ITU-R Rec. P68-8 gives the following expression for the atmospheric noise temperature as seen by the receiving antenna: T atm L T 0 atm m 0 T m is the effective temperature (K) of the atmosphere, a common value is 75 K Based on the atmospheric loss calculated with the algorithms in ITU-R Rec. P676, the above noise temperature equation gives results for: Atmospheric noise temperature as function of frequency of terrestrial paths of different lengths Atmospheric noise temperature as function of frequency of slant paths with different elevation angles Observations: The figures show that for terrestrial path up to 30 km length the noise temperature is quite low (< 0 K) for frequencies less than 0 GHz. The same is true for slant paths with elevation angles less than 0 degrees- At the absorption peaks at 60, 9.75, 83.3, and 35. GHz the noise temperature approaches the effective physical temperature.
Atmospheric noise temperature on terrestrial paths 000 K 00 0 d = 300 km d = 00 km 0 0 0 0 d = 30 km d = 0 km 0 0 0 d = 3 km d = km 0. 0 Slant_path_loss.dsf 0 0 0 p = 03 hpa t = 88 K = 7.5 g/m 3 0 0 00 000 f/ghz
Atmospheric noise temperature on slant paths 000 K 00 0 = 5 o = 7 o = 0 o = 5 o = 0 o = 5 o = 30 o 0 = 35 o p = 03 hpa t = 88 K 0 = 7.5 g/m 3 0 0 0 = 40 o 0 = 45 o = 50 o = 55 o 0 = 60 o 0 0 = 70 o = 80 o = 90 o Slant_path_loss.dsf 0 f/ghz 00 000
Example (replaces slide 43) Wanted signal carrier in an FM-audio receiver at 00 MHz Signals in the band 87.5 08 MHz will pass the receiver input band filter The IF is 0 MHz The local oscillator frequency is 90 MHz Which carrier frequencies are potentially harmful due to receiver nonlinearity and local oscillator harmonic components? kf c +lf c f IF f ci f Input signal Band filter Non-linear amplifier Mixer IF-filter f c f c f BPF Local oscillator f lo BPF f 3f lo 5f lo
Derivation of the expression of critical carrier frequencies (Insert after slide 44) Bif Bif kf i f f, f kf i f f kf i f f c lo if if c lo if c lo if This can be written as four separate equations: fif i flo i flo fif kfc i flo fif fc k k f i f i f f kfc i flo fif fc k k i flo f if fc k if lo lo if
Example: (replaces page ) bo 0.95exp j46 co 0.45exp j50 fc ( 6.8 GHz) 73.5 ps 3 5ns channel bandwidth 40 MHz phasor phase shift over channel bandwidth Path : 40 73.5 0 6 360.06 Path 3: 40 50 3 360 7.0 path 3 phasors at channel extreme frequencies path phasors at channel extreme frequencies resulting phasors at different channel frequencies path phasor equi_3path.dsf
Outage improvement of vertical space diversity with maximum ratio combining (replaces p. 35) For flat fade outage ITU-R Recommendation P.530- gives P out, FF, div P out, FF I sd, FF where the flat fade diversity improvement is.04 I 0.87 0. 0.48 0.( ), exp 0.04 0 FFM V sd FF S f d po S is centre to centre spacing (m), S 3 m, 3m f is the carrier frequency (GHz) d 43 km,40km 5 km,40km d is the hop length (km) p o is the multipath occurrence factor (%) FFM is the flat fade margin V G G (dbi)
The example continues (replaces p. 38) Predict the total outage in a radio link with space diversity in Southern Finland having the following parameters: d = 40 km f c = 6.8 GHz h rx = 00 m h tx = 00 m FFM = 35 db dn = 400 NU/km B MP = B NMP =8 db W MP = W NMP = 40 MHz Signature parameters are determined with = 6.3 ns S = 0 m G = G V = 0 db Calculated before: p o = 77.75 P out,ff = 0.046% = 0.53 P out,sf = 0.04% P out = 0.0484% I sd,ff = 77.0
The flat fade outage probability is (replaces p. 39) P out, FF, div Pout, FF 0.046 % 3.9 0 4 % I 77.0 sd, FF The non-selective correlation coefficient is (using the absolute value of the FF outage probability) k I P 0.875 0.53 sd, FF out, FF 77.0 0.00046 ns which gives r w.70 ns kns 0.9746 k, 0.6.034 0.69 k, ns kns 0.6.034 r 0.69 0.875 0.995 w
Outage improvement with frequency diversity (replaces p. 43) For flat fade outage ITU-R Recommendation P.530- gives P out, FF, div P I out, FF fd, FF where the flat fade diversity improvement is given by I fd, FF 80 f 0 fd f 0.FFM f is the diversity frequency spacing in GHz, if f >0.5GHz, this value is used f is the carrier frequency in GHz, < f < GHz d is the path length in km, 30 < d < 70 km f/f < 0.05 Otherwise the procedure is the same as for space diversity
The example continues (replaces p. 44) Predict the total outage in a radio link with space diversity in Southern Finland having the following parameters: d = 40 km f c = 6.8 GHz h rx = 00 m h tx = 00 m FFM = 35 db dn = 400 NU/km B MP = B NMP =8 db W MP = W NMP = 40 MHz Signature parameters are determined with = 6.3 ns f = 0.6 GHz Calculated before: p o = 77.75 P out,ff = 0.046% = 0.53 P out,sf = 0.04% P out = 0.0484% I fd,ff =.9
The flat fade outage probability is (replaces p. 45) P out, FF, div Pout, FF 0.046 %. 0 3 % I.9 fd, FF The non-selective correlation coefficient is k I P 0.965 0.53 fd, FF out, FF.9 0.00046 ns which gives r w.70 ns kns 0.9746 k, 0.6.034 0.69 k, ns kns 0.6.034 r 0.69 0.965 0.978 w
Outage improvement with combined space and frequency diversity with two receivers (replaces p. 49) Step. The non-selective correlation coefficients are calculated as above for both diversity types: k ns, sd I P sd, FF out, FF k ns, fd where I sd, FF I P fd, FF out, FF P P out, FF out, FF, sd tx f rx tx BU rx f f f Div. comb. space_div.dsf I fd, FF P P out, FF out, FF, fd
The example continues (replaces p. 5) Predict the total outage in a radio link with space diversity in Southern Finland having the following parameters: d = 40 km f c = 6.8 GHz h rx = 00 m h tx = 00 m FFM = 35 db dn = 400 NU/km B MP = B NMP =8 db W MP = W NMP = 40 MHz Signature parameters are determined with = 6.3 ns S = 0 m f = 0.6 GHz Calculated before: p o = 77.75 P out,ff = 0.046% = 0.53 P out,sf = 0.04% P out = 0.0484% I sd,ff = 77.0 I fd,ff = 30.3 k ns,sd = 0.9355 k ns,fd = 0.983
Outage improvement with combined space and frequency diversity with four receivers (replaces p. 57) Step. The non-selective correlation coefficients are calculated as above for both diversity types: k k ns, sd ns, fd where I sd, FF I I P sd, FF out, FF P fd, FF out, FF P P out, FF out, FF, sd BU rx tx f rx3 f tx f BU BU rx4 f f rx f Div. comb. space_div.dsf I fd, FF P P out, FF out, FF, fd
The example continues (replaces p. 60) Predict the total outage in a radio link with space diversity in Southern Finland having the following parameters: d = 40 km f c = 6.8 GHz h rx = 00 m h tx = 00 m FFM = 35 db dn = 400 NU/km B MP = B NMP =8 db W MP = W NMP = 40 MHz Signature parameters are determined with = 6.3 ns S = 0 m f = 0.6 GHz Calculated before: p o = 77.75 P out,ff = 0.046% = 0.53 P out,sf = 0.04% P out = 0.0484% I sd,ff = 77.0 I fd,ff = 30.3 k ns,sd = 0.9355 k ns,fd = 0.983
The example continues (replaces p. 69) Predict the total outage in a radio link with angle diversity in Southern Finland having the following parameters: d = 40 km f c = 6.8 GHz h rx = 00 m h tx = 00 m FFM = 35 db dn = 400 NU/km B MP = B NMP =8 db W MP = W NMP = 40 MHz Signature parameters are determined with = 6.3 ns = 0.5 = 0.5 =.0 G m = 40 NU/km Calculated before: p o = 77.75 P out,ff = 0.046% = 0.53 P out,sf = 0.04% P out = 0.0484% The average angle of arrival is 5 5.89 0 Gmd.89 0 40 40 0.046
Rain attenuation estimation procedure (Add these 6 pages after page 30) h R D: Earth-space path A: frozen precipitation B: rain height C: liquid precipitation (h R h s ) Ls h s Slant_path_loss.dsf L G
Step : Determine the rain height, h R, as given in Recommendation ITU-R P.839. Step : For 5compute the slant-path length, L s, below the rain height from: L s ( h h ) R sin s km For 5, the following formula is used: L s sin ( h h ) R s ( hr hs) Re / sin km If h R h s is less than or equal to zero, the predicted rain attenuation for any time percentage is zero and the following steps are not required.
Step 3: The horizontal projection, L G, of the slant-path length is: L G = L s cos km Step 4: Determine the rainfall rate, R 0.0, exceeded for 0.0% of an average year Step 5: Obtain the specific attenuation, R, using the frequencydependent coefficients given in Recommendation ITU-R P.838 and the rainfall rate, R 0.0, determined from Step 4, by using: R k (R 0.0 ) db/km Step 6: Calculate the horizontal reduction factor, r 0.0, for 0.0% of the time: r 0.0 L 0.78 G R 0.38 e f L G
Step 7: Calculate the vertical adjustment factor, v 0.0, for 0.0% of the time: h tan R hs degrees LG r0.0 L For G r L 0.0 R km else, L cos If 36, 36 R ( hr hs) sin degrees else, 0 degrees ν 0.0 km /( ) L sin 3 e R R 0.45 f Step 8: The effective path length is: L E L R 0.0 km
Step 9: The predicted attenuation exceeded for 0.0% of an average year is obtained from: A 0.0 R L E db Step 0: The estimated attenuation to be exceeded for other percentages of an average year, in the range 0.00% to 5%, is determined from the attenuation to be exceeded for 0.0% for an average year: If p % or 36: 0 If p < % and < 36 and 5: 0.005( 36) otherwise: 0.005( 36) +.8 4.5 sin A p (0.655 0.033ln( p) 0.045ln( A0.0 ) ( p)sin ) p 0.0 db A 0.0
Example R 0.0 = 50 mm/h h s = 0 km = 0-90 = 0 f = 30 GHz k = 0.9, = 0.99 R e = 8500 km h R = 4 km
P5. A radio receiver has a noise figure of 6 db and includes a modem that requires a db signal to noise ratio for proper performance in a 0 khz bandwidth. Determine the equivalent noise temperature and the sensitivity in dbm of the receiver when the antenna sees a 70 K noise temperature. kt 4 0 o W/Hz SOLUTION 0.6 o T F T 0 90 864.5 K rx Trx Tant S k Trx Tant B kto B To 0. 864.5 70 0 4 0 0 4 90.4800 5 W.480 0 mw 6. dbm
P7 c) Now the configuration in the figure below is investigated L, T o ANT. FEED Rx F rx G rx The expression of the total noise temperature is now simplified to: 0 0 T T ( L ) T LT 50 0 90 0 70 totiii a o rx 50 70 3 K A comparison of alternatives II and III shows that the performance of alternative III is T 3 SNR 0lg totiii 0lg.9 db worse. T 63 totii Here it is important that the total noise temperatures are compared in a point where the signal power is independent of the receiver configuration, i.e. in the receiver antenna output.
P43 The downlink characteristics of a GEO-satellite at longitude 0 W are: Transmitter parameters: - frequency GHz, vertical polarization - transmitter power 00 W - antenna feeder loss.5 db - antenna diameter.4 m, efficiency = 0.55. Receiver parameters: - location: () Hanko; 5950' N, 300' E () Utsjoki; 6950' N, 700' E - antenna feeder 0 m, = 30 db/km - other receiver losses 0.5 db a) Determine the receiver G/T=0lg(grx/Ttot) required for a SNR-value 0 db, when the channel bandwidth is 30 MHz. (Clear air is assumed, and the atmospheric loss is obtained from the attached figure. The sky temperature seen by the receiver antenna is assumed to be 0 K). b) Determine the required receiver antenna diameter (=0.55), when the receiver noise temperature is 00 K. c) Calculate the rain attenuation with the rain shower in the figure, when the rain rate is R = 0 mm/h. d) Determine the G/T-degradation caused by the rain shower assuming the rain temperature to be 90 K. How large receiver antenna 3 km diameter is needed to cancel the rain loss? 0 km
SOLUTION a) The signal to noise ratio is 0. 0 SNR prx ptx gtx grx ptx gtx g rx p l l l kt B l l l l kt B grx l l l l T p g tot tx rx o atm tx tx n tx rx ch tot tx rx o atm tot G g 0lg rx SNR Ltx Lrx Lo Latm Gtx Ptx 0lgkB T T tot kb With the given parameters - SNR = 0 db - Ltx =.5 db - Lrx l feed Lother 30 db km 0.0 km 0.5 0.8 db - Lo 9.5 0lg f GHz 0lgd km In the previous problem the distance between the GEO-satellite and the Earth station was derived: cos cos o o d R RR R
where - R is the distance of the GEO-satellite to Earth centre = 4300 km - Ro is the Earth radius = 6370 km - is the longitude difference between the satellite and Earth station - = the latitude of the Earth station in Hanko = 3.0+0.0=33.0, = 59.8 d 4300 4300 6370cos 33 cos59.8 6370 4003km L 9.5 0lg 0lg 4003 06. db o in Utsjoki = 7.0+0.0=37.0, = 69.8 d 4300 4300 6370cos 37 cos69.8 6370 4003 km L 9.5 0lg 0lg4003 06.3 db o The elevation angle of the Earth station antenna is obtained applying the law of cosines to the plane triangle defined by the Earth station, the satellite, and the centre of Earth.
o o R d R dr cos 90 d R dr sin o o z o d R R arcsin dr The atmospheric loss is estimated from the attached figure, where is needed. o R d y x d R
In Hanko: 4003 6370 4300 arcsin 6.6 L atm 0.3 db 4003 6370 In Utsjoki: 4003 6370 4300 arcsin 7.4 L atm 0.5 db 4003 6370 - the satellite transmitter antenna gain is G tx Df.4 0lg 0.55 0lg 0.55 47.0 db c 0.3 - kb 40 0lg 0lg 300 6 53.8 db 90
Now the G/T-values in the two reception locations can be calculated. Hanko: G SNR L tx L rx L o L atm G tx P tx 0lg kb T 0.5 0.8 06. 0.3 47.0 0 53.8 7.9dB K Utsjoki: G SNR L tx L rx L o L atm G tx P tx 0lg kb T 0.5 0.8 06.3 0.5 47.0 0 53.8 8.3dB K b) To get the required receiver antenna gain the total receiver noise temperature at the antenna must be estimated. According to ITU-R Rec. P68-8 0.L atm atm m m T T 0, T 60...80K In Hanko T atm 0.0. In Utsjoki T atm 0.0.46 70 0.7 K 70 0 7.K
Calculation of total noise temperature and receiver antenna diameter Hanko: lrx Tatm Tsky Ttot Trx To lrx lrx lrx 0 0.08.7 0 00 90 0 0.08 0 0.08 0 0.08 00 48.8 0.6 8.3 67.7 K G Grx 0 lgttot 7.9 0 lg 67.7 3. db T D c g 0.3 0 3. g rx rx D c f f 0.55 0.437 m
Utsjoki: lrx Tatm Tsky Ttot Trx To lrx lrx lrx 0 0.08 7. 0 00 90 0 0.08 0 0.08 0 0.08 00 48.8.5 8.3 503.4 K G Grx 0 lgttot 8.3 0 lg 503.4 35.3 db T g rx c g 0.3 0 3.53 rx D D c f f 0.55 0.65 m In clear weather conditions the difference in antenna diameters is not very large.
c) When the rain intensity is 0 mm/h, the characteristic loss at GHz kr 0.0455 0.6 0.938 db/km rain The worst case occurs when the right border of the rain shower is just at the receiving station. The distance travelled through the rain h h, arctan sin l s l h, arctan cos l With the given dimensions of the rain shower the elevation angle corresponding to the discontinuity of the expression is 8.5. The rain attenuation is In Hanko: In Utsjoki: 0 km 3 km 3 Lrain 0.938 9.8 db sin6.6 0 Lrain 0.938 8.9 db cos7.4
d) Now the atmospheric loss is increased with the rain attenuation, which will have impact on both the G/T-requirement and the noise power. Hanko: G SNR L tx L rx L o L atm L rain G tx P tx 0lg kb T 0.5 0.8 06. 0.3 9.8 47.0 0 53.8 7.7 db K l T ( l ) T T T T rx atm rain rain sky tot rx o lrx lrx lrainlrx lrainlrx 0.98 0.08 0 90 0.7 0 00 90 0 0 0 0 0 00 48.8 0.6 6.0 0.9 476.3 K G Grx 0 lgttot 7.7 0 lg 476.3 44.5 db T g rx 0.08 0.08 0.98 0.08 0.980.08 c g 0.3 0 4.45 rx D D.80 m c f f 0.55 T
Utsjoki: G SNR L tx L rx L o L atm L rain G tx P tx 0lg kb T 0.5 0.8 06.3 0.5 8.9 47.0 0 53.8 7. db K l T ( l ) T T T T rx atm rain rain sky tot rx o lrx lrx lrainlrx lrainlrx.89 0.08 0 90 0 7. 0 00 90 0 0 0 0 00 48.8.5 38. 0. 509.6 K 0.08 0.08.890.08.890.08 T G Grx 0 lgttot 7. 0 lg 509.6 54.3 db T D c g 0.3 0 5.43 g rx rx D c f f 0.55 5.57 m
In the investigated rain situation the antenna diameter in Hanko should be increased 4. times but in Utsjoki 8.9 times. However, this is a very rare rain situation and unfavourable for Utsjoki. Why? With a rain rate of 0 mm/h a 0 km large shower is very unlikely to occur. The rain rates for a given occurrence probability is moreover clearly lower in Utsjoki than in Hanko. Compare the result to the results with a 5 km wide rain shower, even if the rain intensity would be 40 mm/h.
0 Earth - space slant path loss A gas /db 5 0 5 0 0. Slant_path_loss.dsf 0 f /GHz 00