789
At a lower altitude, the shuttle will slow down due to friction with earth's atmosphere and it will return to earth. Thus, the spacecraft be in stable if its orbit period is more than 89 min 30 sec.
Solved Problems
--.......
Example 3. In the above question, estimate the linear velocity of the le along its orbit. Solution: The circumference of the orbit is 2na
= 2 n x 6628.14 = 41,645.83
km.
Therefore, the velocity of the shuttle in orbit will be 2na 41,645.83 . Vs = T = 5370.3 km/sec = 7.755 km/sec. j?'xample 1 : Calculate the di ra lusofa~ t . . Solution: The orbit eriod eos atLOnarysatellite. g'lVenby p of a geostationary satell't . I e m second is T2 _ 4n2 a3
--,:z-
or
a3=T2~
where ~ -- K ep 1er 's constant _ 4n2 '. earth (ME) = 3.986 x 105 kIn3/~e~aVltabonal constant
(G) x
Mass of
Now the time . d hours 56' perio of earth rotations is . nun 4.09 sec. or T = 86,164.09 sec. once per sIdereal dayof23 Hence a3 - (86 2 ,164.1) x 3.986 x 105 (4n2) = 7.496 x 1013 kIn3 or a - 42,164 km. ~mple
Example 4.A satellite is rotating in an elliptical orbit with a perigee int where satellite is closest to earth) of 1000 km and an apogee (point re satellite is farthest from earth) of 4000 km. Calculate its orbital iod. Take mean earth radius = 6378.14 km. Solution: The major axis of the elliptical orbit is a straight line een the apogee and perigee, as seen in Fig. 29.1. Hence, for earth dius RE, perigee height hp, and apogee height b«, the major axis will given by 2a = 2R + hp + h« = 2 x 6378.14 + 1000 + 4000 = 17,756.28 km a = 8878.14 km. Satellite
Orbit
2 : The space sh t I
p
1.-------------------2a--------------------~1 Fig. 29.1
The orbit period of a geostationary satellite in seconds is given by 2 3
'f2
= 4n a = 4n2 x (8878.07)3/3.986 x 105sec3 1.1.
= 6.93 x 107 sec2
790
SATELLITE
or
COMMUN
1CAll
T = 8325.1864 sec = 2 hours 18 min 45 . 19 see
Q~
Example 5. A satellite is in . 1 . g~osynchronous.altitude. If the sa~l~;~cu;::; equatorial orbit;;;;;;:-ave sidereal day ~ northh a period of 0n.e the . one solar day ' calculate theera d £us 0 t e orbit. So Iution s Here ,or the 24 hours . Therefore,
bit I . . 1 a penod IS one solar day that' ' ISexactly
T = 24 hours = 86 ,400 sec.
r- =
2 3
a
41t
Solution: The component of velocity an observer in the plane of orbit as the satellite appears over the horizon as will be VTR =vs eos9 ere 9 is the angle between the satellite velocity vector and the tion of the observer at the satellite.
The angle 9 will be
~ a3 =
Example 8. In the above problem, calculate the component of ity towards an observer at an earth station as the satellite appears r the horizon, for an observer who is in the plane of the satellite orbit.
r-~= (86,400)2 x 3.986 x 1005 41t
= 7.537
41t2
x 1013 km3
hence a
42 , 241 k m.
Example 6. In the above q ti . the elJ.uatorof the sub_satelliteue~:~;,. estimate the rate of drift around satellite moving towards the tP m degree per (solar) day. Is the eas or toward the west 'l Solution: The orbit period f h . . longer than a sidereal day by 3 0 .t ~~atelhte is one solar day which is will lead to the drift of the sub- atelli .9 s.ecor 235.9 sec. Clearly, this sa e ite point at a rate of . Draft rate = 360° x 235.9/86400 per day = 0.983° per da Since the earth moves to d y. satellite, the drift ofthe satell7t: frs the east 8:t a faster rate than the on the earth to be towards the w tom earth WIllappear to an observer es. Example 7. A low-earth orb' t lli . . withanaltitudeofl000km At £. s~ e ite zs m a circular polar orbit of 2. 65 GHz. Calculate the' v (a~m£rtter on the satellite has a frequency radius = 6378 km. e ocz y 0 the satellite in orbit. Take earth 4
Solution : Here hei ht f h
. e ISgiven by x (63 78 + 1000)3 5 3.986 x 10 sec
a = s; + h. The period oft~e saOtetlliet ~ate~hte from centre of the earth
= --41t a = 41t2 2 3
~
= 3.978 x 107sec2 or
T = 6306.9 sec The circumference of the orbit will be.2M
6378
Hence the component of satellite velocity toward the observer is VTR
,.,.t), 1-
s,
cos 9 = Re + h = 7378= 0.8645
= 46,357.3
km.
Therefore, the velocity of the satellite in orbit will be vs = Orbit c~rcumference _ 2M Orbital period - T = 7.350 kmlsec.
46357.3
= 63'06.94
= Vs cos9 =
7.350 x 6378 7378
= 6.354 km/sec. Example 9. In the same problem, the satellite carries aKa-band nsmitter at 20.0 GHz. Find the Doppler shift of the received signal at earth station. The earth radius is 6378 km. Discuss the impact of this ·ft on signal bandwidth. Solution: The Doppler shift of the received signal is given by A~-
1..>1 -
here
VTR
A.
=
component of the transmitter velocity towards the receiver. A. = wavelength of the transmitted frequency. For Ka-band transmitter with frequency 20.0 GHz, A. == 0.015 III VTR
Therefore, the Doppler shift at the receiver will be 6(= Vrf).. = 635410.015 = 423.60 kHz Doppler shift at Ka band with a LEO satellite can be very large~ requires a fast frequency tracking receiver. Ka-band LEO satellites all better suited to wideband signals than narrowband voice communications. Example 10. The continental Asia subtends an angle of ap, proximately 6° x 2° when viewed from geostationary orbit. Calculatet~ dimension of a reflector antenna so as to cover half this area with ' circular beam 3° in diameter at 11 GHz. . Solution: We know that the bandwidth of an aperture antenna1ie given as
792 SATELLITE
75}"
=D
93dB
COMMUN
793
IC~lJa..
degrees
'l'he received power at earth station will be
where, }..= wav~length and
= dimension
D
Here
93dB =
of the antenna in metres.
6° - 2° = 4°
Therefore,
D
=
+ GR - BTo - All losses = 18 + 80 + 59.2 - 1 - 195 - 2 - 0.2 - 0.4
PR = PT + GT
=
and }..= 0.0272 m
75}" = 7.5 x 0.0272 _ 4 - 0.51 metres.
93dB
Example 1l.A satellite located at 40,000 km from earth ~ a frequency of 11 GHz and has EIRP of21 dBW Ifth . . operates at has a gain of 50.5 dB, find the receiued power. . e receWLng antenna Solution.
= EIRP + GR - path loss = 21.0 dBW GR = 50.5 dB and h = 40,000
Example 13. In the aboue problem, et in clear air if the noise bandwidth
Solution.
Here, K
= Boltzmann's
find downlink is 27 MHz.
constant
=-
noise power
228.6 dBWlK/Hz
T« = System noise temperature = 75 K = 10 log 75 = 18.8 dBK BN = Noise bandwidth, 27 MHz = 10 log 27 x 106 = 74.8 dBHz
The receiver noise power will be
PR EIRP
Here
-96.4dBW.
NR = kBNTs = (K + BN + Ts) dBW
= -
km
228.6 + 18.8 + 74.3
= -
185.5 dBW.
Example 14. In the aboue problem, find C/ N ratio in receiuer in
The path loss is given by
arair.
Path loss
= (41th/A)2= 20 log
(4n
x 4 x 107) dB-
10 (2.727
x 10- 2)
CIN ratio
- - 205.8 dB
Therefore, by substituting PR
= 21.0 + 50.5 -
=-
188.8 dBW.
t lli . output power is 20W t nde sa e ite, satellite transponder antenna gain on axis'is ';::t~B ;,;utput bac'!off is -1.0 dB, satellite dB. If free space path loss 'at 4 a ~arth station. antenna gain is 59.2 satellite antenna is 20 dB 1 Gl!z lS 195.0 dB, edge of beam loss for losses are 0.4 dB caicul ~;::r alr.atmospheric loss is 0.2 dB and other . ' ae receiued power at earth station.
i
Example 15.In a satellite link, the antenna of the earth station used to receive a signal at 4150 MHz, has diameter of 30 m and an ouerall r/ficiency of 68%. The system noise temperature is 79 K when the antenna POints at the satellite at an eleuation angle of 28°. Find the earth station G~Tratio.
Solution.
For an antenna with circular aperture, the gain is given
SolutIOn. The satellite transponder output power is PT = 20W = 10 log 20 dBW = 18.0 dBW
=
= 1.0 dB on axis = 80 dB
Transponder output·backoff
GT = Satellite antenna gain,
GR = Earth station antenna gain = 59.2 dB Lfs
=
La =
= Lo =
Lair
Free space path loss at 4 GHz
= (PR - NR) dBW
= - 96.4 - ( - 135.5) = 39.1 dBW = 10 Antilog 89.1 W = 12.6 W.
205.8
Example 12. In a C-band GEO
BTo
= ~~
= 195.0 dB
Edge of beam loss for satellite antenna = _ 2.0 dB Clear air atmospheric loss = 0.2 dB Other losses = 0.4 dB
Ga=l1a
Here, l1a = 0.68, D
D)2
n (T
= 30 m and)" =
2n
6 = 0.072 m 4150 x 10
- 1 16 x 106 -- 60 .6 dB Th ere fiore, G -- 0.68 X (n x 280)2 -. (0.072) Ts = 79 K in dB will be Ts
= 10 log 79 = 19 dBik
Therefore, G/T ratio = (Ga - Ts)
dB/k
= 60.6 -
19 = 41.6 dBIk
795 794
SATELLITE
CO
16 1 E temperature to i . n the above problem xample
MMUNICA
"f ' 88 K, calculate tn;:~~S;'~~sTultingthe rain cause~t . sky . value. syste:n ':ofs;t emperature o ution : 0 rzse to S I Ts = 10 log 88 dBK
Here ,
= 19.4
Therefore , G T ratio. = 60.6 - 19 .4 dB K
dBK
rSYM
=~
_ 36m
10
IS given
25.7 M bits/sec.
=2x
e
514 .' e It rates will be . Mbits/sec
rSYM
Example 19 In a satelli : results in a ratio ~f n a satellite link, thermal . . a carrier-to-nois 2~.0 dB. A signal is receiu nolse m an earth station earth station . e rat", of20.0 dB. Find the vae1ue from a transponder of overall (C/N)o atwith tM The overall C-N ratio' (C/N)o
=
. IS given
1 (C/N)up
=
1
by
1
+ (C/N)DN
1125 +1 1120 = 45
= 10 log 45
1
14.5
Example 20• In a Ku-band satellit I a lin'"' gain --gainofofl27 h dB and a nominal out ,e, t he transponder has 5 ut the o~e' satellite's 14.GHz reeei.ft pouier at saturation of W. IfUlt'" 1 t",;.tput yf an uplink trans"!.~;::;;;'"' 26 dB on axis, calcu " sate lite transponder at Ii at guies an output power of ,!he earth station ant ha a requency of 14.45 GHz. Given: betweensag' the ~ln 0f 5~ dB and there is a l.5·diJ ass m the waveguide run enna l The atmosphere loss is 0.5 dB u ransmitter and antenna. nder clear sky conditions.
v!' ro:
PT :::;PR - Gr - Gt + Losses :: _ 127 _ 50 _ 26
+ 207 + 1.5 + 0.5 + 2.0 dBW
:: 7.2 dBIW:: 5.2 W.
_pIe 21. In the above qu.stion, if the rain attenuation is 7 dB ~Ol% of the year, find output power rating required for the transmit· to ,nsu t/wt a lW output can be obtained from the sarellite refor 99.99% of the year if uplink power control. ponder
that Example In the above can be sent18.through thi pro blem, calculate the' . Solutio . F zstransponder with QPSKmaxzmum bit rate n. or QPSK, m -- 2 . Therefore th bi . rSYM
is
NoWuplink power budget is written as PR :: PT + Gr + Gt - Losses dBW
by
m = 1. Therefi ore, the bit rates will b rSYM -
. jng antenna. solution: Here, output of the transponder Pe= 1 W:::; 10 log 1 :: 0 dBW
symbol rate for an RF li k i
1 + a - 1.4 = 25.7m Mbits/sec For BPSK ,
1'"
'lberefore, the received power will be PR :: 0 - 127 dBW:: - 127 dBW
41.2 dBK
Example 17. In t:.,,:pomkr is 36 MHz. ;f:::::~ite link, the bandwidth ~~. r of 0.4, Calculate the rth. stations use RRC filt of satellit. zs transponder with BPs;Faxzmum bit rate that ca ;rs uiith. roll'off Solution : The . . n e sentthrough maximum
Tilt path loss is 207 dB. earth station is located on the - 2 dB contour of the satellite'S
Solution.
With rain attenuation
of 7dB, the transmitted
power will be
PT (rain) :: 7.2 + 7 :: 14.2 dBW. _pIe
22.A sateUire TV distribution system employs a Ku·band satellite with bent pipe transponders for distribution of figital TV signals from an earth station to many receiving stations. The o/aign requires t/wt an overall CIN ratio 0(9.5 dB be available in the 7V receiver to ensure that the video signal on the TV screen is held to an _table leoel: The satellite is located at 73· W. The gain o( the uplink aatenna is 56.2 dB. The required C IN in Ku·band transponder is 30dB, the required overall C I N at earth station is 17 dB and satellite antenna
-"<,,"nary
IUin is 25 dB. Calculate the uplink transmitter power required to achieve (C/N)up:: 30 dB in clear air aiinospheric conditions. Assume free space ~. = 207 dB and other losses = 3 dB. The signOl bandwidth is 43.2 • Hz.Solution,
We first estimate the noise power in the transponder for
.•3.2 MHz bandwidth, Here, k:: Boltzmann's constant:: Ts = 500 K B
=
= 10 log
43.2 MHz
=
z
_228.6 dBW/KIH
500:: 27 dB 6
10 log 43.2 x 10
::
76.4 dBHz
797
796
SATELLITE
COMMUNICA
110••
The transponder noise power is given by NTR = kTs B = k + Ts + B dBW
= -
228.6 + 27 + 76.4 =
-
(14.125 - 0.01) to (14.125 + 0.01) 14.115 MHz to 14.135 GHz.
125.2 dBW
The received power level at the transponder input must be 30 dB greater than the noise power. Therefore, PR
= NTR + 30 dBW = -
95.2 dBW
::/::::it
Ie 25 In a satellite system, the downlink transmissw.n r~te sec. if the required 7nergy p~r bit t? noise power density zs 8 Calculate the required carrter-to-notse ratio. Solution:
Here,
rb
= 60 Mbits/see = 10 log 60 X 106 dB bits/see
Therefore, the required transmitted power for the transponder is PT = PR
where
GSt
+ GSt + GESt
-
Losses
= 77.8 dB Hz
= Satellite antenna gain
GESt = Earth
EblNo=8dB
station antenna gain
The C/N ratio is given by
Substituting, the values PT
(~o) =
= 95.2 + 56.2 + 25 - 207 - 3
(!)
+ rb = (8 + 77.8) = 85.8.
=-73.6dBW. Example 23. The bandwidth of an RF channel in a satellite link is 1.0 MHz. The transmitter and receiver have RRC filter with roll-off of 0.5. Calculate symbol rate for this link. Solution: is given by
The signal bandwidth expressed in terms of symbol rate B, = rSYM(1+ a) Hz
s,
or
rSYM= 1+ a
=
106 1+ 0.5
= 62 dBbits/sec.
Here the signal bandwidth is B.
= 16 X 10 (1 + 0.25) = 20 MHz The frequency range occupied by the RF signal is given by (1 +
a)]
Here, fc = 14.125 MHz. ..
Now
(~o)=(:)+rb
C = 9 + 62 = 671 dB
6
a)] to [fc + rs~
Eb Energy per bit = 9 dB No = noise power density
Substituting above values
= rSYM(1 + a) Hz
~ - rS;M (1 +
= 10 log 1.544 x 106 dB bits/see
= 666.7 Kbits/see.
Example 24. In a Ku-band satellite uplink, the carrier frequency is 14.125 MHz and carries a symbol stream at rate of 16 Msps. If the transmitter and receiver have RRC filter with roll-off of 0.25. Estimate the bandwidth occupied by the RF signal and the frequency range of transmitted signal. Solution:
I 26 A 14-GHz uplink operates with transmission lo~seds Examp e.. .d telliteG/T=8dB/K. The require margins totaling 2As10 dB ~n Fa~~A operation and an earth station . k Eb/NO is 9 dB suming itt &n . . f 42 dB calculate the earth station transmi er link antenna gaui 0 . . 'f baseband signal which has bit rate wer needed for transmtsswn 0 a 1.544 Mbits / sec. Solution: Here the bit rate is rb = 1.544 Mbitslsec
[ 14.125 - ;6 (1 + 0.25)] to [ 14.125 + ;6 (1 + 0.25)]
N
Now EIRP is given by EIRP =
(~o)
+ (~) + Losses - k
= 71 - 8 + 210 - 228.6 = 44.4 dBW
Therefore, the required power of the transmitter will be Pi = EIRp·- GUPa = 44.4 - 42 = 2.4 dBW
= AntiloglO 2.4 = 251.2 dW.
798
SATELLITE
799
COMMUNICA ~lION
Example 27. In the above problem, if the downlink tra;;;;;;--:-rate is rued at 71 dBbits I see, calculate the uplink power i~88l01l required for TDMA operation. reQse Solution: Here, bit rate == 62 dBbits/sec. When TDMA is empl the uplink bit rate increase will be oYed, llrbUP
= 71 -
62 = 9 dB
The EIRP must be increased by 9 dB. Therefore, the earth stat" transmitted power will be IOn PUP
F
Fig. 29.2
Now, the overall noise temperature of this system is given by T 8 = Toa + T LNA +
The EIRP is given by
For a paraboloidal antenna, the isotropic gain is given
by Gi = T](10.472 f D)
= aperture efficiency f = frequency D = diameter of the antenna in metres. Therefore, G; = 0.55 x (10.472 x 12 x 3)2 = 78,168 T]
G;
LNA
(Ta) = 35 K. The
k = - 228.6 dB
GIT = 19.5 dBlK
= Free-space
less + atmospheric absorption loss + antenna pointing loss + receiver feedback losses = 210 + 2 + 2 + 1 = 215 dB Now the elN spectral density ratio is given by
Losses
G
(CINo) = EIRP + T - Losses - k
F= 12 dB
= 50 + 19 -
10 10glO F = 12 10glO F = 1.2 dB , F = 101.2= 15.85. the cable loss (Lc) = 10°·5 = 3.16 and
185 K
EIRP= 50 dBW
i.
Similarly. 5 (GLNA) = 10
G
Example 31. In the link budget of a satellite, the free-space loss at GHz is 210 dB, the antenna pointing loss is 2 dB, and the atmospheric rption is 2 dB. If the receiver G I T ratio is 19 dB IK, receiver feeder BeS are 1 dB and the EIRP is 50 dBW, calculate the carrier-to-noise ctral density ratio. Here
no~sefigure is 12 dB, the cable loss is 5 dB, the LNA gain is 5~ dB, nozse temperature 150 K and the antenna noise temperature ts 35 . Calculate the overall noise temperature of the system.
or
Lc(F-1)To
Solution:
-----------------------------------------------------Example 30. For the system shown in Fig. 29.2. below, if the receiv.r
or
+
105
= 10 log 78,168 = 48.9 dB.
Solution: Here, the antenna noise temperature receiver noise figure is
LNA
+ 3.16 (15.85 -1)'x'290
Example 29. Calculate the gain of a 3-m paraboloidal antenna with an aperture efficiency of 0.55 operating at a frequency of 12 GHz.
Hence
G
105
.
= 7.8 + 50.2 = 58.0 dBW.
where,
(Lc-1)xTo
= 35 + 150 + (3.16 - 1) x 290
EIRP'= log Ps + Ga dBW = log 6 + 50.2
Solution:
Receiver
= (2.4 + 9) dBW = 11.4 dBW.
Example 28. Calculate the EIRP of a satellite downlink whic~ 12 GHz operates with a transmit power of 6 Wand an antenna gain of 50.2 dB. Solution:
Cable (Lc)
LNA
gaill
215 - ( - 286) = 140 dBW.
Example 32. An uplink of a satellite system operates at 14 GHz, and nux density required to saturate the transponder is - 100 dB Wlm2. free-space loss is 200 dB, and the other propagation losses amount
SATELLITE
800
COMMUNICATIOIi
to 3dB. Calculate the earth-station EIRP required for saturation, assu nt. ing clear-s ky co nditi itions. Solution: The effective area of an isotropic antenna (Ao) is given by Ao = - (21.45 + 20 log f) = - (21.45
Example 36. Determine the clear-sky carrier-to-noise ratio for a ellite TV system having worst case EIRP of 51 dBWat the receiver . . The free space loss is 205.34 dB, the nominal usable figure of merit 13.12 dB, and the gaseous attenuation due to atmospheric absorption 0.17 dB. The highest video frequency is 5MHz and peak-to-peak video • nal frequency deviation is 16 MHz. Solution:
dB
== -44.4
The
+ 20 log 14)
801
Here EIRP
(EIRP)sat = '¥STR + Ao + Losses
where '¥STR is the flux density required to saturate the transponder. Therefore, (EIRP)sat
= -
100 - 44.4 + 200
= 55.6 dBW.
Example 33. The EIRP of the Astra 1A satellite is 52 dBW in the main central service area, and that the transponder power is 45 W, calculate the effective isotropic radiated power in watts as seen by the antenna. EIRP = 10 log Ps
Ps
or
=
1O(EffiP/10)
= 158.5
= 51 dBW,
= 10(52110)
kW.
Example 34. An antenna has a noise temperature of 40 K and is matched into a receiver which has a noise temperature of 100 K. Calculate the noise power for a bandwidth of 36 MHz.
Here Ta = 40 K and TR = 100 K. Therefore, total noise of thus explain (antenna = TN = 40 + 100 = 140 K Now the noise power for a bandwidth of BN is
=
6f + 2 fv = 16 + 10 = 26 MHz
k
=
1.38 x 10-
CIN = 51-
+ receiver)
PN=kTNBN = 1.38 x 10- 23
J/K.
Therefore, for BN = 36 MHz PN = 1.38 x 10-
Latm=0.17dB
B
= EIRP
23
- Lfs + GIT - Latm - 10 log (kB)
205.34 + 13.12 - 10 log (1.38 x 10-23 x.26 x 106)
= 51 - 205.34
+ 13.12 - ( - 154.45) - 0.17
Solution:
6
x140 x 36 x 10
= 0.069~
. . rfre~ 35.A satellite link operating at 14 GHz has r~cew~orptiOfl losses 'of2 dB and a free-space loss of202 dB. r,he atmospheric ab he tottP loss is 0.5 dB, and the antenna pointing loss is 1dB. Calculate t link 'loss. efot' 'Solution: The tota11ink loss is the sum of all the losses ..Tb:~c ,tr Losses = (free-space loss + receiver feeder loss + atmospb sorption loss + antenna pointing loss) Example
= 202 + 2 + 0.5 + 1 dB =
205.5 dB.
0.17
The data rate is 24 Kbits/sec. Therefore, time taken for
smitting one bit is tbit =
1 3 see = 0005 sec. 24x 10
In Stop-and- Wait method, we transmit the block of 127 bits and wait acknowledgement. Since the one-way path delay is 240 msec or 0.240 • the two way path delay will be 2 x 0.240 = 0.480 sec. Therefore, sending 127 bits and waiting for acknowledgement takes .480 + 0.005) sec or 0.485 sec. "91I}~oW after 79 blocks one error :::39.5 sec.
23
-
= 13.16.
Example 37. A satellite link is carrying data at rates of24 Kbits Isee hen a block length of 127 bits is used and the one-way path delay is 240 c. A double error detecting code (127, 120) ARQ scheme is used. If 1 every 79 received blocks has an error, calculate the transmission bit for Stop and Wait system.
Solution:
where k
Lfs = 205.34 dB
GIT=13.12dB,
is detected,
that
is, after
Therefore, bit transmission rate will be
•.&. of blocks transmitted
without error and ACK x No. of bits/b1ock Time taken for transmission
x 127 = 7939.5
biIt s/ see
= 2 5 4 biIt s/ sec.
38. Repeat the above calculation for Go-back N ARQ . What should be the capacity of the transmit buffer?
!example
SATELLITE
COMMUNICATION
802
Calculate the noise power in tran . 803 FDMA channels. sponder-L or Lnthe inbound SCPC-
Solution: The time required to transmit 79 blocks each of 127 bi at 24 kbits/sec will be Its
= 79 x 1273 = 0 .418 sec. 24x 10 Now in Go-back-N method, when error is detected in the 79th block the NAK system is sent to request retransmission ofblock 79. Howev r' sending of NAK and retransmission takes two-path delay time , 2 x 24 msec or 0.48 sec. During this period, the number of blocks whi'~
7
arrive at the receiver will be
0.48 sec
0.48
c
Solution: The inbound VSAT l' k 1have a message data rate of64 ~~s/s that pass through transpondererefore, with BPSK modulation (~ :~~ WIth halfra~e FEC. encoding. 64 x 2 or 128 Khits/sec The . b ' t~e transmitted bit rate will ut rate in BPSK is one bit noise andwidth will be 128 KHz ' simce 1 per symbol. The noise power, N», is given by N BN
11.2 Kbits/sec. The required capacity oftransmit buffer should be to hold 91 blocks or 91 x 127 bits or 11,557
==
11,600 bits.
Example 39. Repeat the same calculation for Selective Repeat ARQ system. Solution: In this system, time is lost only in the retransmission of 79 blocks which show an error. Clearly, 78 blocks are without error. Therefore, the rate efficiency of the system will be 78/79 = 0.987. As a result, the bit transmission rate will be rb = 24 x 0.987 Kbits/sec
----------------------------------------~ Example 40. In a typical VSAT system, each VSAT station ~e~ and receives a 64-Kbits / sec data stream to and from the h~b. ~~g~ne data are sent to the hub from the VSATs by the inbound l.mk a v~ahast transponder at a message bit rate of 64 Kbits / see using bm 7J' Pd'T18 shift keying (BPSK) and half rate forward error correction (FEC) CO ~ ' I' 160 ~.z, giving a transmitted bit rate of 128 Kbits / hr. .,tJ ne The occupied RF bandwidth of each VSAT chan rsce (i r tilt corresponding to ideal RRC filters with a = 0.25. Multiple ac kB ~p(Jrt inbound link is by SCPC-FDMA with RF channels spaced 200 z to allow a 40-KHz guard band between channels. setJI Data from the hub station to the VSATs (the outbound link) (J,.6 as a continuous TDM stream of packets using a second transport . d t d otLt1'- • .J BPSK with half rate FEe. ra ecel The VSAT antenna has a d~ameter of Lm an satu e {r "'" of2 W. The transponder noise temperature is 500 K and that a at hub station is 150 K.
c;!e~
..0;
=
10 log 128 x 106 dBHz
= 51.1 dBHz
d Ts = 500 K
= 10 log 500 dBK
NTRI =k
= 27 dBK.
+ Ts +BN-- - 228 .6 + 27 + 51.1 =
-
150.5 dBW.
Exa~ple 41. In the above question I station receiver or in inbound SCPC'ca culate the noise power in the . -FDMACh I The inbound VSAT si 1 anne. an.smitted through t~~: reach the hub station after they are dwidth again will be 128 kH p~nder by the satellite. The noise el is received by a separatZe'IeFcause.at the hub station, each VSAT receiver Here, Ts = 150 K = 21 .8 dBl{, BN= 128 kHz . NH
= 23.70 Kbits/sec.
Watts
k = - 228.6 dBWlKIHz ,
Now
= Transmission time per block = 0.418/79 = 91.
These 91 blocks are discarded when the retransmission of79 blocks starts. The transmission time decreases by 0.48/0.898 = 0.54. Therefore the bit rate falls by 54% of the original value of 24 Kbits/sec, that is:
= kTsBN
= -
228.6 + 21.8 + 51.1
= -
155.7 dBW.
Example 42. In the same question I ponder-2 i.e. in outbound TDM ' ca culate the Noise Power in e bandwidth of 1MHz in the ;;~Tnnels,. assume a starting value I . recewer. utionpasses : Thethrough outboundtrans TDM biIt streaI? from the hub station to SATs L1
transmitting and receivin po~de~-2. Since not all of the VSATs stream on an average has st!r~~~U t~neou~lY, we assume that the VSAT receiver Thi g va ue noise bandwidth ofl MH d . IS correspond t 0 B z data rate of 500 kbits/sec and hsa If-rate a FEC PSKencodin signal with a erefore BN = 1 MHz = 10 log 106 dBHz -- 60 dBH z _~ ansponder-2, Ts = 500 K = 27 dBK erefore , N TR2 = - 228.6 + 27 + 60 . =-
141.6 dBW.
g.
804
SATELLITECOMMUNI C"lIO~
Example 43. In the same question, calculate the nOise~ VSAT receivers, i.e. in outbound TDM Channel.
er In. the
Solution: As already seen, each VSAT in the network receiv outbound TOM stream from the hub station in a noise bandwidthS the MHz. Here Ts = 150 K = 21.8 dBK and BN = 1 MHz = 60 MHz. of 1 NVR
= - 228.6 + 21.8 + 60 = - 146.8 dBW.
805
Example 46. In a VSAT link, the saturated output powe: of t~ llite transponder is 20 W. The VSAT channels are accessing this T/.Sponderby SCPC-FDMA method. If the amplifier in the transponder a backoff of2 dB, calculate the maximum number ofVSAT channels can be handled by the system. Assume each VSAT channel uses IW power on the downlink. F2
--
Example 44. In the same question, calculate the power received the satellite transmitted from the VSAT uplink. Assume as the worst caat that the VSAT is located at the satellite beam's edge-o{-coverage, the~; dB contour of the satellite's uplink (receiving) antenna pattern. The gain of the VSAT transmit antenna is 42 dB and that of satellite receive antenna 30 dB. Assume free path loss of207 dB.
Solution: The power received at the satellite from a single VSAT in dB, is given by I
PR
Here
PT
= PT + GT + GR
- Lp -
other losses
= transmit power = 2W = 10 log 2 = 3 dBW
GT
= gain of the VSAT transmit antenna = 42 dB
GR
= gain of the satellite receive antenna = 30 dB
Lp
= free space path loss at 14 HGz = 207 dB
other losses = satellite antenna edge of beam loss (2 dB) + clear air uplink atmospheric loss due to gases (0.5 dB) + miscellaneous losses of 0.5 dB to account for antenna mispointing feed loss. Therefore, power received by the transponder is PR
= 3 + 42 + 30 - 207 - 2 - 0.5 - 0.5 =-135 dB.
----------------------------------------------------Example 45. In the same question, calculate the uplink inbound C I N ratio is transponder-L,
0;
Solution: Each VSAT has a separate receiver in the hub station noise bandwidth of28 kHz. Since the received power at the transponder input as calculated in preceding example is - 135 dB, and noise po~~o in transponder-1 (in Example 40) is -150.5 dBW, the required CIN ra I will be PTRI (C/N)TRI = -NTRI
= - 135 - (- 150.5) dB = 15.5 dB.
Fig. 29.3
Solution: The saturated transponder output power
= 20 W = 10 log 20 = 13 dBW. With 2 dB backoff, this power becomes (13-2) 11 dBW 10 log 11 12.6 W. Since the power for VSAT channel is lW, the number of nels that can be handled is 12.611 == 12 channels.
=
=
Example 47. Refer to Fig. 29.3. The basis of ~ satel.lite. orbitting und earth is the centripetal force (FI) due to earth sgraoitation acting 'towards the center of the earth balancing the centrifugal force (F2) awng Ot.uayfrom the center. Calculate the centrifugal force for a satellite of rt&a88 IOO kg orbitting with a velocity of 8 km Is at a height of 200 km Gbove the surface of earth. Assume mean radius of earth to be 6370 km. Solution: mV2
Centripetal force = (R + H) here m = mass of satellite V = orbital velocity R = mean radius of earth
806
SATELLITE
807
COMMUNICA ••TIO~
H = height of satellite above surface of earth The centripetal force balances the centrifugal force.
. m y2 100 x (8000)2 Therefore, Centnfugal force = H) = + (6370 + 200) x 103
. . . Apogee + Perigee SemI-maJor axis = 2
IT)
=
108
64 x 3 = 974 Newtons 6570 x 10
Example 48. Determine the orbital velocity of a satellite mo~ a circular orbit at a height of 150 km above the surface of earth giv;~ that gravitation constant, G = 6.67 x 10-11 N-m2Ik!l, mass of earth M = 5.98 x 1024 kg, radius of earth, Re = 6370 km. • Solution: The orbital velocity
(V)
is given by
V = "I!I(R + H) where ~ = GM = 6.67 x 10- 11 x 5.98 x 1024 = 39.8 x 1013 Nm2/kg
= 30000 + 1000 = 15500 km 2 Fig. 29.4 illustrates the point further. Example 50. A satellite moving in an elliptical eccentric orbit has semi-major axis of the orbit equal to 16000 km (Fig. 29.5). If the '6 rence between the apogee and the perigee is 30000 km, determine the e it eccentricity. ../ Solution: Apogee = a (1 + e) Perigee = a (1 - e) ere a = semi-major axis of the ellipse e = orbit eccentricity
R=6370km
I I I I I
H= 150km
V = "39.8
x 1013/(6370 + 150) x 103 = 7.813 kmIs.
Example 49.A satellite in an elliptical orbit has an apogee of30,000 km and a perigee of 1000 km. Determine the semi-major axis of the elliptical orbit.
:@
I:_ -----e----________ tI b
:
__
Earth
I I
Fig. 29.5
Apogee - Perigee = a (1 + e) - a (1- e) = 2ae erEc tricit Apogee - Perigee cen nCI y, e= 2a 30000 - 2 x 16000 30000 = 3200 = 0.93 .
Fig. 29.4
. Example 51 : The farthest and the closest points in a satellite's :ptical eccentric orbit from earth's surface are 30,000 km and 200 km lpectively. Determine the apogee, the perigee and the orbit eccentricity. ume radius of earth to be 6370 km.
808
SATELLITE COMMUNI
809 CA1'Iott
Solution:
E"ample 53 : Satellite-1 in an elliptical orbit has the orbit semi. r axis equal to 18000 km and Satellite-2 in an elliptical orbit has ._major axis equal to 24000 km (Fig. 29.7). Determine the relationbetween their orbital periods.
Apogee = 30000 + 6370 = 36370 Jon Perigee
= 200 + 6370 = 6570 Jon
Eccentricity = Apogee - Perigee 2a where a
= semi-major
axis of the elliptical orbit a = Apogee + Perigee
Also,
2
or
2a Therefore, orbit eccentricity
= Apogee
+ Perigee
= Apogee - Perigee Apogee + Perigee
_ 36370 - 6570 - 36370 + 6570
29800
= 42940 = 0.693.
. ~xample 52 : Refer to Fig. 29.6 showing a satellite m . . elllptlc~l, eccentric orbit. Determine the apogee and peri, a o;.l~g In a? the orbit eccentricity is 0.5. tgee ts ances if Solution:
_
24000km
-' ,
Fig. 29.7
Solution: The orbital time period T
= 21t
(T)
is given by
~a3/1!
here I! = GM G = Earth's gravitational constant M = mass of earth a = semimajor axis of ellipse
-:14000 ,
, ,
km:.-
If (a 1) and (a2) are the values of the semi-major axis of the elliptical its of the satellites-1 and 2, (Tl) and (T2) are the corresponding orbital riods, then Tl=21t~
Fig. 29.6
and
T2=21t~ 312
. The distance from center of ellipse (0) to the centre of earth (c) is gIven. ~y (a x e) where (a) is the semi-major axis and (e) is the eCcentricity. Therefore,
= 14000
axe a
14000
= OT = 28000 Jon
Now apogee
= a(l + e) = 28000(1 + 0.5) = 42000 Ian
Perigee
= a(1 - e) = 28000(1 - 0.5) = 28000 x 0.5 = 14000 Jon.
T2ITI
= (a2lal) 312 = (24000) 18000
= (4/3)312 = 1.54.
Thus orbital period of satellite-2 is 1.54 times the orbital period of tellite-1. Example 54 : Determine the escape velocity for an object to be nehed from surface of earth from a point where earth's radius is 6360 4 (G = 6.67 x Nm2/kg'Z and M = 5.98 x 102 kg).
io:"
Solution:
Escape velocity
= -J2}J1r
811 SATELLITE COMMUNICA1l0ti
solution. t\pogee distance = 35000 + 6360
Here r = 6360 km = 6360 x 103m I! = OM = 6.67 x 10 - 11 x 5.98 X 1024 = 39.8 x 1013 Nm2/kg
= =
-V
2 x 39.8 x 1~13 6360 x 10 11.2 km/s.
km
Semi-major axis of elliptical orbit,
,-------
Therefore, Escape Velocity
= 41360
perigee distance = 500 + 6360 = 6860 km a = 41360 + 6860 = 24110 km
=~
2
6360
Orbital time period, T =
21t
-Va3/1!
= 6.28~,..--(2-4-1-10-X-1-03)3
/Example 55.Calculate the orbital period of a satellite in an~ elliptical orbit shown in Fig. 29.8.
6.67
x 10- 11 x 5.98 X 1024
= 10 hrs 20 minutes I
Velocity at any point on the orbit is given by: V = -J1.l [(21r) - (l/a»)
I
24
---------------------1------~-----
L:=:=:
V = OM = 6.67 x 10 - 11 x 5.98 X 10 = 39.8 X 1013 Nm /kg 2
At apogee point, r = 41360 km. Therefore, 50000k~
V="-~-9.-8X--10-13-[---2---------1----]
•
41360 x 103
Fig. 29.8
Solution.
Semi major axis, a = 50~00 = 25000 km = ,,39.8
Orbital time period, T = 21t and
(rp)
3
24110 X 10
-Va3/1!
I! = OM = 6.67 x 10 - 11 x 5.98 x 1024
T = 2 x 3.14 = 6.28
..y
At perigee point,
3 (25000 x 10 )3 39.8 x 1013
km. Ther~fore, 6860 x 103
= ,,39.8
15625 x 1018 39.8 x 1013
x 6.25 x 103 = 39250 seconds = 10 hours 54 minutes.
=
= 6.28
Example 56 :A satellite moving in a highly eccentric Mo~i~g;1~ having the farthest and the closest points as 35000 km an . 1 time respectively from the surface of the earth. Determine the orbzt~arth'S period and the velocity at the apogee and perigee point. (Assume radius = 6360 km.)
r = 6860
V=,/~3-9.-8-X-10-1-3[----2---------1---]3
..J
-------------------------------------------------
48220 - 41360 ] 3 41360 x 24110 x 10
= - ,r--3-9-.8-x-1-0-1'::"3-X6-86-0-= 523 mls 3 " 41360 x 24110 x 10
are the apogee and perigee distances, then = 39.8 x 1013 Nm2/kg
X 1013 [
~I
x 1013 [
24110 x 10
48220 - 6860 3 ] 6860 x 24110 x 10
13 39.8 x 10 x 41360 = 9.976 kmls. 3 6860 x 24110 x 10
Example 57 : The sum of apogee and perigee distance ora certain iptical satellite orbit is 50000 km and the difference of apogee and ~rigeedistance is 30000 kms. Determine the target eccentricity (e). Solution: If (ra) and
(rp)
are the apogee and perigee distances, then
/
SATELLITE
813
COMMUNIC
A"O~
e = (r'rol -+ rp
_ 21t(R + H) _ 6.28 x 7000 V 7.54
= 30000 = 0 6
rp)
50000
..
= 5830 sec = 1 hour 37 minutes.
flxample
58 : The semi-major
axis and the semi-minor ;-:--ell.iptical satellite orbit are 20,000 km and 16000 km respectivel an mine the apogee and perigee distances. y. eter.
LS;r
Solution:
v» are apogee and perigee distances
If (ra) and .
.
.
semi-major axis
=
respectively, then
Example 60 : A satellite moving in an eccentric elliptical orbit has i-major axis and semi-minor axis of (a) and (b) respectively and an ntricity of 0.6. The satellite takes 3 hrs 10 minutes in moving from B A in the direction shown. What will be the time taken by the satellite move from A to B in the direction shown in Fig. 29.9. y
ra + rp 2
Semi-minor axis = "ra rp ra
+ rp = 20000 km
2
L
x
Therefore, ra + rp = 40000 km
+-------a--------~
"ra x rp = 16000
Therefore, To rp = 256000000 Now ro + rp = 40000 ra x rp = 256000000 Substituting the value of (rp) from (2) in (1)
(1) Selution r According to Kepler's law for elliptical satellite orbits, the line . the satellite and the center ofthe earth spans equal ellipse area equal time.
ra (40000 - ra) = 256000000
or r~ - 40000 ra + 256000000 = 0 r« =
40000 ± ..J16 x 108 - 10.24 x 108 2 40000 ± ..J5.76 x 108 2
= 3.2
X 104,
1.6
X 104
As a first step, we shall determine the area spanned while moving
m B to A. It is given by shaded region and is given by
40000 + 2.4 2
X
104
= 32000 km, 16000 krn
20,000km. rp = 40000 - 32000 = 8000 km. ,
---
.Example 59 : A satellite is moving in a near earth circular orbi~~ of 640 km. Determine
its orbital period.
Solution: Orbital velocity = ~ (RO:H) = ••.139.8
'V
7
=
X 1013 X
106
Area of half of ellipse - Area of MOB 1tab = --2-
ro = 32000 km as it cannot be 16000 km if the semi-major axis is
a distance km.)
Fig. 29.9
(2)
6.67 x 10 -
(Assume
11
x 5.98 x
7000 x = 7.54
kmIs.
103
R ::;6
= ab
-b
x OC
1tab =-2- - bae
(3'i4 - 0.6J
= ab
(rr/2-
e)
= 0.97 ab == ab
The area spanned in moving from A to B. 1tab
= -2-
+ 0.6 ab = ab (rr/2 + 0.6) = 2.2 ab.
The r a t'100 f t h etwoareasls~=2.2. . 2.2 ab
24
!.!!.-
l~~refore, the time taken by satellite to move from A to B should .tunes the time taken by the satellite to move from B to A in the IOnshown. Therefore, 'll Ille taken = 2.2 x 3 hrs 10 min = 7 hrs.
O~
T2) _ 2 n .JaW _ ( al ( Tl - 2n ~a~/Il - a2J T = T (a2) = 10 (24000) 2 1 al 16000 312
. Example ~1 : Fo: an eccentric elliptical satellite o~ apogee .and perigee points at a distance of 50,000 km an t Wzth Qil respectively from the centre. of earth. Determine the semi-! 8.000 kill semi-minor axis and the orbit eccentricity. aJor Cl:cia Solution: ' Apogee distance,
Ta
Perigee distance,
rp = 8000 km
50000
+ 8000
2
-29000
= "=50=-=0-=-00"'-x-80-0-0 = 20000 Orbit eccentricity,
e
= (ra -
rp) ra + rp
Example 63 : A geosynchronous satellite moving in an equatorial ular orbit at a height of 35800 km above the surface of earth gets lined at an angle of 2° due to some reasons. Calculate the maximum viation in latitude, if the maximum deviation in longitude (with erenceto longitude of ascending node) is 0.0175°, determine maximum . :placements in kms caused by latitude and longitude displacements. sume earth's radius = 6364 hm).
km
Solution:
km
Height of orbit = 35800 km
= (50000
- 8000 '\ 50000 + 8000)
Earth's Therefore,
42000 = 58000 = 0.724.
tu
radius
0':
Maximu~ • en by:
latitude
Maximum
= 6364 km
Orbit radius, r
Angle of inclination
E.xalmp!~62.. Satellite-1 and Satellite-2 are orbitting in different e h LPtlC~ its uiitli same perigee but different apogee distances as sown m FLg.29.10. The semi-major axes of the two orbits are 16000 km 24~00 km: If the orbital period of satellite-1 is 10 hours determine t e orbital period of satellite-2. '
z:
312
= 10(1.!?)312= 18.37 hours.
axis, b = "ra x rp
Semi-minor
312
= 50000 km
axis, a = ra + rp 2
Semi-major
815
V£D PROBLEMS
SATELLITE COMMUNICl\ll
= 35800
D", (max)
= ai
from equator
(in krn) due to
( 1~0
J
180
longitude
due to inclination
Amax
(i) is
is given by:
deviation
n
= 1471 km
.
from ascending
node,
i2 22 4 = 228 = 228 = 228 = 0.0175°
\jImax
Maximum
km
where (i) is in deg
= 42164 x 2 x
Maximum
= 42164
= 2°
deviation
displacement
+ 6364
displacement
(DljI)due to
(\jImax)
is given by:
Fig. 29.10
DljI= D",(~::
Solution:
------------------------------------------------------
Orbital period (T) is given by, T = 2n -.Ja3/1l where
a = semi-major Il =GM
Example 64 : Determine the magnitude of velocity impulse needed to correct the inclination of 2° in the satellite orbit of example 63. Solution: The magnitude of velocity impulse is given by:
axis .
(T2) are
If•.~Tl)land respec ive y, then
the orbital periods of satellite-l
) = 1471 x 0.0;75 = 12.9 km.
and sateIIite-2 - r;;-tan -" ~ ;
i
=~
13
39.8 x 10 tan 2° 42164 x 103
= 107
mls.
816
SATELLITECOMMUNICA ••lIO~
817
Example 65 :Ageosynchronous satellite orbitting at 4216;;;;-earth's center has a circular and equatorial orbit. The orbit gets inc{:OTn due to some reason and it is observed that the maximum displace::::d due to latitude deviation in 500 km. Determine angle of inclinatio ~t between the new orbital plane and the equatorial plane. n Ii) Solution: ~
Maximum
displacement
D').. (max) due to latitude
deviation
is mv
~~
DJ...(max) = r A.max where A.max= maximum
latitude
deviation
= i (angle of inclination)
DJ...(max) = r x i
or
i = DJ...(max)/r = 4~~~4 = 0.012 rad
Therefore,
= 0.68
deg.
Fig. 29.11
Example 66 : A geostationary satellite moving in an equatorial circular orbit is at a height of35786 km from earth's surface. Ifthe earth's radius is taken as 6378 km, determine the theoretical maximum coverage angle. Also determine the maximum slant range.
Solution: The maximum
coverage angle, (2
2amax = 2 sin - 1 [(
Re~
H) cos Emin]
Solution: For theoretical Maximum
maximum
coverage angle, elevation
coverage angle, 2amax
where Re
= earth's
radius
H
= height
of satellite
= 2 sin -1Re~
above earth's
angle, E
H cos
= o.
re
Emin = Minimum elevation angle
Emin)
2 amax
. -il(
= 2 sin
6378
~ 6378 + 35786
} cos
50]
= 2 sin - 1[0.1512 x 0.996] = 2 sin -10.1506 = 2 x 8.66 = 17.32°.
surface
2 amax = 2 Si{ 3578663~~378 cos 00) = 17.4° or
Example 68 : Refer to Fig. 29.12 showing a geostationary satellite itting earth. Calculate the angle (6) subtended by the arc of the llite's footprint at the center of earth.
amax = 8.7° If D is the maximum slant range, then
D2 =R; + (Re +H)2 - 2Re (Re +H) x sin
.
[E + sin - 1(&~ H
CDS
Ell
= (6378)2 + (42164)2 - 2 x 6378 x 42164 x sin 8.70
= 40678884 + 1777802896
- 537843984
x 0.1512
D =41679
km.
Therefore
62 = 90° - a2 - E2 6 = (90 - a1 - E1) + (90 - a2 - E2)
= 180
- (a1
+ a2) - (E1 + E2)
= 175° - (a1 + a2)
= 1737139041 or
6 = 61 + 62 61 = 90° - a1 - E1
.------:
----------------------ngleand Example 67. What would be the new maximum coverage a t zero the slant range if the minimum possible elevation angle is 5° and no as in Example-66 (Refer to Fig. 29.11).
. - 1
ci = sm
6378 cos 5° 6378 + 35786
. _ 1 6378 cos 5° _ 8 660 =sm 42164 -.
(E1 = 5°) (E2 = 0°)
