Solved Problems--.......

j?'xample 1 : Calculate the dira 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 1 ' 4n2- ep er s constant _ '.earth (ME) = 3.986 x 105 kIn3/~e~aVltabonal constant (G) x Mass of

Now the time . dhours 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 2 : The space sh t I

789

At a lower altitude, the shuttle will slow down due to friction withearth's atmosphere and it will return to earth. Thus, the spacecraftbe in stable if its orbit period is more than 89 min 30 sec.

Example 3. In the above question, estimate the linear velocity of thele along its orbit.

Solution: The circumference of the orbit is2na = 2 n x 6628.14 = 41,645.83 km.

Therefore, the velocity of the shuttle in orbit will be2na 41,645.83 .

Vs = T = 5370.3 km/sec = 7.755 km/sec.

Example 4.A satellite is rotating in an elliptical orbit with aperigeeint where satellite is closest to earth) of 1000 km and an apogee (pointre satellite is farthest from earth) of 4000 km. Calculate its orbitaliod. Take mean earth radius = 6378.14 km.Solution: The major axis of the elliptical orbit is a straight lineeen 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 willgiven by

2a = 2R + hp + h«= 2 x 6378.14 + 1000 + 4000 = 17,756.28 km

a = 8878.14 km.

Orbit Satellite

p

1.-------------------2a--------------------~1Fig. 29.1

The orbit period of a geostationary satellite in seconds is given by2 3

'f2 = 4n a = 4n2 x (8878.07)3/3.986 x 105sec31.1.

= 6.93 x 107 sec2

790 SATELLITE COMMUN 1CAll

T = 8325.1864 sec = 2 hours 18 min 45 19 Q~. see

Example 5. A satellite is in . 1 .g~osynchronous.altitude. If the sa~l~;~cu;::; equatorial orbit;;;;;;:--sidereal day one solar day calculate thee d ~ northhave

a period of the

S I. ' ra £us 0 t e orbit. 0n.e

o ution s Here the bit I . .24 hours ,or 1 a penod ISone solar day that'. ' ISexactly

Therefore, T = 24 hours = 86 400, sec.

r- = 41t2a3

~a3 = r-~= (86,400)2 x 3.986 x 1005

41t 41t2

= 7.537 x 1013 km3 hence a 42 241 k, m.

or

Example 6. In the above q ti .the elJ.uatorof the sub_satelliteue~:~;,. estimate the rate of drift aroundsatellite moving towards the tP m degree per (solar) day. Is theeas 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 iswill lead to the drift of the sub- atelli .9 s.ecor 235.9 sec. Clearly, thissa e ite point at a rate of

. Draft rate = 360° x 235.9/86400 per day = 0.983° per daSince the earth moves to d y.

satellite, the drift ofthe satell7t: frs the east 8:t a faster rate than theon the earth to be towards the w tom earth WIllappear to an observeres.

Example 7. A low-earth orb' 4 t lli . .withanaltitudeofl000km At £. s~ e ite zs m a circular polar orbitof 2. 65 GHz. Calculate the' v (a~m£rtter on the satellite has a frequencyradius = 6378 km. e ocz y 0 the satellite in orbit. Take earth

Solution : Here hei ht f h .a =s; + h. The period oft~e saOtetlliet~ate~hte from centre of the earthe ISgiven by,.,.t), 41t2a3 (63 31- = -- = 41t2 x 78 + 1000) 5~ 3.986 x 10 sec

= 3.978 x 107sec2

T = 6306.9 secThe circumference of the orbit will be.2M = 46,357.3 km.

Therefore, the velocity of the satellite in orbit will bevs = Orbit c~rcumference _ 2M 46357.3

Orbital period - T = 63'06.94= 7.350 kmlsec.

or

Example 8. In the above problem, calculate the component ofity 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.Solution: The component of velocity an observer in the plane oforbit as the satellite appears over the horizon as will be

VTR =vs eos9ere 9 is the angle between the satellite velocity vector and the

tion of the observer at the satellite.

The angle 9 will bes, 6378

cos 9 = Re + h = 7378= 0.8645

Hence the component of satellite velocity toward the observer is7.350 x 6378

VTR = Vs cos9 = 7378

= 6.354 km/sec.

Example 9. In the same problem, the satellite carries aKa-bandnsmitter at 20.0 GHz. Find the Doppler shift of the received signal atearth 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~- VTR1..>1 - A.

here VTR = component of the transmitter velocity towards thereceiver.

A. = wavelength of the transmitted frequency.For Ka-band transmitter with frequency 20.0 GHz, A. == 0.015 III

Therefore, the Doppler shift at the receiver will be6(= 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 allbetter suited to wideband signals than narrowband voice communica-tions.

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 11GHz.

. Solution: We know that the bandwidth of an aperture antenna1ie

given as

792SATELLITE COMMUN

IC~lJa..75}"

93dB = D degrees

where, }..= wav~length

and D = dimension of the antenna in metres.

Here 93dB = 6° - 2° = 4° and }..= 0.0272 m

Therefore, D = 75}"= 7.5 x 0.0272 _93dB 4 - 0.51 metres.

Example 1l.A satellite located at 40,000 km from earth ~a frequency of 11GHz and has EIRP of21 dBW Ifth . . operates athas a gain of 50.5 dB, find the receiued power. . e receWLng antenna

Solution.

PR = EIRP + GR - path lossHere EIRP = 21.0 dBW

GR = 50.5 dB and h = 40,000 kmThe path loss is given by

Path loss = (41th/A)2= 20 log (4n x 4 x 107) dB-10 (2.727 x 10- 2) - - 205.8 dB

Therefore, by substituting

PR = 21.0 + 50.5 - 205.8= - 188.8 dBW.

Example 12. In a C-band GEO t lli .output power is 20W t nde sa e ite, satellite transponderantenna gain on axis'is ';::t~B ;,;utput bac'!off is -1.0 dB, satellitedB. If free space path loss 'at 4 a ~arth station. antenna gain is 59.2satellite antenna is 20 dB 1 Gl!z lS 195.0 dB, edge of beam loss forlosses are 0.4 dB caicul i ~;::r alr.atmospheric loss is 0.2 dB and other

. ' a e receiued power at earth station.SolutIOn. The satellite transponder output power isPT = 20W = 10 log 20 dBW = 18.0 dBW

BTo = Transponder output·backoff = 1.0 dBGT = Satellite antenna gain, on axis = 80 dBGR = Earth station antenna gain = 59.2 dBLfs = Free space path loss at 4 GHz = 195.0 dBLa = Edge of beam loss for satellite antenna = _ 2.0 dB

Lair = Clear air atmospheric loss = 0.2 dBLo = Other losses = 0.4 dB

793

'l'he received power at earth station will bePR = PT + GT + GR - BTo - All losses

= 18 + 80 + 59.2 - 1 - 195 - 2 - 0.2 - 0.4= -96.4dBW.

Example 13. In the aboue problem, find downlink noise poweret in clear air if the noise bandwidth is 27 MHz.

Solution. Here, K = Boltzmann's constant = - 228.6 dBWlK/HzT« = System noise temperature = 75 K = 10 log 75 = 18.8 dBK

BN = Noise bandwidth, 27 MHz = 10 log 27 x 106 = 74.8 dBHzThe receiver noise power will be

NR = kBNTs = (K + BN + Ts) dBW= - 228.6 + 18.8 + 74.3 = - 185.5 dBW.

Example 14. In the aboue problem, find C/N ratio in receiuer inarair.

CIN ratio = ~~ = (PR - NR) dBW

= - 96.4 - ( - 135.5) = 39.1 dBW= 10 Antilog 89.1 W = 12.6 W.

Example 15.In a satellite link, the antenna of the earth station usedto receive a signal at 4150 MHz, has diameter of 30 m and an ouerallr/ficiency of 68%. The system noise temperature is 79 K when the antennaPOints at the satellite at an eleuation angle of 28°. Find the earth stationG~Tratio.

Solution. For an antenna with circular aperture, the gain is given

(n D)2

Ga=l1a TH 2nere, l1a = 0.68, D = 30 m and)" = 6 = 0.072 m

4150 x 10

Th fi G - 0.68 X (n x 80)2 - 1 16 106 - 60 6 dBere ore, - 2 -. x - .(0.072)

Ts = 79 K in dB will beTs = 10 log 79 = 19 dBik

Therefore, G/T ratio= (Ga - Ts) dB/k = 60.6 - 19 = 41.6 dBIk

794

E

SATELLITE CO

xample 16 1 MMUNICA

temperature to i . n the above problem "f '88 K, calculate tn;:~~S;'~~sTultingthe syste:n ':ofs;t rain cause~

S I

. value. emperature t . skyo ution : 0 rzse to

Here T, s = 10 log 88 dBK = 19.4 dBK

Therefore G ., T ratio = 60.6 - 19 4 dB. K 41.2 dBK

Example 17. Int:.,,:pomkr is 36MHz. ;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'offSolution : The . . n e sentthrough

maximum symbol rate for an RF li k irSYM = ~ _ 36m 10 IS given by

1 + a - 1.4 = 25.7m Mbits/sec

For BPSK , m = 1 Therefi. ore, the bit rates will brSYM - 25.7 M bits/sec. e

that Example 18. In the above b .can be sent through thi pro lem, calculate the'Solutio . F zstransponder with QPSKmaxzmum bit raten. or QPSK, m - 2 .- . Therefore th bi

rSYM = 2 x rSYM 514 .' e It rates will be. Mbits/sec

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 stationearth station e rat", of20.0 dB. Find the e1 from a transponder with. va ue of overall (C/N)o at tM

The overall C-N ratio' .IS given by

(C/N)o = 11 1

(C/N)up + (C/N)DN= 11125 + 1120 = 45 = 10 log 45 14.51

Example20 I Igain ofl27 dB • n a Ku-band satellit h ---gainof h and a nominal out ,e, t e transponder has a lin'"'the o~e' satellite's 14.GHz reeei.ftut pouier at saturation of

5W. If t'"

1 v!' ro:t",;.tput yf an uplink trans"!.~;::;;;'"' 26dB on axis, calcuUl

"sate lite transponder at Ii at guies an output power of

l

,!he earth station ant ha a requency of 14.45 GHz. Given:ass m the enna sag' fwaveguide run between the ~ln 0 5~dB and there is a l.5·diJ

The atmosphere loss is 0.5 dB u ransmitter and antenna.nder clear sky conditions.

795

Tilt path loss is 207 dB.1'" earth station is located on the - 2 dB contour of the satellite'S

. jng antenna.solution: Here, output of the transponder is

Pe= 1 W:::; 10 log 1 :: 0 dBW

'lberefore, the received power will bePR :: 0 - 127 dBW:: - 127 dBW

NoWuplink power budget is written asPR :: PT + Gr + Gt - Losses dBW

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

ret/wt a lW output can be obtained from the sarellite

ponder for 99.99% of the year if uplink power control.

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-"<,,"nary satellite with bent pipe transponders for distribution offigital TV signals from an earth station to many receiving stations. Theo/aign requires t/wt an overall CIN ratio 0(9.5 dB be available in the7V 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 uplinkaatenna is 56.2 dB. The required CIN in Ku·band transponder is 30dB,the required overall C I N at earth station is 17 dB and satellite antenna

IUin is 25 dB.Calculate the uplink transmitter power required to achieve(C/N)up:: 30 dB in clear airaiinospheric 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:: _228.6 dBW/KIH

z

Ts = 500 K = 10 log 500:: 27 dBB = 43.2 MHz = 10 log 43.2 x 106

:: 76.4 dBHz

796 SATELLITE COMMUNICA110••

The transponder noise power is given byNTR = kTs B = k + Ts + B dBW

= - 228.6 + 27 + 76.4 = - 125.2 dBWThe received power level at the transponder input must be 30 dB

greater than the noise power. Therefore,PR = NTR + 30 dBW = - 95.2 dBW

Therefore, the required transmitted power for the transponder isPT = PR + GSt + GESt - Losses

where GSt = Satellite antenna gainGESt = Earth station antenna gain

Substituting, the valuesPT = 95.2 + 56.2 + 25 - 207 - 3

=-73.6dBW.

Example 23. The bandwidth of an RF channel in a satellite link is1.0 MHz. The transmitter and receiver have RRC filter with roll-off of0.5. Calculate symbol rate for this link.

Solution: The signal bandwidth expressed in terms of symbol rateis given by

or

B, = rSYM(1+ a) Hz

s, 106

rSYM= 1+ a = 1+ 0.5 = 666.7 Kbits/see.

Example 24. In a Ku-band satellite uplink, the carrier frequency is14.125 MHz and carries a symbol stream at rate of 16 Msps. If thetransmitter and receiver have RRC filter with roll-off of 0.25. Estimatethe bandwidth occupied by the RF signal and the frequency range oftransmitted signal.

Solution: Here the signal bandwidth isB. = rSYM(1+ a) Hz

= 16 X 106 (1 + 0.25) = 20 MHzThe frequency range occupied by the RF signal is given by

~ - rS;M (1+ a)] to [fc + rs~ (1+ a)]Here, fc = 14.125 MHz.

. . [ 14.125 - ;6 (1 + 0.25)] to [ 14.125 + ;6 (1 + 0.25)]

(14.125 - 0.01) to (14.125 + 0.01)14.115 MHz to 14.135 GHz.

Ie 25 In a satellite system, the downlink transmissw.n r~te::/::::it sec. if the required 7nergy p~r bit t? noise power density zs 8Calculate the required carrter-to-notse ratio.

Solution: Here, rb = 60 Mbits/see= 10 log 60 X 106 dB bits/see

= 77.8 dB Hz

EblNo=8dB

The C/N ratio is given by

(~o) =(!)+ rb = (8 + 77.8) = 85.8.

I 26 A 14-GHz uplink operates with transmission lo~sedsExamp e.. . d telliteG/T=8dB/K. The requiremargins totaling 2As10dB ~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 erlink antenna gaui 0 . . 'f baseband signal which has bit ratewer needed for transmtsswn 0 a1.544 Mbits / sec.

Solution: Here the bit rate is rb = 1.544 Mbitslsec

= 10 log 1.544 x 106 dB bits/see

= 62 dBbits/sec.

Eb Energy per bit = 9 dBNo = noise power density

Now (~o)=(:)+rb

Substituting above values

C = 9 + 62 = 671 dBN

Now EIRP is given by

EIRP = (~o)+ (~) + Losses - k

= 71 - 8 + 210 - 228.6 = 44.4 dBWTherefore, the required power of the transmitter will be

Pi = EIRp·- GUPa = 44.4 - 42 = 2.4 dBW= AntiloglO 2.4 = 251.2 dW.

797

SATELLITE 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~88l01lrequired for TDMA operation. reQse

Solution: Here, bit rate == 62 dBbits/sec. When TDMA is emplthe uplink bit rate increase will be oYed,

llrbUP = 71 - 62 = 9 dBThe EIRP must be increased by 9 dB. Therefore, the earth stat"

transmitted power will be IOn

PUP = (2.4 + 9) dBW = 11.4 dBW.

798

Example 28. Calculate the EIRP of a satellite downlink whic~12 GHz operates with a transmit power of 6 Wand an antenna gain of50.2 dB.

Solution: The EIRP is given byEIRP'= log Ps + Ga dBW = log 6 + 50.2

= 7.8 + 50.2 = 58.0 dBW.

Example 29. Calculate the gain of a 3-m paraboloidal antenna withan aperture efficiency of 0.55 operating at a frequency of 12 GHz.

Solution: For a paraboloidal antenna, the isotropic gain is givenby

Gi = T](10.472 f D)

where, T] = aperture efficiencyf = frequency

D = diameter of the antenna in metres.

Therefore, G; = 0.55 x (10.472 x 12 x 3)2 = 78,168

Hence G; = 10 log 78,168 = 48.9 dB.------------------------------------------------------

Example 30. For the system shown in Fig. 29.2. below, if the receiv.rno~sefigure is 12 dB, the cable loss is 5 dB, the LNA gain is 5~ dB, i.nozse temperature 150 K and the antenna noise temperature ts 35 .Calculate the overall noise temperature of the system.

Solution: Here, the antenna noise temperature (Ta) = 35 K. Thereceiver noise figure is

F= 12 dBor 10 10glO F = 12

10glO F = 1.2 dBor , F = 101.2= 15.85.

Similarly. the cable loss (Lc) = 10°·5= 3.16(GLNA) = 105

gailland LNA

799

Cable (Lc) ReceiverF

Fig. 29.2Now, the overall noise temperature of this system is given by

T To T (Lc-1)xTo Lc(F-1)To8 = a + LNA + G + G

LNA LNA

= 35 + 150 + (3.16 - 1) x 290. 105

+ 3.16 (15.85 -1)'x'290 185 K105

Example 31. In the link budget of a satellite, the free-space loss atGHz is 210 dB, the antenna pointing loss is 2 dB, and the atmospheric

rption is 2 dB. If the receiver GI T ratio is 19 dB IK, receiver feederBeS are 1dB and the EIRP is 50 dBW, calculate the carrier-to-noisectral density ratio.Solution:Here k = - 228.6 dB

EIRP= 50 dBWGIT = 19.5 dBlK

Losses = Free-space less + atmospheric absorption loss + antennapointing loss + receiver feedback losses

= 210 + 2 + 2 + 1 = 215 dBNow the elN spectral density ratio is given by

G(CINo) = EIRP + T - Losses - k

= 50 + 19 - 215 - ( - 286) = 140 dBW.

Example 32. An uplink of a satellite system operates at 14 GHz, andnux density required to saturate the transponder is - 100 dB Wlm2.free-space loss is 200 dB, and the other propagation losses amount

800 SATELLITE COMMUNICATIOIi

to 3dB. Calculate the earth-station EIRP required for saturation, assuk nditi nt.ing clear-s y co itions.

Solution: The effective area of an isotropic antenna (Ao) is givenby

Ao = - (21.45 + 20 log f)= - (21.45 + 20 log 14)

== -44.4 dBThe (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 themain central service area, and that the transponder power is 45 W,calculate the effective isotropic radiated power in watts as seen by theantenna.

or

EIRP = 10 log PsPs = 1O(EffiP/10) = 10(52110)

= 158.5 kW.

Example 34. An antenna has a noise temperature of 40 K and ismatched into a receiver which has a noise temperature of 100 K. Calculatethe noise power for a bandwidth of 36 MHz.

Solution: Here Ta = 40 K and TR = 100 K.Therefore, total noise of thus explain (antenna + receiver)

= TN = 40 + 100 = 140 KNow the noise power for a bandwidth ofBN is

PN=kTNBN

where k = 1.38 x 10- 23 J/K.

Therefore, for BN = 36 MHzPN = 1.38 x 10- 23 x140 x 36 x 106 = 0.069~

. . rfre~Example 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 tottPloss is 0.5 dB, and the antenna pointing loss is 1dB. Calculate tlink 'loss. efot'

'Solution: The tota11ink loss is the sum of all the losses ..Tb:~c ,trLosses = (free-space loss + receiver feeder loss + atmospb

sorption loss + antenna pointing loss)= 202 + 2 + 0.5 + 1 dB = 205.5 dB.

801

Example 36. Determine the clear-sky carrier-to-noise ratio for aellite TV system having worst case EIRP of 51 dBWat the receiver

. . The free space loss is 205.34 dB, the nominal usable figure of merit13.12 dB, and the gaseous attenuation due to atmospheric absorption0.17 dB. The highest video frequency is 5MHz and peak-to-peak video

• nal frequency deviation is 16MHz.Solution:

Here EIRP = 51 dBW, Lfs = 205.34 dBGIT=13.12dB, Latm=0.17dB

B = 6f + 2 fv = 16 + 10 = 26 MHz

k = 1.38 x 10- 23

CIN = EIRP - Lfs + GIT - Latm - 10 log (kB)

= 51- 205.34 + 13.12 - 10 log (1.38 x 10-23 x.26 x 106) - 0.17

= 51 - 205.34 + 13.12 - ( - 154.45) - 0.17 = 13.16.

Example 37.A satellite link is carrying data at rates of24 Kbits Iseehen 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 1every 79 received blocks has an error, calculate the transmission bit

for Stop and Wait system.Solution: 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 waitacknowledgement. 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 is detected, that is, after:::39.5 sec.

Therefore, bit transmission rate will be

•.&. of blocks transmitted without error and ACK x No. of bits/b1ockTime taken for transmission

79 x 127 bi s/ 2 4 bi s/= 39.5 It see = 5 It sec.

!example 38. Repeat the above calculation for Go-back N ARQ. What should be the capacity of the transmit buffer?

SATELLITE COMMUNICATION

Solution: The time required to transmit 79 blocks each of 127 biat 24 kbits/sec will be Its

= 79 x 127 = 0 4183 . sec.24x 10

802

Now in Go-back-N method, when error is detected in the 79th blockthe NAK system is sent to request retransmission ofblock 79. Howev 'sending of NAK and retransmission takes two-path delay time 7

r,

2 x 24 msec or 0.48 sec. During this period, the number ofblocks whi'~arrive at the receiver will be c

0.48 sec 0.48= Transmission time per block = 0.418/79 = 91.

These 91 blocks are discarded when the retransmission of79 blocksstarts. The transmission time decreases by 0.48/0.898 = 0.54. Thereforethe bit rate falls by 54% of the original value of 24 Kbits/sec, that is:

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 of79 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 aresult, the bit transmission rate will be

rb = 24 x 0.987 Kbits/sec= 23.70 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~nedata are sent to the hub from the VSATs by the inbound l.mk v~ahasttransponder at a message bit rate of 64 Kbits / see using bm

a7J' Pd'T18

shift keying (BPSK) and half rate forward error correction (FEC) CO ~ '

giving a transmitted bit rate of 128Kbits / hr. .,tJI' 160 ~.z,The occupied RF bandwidth of each VSAT channe

rs (i r tiltcorresponding to ideal RRC filters with a = 0.25. Multiple ac

cekB ~p(Jrt

inbound link is by SCPC-FDMA with RF channels spaced 200 zto allow a 40-KHz guard band between channels. setJI

Data from the hub station to the VSATs (the outbound link) c;!e~(J,.6as a continuous TDM stream of packets using a second transportBPSK with half rate FEe. ..0;. d t d otLt1'- • .J

The VSAT antenna has a d~ameter of Lm an satura

e {recel

"'"of2 W. The transponder noise temperature is 500 K and that aat hub station is 150 K.

Calculate the noise power in tran . 803FDMA channels. sponder-L or Lnthe inbound SCPC-

Solution: The inbound VSAT l' k1have a message data rate of64 ~~s/s that pass through transponder-erefore, with BPSK modulation (~ :~~ WIthhalfra~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 si1 per symbol. ' mce

The noise power, N», is given byN = kTsBN Wattsk = - 228.6 dBWlKIHz,

BN = 10 log 128 x 106 dBHz= 51.1 dBHz

d Ts = 500 K = 10 log 500 dBK = 27 dBK.

NTRI =k + Ts +BN- 228- - .6 + 27 + 51.1 = - 150.5 dBW.

Now

Exa~ple 41. In the above question Istation receiver or in inbound SCPC'ca culate the noise power in the. -FDMACh IThe inbound VSAT si 1 anne.an.smitted through t~~: reach the hub station after they aredwidth again will be 128kH p~nder by the satellite. The noise

el is received by a separatZe'IeFcause.at the hub station, each VSATreceiverHere, Ts = 150 K = 21 8 dBl{, B .. N= 128 kHz

NH = - 228.6 + 21.8 + 51.1 = - 155.7 dBW.

Example 42. In the same question Iponder-2 i.e. in outbound TDM ' ca culate the Noise Power ine bandwidth of 1MHz in the ;;~Tnnels,. assume a starting valueI . L1 recewer.ution : The outbound TDM biSATs passes through trans It streaI? from the hub station totransmitting 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 theVSAT receiver Thi g va ue noise bandwidth ofl MHd . IS correspond t B z

data rate of 500 kbits/sec and hs If0 a PSK signal with aa -rate FEC encodin

erefore BN = 1 MHz = 10 log 106 dBHz - 60 dBH g._~ - zansponder-2, Ts = 500 K = 27 dBK

erefore N ., TR2 = - 228.6 + 27 + 60= - 141.6 dBW.

804 SATELLITECOMMUNIC"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 receivoutbound TOM stream from the hub station in a noise bandwidthS theMHz. Here Ts = 150 K = 21.8 dBK and BN = 1 MHz = 60 MHz. of 1

NVR = - 228.6 + 21.8 + 60 = - 146.8 dBW. --Example 44. In the same question, calculate the power receivedthe satellite transmitted from the VSAT uplink. Assume as the worst caatthat the VSAT is located at the satellite beam's edge-o{-coverage, the~;dB contour of the satellite's uplink (receiving) antenna pattern. The gainof the VSAT transmit antenna is 42 dB and that of satellite receiveantenna 30 dB. Assume free path loss of207 dB.

Solution: The power received at the satellite from a single VSATin dB, is given by I

PR = PT + GT + GR - Lp - other lossesHere PT = transmit power = 2W = 10 log 2 = 3 dBW

GT = gain of the VSAT transmit antenna = 42 dBGR = gain of the satellite receive antenna = 30 dBLp = free space path loss at 14 HGz = 207 dB

other losses = satellite antenna edge of beam loss (2 dB) + clear airuplink atmospheric loss due to gases (0.5 dB) + miscel-laneous losses of 0.5 dB to account for antenna mispoint-ing feed loss.

Therefore, power received by the transponder isPR = 3 + 42 + 30 - 207 - 2 - 0.5 - 0.5

=-135 dB.-----------------------------------------------------

Example 45. In the same question, calculate the uplink inboundC IN ratio is transponder-L,

Solution: Each VSAT has a separate receiver in the hub station 0;noise bandwidth of28 kHz. Since the received power at the transponderinput as calculated in preceding example is - 135 dB, and noise po~~oin 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.

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 transpondera backoff of2 dB, calculate the maximum number ofVSAT channelscan be handled by the system. Assume each VSAT channel uses IW

power on the downlink. F2

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 1112.6 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. orbittingund earth is the centripetal force (FI) due to earth sgraoitation acting

'towards the center of the earth balancing the centrifugal force (F2) awngOt.uayfrom the center. Calculate the centrifugal force for a satellite ofrt&a88 IOO kg orbitting with a velocity of 8 km Is at a height of 200 kmGbove the surface of earth. Assume mean radius of earth to be 6370 km.

Solution:mV2

Centripetal force = (R +H)

here m = mass of satelliteV = orbital velocityR = mean radius of earth

806 SATELLITE COMMUNICA••TIO~

H = height of satellite above surface of earthThe centripetal force balances the centrifugal force.

. my2 100 x (8000)2Therefore, Centnfugal force = IT) H) = -

+ (6370 + 200) x 103

64 x 108= 3 = 974 Newtons

6570 x 10

. . . Apogee + PerigeeSemI-maJor axis = 2

= 30000 + 1000 = 15500 km2

Fig. 29.4 illustrates the point further.

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 earthM = 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/kgR=6370kmH= 150km

V = "39.8 x 1013/(6370 + 150) x 103 = 7.813 kmIs.

Example 50. A satellite moving in an elliptical eccentric orbit hassemi-major axis of the orbit equal to 16000 km (Fig. 29.5). If the

'6erence between the apogee and the perigee is 30000 km, determine theit eccentricity. ../Solution:

Apogee = a (1 + e)Perigee = a (1 - e)

ere a = semi-major axis of the ellipsee = orbit eccentricity

IIIII

b :@I I e----- __________t :_ -----: EarthII

Example 49.A satellite in an elliptical orbit has an apogee of30,000km and a perigee of 1000 km. Determine the semi-major axis of theelliptical orbit.

Fig. 29.5

Apogee - Perigee = a (1 + e) - a (1- e) = 2aeerEc tricit Apogee - Perigeecen nCI y, e = 2a

30000- 2 x 16000

30000= 3200 = 0.93 .

. 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 kmlpectively. Determine the apogee, the perigee and the orbit eccentricity.ume radius of earth to be 6370 km.Fig. 29.4

807

808 SATELLITE COMMUNICA1'Iott

Solution:Apogee = 30000 + 6370 = 36370 Jon

Perigee = 200 + 6370 = 6570 Jon

Eccentricity = Apogee - Perigee2a

where a = semi-major axis of the elliptical orbit

Also, a = Apogee + Perigee2

2a = Apogee + Perigee

Therefore, orbit eccentricity = Apogee - PerigeeApogee + Perigee

_ 36370 - 6570 29800- 36370 + 6570 = 42940 = 0.693.

or

. ~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:

,-:14000 km:.-, ,

Fig. 29.6

. The distance from center of ellipse (0) to the centre of earth (c) isgIven. ~y (a x e) where (a) is the semi-major axis and (e) is the eC-centricity.

Therefore,

Now apogeePerigee

axe = 1400014000

a = OT = 28000 Jon

= a(l + e) = 28000(1 + 0.5) = 42000 Ian

= a(1 - e) = 28000(1 - 0.5)= 28000 x 0.5 = 14000 Jon.

809

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 relation-between their orbital periods.

_ 24000km-',Fig. 29.7

Solution:The orbital time period (T) is given by

T = 21t ~a3/1!

here I!= GMG = Earth's gravitational constantM = mass of eartha = semimajor axis of ellipse

If (a1) and (a2) are the values of the semi-major axis of the ellipticalits ofthe satellites-1 and 2, (Tl) and (T2) are the corresponding orbital

riods, thenTl=21t~ and T2=21t~

312312 (24000) 312T2ITI = (a2lal) = 18000 = (4/3) = 1.54.

Thus orbital period of satellite-2 is 1.54 times the orbital period oftellite-1.

Example 54 : Determine the escape velocity for an object to benehed from surface of earth from a point where earth's radius is 6360(G = 6.67 x io:" Nm2/kg'Z and M = 5.98 x 1024kg).

Solution:Escape velocity = -J2}J1r

SATELLITE COMMUNICA1l0ti

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,-------

Therefore, Escape Velocity = -V2 x 39.8 x 1~13 = ~6360 x 10 6360

= 11.2 km/s.

/Example 55.Calculate the orbital period of a satellite in an~elliptical orbit shown in Fig. 29.8.

II

---------------------1------~-----

L:=:=: 50000k~

Fig. 29.8

Solution. Semi major axis, a = 50~00 = 25000 km

Orbital time period, T = 21t -Va3/1!and (rp) are the apogee and perigee distances, then

I! = OM = 6.67 x 10 - 11x 5.98 x 1024

= 39.8 x 1013 Nm2/kg

T = 2 x 3.14 ..J(25000 x 103)339.8 x 1013

= 6.28 ..y 15625 x 1018

39.8 x 1013

= 6.28 x 6.25 x 103 = 39250 seconds= 10 hours 54 minutes.

•

-------------------------------------------------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 timerespectively from the surface of the earth. Determine the orbzt~arth'Speriod and the velocity at the apogee and perigee point. (Assumeradius = 6360 km.)

811

solution.t\pogee distance = 35000 + 6360 = 41360 kmperigee distance = 500 + 6360 = 6860 kmSemi-major axis of elliptical orbit,

a = 41360 + 6860 = 24110 km2

Orbital time period, T = 21t -Va3/1!= 6.28~,..--(2-4-1-10-X-1-03)3

6.67 x 10- 11 x 5.98 X 1024

= 10 hrs 20 minutes

Velocity at any point on the orbit is given by:V = -J1.l [(21r) - (l/a»)

V = OM = 6.67 x 10 - 11 x 5.98 X 1024

= 39.8 X 1013 Nm2/kg

At apogee point, r = 41360 km. Therefore,

V="-~-9.-8X--10-13-[---2---------1----]41360 x 103 24110 X 10

3

= ,,39.8 X 1013 [ 48220 - 41360 ]41360 x 24110 x 103

= - ,r--3-9-.8-x-1-0-1'::"3-X6-86-0-=523 mls

" 41360 x 24110 x 103

At perigee point, r = 6860 km. Ther~fore,V=,/~3-9.-8-X-10-1-3[----2---------1---]-

6860 x 103 24110 x 103

= ,,39.8 x 1013 [ 48220 - 6860 ]6860 x 24110 x 103

= ~I 39.8 x 1013 x 41360 = 9.976 kmls.6860 x 24110 x 103

Example 57 : The sum of apogee and perigee distance ora certainiptical 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 COMMUNIC A"O~e = (r'l - rp) = 30000 = 0 6

ro + rp 50000 ..

flxample 58 : The semi-major axis and the semi-minor ;-:---ell.iptical satellite orbit are 20,000 km and 16000 km respectivel LS;r anmine the apogee and perigee distances. y. eter.

Solution:

If (ra) and v» are apogee and perigee distances respectively, then. . . ra + rpsemi-major axis =

2

Semi-minor axis = "ra rp

ra + rp = 20000 km2

Therefore, ra + rp = 40000 km

"ra x rp = 16000Therefore, To rp = 256000000Now ro + rp = 40000 (1)

ra x rp = 256000000 (2)Substituting the value of (rp) from (2) in (1)

ra (40000 - ra) = 256000000

or r~ - 40000 ra + 256000000 = 0

40000 ± ..J16 x 108 - 10.24 x 108r« = 2

40000 ± ..J5.76 x 108 40000 + 2.4 X 104

2 2

= 3.2 X 104, 1.6 X 104 = 32000 km, 16000 krnro = 32000 km as it cannot be 16000 km if the semi-major axis is

20,000km.rp = 40000 - 32000 = 8000 km.

, ---.Example 59 :A satellite is moving in a near earth circular orbi~~

a distance of 640 km. Determine its orbital period. (Assume R ::;6km.)

Solution:Orbital velocity = ~ (RO:H) =

= ••.139.8 X 1013'V 7 X 106 = 7.54 kmIs.

246.67 x 10 - 11 x 5.98 x !.!!.-7000 x 103

813

_ 21t(R + H) _ 6.28 x 7000- V - 7.54

= 5830 sec = 1 hour 37 minutes.

Example 60 :A satellite moving in an eccentric elliptical orbit hasi-major axis and semi-minor axis of (a) and (b) respectively and anntricity 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 satellitemove from A to B in the direction shown in Fig. 29.9.

y

Lx+-------a--------~

Fig. 29.9

Selution rAccording to Kepler's law for elliptical satellite orbits, the line. the satellite and the center ofthe earth spans equal ellipse area

equal time.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 byArea of half of ellipse - Area of MOB

1tab 1tab= --2- - b x OC =-2- - bae = ab (rr/2- e)

= ab (3'i4 - 0.6J = 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 t' f h . 2.2 aba 100 t etwoareasls~=2.2.

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 theIOnshown. Therefore,

'llIlle taken = 2.2 x 3 hrs 10 min = 7 hrs.

SATELLITE COMMUNICl\llO~

. 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:ciaSolution: '

Apogee distance, Ta = 50000 km

Perigee distance, rp = 8000 km

Semi-major axis, a = ra + rp2

50000 + 80002 -29000 km

Semi-minor axis, b = "ra x rp= "=50=-=0-=-00"'-x-80-0-0= 20000 km

Orbit eccentricity, e = (ra - rp) = (50000 - 8000 '\ra + rp 50000 + 8000)

42000= 58000 = 0.724.

tu E.xalmp!~62.. Satellite-1 and Satellite-2 are orbitting in differentehLPtlC~ 0': its uiitli same perigee but different apogee distances assown m FLg.29.10. The semi-major axes of the two orbits are 16000 kmz: 24~00 km: If the orbital period of satellite-1 is 10 hours determinet e orbital period of satellite-2. '

Fig. 29.10Solution:

Orbital period (T) is given by, T = 2n -.Ja3/1l

where a = semi-major axisIl =GM .

If•.~Tl)land (T2) are the orbital periods of satellite-l and sateIIite-2respec ive y, then

V£D PROBLEMS

(T2) _ 2 n .JaW _ (a2J312

Tl - 2n ~a~/Il - al312 312

T = T (a2) = 10 (24000)2 1 al 16000

= 10(1.!?)312= 18.37 hours.

815

Example 63 : A geosynchronous satellite moving in an equatorialular orbit at a height of 35800 km above the surface of earth getslined at an angle of 2° due to some reasons. Calculate the maximumviation in latitude, if the maximum deviation in longitude (witherenceto longitude of ascending node) is 0.0175°, determine maximum

. :placements in kms caused by latitude and longitude displacements.sume earth's radius = 6364 hm).Solution:

Height of orbit = 35800 km

Earth's radius = 6364 km

Therefore, Orbit radius, r = 35800 + 6364 = 42164 km

Angle of inclination = 2°Maximu~ latitude deviation from equator due to inclination (i) is

• en by:

Maximum displacement (in krn) due to Amax is given by:

D", (max) = ai ( 1~0J where (i) is in deg

= 42164 x 2 x n = 1471 km180 .

Maximum longitude deviation from ascending node,

i2 22 4\jImax = 228 = 228 = 228 = 0.0175°

Maximum displacement (DljI)due to (\jImax) is given by:

DljI= D",(~:: ) = 1471 x 0.0;75 = 12.9 km.

------------------------------------------------------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:

- r;;-tan i = ~ 39.8 x 1013

tan 2° = 107 mls.-" ~ ; 42164 x 103

816 SATELLITECOMMUNICA••lIO~

Example 65 :Ageosynchronous satellite orbitting at 4216;;;;--earth's center has a circular and equatorial orbit. The orbit gets inc{:OTndue to some reason and it is observed that the maximum displace::::ddue to latitude deviation in 500 km. Determine angle of inclinatio ~tbetween 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)

or DJ...(max) = r x i

Therefore, i = DJ...(max)/r = 4~~~4 = 0.012 rad = 0.68 deg.

Example 66 : A geostationary satellite moving in an equatorialcircular orbit is at a height of35786 km from earth's surface. Ifthe earth'sradius is taken as 6378 km, determine the theoretical maximum coverageangle. Also determine the maximum slant range.

Solution:For theoretical maximum coverage angle, elevation angle, E = o.

Maximum coverage angle, 2amax = 2 sin -1Re~ H cos Emin)

where Re = earth's radius

or

H = height of satellite above earth's surface

2 amax = 2 Si{ 3578663~~378 cos 00) = 17.4°

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

= 1737139041

D =41679 km.or .------:----------------------ngleand

Example 67. What would be the new maximum coverage a t zerothe slant range if the minimum possible elevation angle is 5° and noas in Example-66 (Refer to Fig. 29.11).

817

Fig. 29.11

Solution:The maximum coverage angle, (2<Xma.x)is given by

2amax = 2 sin - 1[( Re~ H) cos Emin]

re Emin = Minimum elevation angle

. - il( 6378 } cos 50]2 amax = 2 sin ~ 6378 + 35786

= 2 sin - 1[0.1512 x 0.996]

= 2 sin -10.1506 = 2 x 8.66 = 17.32°.

Example 68 : Refer to Fig. 29.12 showing a geostationary satelliteitting earth. Calculate the angle (6) subtended by the arc of thellite's footprint at the center of earth.

Therefore

6 = 61 + 6261 = 90° - a1 - E162 = 90° - a2 - E26 = (90 - a1 - E1) + (90 - a2 - E2)

= 180 - (a1 + a2) - (E1 + E2)= 175° - (a1 + a2)

. - 1 6378 cos 5°ci = sm 6378 + 35786

. _ 1 6378 cos 5° _ 8 660=sm 42164 -.

(E1 = 5°)

(E2 = 0°)