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					Wireless Networks (CS698T) Course: Prof. Bhaskaran Raman Instructor: Lecture Number: 03 04-AUG-2005 Lecture Date: Prateek Bhansali (Y3228) Scribe:

Table of Contents
I. II. III. IV. Antenna Parameters Frii’s Free Space Equation Link –Budget Calculation Path loss Models

Antenna Parameters
The fundamental characteristics of an antenna are its gain and half power beamwidth Antenna Gain :The gain is ratio of transmit/receive power in a particular direction, to that of an isotropic antenna . The actual gain is given by what is specified on radiation pattern diagram plus what is specified by Vendor. Half power beamwidth: The half power beamwidth is the angular separation between the half power points on the antenna radiation pattern, where the gain is one half the maximum value. Antenna Polarization
What is polarization?

Polarization of a wave is the direction of the electric field vector Various types of polarization are possible such as circular,elliptical and linear. The orientation of the antenna will change the orientation of the signal and lead to power loss . The transmitting ( Tx ) and receiving ( R x )antennas should be both polarized either horizontally or vertically. 45 deg mismatch will lead to 3dB loss 90 deg mismatch will lead up to 20 dB loss To gain more insight of antenna parameters visit

Frii’s Free Space Equation
The free space propagation model is used to predict received signal strength when the transmitter and receiver have a clear ,unobstructed line-of-sight path between them. The free space power received by receiver antenna which is separated from a radiating transmitter antenna by distance d , is given by Friis free space equation

Pt Gt Gr λ2 Pr = ( 4Π d ) 2
c2 = PGt Gr ( f 4Π d ) 2

Pr =Received Power Pt = Transmitted Power Gt =Transmitter antenna gain Gr =Receiver antenna gain λ =wavelength d=distance between transmitter and receiver f= carrier frequency c= speed of light,3x10 8 m/s

Which is equivalent to

( Pr ) dBm = ( Pt ) dBm + (Gt ) dB + (Gr ) dB − 32.5 − 20 log10 f − 20 log10 d
where f is in MHz and d is in Km

(P )dBm + (Gt )dB t

is collectively known as Effective Isotopic Radiated Power

The Effective Isotropic Radiated Power (EIRP) is defined as the maximum radiated power available from a transmitter to the receiver in the direction of maximum antenna gain ,as compared to isotropic radiator. An isotropic antenna radiates the signal equally in all directions with unity gain and is used to reference antenna gains in wireless systems .It is important in the sense that Power Output rules for ISM bands are specified in terms of EIRP and the operator has to stay within this power limits. Problem : Calculate Pr , given Pt = 50mW ,carrier frequency=2.4 GHz, d=2Km for following cases: Case I : Assuming transmitter and receiver antenna to be isotropic . Case II: Transmitter and receiver antennas with Gt = 24dBi, Gr = 24dBi.

Solution: Case I : Since the antennas are give to be isotropic ,they have unity gain. Using

( Pr ) dBm = ( Pt ) dBm + (Gt ) dB + (Gr ) dB − 32.5 − 20 log10 f − 20 log10 d
With (Gt ) dB = (Gr ) dB = log1 = 0 (also f in MHz and d in Km ) We get Path loss = 32.5 +20log2400+ 20 log2 =106 dB 50mW => Pr =10 log -106 1mW =17-106 = -89dBm

Case II: Now the antennas are not isotropic ,so we have to take in account of antenna gains ,which merely leads to addition of (24+24) dBm in above result
∴ With Gt = 24dBi, Gr = 24dBi.

=> Pr =-89dBm+24dBm+24dBm= -41dBm

Is Friis free space equation always valid?
Certainly the model is not valid for d=o. The Friis free space equation is only valid ,for estimation of Pr , for values of d which are in the far-field of the transmitting antenna .The Far-field of transmitting antenna is defined as the region beyond the far-field distance d f ,which is related to largest linear dimension of the transmitter antenna aperture, D and the carrier wavelength, λ by following relation: 2D 2 d > df =


Additionally ,

d f >>D and d f >> λ Near field Region:-It is defined as the part of space between the antenna and far field region.
Problem: What is Near Field of the Parabolic Grid Antenna with D=1 m and carrier frequency , f=2.44 GHz? Solution: We know ,

d = Inserting values, D=1 , λ = d=16 m

2D 2

3 * 10 8 c =12.3 cm = f 2.44 * 10 9

What is Link –Budget Calculation?
A link budget is a rough calculation of all known elements of the link to determine if the signal will have the proper strength when it reaches the other end of the link. To make this calculation, various details are required like total length of transmission cable and loss per unit length at the specified frequency as there will always be some loss of signal strength through the cables. For example : loss through WBC400 is 6dB/100ft.

Path Loss Models

Free Space path loss model is idealistic based on one unobstructed path from transmitter to receiver and it states that power falls off proportional to d² . In reality there are many more path losses and several models which may vary in their complexity and accuracy are proposed to explain those losses .

Ground Reflection (two ray) model

It models path loss for one unobstructed path and one ground (or reflected) bounce according to which power falls off proportional to d 4 .

Knife Edge diffraction model
If the direct line-of-sight is obstructed by a single object ( of height hm), such as a mountain or building, the attenuation caused by diffraction over such an object can be estimated by treating the obstruction as a diffracting knife-edge.

Figure: Path profile model for (single) knife edge diffraction These models are not generally accurate for cities or indoors.

Indoor Propagation models

The indoor signal propagation differs from an outdoor case particularly in distances and in variability of the environment , so these models are site specific and they have to take account of partition losses and surrounding objects which causes multipath propagation.

(directly quoted from

The most appropriate path loss model depends on the location of the receiving antenna. For the example above at:
• • • • •

location 1, free space loss is likely to give an accurate estimate of path loss. location 2, a strong line-of-sight is present, but ground reflections can significantly influence path loss. The plane earth loss model appears appropriate. location 3, plane earth loss needs to be corrected for significant diffraction losses, caused by trees cutting into the direct line of sight. location 4, a simple diffraction model is likely to give an accurate estimate of path loss. location 5, loss prediction fairly difficult and unreliable since multiple diffraction is involved.

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