A SIMPLE APPROACH TO RADIO TRANSMISSIONS. Alessandro Iscra and Maria Teresa Quaglini, IIS Maserati, Via Mussini 22, 27058 Voghera (P avia, Italy ) Tel . : +39- 0383-43644, Fax: +39- 0383-62862, e-mail: firstname.lastname@example.org Translation: Fabrizia Rolla July 2003.
Alessandro Iscra and Maria Teresa Quaglini,
IIS Maserati, Via Mussini 22, 27058 Voghera (Pavia, Italy)
Tel.: +39-0383-43644, Fax: +39-0383-62862, e-mail: email@example.com
Translation: Fabrizia Rolla
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.
A bulb in the deep space
Let’s measure the el. pollution
A bulb and an antenna
The electromagnetic pollution
A transmission in the space
Transmitting by a mobile phone
Terrestrial… …radio links
A brief analysis of the results
A 9 km radio link…
A 55 km radio link…
Let’s turn on a bulb, placed in the space, that radiates a power P = 10 W evenly.
Let’s imagine a sphere around the bulb, with radiusR = 100 meters.
The area of the spherical surface isA = 4pR2 = 125600 m2.
In the empty space, energy diffuses without any form of leak phenomena.
Each point of this surface is crossed by a power density
S = P/A =
10 W / 125600 m2 =
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.
R = 100 m
We have introduced thepower density, that is defined asS = P/A.
Where P is the power that crosses vertically a surface and A is the area of the surface itself.
A = 125600 m2
The bulb approximates an isotropic (i.e. that radiates evenly in all directions) and punctiform source.
S = 79.6mW/m2
Which form of energy does the bulb radiate?
Frequencies of light emitted by a bulb are extremely high (1014Hz) and are distributed over a wide spectrum.
The bulb generates visible radiation, i.e. electromagnetic waves with an extremely high frequency.
Energy radiated by an antennahas (almost)a single frequency,collocated in the radio wave region of electromagnetic spectrum (104..1011 Hz) and it is coherent (as laser light).
Its wavelength is:
l = c/f (c = speed of light).
Is it possible to generate electromagnetic waves with other methods?
Yes, by applying a high frequency alternate voltage to an antenna, for instance.
Radiation emitted by an antenna is polarized, i.e. electric and magnetic vectors oscillate along well defined directions.
Which are the main differences between energy radiated by a bulb and energy radiated by an antenna?
An antenna is not an isotropic source: energy is radiated not evenly.
Let’s stay now in a place at
Rdistance from a trnsmitting antenna thatradiatesP watts 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).
With an isotropic antenna, the power density will be:
S = P/(4pR2).
The capture phenomena is intuitive in dish antennas:ifARis the area of the cross-section of the dish, the received power is:
PR = AR S.
Since an antenna is not isotropic,a factor Gis introduced, defined as antenna gain, so:
S = PG/(4pR2).
S = PG/(4pR2)
Each antenna has a “capture-area”. Moreover, each transmitting antenna can be a receiveing one.
Usually, we are interested inGreferred to the direction along which the antenna transmits the maximum amount of power.
PR = AR S
The capture-area and the gain are related by the formula: AR = l2GR/(4p).
AR = l2GR/(4p)
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.
Radio links “a” e “b” operate in line of sight and free-space conditions.
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.
airplanes or satellites
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).
Radio link “d” operates in line of sight conditions, but, probably, notin free-space conditions: the edge between the two antennas, will corrupt the received power (compared to the free-space conditions).
We have tested a radio link between theSecondary School “Maserati”of Voghera (Pavia, Italy), and a house placed in Ca’ Mori, a hillabove 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 wasR = 9 km.
The frequency used wasf = 433.3 MHz, (then, the wavelength wasl = c/f = 0.69 m). The gains of the two (identical) antennas wereG = GR = 14.5.
R = 9 km
Consequently, the capture area of the receiving antenna was: AR = GRl2/(4p) = 14.50.692/(4p) = 0.55 m2.
The power of the transmitter, placed in Cà Mori, was4.2 W, but only2.25 W were radiated by the antenna, due to the power dissipation of the cable placed between the transmitterand the antenna. Then, P = 2.25 W.
The estimated power density on“Maserati”was:
S = PG/(4pR2) = 2.2514.5/(4p90002) = 32.110-9 W/m2.
Consequently, the estimated received power was PR = ARS = 0.5532.110-9 = 17.6 nW, but we expected to measure 9.44 nW, according to receiving cable loss.
We measuredPR,MEAS= 10.5 nW, a good result.
We tested another link, between“Maserati”in Voghera, and apoint 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.
Also in this case we usedf = 433.3 MHz,l = c/f = 0.69 m,and two identical antennas withG = GR = 14.5, then AR = 0.55 m2.
By transmittingP = 2.25 W from Castelrocchero, the estimated power density on“Maserati”was:
S = PG/(4pR2) = 2.2514.5/(4p550002) = 85910-12 W/m2.
The estimated received power was:
PR = ARS = 0.5585910-12 = 472 pW,
but we expected to measure 253 pW,according to power loss associated to the receiving cable.
R = 55 km
PR,MIS= 28.1 pW, about one ninth of the whole expected value.
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.
d - R/2 [km]
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 operatesboth 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 lineconnecting 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 with0.1 pW; an FM receiver withfew pW.
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 or1800 MHz);
-the phone distance from the base station;
The gain of the antenna built inside a mobile phone has about an unitary gain.
P = ?
The base station has an antenna with GR20, and can receive well with PR = 1 pW.
Then, with a frequency of1800 MHz (l= 0.17 m), and with a distance R = 500 m, in free-space,P = PR (4pR/l)2/(GGR) = 68.2mW (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.
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/m2, that should be reduced to0.1 W/m2 in commonly crowded areas.
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 = HR0whereR0 = 377 V/A = 377W.
Moreover, S = EH = E2/ R0 = H2R0 , then: E = (S R0).
At the value of1 W/m2, are associated: E = 20 V/m, H = 0.05 A/m.
At the value of0.1 W/m2, are associated: E = 6 V/m, H = 0.016 A/m.
By using the above formulas, we are able to determine values of E and H atan adequate distance r from the transmitting antenna:
E = ?, H = ?
S = PG/(4pr2), E = (1/r)[PGR0/(4p)] , H = E/R0
The product PG is defined as EIRP (Equivalent Isotropic Radiated Power).