Introduction to Radiowave Propagation
Introduction to Radiowave PropagationDr Costas ConstantinouSchool of Electronic, Electrical & Computer EngineeringUniversity of BirminghamW: www.eee.bham.ac.uk/ConstantinouCC/E: [email protected] an overview, see Chapters 1 4 of L.W. Barclay (Ed.), Propagation of Radiowaves, 2nd Ed., London: The IEE, 2003The main textbook supporting these lectures is: R.E. Collin, Antennas and Radiowave Propagation, New York: McGraw-Hill, 19852Introduction (cont.)Simple free-space propagation occurs only rarelyFor most radio links we need to study the influence of the presence of the earth, buildings, vegetation, the atmosphere, hydrometeors and the ionosphereIn this lectures we will concentrate on simple terrestrial propagation models only3Radio SpectrumSymbolFrequency rangeWavelength, CommentsELF< 300 Hz> 1000 kmEarth-ionosphere waveguide propagationULF300 Hz 3 kHz1000 100 kmVLF3 kHz 30 kHz100 10 kmLF30 300 kHz10 1 kmGround wave propagationMF300 kHz 3 MHz1 km 100 mHF3 30 MHz100 10 mIonospheric sky-wave propagationVHF30 300 MHz10 1 mSpace waves, scattering by objects similarly sized to, or bigger than, a free-space wavelength, increasingly affected by tropospheric phenomenaUHF300 MHz 3 GHz1 m 100 mmSHF3 30 GHz100 10 mmEHF30 300 GHz10 1 mm
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Electromagnetic wavesSpherical wavesIntensity (time-average)Conservation of energy; the inverse square law
5Electromagnetic wavesConservation of energy; the inverse square lawEnergy cannot flow perpendicularly to, but flows along light rays
6Sphere vs SteradianThe surface area of a sphere is 4r2,The surface area of a steradian is just r2.
A Radian "cuts out" a length of a circle's circumference equal to the radius
A Steradian "cuts out" an area of a sphereequal to (radius)2.Free-space propagationTransmitted powerEIRP (equivalent isotropically radiated power)Power density at receiver
Received power
Friis power transmission formula
TxRxR8Free-space propagation (cont.)Taking logarithms gives
where is the free-space path loss, measured in decibels
Maths reminder
9Basic calculationsExample: Two vertical dipoles, each with gain 2dBi, separated in free space by 100m, the transmitting one radiating a power of 10mW at 2.4GHz
This corresponds to 0.4nW (or an electric field strength of 0.12mVm-1)The important quantity though is the signal to noise ratio at the receiver. In most instances antenna noise is dominated by electronic equipment thermal noise, given by where is Boltzmans constant, B is the receiver bandwidth and T is the room temperature in Kelvin
10Basic calculations (cont.)The noise power output by a receiver with a Noise Figure F = 10dB, and bandwidth B = 200kHz at room temperature (T = 300K) is calculated as follows
Thus the signal to noise ratio (SNR) is given by
11Basic calculations (cont.)
12Propagation over a flat earthThe two ray model (homogeneous ground)
Valid in the VHF, band and above (i.e. f 30MHz where ground/surface wave effects are negligible)Valid for flat ground (i.e. r.m.s. roughness z < , typically f 30GHz)Valid for short ranges where the earths curvature is negligible (i.e. d < 1030 km, depending on atmospheric conditions)zhthrdr1r2air, e0, m0ground, er, m0, sTxRxPx13Propagation over flat earthThe path difference between the direct and ground-reflected paths isand this corresponds to a phase differenceThe total electric field at the receiver is given by
The angles and are the elevation and azimuth angles of the direct and ground reflected paths measured from the boresight of the transmitting antenna radiation pattern
14Reflection of plane wavesReflection coefficient is a tensor
The reflection coefficient can be resolved into two canonical polarisations, TE and TM and has both a magnitude and phase
Plane of incidence1516Mobihoc '03 Radio Channel Modelling TutorialReflection of plane wavesTypical reflection coefficients for ground as a function of the grazing angle (complement of the angle of incidence). In this instance,
Pseudo-Brewster angle16Propagation over flat earthThis expression can be simplified considerably for vertical and horizontal polarisations for large ranges d >> ht, hr, l,
17Propagation over flat earthThere are two sets of ranges to consider, separated by a breakpoint
18Propagation over flat earthThus there are two simple propagation path loss laws
where l is a rapidly varying (fading) term over distances of the scale of a wavelength, and
This simplifies to
The total path loss (free space loss + excess path loss) is independent of frequency and shows that height increases the received signal power (antenna height gain) and that the received power falls as d-4 not d-2
19Propagation over flat earth
Typical ground (earth), withr = 15= 0.005Sm-1ht = 20m andhr = 2mdeep fade1/d2 power law regime (d < dc)1/d4 power law regime (d > dc)20Propagation over flat earthWhen ht = 0 or hr = 0
This implies that no communication is possible for ground based antennas (not quite true in practice)
Furthermore, for perfectly conducting ground and vertical polarisation at grazing incidence,
21Propagation over flat earthProblem: A boat has an elevated antenna mounted on a mast at height ht above a highly conducting perfectly flat sea. If the radiation pattern of the antenna approximates that of a vertically polarised current element, i.e. , determine the in-situ radiation pattern of the antenna and in particular the radiation pattern nulls as a function of the elevation angle above the horizon.
Answer:
22Path clearance on LOS pathsAssume that in the worst case scenario we get the strongest possible scattering from the sub-path obstacle: specular reflection at grazing incidencehtdr0r1TxRxPqhcr01r02hrhr11r22d1d223Path clearance on LOS pathsThe electrical path difference between the direct and scattered rays from the top of the obstacle is,
Since typically
24Path clearance on LOS pathsAdditionally, comparing similar parallelograms gives,
Under the assumptions made, the direct and scattered waves have similar magnitudes and differ in phase by p due to the grazing incidence reflectionIf the electrical path difference is p this corresponds to a first Fresnel zone path clearance
Problem: Verify that the breakpoint distance in the two ray model corresponds to the point at which the first Fresnel zone touches the ground
25Site shieldingWe consider the two-dimensional problem of site shielding by an obstacle in the line-of-sight path for simplicity (rigorous diffraction theory is beyond the scope of these introductory lectures)We invoke the Huygens-Fresnel principle to describe wave propagation:Every point on a primary wavefront serves as the source of spherical secondary wavelets such that the primary wavefront at some later time is the envelope of these wavelets. Moreover, the wavelets advance with a speed andfrequency equal to that of the primary wave at each point in space. Huygens's principle was slightly modified by Fresnel to explain why no back wave was formed, and Kirchhoff demonstrated that the principle could be derived from the wave equationSite shielding
TRPd1d2d1d1r = d2 + dPOobservation planeperfectly absorbing knife-edgeduu0 (u0 > 0 path obstraction) (u0 < 0 path clearance)uaSite shieldingSite sheildingThe Kirchhoff integral describing the summing of secondary wavefronts in the Huygens-Fresnel principle yields the field at the receiver
where k1 describes the transmitter power, polarisation and radiation pattern, f(r) describes the amplitude spreading factor for the secondary waves (2D cylindrical wave f(r) = r1/2, 3D spherical wave f(r) = r) and u1 is a large positive value of u to describe a distant upper bound on the wavefront
Site shieldingStationary phase arguments (since the exponent is oscillatory, especially for high frequencies) show that only the fields in the vicinity of the point O contribute significantly to the field at RIf point O is obstructed by the knife-edge, then only the fields in the vicinity of the tip of the knife-edge contribute significantly to the field at RUsing the cosine rule on the triangle TPR, gives
Site shieldingIf we assume that d1, d2 >> l, u (stationary phase and far-field approximations), then u/d1, a
