Introduction to Radiowave Propagation Dr Costas Constantinou School of Electronic, Electrical & Computer Engineering University of Birmingham W: www.eee.bham.ac.uk/ConstantinouCC/ E: c.constantinou@bham.ac.uk

Introduction For an overview, see Chapters 1 – 4 of L.W. Barclay (Ed.), Propagation of Radiowaves, 2nd Ed., London: The IEE, 2003 The main textbook supporting these lectures is: R.E. Collin, Antennas and Radiowave Propagation, New York: McGraw-Hill, 1985

Introduction (cont.) Simple free-space propagation occurs only rarely For most radio links we need to study the influence of the presence of the earth, buildings, vegetation, the atmosphere, hydrometeors and the ionosphere In this lectures we will concentrate on simple terrestrial propagation models only

Radio Spectrum Symbol Frequency range Wavelength,  Comments ELF < 300 Hz > 1000 km Earth-ionosphere waveguide propagation ULF 300 Hz – 3 kHz 1000 – 100 km VLF 3 kHz – 30 kHz 100 – 10 km LF 30 – 300 kHz 10 – 1 km Ground wave propagation MF 300 kHz – 3 MHz 1 km – 100 m HF 3 – 30 MHz 100 – 10 m Ionospheric sky-wave propagation VHF 30 – 300 MHz 10 – 1 m Space waves, scattering by objects similarly sized to, or bigger than, a free-space wavelength, increasingly affected by tropospheric phenomena UHF 300 MHz – 3 GHz 1 m – 100 mm SHF 3 – 30 GHz 100 – 10 mm EHF 30 – 300 GHz 10 – 1 mm

Electromagnetic waves Spherical waves Intensity (time-average) Conservation of energy; the inverse square law

Electromagnetic waves Conservation of energy; the inverse square law Energy cannot flow perpendicularly to, but flows along “light rays”

Free-space propagation Tx Rx R Transmitted power EIPR (equivalent isotropically radiated power) Power density at receiver Received power Friis power transmission formula

Free-space propagation (cont.) Taking logarithms gives where is the free-space path loss, measured in decibels Maths reminder

Basic calculations Example: 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 Boltzman’s constant, B is the receiver bandwidth and T is the room temperature in Kelvin

Basic 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

Basic calculations (cont.)

Propagation over a flat earth The 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 earth’s curvature is negligible (i.e. d < 10–30 km, depending on atmospheric conditions) z ht hr d r1 r2 air, e0, m0 ground, er, m0, s Tx Rx P  x

Propagation over flat earth The path difference between the direct and ground-reflected paths is and this corresponds to a phase difference The 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

Reflection of plane waves Reflection 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 incidence

Reflection of plane waves Typical reflection coefficients for ground as a function of the grazing angle (complement of the angle of incidence). In this instance, Pseudo-Brewster angle Mobihoc '03 Radio Channel Modelling Tutorial

Propagation over flat earth This expression can be simplified considerably for vertical and horizontal polarisations for large ranges d >> ht, hr, l,

Propagation over flat earth There are two sets of ranges to consider, separated by a breakpoint

Propagation over flat earth Thus 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

Propagation over flat earth Typical ground (earth), with r = 15 = 0.005Sm-1 ht = 20m and hr = 2m 1/d4 power law regime (d > dc) 1/d2 power law regime (d < dc) deep fade

Propagation over flat earth When 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,

Propagation over flat earth Problem: 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:

Path clearance on LOS paths ht d r0 r1 Tx Rx P q hc r01 r02 hr h r11 r22 d1 d2 Assume that in the worst case scenario we get the strongest possible scattering from the sub-path obstacle: specular reflection at grazing incidence

Path clearance on LOS paths The electrical path difference between the direct and scattered rays from the top of the obstacle is, Since typically

Path clearance on LOS paths Additionally, 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 reflection If 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

Site shielding We 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 equation

Site shielding

Site shielding T R P d1 d2 r = d2 + d O du observation plane perfectly absorbing knife-edge du u0 (u0 > 0  path obstraction) (u0 < 0  path clearance) u a Site shielding

Site sheilding The 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 shielding Stationary 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 R If 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 R Using the cosine rule on the triangle TPR, gives

Site shielding If we assume that d1, d2 >> l, u (stationary phase and far-field approximations), then u/d1, a << 1 and d 2 << d Thus, using stationary phase arguments, we may only keep the fast varying exponential term inside the Kirchhoff integral and evaluate the slowly varying f(r) term at the stationary phase point O, to give,

Site shielding Since , we make the substitution which simplifies the integral to the form, where we have used the stationary phase argument to make the upper limit  Using the definition of the complex Fresnel integral,

Site shielding To determine k3 we let n  – and use F(–)= – F() and the fact that in this case we have free-space propagation (i.e. E(R) = E0(R)) , to get,

Site shielding Therefore, where, The path-gain factor, F, is given by, Useful engineering approximations:

Site shielding

Multipath propagation Mobile radio channels are predominantly in the VHF and UHF bands VHF band (30 MHz  f  300 MHz, or 1 m    10 m) UHF band (300 MHz  f  3 GHz, or 10 cm    1 m) In an outdoor environment electromagnetic signals can travel from the transmitter to the receiver along many paths Reflection Diffraction Transmission Scattering

Multipath propagation Narrowband signal (continuous wave – CW) envelope Area mean or path loss (deterministic or empirical) Fast or multipath fading (statistical) Local mean, or shadowing, or slow fading (deterministic or statistical)

Multipath propagation The total signal consists of many components Each component corresponds to a signal which has a variable amplitude and phase The power received varies rapidly as the component phasors add with rapidly changing phases Averaging the phase angles results in the local mean signal over areas of the order of  10l2 Averaging the length (i.e. power) over many locations/obstructions results in the area mean The signals at the receiver can be expressed in terms of delay, and depend on polarisation, angle of arrival, Doppler shift, etc.

Area mean models We will only cover the Hata-Okumura model, which derives from extensive measurements made by Okumura in 1968 in and around Tokyo between 200 MHz and 2 GHz The measurements were approximated in a set of simple median path loss formulae by Hata The model has been standardised by the ITU as recommendation ITU-R P.529-2

Area mean models The model applies to three clutter and terrain categories Urban area: built-up city or large town with large buildings and houses with two or more storeys, or larger villages with closely built houses and tall, thickly grown trees Suburban area: village or highway scattered with trees and houses, some obstacles being near the mobile, but not very congested Open area: open space, no tall trees or buildings in path, plot of land cleared for 300 – 400 m ahead, e.g. farmland, rice fields, open fields

Area mean models where

Area mean models The Hata-Okumura model is only valid for: Carrier frequencies: 150 MHz  fc  1500 MHz Base station/transmitter heights: 30 m  hb  200 m Mobile station/receiver heights: 1 m  hm  10 m Communication range: R > 1 km A large city is defined as having an average building height in excess of 15 m

Local mean model The departure of the local mean power from the area mean prediction, or equivalently the deviation of the area mean model is described by a log-normal distribution In the same manner that the theorem of large numbers states that the probability density function of the sum of many random processes obeys a normal distribution, the product of a large number of random processes obeys a log-normal distribution Here the product characterises the many cascaded interactions of electromagnetic waves in reaching the receiver The theoretical basis for this model is questionable over short-ranges, but it is the best available that fits observations

Local mean model Working in logarithmic units (decibels, dB), the total path loss is given by where Xs is a random variable obeying a lognormal distribution with standard deviation s (again measured in dB) If x is measured in linear units (e.g. Volts) where mx is the mean value of the signal given by the area mean model

Local mean model Cumulative probability density function This can be used to calculate the probability that the signal-to-noise ratio will never be lower than a desired threshold value. This is called an outage calculation Typical values of dB = 10 dB are encountered in urban outdoor environments, with a de-correlation distance between 20 – 80 m with a median value of 40 m

Fast fading models Constructive and destructive interference Im Re Constructive and destructive interference In spatial domain In frequency domain In time domain (scatterers, tx and rx in relative motion) Azimuth dependent Doppler shifts Each multipath component travels corresponds to a different path length. Plot of power carried by each component against delay is called the power delay profile (PDP )of the channel. 2nd central moment of PDP is called the delay spread  t P

Fast fading models The relation of the radio system channel bandwidth Bch to the delay spread d is very important Narrowband channel (flat fading, negligible inter-symbol interference (ISI), diversity antennas useful) Wideband channel (frequency selective fading, need equalisation (RAKE receiver) or spread spectrum techniques (W-CDMA, OFDM, etc.) to avoid/limit ISI) Fast fading refers to very rapid variations in signal strength (20 to in excess of 50 dB in magnitude) typically in an analogue narrowband channel Dominant LOS component  Rician fading NLOS components of similar magnitude  Rayleigh fading

Fast fading models Working in logarithmic units (decibels, dB), the total path loss is given by where Y is random variable which describes the fast fading and it obeys the distribution for Rayleigh fading, where the mean value of Y is

Fast fading models For Rician fading where ys is the amplitude of the dominant (LOS) component with power . The ratio is called the Rician K-factor. The mean value of Y is The Rician K-factor can vary considerably across small areas in indoor environments

Fading models Similar but much more complicated outage calculations E.g. Rayleigh and log-normal distributions combine to give a Suzuki distribution The spatial distribution of fades is such that the “length” of a fade depends on the number of dB below the local mean signal we are concerned with Fade depth (dB) Average fade length () 0.479 -10 0.108 -20 0.033 -30 0.010

Tropospheric propagation Over long-distances, more than a few tens of km, and heights of up to 10 km above the earth’s surface, clear air effects in the troposphere become non-negligible The dielectric constant of the air at the earth’s surface of (approx.) 1.0003 falls to 1.0000 at great heights where the density of the air tends to zero A consequence of Snell’s law of refraction is that radiowaves follow curved, rather than straight-line trajectories

Tropospheric propagation B B O d   d dh n + dn n The variation of the ray curvature with refractive index is derived: AA: wavefront at time t BB: wavefront at time t + dt AB and AB: rays normal to the wavefronts  : radius of curvature of AB

Tropospheric propagation Retaining only terms which are correct to first order in small quantities, But this is the curvature, C, of the ray AB, by definition. Furthermore, For rays propagating along the earth’s surface b is very small and we may take cosb = 1. Moreover, n–1 1.

Tropospheric propagation If n = constant, dn/dh = 0  C = 0 and the ray has zero curvature, i.e. the ray path is a straight line A ray propagating horizontally above the earth must have a curvature C = (earth’s radius)–1 = a–1 in order to remain parallel with the earth’s surface. But its actual curvature is given by C and not C. The difference between the two curvatures gives the curvature of an equivalent earth for which dn/dh = 0 and which has an effective radius ae,

Tropospheric propagation k is known as the k-factor for the earth Typically, dn/dh –0.03910–6 m–1  1/(25,600 km) Therefore, The k-factor of the earth is k = 4/3 The effective radius of the earth is ae = 4a/3 These values are used in the standard earth model which explains why the radio horizon is bigger than the radio horizon

Tropospheric propagation Problem: Find the radio horizon of an elevated antenna at a height ht above the earth Answer: