Antennas: Problems and exercises: Answers

Transcription

1 adio echnology Metropolia/A. Koivumäki Antennas: Problems and exercises: Answers 1. he maximum transmit power of a.4 GHz WLAN base station is 13 dbm and the gain of the transmit antenna is 3.5 dbi. Find out, what is the electric field strength at these distances from the base station: a) 1 meter b) 1 meters c) 1 meters (We assume free space propagation, which of course is often not the case in reality in WLAN's.) P Solution: Power density at distance r from transmit antenna: S = G r lectric field strength from power density: = Z S, where Z = 377 Ω. Now G = 1 3.5/1 ja P = 1 13/1 mw. We get: a) = 1.64 V/m b) =.164 V/m c) = 16.4 mv/m. Compare the results you got in exercise 1 to the "safe field strength value" given by e.g. the UK Health Protection Agency/adiation Protection Division in the document ebfile/hpaweb_c/ From there you can find the recommendation that the field strength exposure at.4 GHz should not exceed 61 V/m ("exposure guideline for general public"). How many db is the difference between the calculated WLAN field strengths and the given maximum exposure value? Solution: he value 61 V/m given above comes from table A4 on page 34 of the document. In the table, also the value of the corresponding power density is given, it is 1 W/m. From this value we can conclude that 61 V/m in the table is the rms value of the field strength. How can we conclude that? By noticing 61 that 1. his means that power density given in the table is calculated using the equation 377 S = instead of the equation S = which we have used e.g. in the previous exercise. Our Z Z equation S = assumes that the value of is the peak value of the field strength. So, in order Z to compare the safety limit given in the table to the value we calculated in the previous exercise, we have to convert the rms value 61 V/m into peak value, which of course is 61 V/m = 86 V/m. After this we can answer the question: he field strength calculated at a distance of 1 m in the previous exercise is V/m. his is log db = 34 db below the safety limit given in the document Doesn't sound very dangerous. 3. he maximum transmit power of a GSM phone in a 9 MHz network must be somewhere between 1 W... 4 W. Let's assume that in your phone this maximum power is 1.5 W. he gain of the phone antenna is about dbi. How near the antenna must you be in order that the field strength of the antenna radiation exceeds the maximum 9 MHz exposure limit of 41 V/m?

2 (he answer is rather surprising, and maybe a bit worrying, but fortunately the reality is quite a bit more complicated, and not so worrying.) Solution: Again, the safety limit 41 V/m is an rms value, so the corresponding power density is W P S = = From equation S = G Z we must now solve distance r, and we get m r. G P W r = = = 1cm. S 4.46 W/m So, if the antenna of the phone is closer than 1 cm, field strength (rms value) will exceed the value 41 V/m. However, this kind of simple calculation does not tell very much about the safety or unsafety of mobile phone use. eality is quite a lot more complicated. 4. A 36 GHz link hop of 4 km hop length uses 5 centimeter reflector antennas. What must be the transmit power in order that the radio link connection functions also during heavy rain (5 mm/h)? Functioning of the connection requires a received signal power of 5 dbm. Solution: Link budget equation: P = P + G L + G. Now transmit power P is unknown, we get it with P = P G + L G. c Values : Wavelength: = f m/s = = Hz.833 m π π.5 m Gain of antennas: G = = 1 log η D G db = 1 log.7 db = db.833 m (Surface efficiencies of the antennas were not given, here η =.7 is used for both antennas.) Free space loss: 3 r r 4 1 m L f = 1 log db = log db = log db = dB.8333 m ain causes medium loss. From figure 1.3 of the textbook we see that heavy rain of 5 mm/h causes an attenuation of about 1 db/km at 36 GHz:

3 So, total medium loss caused by rain in a 4 km link hop is L m = 4 db, which means that the total path loss is now L = L f + L m 176 db. equired transmit power is P = P G + L G = 38.1 dbm = 6.4 W. 5. A radio amateur has a 1 GHz db transmitter with a transmit power of 1 mw and an antenna with 5 db gain. he antenna is directed to the north, the radiation pattern of the antenna is in the figure. a) How much is the power density and field strength at a distance of 5 km in the north? b) How much is the power density and field strength at a distance of 5 km in the east and in the west? c) At a certain distance in the north, it is possible to receive the transmission with a 5 cm reflector antenna. How large an antenna is needed to receive the transmission at Angle (degrees) the same distance in the east and in the west? (Free space propagation is assumed, attenuation of the atmosphere is assumed to be db.).5 G P 1 1 mw Solution: a) S = = = 1.1nW/m r 5 m S = = Z S = 377 Ω 1.1 nw/m =.8 mv/m. Z b) We see from the radiation pattern that in the west power density and field strength are 8 db smaller than in the north. So, in the west they are 16. pw/m and 11 µv/m. In the east they are 31.5 db smaller than in the north, or 7.1 pw/m and 73.3 µv/m. c) he signal power received by an antenna is given by P = Ae S, where A e is the effective area (or capture area) of the receive antenna. So, in order to receive power P from an P electromagnetic field of power density S, an antenna with effective area Ae = is needed. S 1.1 nw/m Because the ratio of power densities in the north and in the west is = 631, the 16. pw/m effective area of the antenna needed in the west is 631 times the effective area of the antenna needed in the north. For a reflector antenna, effective area is given by A e = η A, where η is the aperture efficiency (or surface efficiency) of the antenna and A is the physical area of the reflector. If we assume that the reflector rim is a circle of diameter D, then A = πd. So, in 4 order to have a reflector antenna with an effective area 631 times as large as that of the.5 meter antenna used in the north, a reflector with a physical area 631 times as large as the reflector used in the north is needed. his means that the diameter of the reflector needed in the west must be 631 times as large as the diameter of the reflector used in the north, or m = 1.6 m. In a similar way we can calculate that a reflector with a diameter of 18.8 m is needed in the east.

4 6. You have to build a reflector antenna with a gain of 48 db at 18 GHz. What must be the diameter of the reflector? A couple of points to be taken into account: You use a corrugated horn as the feed antenna of the reflector antenna. With good design, using a corrugated feed horn, you can achieve a surface efficiency of 78 % with an ideal parabolic reflector. However, it is not possible to manufacture an ideal parabolic reflector. If we assume that the average deviation of the reflector surface from an ideal parabolic form is.4 mm, it can be calculated that the gain of the antenna is decreased by.5 db (compared to using an ideal paraboloid). Solution: Gain of reflector antenna: π G = 1 log η D dbi. 1 G /1 Now G = 48 db, η =.78, = mm, so we can solve D = = 1.51 m π η his would be the size of an ideal reflector. However, with a real life reflector the actual gain of the antenna would be 47.5 db. So the actual aperture efficiency (=surface efficiency) of the G /1 1 antenna will be η = =.694. Using this value for η instead of.78, we find the actual πd reflector size required for 48 db gain: D = 1.6 m. 7. Frequency is 1 GHz. A pyramidal horn with aperture dimensions a 1 = cm, b 1 = 15 cm is attached to a rectangular waveguide (size: a= 19 mm, b = 9.5 mm). What is the gain of this horn antenna, if the length of the horn L is a) L = 5 cm b) L = 5 cm he gain is found by using equation (9.51) and figure 9.4 of the textbook. Solution: quation (9.51) gives the directivity of a pyramidal horn antenna: Numerical values are obtained from figure 9.4: πd DH D 3ab

5 xplanation: f = 1 GHz = 5 mm a 1 / = 8, b 1 / = 6 a) L/ = 1 red horizontal lines give D = 7 and DH a b πd DH π D DH D = = 84.8 = 19.3 db 3ab 3 a b = 3 b) L/ = green horizontal lines give D = 5 and DH a b πd DH π D DH D = = 36.7 = 5.1dB 3ab 3 a b = 64 Directivities were calculated, but because gain = directivity minus losses, gain values are quite accurately the same, because losses in a horn antenna are very small.