”

Ω fvb

b b

= − FHGIKJ

L

NM

O

QP

≅ <<

2 12

412

π φ

π φ φ

cos

• Let us compute the amount of power received in an area A.

4

• P w

A

d

Pg A

d

A l z

t l z

≅

=

( , )

( , )

φ φ

φ φπ

2

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(d is large)

gw

Pl zl z

t( , )

( , )φ φ φ φ

π=

F

HGG

I

KJJ

4

• P

Pg

dAA

t l z≅ ( , )φ φπ4 2

• One also has to add “ radiation efficiency” of the antenna

A

Ωa

d

solid angle

5

P

P g

dAA

t r l z≅ ρ φ φπ( , )

4 2 Area

Power/Area

6

power/area = flux density

• Let us define P gt r l zρ φ φ( , ) as EIRP.

EIRP P gt l z r= ( , )φ φ ρ Effective Isotropic Radiated Power

• P

EIRP

dAA =

4 2π

• A may now correspond to the “area” of a receiving

antenna

• Ω fv

ap aA A= =λ

ρλ2 2

(True for all antennae)

Area of the sphere

aperture loss factor antenna “area”

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PP g

d

g

At l z r

fv

fvr

= ×

=

( , )φ φ ρπ

λ

π4

4

2

2

Ω

Ω

• Let the receiving antenna be pointed directly at the

receiving antenna. • Note that apreture ineffiency reduces the effective

antenna size, which in turn leads to a smaller antenna gain at the receiver

• Also, let g gt l z= max ( , )φ φl q occur at φ l and φ z that is

identical to the receiving antenna’s azimuth and elevation angles.

P dP g g

drt t r r( )( )

= λ ρπ

2

2 24

gain of the receiving antenna

received power

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P d P g gdr t t r r( ) = F

HGIKJ

ρ λπ4

2

•

10 10 10 10

10 10

10 10 10 10

10 10

log log log log

log log

P d P g g

L

P dB P dB g dB g dB

dB L dB

r t t r

r P

r t t r

r p

b g

b g

b g b g b g b g

b gb g b g

= + +

+ +

= + +

+ +

ρ

ρ

• L

dP = FHG

IKJ

λπ4

2

is known as the “propagation loss”

• In practice, one uses ERP (effective radiated power)

instead of EIRP ERP = Effective Radiated Power

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ERP = the maximum radiated power as compared to a half-wave dipole antenna • Dipole antenna has more “directivity”

⇒ >=

g g

g or dB

da isotropic

da 164 215. ( . )

For the same transmission system

ERP EIRP

ERP EIRPdB dB

≅ ⇒

= −164

215.

.b g b g

• In arriving at our equations, we have assumed that the

receiver is at the “ far-field” of the antenna. This means that d > df (far-field distance or Fraunhofer distance).

• Furthermore, in order to be in the far field, D (the largest physical linear dimension) of the antenna << df

•

dD

and d

f

f

=

>>

2 2

λλ

10

• Note that, for d~0, the previous equations fail • For

P d P dd

d

for d d

where d d

r r

f

b g b g= FHGIKJ

>≥

00

2

0

0 .

• Our previous equations are valid for d d f≥ .

• Pr(d0) is usually known (the received power at a

designated point). Then, P d P d dd d d dr r fa f b g c h= F

HIK > ≥0

02

0

• units: one may use dBm or dBW for P.

Pr logPr( )

a fdBmwatts

watts= F

HIK−10

1010 3

11

• Note that the above results are only valid for the “ free-space” propagation.

• The other key issue in analyzing performance is the

equivalent noise power spectrum level N 0 • In general, this paramter is impacted by the input noise

power to the antenna and the receiver-induced noise (primarily caused by thermal-induced random emissions in lossy devices)

• Let us assume we have a resistive element with resistance R

• We then can show that the two-sided voltage spectrum density of the noise generated by the resistor is

S f kT Rv ( ) = 2 0

where k is the Boltzmann’s constant, T0 denotes the device temperature in Kelvin, and R is the resistance of the device in ohms. Assuming that we are interested in a bandwidth of B Hz (one-sided), the rms value of the noise-induced voltage in R is

V kT BRrms2

04= • The equivalent noise power for this rms noise voltage

is

PV

RkT Brms= =

2

04

• Given that a perfect matching of this resistor to a load implies that one must use a load with resistance R, the

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maximum noise power that can be delivered to a matched load is ¼ of the above

P kT Bout = 0 • Hence, one can define the equivalent one-sided noise

power spectrum level as

PN

B kT B N kTout = = → =00 0 02

2( )

• Hence, by defining a “ temperature” for a resistor, one can compute the equivalent power spectrum level of the “white” noise that is created by the noisy resistor

• The noise from the antenna due to background radiation may also be modeled as a white noise, and hence one may define a “ temperature” for the antenna as well (let us define that as Tant )

• For more complicated devices (such as amplifiers, mixers, etc), one may define noise figure as

Fs n

s ni i= =Input SNR

Output SNR

/

/0 0

where the input noise is assumed to be due to a source at a predefine temperature (typically room temperature T0 295= K) • This implies that for active elements with a gain of G

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s Gs

n n G kT BG

n kT B

FT

TT F T

i

i eq

i

eqeq

0

0

0

001 1

== +

=

= + → = −b g

• This implies that the device may be replaced by a noise-less device and a noise source at its input having a noise temprature of Teq (and hence a power spectrum density of kTeq )

• For lossy devices (such as transmission lines) at

temperature T and the loss factor L (L<1),

Noise Figure F Gain =G

Noiseless Gain= G

Noise source with Temperature eqT

14

s Ls

n n L kT BL

n kT B n

T

T LT F T F

L

i

i eq

i

eqeq

0

0

0 0

00

11 1

1

== +

= =

= − → = − → =b g

• Hence, the noise figure of a transmission line is the inverse of its attenuation factor

• Finally, if a series of systems are cascaded, then one may find an equivalent noise figure, and hence temperature for the entire chain

F G1 1,

F G2 2, F Gn n, …

F G G Geq n, ...1 2

Antenna

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• In that case,

F FF

G

F

G G

F

G G G Geqn

n

= + − + − + + −

−1

2

1

3

1 2 1 2 3 1

1 1 1...

...

Which implies that in order to achieve the lowest overall noise figure, one must use the device with the largest gain and the lowest noise figure as close as possible to the antenna (LNA) • Note that only the first two stages will determine the

entire noise level of the receiver • Therefore, the noise power spectrum level that you

may now use to compute the error rate of your digital communications system is given by

N kT

T T F T

tot

tot ant eq

0

01

=

= + −d i