Terrain Reflection and Diffraction, Part One 1 UHF and VHF paths near the ground 2 Propagation over a plane Earth 3 Fresnel zones Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 1 / 52 I. UHF and VHF paths near the ground In many applications communications are influenced by reflections from the Earth surface Particular problem when antenna gains are small and receivers are near the ground In Chapter 7 we will study the direct + Earth reflections mechanism and also terrain diffraction The Earth surface also plays a role in groundwave propagation (Chapter 9) but this applies mainly at lower than VHF frequencies Since ionospheric propagation is also negligible for VHF and higher freqencies, direct + Earth reflections and terrain diffraction are the main mechanism for these frequencies over medium distances Tropospheric scatter applies at very long distances Consideration of \multipath environments" in Chapter 8 Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 2 / 52 We will learn basic methods for predicting propagation over a specified terrain path Primarily applied to assess antenna heights needed to maintain communications for various values of κ Used in designing line-of-sight UHF and SHF point-to-point links Numerical methods are also now being applied here; we will learn a little about these as well Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 3 / 52 II. Propagation over a plane Earth Begin by assuming the Earth surface to be planar (valid approximation for short distances and smooth terrain), so that 0 00 R2 = R2 + R2 (1) tan 2 = (h1 + h2)=d (2) tan 1 = (h1 − h2)=d (3) d = R1 cos 1 = R2 cos 2 (4) We know the power density in the direct ray in free space is PT GT jSdir j = 2 (5) 4πR1 from which we can find p dir 60P G jE j = T T (6) R p 1 60P G ≈ T T (7) d Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 4 / 52 Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 5 / 52 We can re-write the direct field strength as dir E jE j = 0 (8) d where p E0 = α 2PT (9) and p α = 30GT (10) E0 is known as the \unattenuated field intensity at unit distance" Antenna type α for d in km α for d in mi isotropic 173 108 short dipole 212 132 λ/2 dipole 222 138 short monopole* 150 93.2 λ/4 monopole* 157 97.5 Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 6 / 52 Table: Values for α in eqn (7.11) with E0 in millivolts/m, note that * antennas are fed against ground, to find unattenuated field intensity in presence of ground use 2E0 The magnitude of the reflected ray can be found from ref E jE j = j 0 Γj (11) d and the total field should be found by adding the direct and reflected fields in the proper phase relationship tot E0 jE j = e−jk0R1 + Γe−jk0R2 ; (12) d Since distances are likely to be large in terms of λ, it is better to use tot E0 jE j = 1 + Γe−jk0(R2−R1) (13) d Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 7 / 52 Examining the geometry, we can find 2 21=2 R2 = d + (h1 + h2) (14) 1 ≈ d + (h + h )2 (15) 2d 1 2 and 1 R ≈ d + (h − h )2 (16) 1 2d 1 2 so that 2h h R − R ≈ 1 2 ; (17) 2 1 d if we are near grazing, to obtain tot E0 jE j = 1 + Γe−j2k0h1h2=d (18) d dkm If jΓj ≈ 1 and h1h2 < 12 , we obtain fGHz tot E 4πh h jE j = 0 1 2 (19) d λd Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 8 / 52 Example: 1 GHz source 10 m above a perfectly conducting plane, plot received power as a function of height relative to free space for varying receiver heights at 1 km range. PEC, horizontal polarization: Γ = −1 tot E0 −j2k0h1h2=d Thus, jE j = d 1 − e This will be zero when 2k0h1h2=d = 2nπ Solve to find h2 = 15n m Field strength relative to free space is given by term in brackets Maximum value: 6 dB, minimum: -1 dB Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 9 / 52 1 GHz, 1 km distance, Transmitter height 10 m 40 35 30 25 Γ=−1 Γ=−0.5 20 Low antenna 15 Receiver Height (m) 10 5 0 −20 −15 −10 −5 0 5 10 Received power relative to free space (dB) Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 10 / 52 1 GHz, Transmitter/Receiver heights 10 m 10 0 −10 Γ=−1 −20 Γ=−0.5 Low antenna −30 −40 200 400 600 800 1000 1200 1400 1600 1800 2000 Relative Received Power (dB) Distance (m) Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 11 / 52 III. Fresnel zones Fresnel zones are defined as the locus of points around a direct ray that produce a specified phase delay if a reflection occurs there Convention is to define the nth Fresnel zone as the set of points at which reflection would produce an excess path length of nλ/2 Analytic geometry shows that these are ellipsoids of revolution centered on the direct ray, axes increase with increasing n When considered in a two dimensional plane, these become ellipses Since Γ ≈ −1 at grazing incidence, a signal reflected from the first Fresnel zone will be received in phase with the direct signal Thus, odd Fresnel zone reflections produce constructive interference, even Fresnal zone reflections produce destructive interference Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 12 / 52 2nd Fresnel zone 1st Fresnel zone Direct Ray Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 13 / 52 To derive the Fresnel zone radii at a specified distance d1 from one end of the path use r1 + r2 = d1 + d2 + nλ/2 (20) q q 2 2 2 2 d1 + Fn + d2 + Fn = d1 + d2 + nλ/2 (21) Approximating yields 2 2 Fn Fn d1 + + d2 + = d1 + d2 + nλ/2 (22) 2d1 2d2 and 2 Fn 1 1 2 d1 + d2 + = Fn = nλ/2 (23) 2 d1 d2 2d1d2 so r nλd d F = 1 2 (24) n D s nd d = 17:3 1;km 2;km (25) fGHz Dkm with Fn in m Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 14 / 52 C r1 r2 Fn B A d1 O d2 Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 15 / 52 0 0.6 F 1 F 1 20 F 2 F 3 F Fresnel Zone Size (m) 4 40 0 5 10 15 20 Distance (km) Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 16 / 52 Revisit our planar Earth direct plus reflected model in terms of Fresnel zones: transmit/receive heights h1 & h2, distance d First, write an equation for a line between transmitter image and receiver; intersection with plane is the reflection point Next, write an equation for a line between transmitter and receiver (the \direct path"); we can determine the height of this line above the reflection point As derived in Section 7.3.1 of the book, this clearance h divided by the first Fresnel zone radius at the reflection point F1 is r h h h = 2 1 2 (26) F1 λd When h=pF1 is unity, constructive interference occurs, when h=F1 = 2 destructive interference occurs, etc. 2 Also it turns out that π h is the phase difference between the F1 direct and reflected rays at the receiver As the receiver height increases, the lowest height at which a free space equivalent power occurs is when h=F1 = 0:577 Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 17 / 52 Terrain Reflection and Diffraction, Part Two 1 Path profile construction 2 Calculation of Earth \bulge" 3 Height gain curves 4 Obstructed region 5 Reflection region Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 18 / 52 I. Path profile construction Predicting propagation over a particular path requires knowledge of the terrain between Tx and Rx Such information can be obtained from topographical maps or from mapping agencies, although accuracy may be questionable Ground or aerial surveys can be used to obtain more accurate information, but these can get expensive On longer paths, Earth curvature and refraction become more important and we need to consider these Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 19 / 52 We'll use our Chapter 6 models: either straight rays-modified Earth radius or curved rays-flat Earth The former is more useful if we want to consider several antenna heights for a given κ value, the latter if we want to consider several different refractive conditions for fixed antennas Note when we draw paths the horizontal scale will usually be much larger than the vertical; this makes angles appear incorrect Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 20 / 52 II. Calculation of Earth \bulge" Using the figure and a theorem for chords in a circle, we have AD · DC = BD · DE (27) AD ≈ d1 (28) DC ≈ d2 (29) BD ≈ b (30) ED ≈ 2Re = 2κa (31) so that d1d2 = (2b)(κa) (32) and d d b = 1 2 (33) 2κa Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 21 / 52 d2 B d1 b= BD A C AD D DC Re - b= DO Re= AO Re= CO O Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 22 / 52 If we use d in km and b in m we find d d b = 1 2 (34) 12:74κ d2 Maximum value occurs at center of path, where b = 4(12:7κ) Note that the Earth bulge becomes larger as the antennas are spread further apart Note this \bulge" is not relative to the horizon, it is relative to the direct ray (through the Earth) between Tx and Rx This occurs because as Rx moves further from Tx it decreases in height so that it appears the Earth surface is \bulging" Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 23 / 52 For ground based links: We want to minimize costs so we should minimize the number of stations by spacing them far apart If they are too far apart the Earth \bulge" will block the line of sight under certain refractive index variations Even if we are not blocked, reflections from the ground can produce nulls in our pattern that change as refractive indices vary This would require a reflection coefficient magnitude near unity The aim of microwave link design is to assure proper path clearance and to avoid interference due to reflections by choosing paths and antenna heights Levis, Johnson, Teixeira (ESL/OSU) Radiowave Propagation August 17, 2018 24 / 52 Distance to the \radio horizon": Given the Earth bulge, we can find the distance to the \radio horizon" by drawing lines between specified transmitters and receivers and trying to determine whether the Earth bulge is intersected Derivation in Sect.