free space radio wave propagation
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Free Space Radio Wave Propagation
RKTiwary
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propagation effects Reflection : EM waves impinge on objects
which are much greater than the wavelength of the traveling wave
diffraction: surface having sharp irregularities.
scattering: occurs when the medium through which the wave is traveling contains objects
which are much smaller than the wavelength of the EM wave.All are due to the presence of buildings, mountains and other such obstructions
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Friis free space equation is given byPr(d) = PtGtGrλ2/(4π)2d2L
transmitted power : Pt
transmitter antenna gain: Gt
Reciever antenna gain : Gr
Wavelength : λ Tx-Rx separation :d system loss factor : L
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The gain of the antenna is related to the effective aperture of the antenna which in turn is dependent upon the physical size of the antenna as given below
The path loss, representing the attenuation
The fields of an antenna can broadly be classified in two regions, I. the far field and II. the near field.
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The far eld region is also termed as Fraunhofer region and the Friis equation holds in this region. Hence, the Friis equation is used only beyond the far field distance, df , which is dependent upon the largest dimension of the antenna as
Also we can see that the Friis equation is not dened for d=0. For this reason, we use a close in distance, do, as a reference point. The power received, Pr(d), is then given by:
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Free space propagation model, showing the near and far elds.
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Eg: Find the far field distance for a circular antenna with maximum dimension of 1 m and operating frequency of 900 MHz. Solution: Since the operating frequency f = 900 Mhz, the
wavelength= c/f2
Thus, with the largest dimension of the antenna, D=1m, the far eld distance is
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Eg2:A unit gain antenna with a maximum dimension of 1 m produces 50 W power at 900 MHz. Find (i) the transmit power in dBm and dB, (ii) the received power at a free space distance of 5 m and 100 m.Solution:(i) Tx power = 10log(50) = 17 dB = (17+30) dBm = 47 dBm(ii) df = 2XD2/λ = 2X12/(1/3) = 6mThus the received power at 5 m can not be calculated using free space distance formula.At 100 m
= 3.5x10-3mW PR(dBm) = 10logPr(mW) = -24.5dBm
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Basic Methods of PropagationReflection, diffraction and scattering are the three fundamental phenomena that causes signal propagation in a mobile communication system, apart from LOS communication.
Reflection: Reflection occurs when an electromagnetic wave falls on an object, which has very large dimensions as compared to the wavelength of the propagating wave. Example: the earth, buildings and walls
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Properties of EM wave incident: dielectric, some energy is reflected back and some energy
is transmitted. perfect conductor, all energy is reflected back to the first
medium Amount of energy reflected depends on the polarization of
the e.m. wave Brewster's angle:
Angle arises in parallel polarization, when no reflection occurs in the medium of origin
Reflection coeficient is equal to zero.
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• Large-scale fading, due to path loss of signal as a function of distance and shadowing by large objects such as buildings and hills. This occurs as the mobile moves through a distance of the order of the cell size, and is typically frequency independent.• Small-scale fading, due to the constructive and destructive interference of the multiple signal paths between the transmitter and receiver. This occurs at the spatial scale of the order of the carrier wavelength, and is frequency dependent
Variations of the channel strength over time and over frequency
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Channel quality varies over multiple time-scales. At a slow scale, channel varies due to large-scale fading effects. At afast scale, channel varies due to multipath effects.
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Free space, fixed transmit and receive antennas
In response to a transmitted sinusoid cos 2πft, we can express the electric far field at time t as
(r, θ, ψ) represents the point u in space r is the distance from the transmit antenna to u &(θ, ψ) represents the vertical and horizontal angles from the antenna to u respectively.αs (θ, ψ,ƒ) is the radiation pattern of the sending antenna at frequency f in the direction (θ, ψ)
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As the distance r increases, the electric field decreases as r−1 and thus the power per square meter in the free space wave decreases as r−2. In concentric spheres of increasing radius r around the antenna, the total power radiated through the sphere remains constant, but the surface area increases as r2. Thus, the power per unit area must decrease as r−2. Not true always when obstruction
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For fixed receive antenna at the location u = (r, θ, ψ)
Where α (θ, ψ,ƒ) is the product of the antenna patterns of transmit and receive antennas in the given direction.Now suppose, for the given u, that we define
We then have H(f) is the system function for an LTI (linear time-invariant) channel
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Free space, moving antennaLet receive antenna is at a moving location described as
describe the free space electric field at the moving point u(t) (for the moment with no receive antenna), we have
we can rewriteThus, the sinusoid at frequency f has been converted to a sinusoid of frequencythere has been a Doppler shift of
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Intuitively, each successive crest in the transmittedsinusoid has to travel a little further before it gets observed at the moving observation point.If the antenna is now placed at u(t),
This channel cannot be represented as an LTI channel
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Reflecting wall, fixed antenna
Illustration of a direct path and a reflected path
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We assume that in the absence of the receive antenna, the electromagnetic field at the point where the receive antenna will be placed is the sum of the free space field coming from the transmit antenna plus a reflected wave coming from the wall.Approximate the solution of Maxwell’s equations by amethod called ray tracing.
Sum of the free space wave from the transmitter andthe reflected free space waves from each of the reflecting obstacles
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For both the direct and the reflected wave, and assuming the same antenna gain for both waves, we get
The received signal is a superposition of two waves, both of frequency f.The phase difference between the two waves is
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When : Δθ = n2π The two waves add constructively,and the received signal is strong & When : Δθ = (2n+1) π Two waves add destructively, and the received signal is weak As a function of r, this translates into a spatial pattern of constructive and destructive interference of the waves The distance from a peak to a valley is called the
coherence distance: Where
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For distance much smaller than the received signal at a particular time does not change
appreciably The quantity is called the delay spread of the channel:
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Illustration of a direct path and a reflected path.
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Reflecting wall, moving antenna
Doppler shift
second is a sinusoid at frequencya Doppler shift
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Doppler spread
This is the product of two sinusoids, one at the input frequency f, which is typically of the order of GHz, and the other one at
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The received waveform oscillating at frequency f with a slowly varying envelope at frequency Ds/2.
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multipath propagationAt the MS, plane waves arrive from many different directions and with different delays, as shown in Fig. and this property is called multipath propagation
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Reflection from smooth surface