Satellite Communication Dc Agrawal

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

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