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A SIMPLE APPROACH TO RADIO TRANSMISSIONS Alessandro Iscra and Maria Teresa Quaglini, IIS Maserati, Via Mussini 22, 27058 Voghera (Pavia, Italy) Tel.: +39-0383-43644,

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Presentation on theme: "A SIMPLE APPROACH TO RADIO TRANSMISSIONS Alessandro Iscra and Maria Teresa Quaglini, IIS Maserati, Via Mussini 22, 27058 Voghera (Pavia, Italy) Tel.: +39-0383-43644,"— Presentation transcript:

1 A SIMPLE APPROACH TO RADIO TRANSMISSIONS Alessandro Iscra and Maria Teresa Quaglini, IIS Maserati, Via Mussini 22, Voghera (Pavia, Italy) Tel.: , Fax: , Translation: Fabrizia Rolla July 2003 A bulb and an antenna A transmission in the space Terrestrial… …radio links A 9 km radio link…A 55 km radio link… A brief analysis of the results Transmitting by a mobile phone The electromagnetic pollution Let’s measure the el. pollutionA bulb in the deep space This presentation is related to a project (“An introduction to radiocommunications systems and electromagnetic hazards”) that was developed in our school. This work was originally designed for students of technical secondary schools. During an Italian exhibit, (“The Science Day”, Pavia, Italy, May 2003) many young people were interested to radiocommunications topics. This interest is stimulated by the great development of modern wireless system, such as mobile phones, satellite television receivers, wireless computer networks, etc. This presentation summarizes a work realized into our school and also demontrates that basics of radiocommunications can be shown in very simple ways (i.e. by comparing an antenna with a common bulbe). We suggest to follow the sequence of slides here traced.

2 Let’s turn on a bulb, placed in the space, that radiates a power P = 10 W evenly. In the empty space, energy diffuses without any form of leak phenomena. A bulb in the deep space Let’s imagine a sphere around the bulb, with radius R = 100 meters. R = 100 m If we see the bulb from an adequate distance, it will appear as a small point. In this case, the bulb seems a punctiform power source. The bulb approximates an isotropic (i.e. that radiates evenly in all directions) and punctiform source. The area of the spherical surface is A = 4  R 2 = m 2. A = m 2 Each point of this surface is crossed by a power density S = P/A = 10 W / m 2 = 79.6  W/m 2. S = 79.6  W/m 2 We have introduced the power density, that is defined as S = P/A. Where P is the power that crosses vertically a surface and A is the area of the surface itself.

3 ~ Which form of energy does the bulb radiate? The bulb generates visible radiation, i.e. electromagnetic waves with an extremely high frequency. Is it possible to generate electromagnetic waves with other methods? Yes, by applying a high frequency alternate voltage to an antenna, for instance. Which are the main differences between energy radiated by a bulb and energy radiated by an antenna? A bulb and an antenna Frequencies of light emitted by a bulb are extremely high (10 14 Hz) and are distributed over a wide spectrum. Energy radiated by an antenna has (almost) a single frequency, collocated in the radio wave region of electromagnetic spectrum ( Hz) and it is coherent (as laser light). Its wavelength is:  = c/f (c = speed of light). Radiation emitted by an antenna is polarized, i.e. electric and magnetic vectors oscillate along well defined directions. An antenna is not an isotropic source: energy is radiated not evenly. ~ E H

4 A transmission in the space Let’s stay now in a place at R distance from a trnsmitting antenna that radiates P watts power. R With an isotropic antenna, the power density will be: S = P/(4  R 2 ). Since an antenna is not isotropic, a factor G is introduced, defined as antenna gain, so: S = P  G/(4  R 2 ). Usually, we are interested in G referred to the direction along which the antenna transmits the maximum amount of power. How does a receiving antenna work? A receiving antenna “captures” a small amount of power crossing around itself. The captured power is transmitted to the radio receiver (i.e. by a cable). The capture phenomena is intuitive in dish antennas: if A R is the area of the cross- section of the dish, the received power is: P R = A R S. Each antenna has a “capture-area”. Moreover, each transmitting antenna can be a receiveing one. The capture-area and the gain are related by the formula: A R = 2 G R /(4  ). P R = A R S A R = 2 G R /(4  ) S = P  G/(4  R 2 )

5 Terrestrial (and satellite) radio links Performances of terrestrial radio links are corrupted by obstacles placed between the transmitting and the receiving antennas, by refraction, reflection and leack phenomena due to atmosphere and various obstacles. These effects are almost negligible (except absorbtion of microwaves due to rain) if obstacles are adequately far from the line joining the transmitting and receiving antennas ( “a” and “b” links). a b towards airplanes or satellites Radio links “a” e “b” operate in line of sight and free- space conditions. d Radio link “c”, used for transmission to a village, probably works without both line of sight and free-space conditions. The simple formulas used for free-space conditions are absolutely not valid in this case. c Radio link “d” operates in line of sight conditions, but, probably, not in free-space conditions: the edge between the two antennas, will corrupt the received power (compared to the free-space conditions).

6 A 9 km radio link operating in line of sight and free-space conditions We have tested a radio link between the Secondary School “Maserati” of Voghera (Pavia, Italy), and a house placed in Ca’ Mori, a hill above Salice Terme (Pavia, Italy). Antennas on “Maserati” and in Ca’Mori were placed respectively at an altidude of 114 m and 255 m above the sea-level. The distance between antennas was R = 9 km. Maserati 114 m Cà Mori 255 m The frequency used was f = MHz, (then, the wavelength was = c/f = 0.69 m). The gains of the two (identical) antennas were G = G R = The power of the transmitter, placed in Cà Mori, was 4.2 W, but only 2.25 W were radiated by the antenna, due to the power dissipation of the cable placed between the transmitter and the antenna. Then, P = 2.25 W. R = 9 km The estimated power density on “Maserati” was: S = P  G/(4  R 2 ) = 2.25  14.5/(4  ) = 32.1  W/m 2. Consequently, the capture area of the receiving antenna was: A R = G R 2 /(4  ) = 14.5  /(4  ) = 0.55 m 2. Consequently, the estimated received power was P R = A R S = 0.55  32.1  = 17.6 nW, but we expected to measure 9.44 nW, according to receiving cable loss. We measured P R,MEAS = 10.5 nW, a good result. P=2.25W S=32.1 nW/m 2 P R =9.44 nW P R,MEAS =10.5 nW

7 A 55 km radio link, operating only in line of sight conditions We tested another link, between “Maserati” in Voghera, and a point placed in Castelrocchero (Asti, Italy). Antennas on “Maserati” and in Castelrocchero were placed respectively at an altitude of 111 m and 405 m above the sea-level. The distance between antennas was R = 55 km. Maserati 111 m Castelrocchero 405 m R = 55 km Also in this case we used f = MHz, = c/f = 0.69 m, and two identical antennas with G = G R = 14.5, then A R = 0.55 m 2. By transmitting P = 2.25 W from Castelrocchero, the estimated power density on “Maserati” was: S = P  G/(4  R 2 ) = 2.25  14.5/(4  ) = 859  W/m 2. The estimated received power was: P R = A R S = 0.55  859  = 472 pW, but we expected to measure 253 pW, according to power loss associated to the receiving cable. We measured: P R,MIS = 28.1 pW, about one ninth of the whole expected value. P=2.25W S=859 pW/m 2 P R =253 pW P R,MIS =28.1 pW Note: In radio engineering, and in absence of free-space conditions, considerable differences between the estimated (by formulas valid in free-space) and measured values are normal and predictable.

8 A brief analysis of the results As you can see by the path profile (obtained by a map), the link between Salice Terme and Voghera operates both in line of sight and in free-space conditions: all obstacles are deeply far from the line traced between the transmitter antenna and the receiver one. By observing the path profile associated with the link between Castelrocchero – Voghera you can see that the line connecting the two points does not intercept any obstacle (line of sight conditions verified), but near “Maserati” this line is closed to the ground (and buildings): the free-space conditions are not verified. Is it possible to objectively establish if the free-space conditions are verified or not? Yes, but the method is quite complex and cannot be explained in this simple presentation. Can a mobile phone or a radio receiver work with the very low power values measured? Yes (in many cases): a mobile phone works well with 0.1 pW; an FM receiver with few pW d-R/2 [km] y [m] Maserati Cà Mori Ground (sea-level) Path profile

9 Transmitting by a mobile phone How many power does a GSM mobile phone need to be received from a base terrestrial station? It depends on many factors, such as: -the frequency (900 or 1800 MHz); -the phone distance from the base station; -the obstacles. P = ? The gain of the antenna built inside a mobile phone has about an unitary gain. The base station has an antenna with G R  20, and can receive well with P R = 1 pW. Then, with a frequency of 1800 MHz ( = 0.17 m), and with a distance R = 500 m, in free- space, P = P R (4  R/ ) 2 /(GG R ) = 68.2  W (a very low value!) is adequate. Due to obstacles and interference at the base stations, a mobile phone needs to transmit a greater power: from tenths of milliwatts to few watts. This power is controlled from the base station.

10 The electromagnetic pollution The electromagnetic energy radiated by an antenna crosses trough our bodies. Which are its effects? A very high power density (i.e. when a mobile phone is placed closed to our head) causes thermal (and, probably, biological) effects. A lower power density may cause biological effects, that many researchers are currently sudying. For frequencies between 3 MHz and 3 GHz, Italian laws limit the human exposure to the maximum power density of 1 W/m 2, that should be reduced to 0.1 W/m 2 in commonly crowded areas.

11 Let’s measure the electromagnetic pullution Electromagnetic waves are composed by an electric field vector and a magnetic field vector. At a great distance frome the transmitting antenna the two vectors oscillate along directions that are mutually orthogonal. In the empty space and, approximately, on the air, the strengths of the two vectors are related by the following formula: E = HR 0 where R 0 = 377 V/A = 377 . Moreover, S = EH = E 2 / R 0 = H 2 R 0, then: E =  (S R 0 ). E [V/m] H [A/m] S [W/m 2 ] Direction of propagation At the value of 1 W/m 2, are associated: E = 20 V/m, H = 0.05 A/m. At the value of 0.1 W/m 2, are associated: E = 6 V/m, H = A/m. By using the above formulas, we are able to determine values of E and H at an adequate distance r from the transmitting antenna: S = PG/(4  r 2 ), E = (1/r)  [PGR 0 /(4  )], H = E/R 0 The product PG is defined as EIRP (Equivalent Isotropic Radiated Power). E = ?, H = ? r


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