Radio Technology Metropolia/A. Koivumäki Propagation of radio waves, exercises: Solutions Solutions also done in Excel file http://users.metropolia.fi/~koiva/K2016/RadTech/propagation.exercises.answers.xls 1. Link budget equation: P ⋅ G ⋅ GR PR = T T L Here PT is transmit power, GT is gain of transmit antenna, GR is gain of receive antenna and L is path loss of the radio link. Link budget equation in dB: PR = PT + GT − L + G R Here, unit of power is dBm or dBW, unit of gain and loss is dB. These are the given values: • • 20 mW PT = 20 mW. In decibel milliwatts: PT = 10 ⋅ log dBm = 13.0 dBm 1 mW GT = GR = 25 dB So, we can solve the measured path loss from the link budget equation ⇒ L = 154 dB. Every radio link has at least the path loss given by the free space loss: 2 4π ⋅ r 2 4π ⋅ r = ⋅ , or 10 log L Lf = dB f λ λ where r is length of the link = 1.5 km and λ is wavelength = c0 3 ⋅108 m = = 5 mm. f 60 ⋅10 9 Hz So now 4π ⋅ 1500 m 2 L f = 10 ⋅ log dB =131.5 dB 0.005 m So, the measured path loss (154 dB) is 22.5 dB larger than the free space loss of the radio link. This means that there has to be something between the two antennas causing extra loss on the radio path. When the frequency is 60 GHz, this "something" is the oxygen of the atmosphere. Oxygen molecules resonate at 60 GHz, and that's why they absorb part of the energy of 60 GHz radio waves. Because the path length is 1.5 km, the extra loss can be said to be 15 dB/km. If we look at figure 10.2 in the textbook (the upper curve giving clear atmosphere attenuation at sea level), we see the same 15 dB/km at 60 GHz. 2. (This matter is discussed in chapter 10.4 of the textbook.). If an "over-the-horizon" radio link is required, the antenna masts have to be at least so high, that there is a line-of-sight from one mast to the other. This means that when looking from the top of each mast, the horizon has to be halfway between the masts. In this case this means that the horizon has to be at a distance of 22.5 km from each mast. This figure helps: The radio waves propagate along the dashed line. The required mast height h0 is given by equation (10.13) in the textbook: r2 h0 = H 2 ⋅ KR In this case the distance of the radio horizon is rH = r/2 = 22.5 km, so we get r2 ( r / 2) 2 r2 45000 2 m 2 = = = 28.9 m h0 = H = 2 ⋅ KR 2 ⋅ KR 8 ⋅ KR 8 ⋅ 8760000 m However, a mast height 28.9 m is not high enough for a 6 GHz radio link, because part of the Earth will be inside the first Fresnel ellipsoid: In order that the ellipsoid be free of obstacles, the antenna masts must be so high, that the whole ellipsoid is raised above the Earth: The extra height (needed in addition to h0) is denoted by h1. The equation of the first Fresnel ellipsoid is given by equation (10.12) of the textbook: λr1 r2 hF = r1 + r2 If the scale of the situation would be like that in the figure above, quite complicated geometrical and trigonometrical calculations would be needed to calculate at first r1 and r2 and then h1. But of course the figure is not drawn in a correct scale. In reality, distances r, r1 and r2 are all very much larger that heights h0, h1 and hF. So, quite accurately: r1 = r2 = r / 2 = 22.5 km c0 3 ⋅108 m/s = = 0.05 m , so the thickness of the Fresnel ellipsoid in the f 6 ⋅10 9 Hz middle of the radio path (where the ellipsoid is allowed to touch the Earth) is Wavelength is now λ = hF = λr1r2 = λ ⋅ ( r / 2) 2 = λr = λr = 23.7 m . r1 + r2 r 4 2 Because of the scale of things, quite accurately h1 = hF, so we get the total height of the masts: h = h0 + h1 = 52.6 m. However, because there is a 30 meter hill halfway, we need 30 meters more of mast height, as can be seen from the next figure. So, the total mast height must be h = h0 + h1 + h2 = 82.6 m. r = 45 km r1 h r2 F h1 h 2 = 30 m Hill h0 3. The calculations needed are very similar to those in exercise 2. The mast height h0 needed to raise the line-of-sight between the antennas just above the horizon is r2 . The extra mast height h1 needed to raise the first Fresnel ellipsoid above the Earth h0 = 8 ⋅ KR λr was derived in exercise 2: h1 = hF = . 2 Now we have masts with height h = h0 + h1 = 50 m. To get distance r we have to solve the r2 λr equation + = 50 m 8 ⋅ KR 2 This leads to a fourth-order equation which cannot be solved in a closed form. Numerical solution is needed, e.g. Excel can be used. The results: f/GHz 1 10 40 r/km 23.6 46.8 53.0 4. Situation is like this:: Chapter 10.5 of the textbook gives this kind of information about the situation: If the radio waves used in this kind of radio link are horizontally polarized, the two signals (direct signal and reflected signal) going to the receive antenna will have a phase difference of 180° if the difference of path lengths r0 and r1+r2 is equal to a whole number of wavelengths. I.e. if r1 + r2 − r0 = n ⋅ λ . And we know that if the phase difference of two signals of equal amplitude and equal frequency is 180°, the sum of the two signals is zero. So, if this is the case, no signal is received if the previous equation is true. Looking at the geometry of the situation we get the condition for this worst-case scenario as follows: 2h h If 1 2 = n ⋅ λ , received signal = 0. d 2h h 2 ⋅ 80 m ⋅ 60 m = 1 .2 m = 1 ⋅ λ Now wavelength is λ = 1.2 m, and 1 2 = d 8000 m So, we now do have a situation where ground reflection ruins reception because the reflected wave cancels the direct signal. We see from figure 10.11b that moving the receive antenna away from this kind of worst-case height will raise the received field strength value from zero. From the text of the book we see that the best situation happens when 2h1h2 λ = (2n + 1) ⋅ d 2 If this is the case, the two signals will have the same phase angle, and the total signal strength will be twice as much as the level of the direct signal. Using the previous equation, we find that the optimum situation is now reached by moving the receive antenna from its original height of 60 m to a height of 30 m or 90 m. But we can also calculate that a much smaller movement of the antenna may give satisfactory results, because, as figure 10.11b shows, the total field strength rises quite steeply, when the antenna is moved even a little away from the worst height. E.g. we can find that by moving the receive antenna 2 m upwards the total field strength rises from zero to about E0/5, where E0 is the field strength of the direct signal. 5. We have to calculate 12 values of free space loss using the equation 4π ⋅ r 2 Lv = 10 ⋅ log dB λ and 12 path losses given by Okumura-Hata model. Of course, e.g. Excel is a good tool for making these calculations. Here are the results: r/km 1 5 10 20 f = 450 MHz f = 900 MHz f = 1500 MHz Lfree/dB LO-H/dB Diff./dB Lfree /dB LO-H/dB Diff./dB Lfree /dB LO-H/dB Diff./dB 85.5 118.5 33.0 91.5 126.4 34.9 96.0 132.2 36.3 99.5 143.2 43.7 105.5 151.0 45.5 109.9 156.8 46.9 105.5 153.8 48.3 111.5 161.6 50.1 116.0 167.4 51.5 111.5 164.4 52.8 117.5 172.2 54.7 122.0 178.1 56.1 An example of calculations for f = 900 MHz and r = 10 km: 2 4π ⋅ r 2 4π ⋅ 10000 m dB = 111.5 dB L free = 10 ⋅ log dB = 10 ⋅ log 8 6 3 ⋅ 10 / 900 ⋅ 10 m λ LO − H = 69.55 + 26.16 ⋅ log(900 ) − 13.82 ⋅ log(30 ) + [44.9 − 6.55 ⋅ log(30 )]⋅ log(10 ) dB =161.6 dB ( ) 6. Let's calculate field strength at a distance of 1 km from the transmitter. Transmit power is PT = 1000 W, transmit antenna gain is GT = 0 dB, or GT = 1. So, power density at r = 1 km is G ⋅P 1 ⋅1000 W = 79.6 µW/m 2 S = T 2t = 4πr 4π ⋅10 6 m 2 E2 Field strength and power density have this dependency: S = , so we get field strength: 2Z 0 E = 2 Z 0 S = 2 ⋅ 377 Ω ⋅ 79.6 µW/m 2 = 0.245 V/m . In decibel units (dBµV/m): 0.245 V/m dBµV/m = 107.8 dBµV/m , which is in good agreement with the value E = 20 ⋅ log 1 µV/m given by figure 10.17. When distance increases 10-fold, power density S = GT ⋅ Pt decreases 100-fold, and field 4πr 2 2Z 0 GT PT S 1 ⋅ decreases 10-fold, or 20 dB. The dashed line in figure r 4π 10.7 also goes down 20 dB for each 10-increase of distance. So, the statement of the exercise really is true. strength E = 2 Z 0 S = 7. See Excel file. 8. See Excel file. 9. After the new building is built the situation looks like this: Base station antenna Your antenna To get rid of the extra attenuation, you must raise your antenna so hight that the whole of the 1st Fresnel ellipsoid rises above the new building halfway between the antennas, where the ellipsoid λr1 r2 is at its thickest. The half-thickness of the ellipsoid is given by hF = . r1 + r2 Now r1 = r2 = r/2 = 2 km. (Here r = 4 km is the distance between the antennas.) We get hF = λr1r2 = λ ⋅ ( r / 2) 2 = λr = λr = 9.26 m. (Wavelength is λ = 0.0857 m.) 4 2 r1 + r2 r So, your antenna must be lifted so much that the line between your antenna and base station antenna rises 9.26 m above the new building. This is what must be achieved:: Base station antenna New line-of-sight New place of your antenna h F Original line-of-sight Original place of your antenna To raise the midpoint of the line by hF = 9.26 m, the left end of the line (i.e. place of your antenna) must of course be raised by 2hF = 18.52 m. 10. The "Fresnel Zone Clearance and Antenna Height Calculator" at http://www.afar.net/fresnel-zone-calculator/ works a bit differently than our Excel calculations. Let's use this calculator to solve our propagation exercise 2: Find the minimum antenna height needed in a link hop between Helsinki and Lohja. Hop length = 45 km, there is a hill of 30 m height halfway between the antenna towers. Frequency is 6 GHz. The calculation sheet for this exercise is below. Some explanations: • "k factor" is the same as K in KR = 8760 km. We have used K = 1.375, but the Excel sheet only accepts values with two decimal places. We use the value 1.38, so now KR = 8791 km. • "Percent of 1st Fresnel Zone (typical 60%)" means that here it is allowed that there are obstacles inside the 1st Fresnel ellipsoid. The %-value of e.g. 60% would mean that the distance from the line-of-sight between antennas to an obstacle below line-of-sight is allowed to be 0.60⋅hF. In our own calculations we have used 100%, and are using it now. • This calculator does not find the mast heights needed, but instead calculates the "Clearance between Earth and Fresnel Zone" for given mast heights. Because in our exercise we have a 30 m hill in the middle of the link hop, we must find antenna heights at site 1 and site 2 so that clearance is 30 m at distance 22.5 km from site 1. This has been done below. Result: Antenna heights must be 82 m. 11. (No answer here.)
