w. tomasi - pp

Eb = 52 x !iNo N fb

The E,jNo ratio is the product of the carrier-to-noise ratio (GIN) and the noise band-width-to-bit rate ratio (B/fb).Expressed as a log,

Eb (dB) = G (dB) +!i (dB)No N fb

The energy per bit (Eb) will remain constant as long as the total wideband carrierpower (C) and the transmission rate (bps) remain unchanged. Also, the noise density (No)will remain constant as long as the noise temperature remains constant. The followingconclusion can be made: For a given carrier power, bit rate, and noise temperature, theE,jNo ratio will ~main constant regardless of the encoding technique, modulation scheme,or bandwidth used.

Figure 7-17 graphically illustrates the relationship between an expected probabilityof error P(e) and the minimum GIN ratio required to achieve the P(e). The GINspecified isfor the minimum double-sided Nyquist bandwidth. Figure 7-18 graphically illustrates therelationship between an expected P(e) and the minimum E,jNo ratio required to achievethat P(e).

A P(e) of 10-5 (1/105) indicates a probability that 1 bit will be in error for every100,000 bits transmitted. P(e) is analogous to the bit error rate (BER).

(7-6)

EXAMPLE7-6

A coherent binary phase-shift-keyed (BPSK) transmitter operates at a bit rate of 20 Mbps. Fora probability of error P(e) of 10-4:(a) Determine the minimum theoretical CIN and EJNo ratios for a receiver bandwidth equal

to the minimum double-sided Nyquist bandwidth.(b) Determine the CIN if the noise is measured at a point prior to the bandpass filter, where

the bandwidth is equal to twice the Nyquist bandwidth.(c) Determine the CIN if the noise is measured at a point prior to the bandpass filter where

the bandwidth is equal to three times the Nyquist bandwidth.

Solution (a) With BPSK, the minimum bandwidth is equal to the bit rate, 20 MHz. FromFigure 7-17, the minimum ClNis 8.8 dB. Substituting into Equation 7-6 gives us

Eb (dB) = C (dB) + !i (dB)No N fb

20 X 106= 8.8 dB + 10 log 0 062 XI

= 8.8 dB + 0 dB = 8.8 dB

Note: The minimum E~o equals the minimum CIN when the receiver noise bandwidthequals the minimum Nyquist bandwidth. The minimum E~o of 8.8 can be verified from Fig-ure 7-18.

What effect does increasing the noise bandwidth have on the minimum CIN and E~oratios? The wideband carrier power is totally independent of the noise bandwidth. Similarly,an increase in the bandwidth causes a corresponding increase in the noise power. Conse-quently, a decrease in CIN is realized that is directly proportional to the increase in the noisebandwidth. Eb is dependent on the wideband carrier power and the bit rate only. Therefore, Ebis unaffected by an increase in the noise bandwidth. No is the noise power normalized to aI-Hz bandwidth and, consequently, is also unaffected by an increase in the noise bandwidth.

Satellite System Parameters 287

P(e)

~!/)C-eo

~!/)C-~

~C-<{cb~

~!/)C-cb

~!/)C-oo

10-3I I

\\\\\\\\\\\\\\\\\\\,,,\,

8-PSK\, I"

v160AM i, I, III, :, II, I, :, I, :, :

~, :I I

:' :" I:, ::, :I , I: , ::' :" [:, ::' ::' :I, " I

16-PSK

10-4

10-6

10-6

10-7

10-6

10-9

10-10

6C/N (dB)

168 10 12 14 18 20 22 24 26

Figure 7-17 Pre) performance of M-ary PSK, GAM, QPR, and M-ary APK coherentsystems. The rms cm is specified in the double-sided Nyquist bandwidth.

(b) Since Ei/No is independent of bandwidth, measuring the C/N at a point in the receiverwhere the bandwidth is equal to twice the minimum Nyquist bandwidth has absolutely noeffect on Ei/No. Therefore, Ei/No becomes the constant in Equation 7-6 and is used tosolve for the new value of C/N. Rearranging Equation 7-6 and using the calculated Ei/Noratio, we have

288 Satellite CommunicationsChap. 7

,I,:

10-1

10-<3

I:

II

II

I'

II~

.il

ll

II:

I

I~

II

1111"1

10-2

10-3

~e 10-4~

.....0

,£:0co

"2 10-5c...

'II

10-<3

7 8 9 10 11 12 13 14

EbiNo (dB)

15 16 17 18 19

'II

II,

~II

III

Iii

II

II

10-7

Figure 7 -18 Probability of error P(e) versus EblNo ratio for various digitalmodulation schemes.

C (dB) = Eb (dB) - !i (dB)N No fb

40 X 106

= 8.8dB - 10 log 20 X 106

= 8.8 dB - 10 log 2

= 8.8 dB - 3 dB = 5.8 dB

(c) Measuring the CIN ratio at a point in the receiver where the bandwidth equals three timesthe minimum bandwidth yields the following results for CIN.

£ = Eb - 10 1 60 X 106N No og 20 X 106

= 8.8dB - 10 log 3

= 4.03 dB

The CIN ratios of 8.8, 5.8, and 4.03 dB indicate the CIN ratios that would be measuredat the three specified points in the receiver to achieve the desired minimum E~o and Pee).


Because Et/No cannot be directly measured to determine the Et/No ratio, the widebandcarrier-to-noise ratio is measured and then substituted into Equation 7-6. Consequently, toaccurately determine the Et/No ratio, the noise bandwidth of the receiver must be known.

EXAMPLE 7-7

A coherent 8-PSK transmitter operates at a bit rate of 90 Mbps. For a probability of error of10-5:

(a) Determine the minimum theoretical CIN and Et/No ratios for a receiver bandwidth equalto the minimum double-sided Nyquist bandwidth.

(b) Determine the CIN if the noise is measured at a point prior to the bandpass filter wherethe bandwidth is equal to twice the Nyquist bandwidth.

(c) Determi~e the CIN if the noise is measured at a point prior to the bandpass filter wherethe bandwidth is equal to three times the Nyquist bandwidth.

Solution (a) 8-PSK has a bandwidth efficiency of 3 bps/Hz and, consequently, requires aminimum bandwidth of one-third the bit rate or 30 MHz. Froin Figure 7-17, the minimumCIN is 18.5 dB. Substituting into Equation 7-6, we obtain

Eb (dB) = 18.5 dB + 10 log 30 MHzNo 90 Mbps

= 18.5dB + (-4.8 dB) = 13.7 dB

(b) Rearranging Equation 7-6 and substituting for Et/No yields

C (dB) = 137 (dB) - 101 60 MHzN . og 90 Mbps

= 13.7dB - (- 1.77dB) = 15.47dB

(c) Again, rearranging Equation 7-6 and substituting for E,jNo gives us

52(dB = 13.7 (dB) - 101 90 MHzN og 90Mbps

= 13.7dB - 0 dB = 13.7 dB

It should be evident from Examples 7-6 and 7-7 that the Et/No and CIN ratios are equalonly when the noise bandwidth is equal to the bit rate. Also, as the bandwidth at the point ofmeasurement increases, the CIN decreases.

When the modulation scheme, bit rate, bandwidth, and CIN ratios of two digital radiosystems are different it is often difficult to determine which system has the lower probabilityof error. Because Et/No is independent of bit rate, bandwidth, and modulation scheme; it is aconvenient common denominator to use for comparing the probability of error performanceof two digital radio systems.

EXAMPLE7-8

Compare the performance characteristics of the two digital systems listed below, and deter-mine which system has the lower probability of error.

Solution Substituting into Equation 7-6 for the QPSK system gives us

290 Chap. 7 Satellite Communications

QPSK 8-PSK

Bit rate 4DMbps 60 MbpsBandwidth 1.5 X minimum 2 X minimumCIN 10.75 dB 13.76 dB

E C B-1L (dB) = - (dB) + 10 log 7No N Jb

= 10.75 dB + 10 log 1.5 X 20 MHz40 Mbps

= 10.75 dB + (-1.25 dB)

= 9.5 dB

From Figure 7-18, the Pee) is 10-4.Substituting into Equation 7-6 for the 8-PSK system gives us

~ Eb (dB) = 13.76 dB + 10 10 2 X 20 MHzNo g 60Mbps

= 13.76dB + (-1.76 dB)

= 12dB

From Figure 7-18, the Pee) is 10-3.Although the QPSK system has a lower C/N and E~o ratio, the Pee) of the QPSK sys-

tem is 10 times lower (better) than the 8-PSK system.

Gain-to-Equivalent Noise Temperature Ratio

Essentially, gain-to-equivalent noise temperature ratio (GITe) is a figure of merit used torepresent the quality of a satellite or an earth station receiver. The GITeof a receiver is theratio of the receive antenna gain to the equivalent noise temperature (Te) of the receiver.Because of the extremely small receive carrier powers typically experienced with satellitesystems, very often an LNA is physically located at the feedpoint of the antenna. Whenthis is the case, GITe is a ratio of the gain of the receiving antenna plus the gain of theLNA to the equivalent noise temperature. Mathematically, gain-to-equivalent noise tem-perature ratio is

Q = Ar + A(LNA)Te Te

(7-7)

Expressed in logs, we have

Q (dBK-1) = Ar (dB) + A(LNA)(dB) - Te (dBK)Te

GITeis a very useful parameter for determining the EtfNo and C/N ratios at the satel-lite transponder and earth station receivers. GITeis essentially the only parameter requiredat a satellite or an earth station receiver when completing a link budget.

(7-8)

EXAMPLE7-9

For a satellite transponder with a receiver antenna gain of 22 dB, an LNA gain of 10 dB, andan equivalent noise temperature of 22 dBK; determine the GlTe figure of merit.

Solution Substituting into Equation 7-8 yields

Q (dBK-1) = 22 dB + 10 dB - 22 dBKTe

= 10dBK-1


SATELLITE SYSTEM LINK EQQAnONS

The error performance of a digital satellite system is quite predictable. Figure 7-19 showsa simplifiedblock diagram of a digital satellite system and identifies the various gains andlosses that may affect the system performance. When evaluating the performance of a dig-ital satellite system, the uplink and downlink parameters are first considered separately,then the overall performance is determined by combining them in the appropriate manner.Keep in mind, a digital microwave or satellite radio simply means the original and demod-ulated baseband signals are digital in nature. The RF portion of the radio is analog; that is,FSK, PSK, QAM, or some other higher-level modulation riding on an analog microwavecarner.

~

C'

ArAt, Pr

HPAPt

Lbo

Lf LfPr Et/No

Lb LNA C/NGrre

Lp

Earth stationtransmitter

Earth stationreceiver

Figure 7-19 Overall satellite system showing the gains and losses incurred in both theuplink and downlink sections. HPA, High-power amplifier; Pto HPA output power; Lbo,back-off loss; Lf, feeder loss; Lb, branching loss; At, transmit antenna gain; Prototalradiated power = Pt - Lbo - Lb - L,; EIRP, effective isotropic radiated power = PrAt;Lu, additional uplink losses due to atmosphere; Lp, path loss; An receive antenna gain;GITe, gain-to-equivalent noise ratio; Ld, additional downlink losses due to atmosphere;LNA, low-noise amplifier; CITe, carrier-to-equivalent noise ratio; CINo, carrier-to-noisedensity ratio; EblNo, energy of bit-to-noise density ratio; CIN, carrier-to-noise ratio.

LINK EQQATIONS

The following link equations are used to separately analyze the uplink and the downlinksections of a single radio-frequency carrier satellite system. These equations consider onlythe ideal gains and losses and effects of thermal noise associated with the earth stationtransmitter, earth station receiver, and the satellite transponder. The nonideal aspects ofthe system are discussed later in this chapter.

QpIink Equation

292

C AtPrCLpLu)Ar AtPr(LpLu) G-- - x-No - KTe K Te

Chap. 7 Satellite Communications

Ar Lf I Lb Lb I Lf AtGrre

LNA Lbo HPAC'1

I

Satellitetransponder

where Ld and Lu are the additional uplink and downlink atmospheric losses, respectively.The uplink and downlink signals must pass through Earth's atmosphere, where they arepartially absorbed by the moisture, oxygen, and particulates in the air. Depending on theelevation angle, the distance the RF signal travels through the atmosphere varies from oneearth station to another. Because Lp, Lw and Ld represent losses, they are decimal valuesless than 1. Grre is the receive antenna gain plus the gain of the LNA divided by theequivalent input noise temperature.

Expressed as a log,

C

(41TD

) (G

)- = 10 log At?r - 20 log - + 10 log - - 10 log Lu - 10 log K~ A ~

~~~~~

~ EIRP free-space + satellite - additional - Boltzmann's- earth path loss G/Te atmospheric constantstation losses

= EIRP (dBW) - Lp (dB) + Q (dBK-1) - Lu (dB) - K(dBWK)Te

Downlink Equation

f.. = At? r(LpLd)Ar = At? r(LvLd) x QNo KTe K Te

Expressed as a log

C

(41TD

) (G

)- = 10 log At?r - 20 log - + 10 log - - 10log Ld - 10log K~ A ~

~~~~~

EIRP free-space + satellite - additional - Boltzmann'ssatellite path loss G/Te atmospheric constant

losses

= EIRP (dEW) - Lp (dB) + Q (dBK-1) - Ld (dB) - K (dBWK)Te

LINK BUDGET

Table 7-4 lists the system parameters for three typical satellite communication systems.The systems and their parameters are not necessarily for an existing or future system; theyare hypothetical examples only. The system parameters are used to construct a link bud-get. A link budget identifies the system parameters and is used to determine the projectedC/N and Ei/No ratios at both the satellite and earth station receivers for a given modulationscheme and desired Pee).

EXAMPLE 7 -1 0,I,Complete the link budget for a satellite system with the following parameters."I,'

Uplink

1. Earth station transmitter output powerat saturation, 2000 W

33 dBW

Link Budget 293

1'1!::::,~~Ii'I'::iii:::ij:I:':f ~ ,'f_1 III,::,. " :"" """~lII,:I::IIIII~:liil:'IIIII;I:'FI;I!11 i::I:Hl:ll>lllm:lliIIilliliilllli

r.i111__""., " ",,_~ILill':II~::"~"'fi' I

294

TABLE 7-4 SYSTEM PARAMETERS FOR THREE HYPOTHETICALSATELLITE SYSTEMS

'"" 0-~~...."0>'0.. 0 -

~ u cS -'" .8" 'g ..~ ""5>. -"0

CI) N 0::c sO~::tCl)'0 A..

CI

'"" 0-01).<:;~::E>000\.. <.) -

~-",cS 'g .gB ,,~'" - '">.N"O

CI)::CO0 SN~-CI)~A..-""

'"~2'~::E"0>N0-

oo <.) -u-"'cS 'g .g" " ,.~ -"5>.N"OCI)::Co

0 s8~<TA..-, 00

~ Uplink

Transmitter output power (saturation, dBW)Earth station back-off loss (dB)

Earth station branching and feeder loss (dB)

Additional atmospheric (dB)

Earth station antenna gain (dB)

Free-space path loss (dB)Satellite receive antenna gain (dB)

Satellite branching and feeder loss (dB)

Satellite equivalent noise temperature (K)Satellite GlTe (dBK-1)

Downlink

Transmitter output power (saturation, dBW)Satellite back -off loss (dB)

Satellite branching and feeder loss (dB)

Additional atmospheric loss (dB)

Satellite antenna gain (dB)

Free-space path loss (dB)Earth station receive antenna gain (dB)

Earth station branching and feeder loss (dB)

Earth station equivalent noise temperature (K)Earth station GlTe (dBK-1)

35230.6

5520020

11000-10

180.510.8

1619751

3250

27

2. Earth station back-off loss

3. Earth station branching and feeder losses

4. Earth station transmit antenna gain (fromFigure 7-20, 15 mat 14GHz)

5. Additional uplink atmospheric losses

6. Free-space path loss (from Figure 7-21,at 14 GHz)

7. Satellite receiver GtTeratio

8. Satellite branching and feeder losses9. Bit rate

10. Modulation scheme

Downlink

1. Satellite transmitter output power atsaturation 10 W

2. Satellite back-off loss

3. Satellite branching and feeder losses


25 332 33 40.4 0.6

45 64208 206.545 23.7

1 0800 800

16 -5.3

20 100.2 0.11 0.51.4 0.4

44 30.8206 205.644 62

3 01000 270

14 37.7

3 dB

4dB

64 dB

0.6 dB

206.5 dB

-5.3 dBK-1

OdB

120 Mbps8-PSK

10 dBW

0.1 dB

0.5 dB

Link Budget

60

50

~ 40c:.~C>.,c:c:2~ 30

10

Figure 7-20 Antenna gain based onthe gain equation for a parabolicantenna:

A (db) = 10 log 1'] ('IT 0/'11.)2

0

where 0 is the antenna diameter, '11.=the wavelength, and 1']= the antennaefficiency. Here 1']= 0.55. To correctfor a 100% efficient antenna, add2.66 dB to the value.

2 3 4 5 10 150.5

Antenna diameter (m)

4. Satellite transmit antenna gain (fromFigure 7-20, 0.37 mat 12 GHz)

5. Additional downlink atmospheric losses

6. Free-space path loss (from Figure 7-21,at 12 GHz)

7. Earth station receive antenna gain (15 m,12 GHz)


9. Earth station equivalent noise temperature10. Earth station GlTe ratioII. Bit rate12. Modulation scheme

Solution Uplink budget: Expressed as a log,

EIRP(earthstation)= Pt + At - Lbo - Lbf

30.8 dB

0.4 dB

205.6 dB

62 dB

OdB

270K

37.7 dBK-1

120 Mbps8-PSK

Carrier power density at the satellite antenna:

= 33 dBW + 64 dB - 3 dB - 4 dB = 90 dBW

295

1

i:D~'".2 202.;;coCo

~ 200coCo'".,~ 198u.

210

208

206

204

Elevation anglecorrection:

Angle90°45°0°

+dB

00.441.33

196

194

192

3 4 5 6 7 8 9 10 11 12 13 14 15

Frequency (GHz)

Figure 7-21 Free-space path 1055 (Lp) determined from Lp = 183.5 + 20 logf(GHz}, elevation angle = 90°, and distance = 35,930 km.

C' = EIRP (earth station) - Lp - Lu

= 90 dBw - 206.5 dB - 0.6 dB = -117.1 dBW

C/No at the satellite:

Thus

~=~=~XJ...No KIe Te K

C Gwhere - = C' X -

Te Te

C G 1-=C'x-x-No Te K

Expressed as a log,

Thus

296

~ (dB) = C' (dBW) + Q (dBK-1) - 10 log (1.38 X 10-23)No Te

~ = -117.1 dBW + (-5.3 dBK-1) - (-228.6dBWK) = 106.2dBNo

Eb (dB) = C/fb (dB) = ~ (dB) - 10 logfbNo No No

Eb = 106.2 dB - 10 (log 120 X 106)= 25.4 dBNo


and for a minimum bandwidth system,

C = Eb - .!i = 25.4 - 10 10 40 X 106 = 30.2 dBN No fb g 120 X 106

Downlink budget: Expressed as a log,

EIRP (satellite transponder) = Pt + At - Lbo - Lbf

= 10 dBW + 30.8 dB - 0.1 dB - 0.5 dB

= 40.2 dBW

Carrier power density at earth station antenna:

~C' = EIRP (dBW) - Lp (dB) - Ld (dB)

= 40.2 dBW - 205.6 dB - 0.4 dB = -165.8 dBW

C/No at the earth station receiver:

~=-.£=~X~No KTe Te K

C Gwhere - = C' X -

Te Te

Thus

~=C'X.QX~No Te K

Expressed as a log,

~ (dB) = C' (dBW) + .Q (dBK-1) - 10 log (1.38 X 10-23)No Te

= -165.8 dBW + (37.7 dBK-1) - (-228.6 dBWK) = 100.5 dB

An alternative method of solving for C/No is

~ (dB) = C' (dBW)+ Ar (dB) - Te(dBK-1) - K (dBWK)No

= -165.8 dBW + 62 dB - 10 log 270 - (-228.6 dBWK)

~ = -165.8 dBW + 62 dB - 24.3 dBK-1 + 228.6 dBWK = 100.5 dBNo

E C!L (dB) = - (dB) - 10logfb

No No

= 100.5 dB - 10 log (120 X 106)

= 100.5dB - 80.8dB = 19.7dB

and for a minimum bandwidth system,

C = Eb - .!i = 19.7 - 10 1 40 X 106 = 24.5 dBN No fb og 120 X 106

i

I

!

t

r

II

I

.I!I

il

i

II

I~

~

With careful analysis and a little algebra, it can be shown that the overall energy of bit-to-noise density ratio (Et/No), which includes the combined effects of the uplink ratio (Et/No)uand the downlink ratio (E,/No)d, is a standard product over the sum relationship and isexpressed mathematically as

Eb (Et/NoL (Et/NO)d

No (overall) = (Et/NoL + (Et/NO)d

Link Budget

(7-9)

297

J

where all E,jNo ratios are in absolute values. For Example 7-10, the overall E,jNo ratio is

Eb (overall) = (346.7)(93.3) = 73.5No 346.7+ 93.3

= 10 log 73.5 = 18.7dB

As with all product-over-sum relationships, the smaller of the two numbers domi-nates. If one number is substantially smaller than the other, the overall result is approxi-mately equal to the smaller of the two numbers.

The system parameters used for Example 7-10 were taken from system C in Table7-4. A complete link budget for the system is shown in Table 7-5.

1

TABLE 7-5 LINK BUDGET FOR EXAMPLE7-10

Uplink

1. Earth station transmitter output power atsaturation, 2000 W

2. Earth station back -offloss


4. Earth station transmit antenna gain5. Earth station EIRP

6. Additional uplink atmospheric losses

7. Free-space path loss8. Carrier power density at satellite

9. Satellite branching and feeder losses10. Satellite Glfe ratio

11. Satellite Clfe ratio

12. Satellite C/No ratio13. Satellite C/N ratio

14. Satellite Ei/No ratio15. Bit rate16. Modulation scheme

Downlink

1. Satellite transmitter output power atsaturation, 10 W

2. Satellite back-off loss

3. Satellite branching and feeder losses

4. Satellite transmit antenna gain5. Satellite EIRP

6. Additional downlink atmospheric losses

7. Free-space path loss8. Earth station receive antenna gain

9. Earth station equivalent noise temperature

10. Earth station branching and feeder losses11. Earth station G/Te ratio

12. Carrier power density at earth station13. Earth station C/Te ratio

14. Earth station C/No ratio15. Earth station C/N ratio

16. Earth station Ei/No ratio17. Bit rate

18. Modulation scheme

33 dBW

3dB4dB

64 dB90 dBW

0.6 dB206.5 dB

-117.1 dBWOdB

-5.3 dBK-1-122.4 dBWK-1

106.2 dB30.2 dB25.4 dB

120 Mbps8-PSK

10 dBW

0.1 dB0.5 dB

30.8 dB40.2 dBW

0.4 dB205.6 dB

62 dB270 K

OdB37.7 dBK-1

-165.8 dBW .-128.1 dBWK-1

100.5 dB24.5 dB19.7 dB

120 Mbps8-PSK


NONIDEAL SYSTEM PARAMETERS

Additional nonideal parameters include the following impairments: AMlAM conversionand AMIPM conversion, which result from nonlinear amplification in HPAs and limiters;pointing error, which occurs when the earth station and satellite antennas are not exactlyaligned; phase jitter, which results from imperfect carrier recovery in receivers; nonidealfiltering, due to the imperfections introduced in bandpass filters; timing error, due toimperfect clock recovery in receivers; andfrequency translation errors introduced in thesatellite transponders. The degradation caused by the preceding impairments effectivelyreduces the Ei/No ratios determined in the link budget calculations. Consequently, theyhave to be included in the link budget as equivalent losses. An in-depth coverage of thenonideal par~eters is beyond the intent of this text.

QUESTIONS

7-1. Briefly describe a satellite.

7-2. What is a passive satellite? An active satellite?

7-3. Contrast nonsynchronous and synchronous satellites.

7-4. Defineprograde and retrograde.

7-5. Define apogee and perigee.

7-6. Briefly explain the characteristics of low-, medium-, and high-altitude satellite orbits.

7-7. Explain equatorial, polar, and inclined orbits.

7-8. Contrast the advantages and disadvantages of geosynchronous satellites.

7-9. Define look angles, angle of elevation, and azimuth.

7-10. Define satellite spatial separation and list its restrictions.

7-11. Describe a "footprint."

7-12. Describe spot, zonal, and earth coverage radiation patterns.

7-13. Explain reuse.

7-14. Briefly describe the functional characteristics of an uplink, a transponder, and a downlinkmodel for a satellite system.

7-15. Define back-off loss and its relationship to saturated and transmit power.

7-16. Define bit energy.

7-17. Define effective isotropic radiated power.7-18. Define equivalent noise temperature.

7-19. Define noise density.

7-20. Define carrier-to-noise density ratio and energy of bit-to-noise density ratio.

7-21. Define gain-to-equivalent noise temperature ratio.

7-22. Describe what a satellite link budget is and how it is used.

PROBLEMS

7-1. An earth station is located at Houston, Texas, which has a longitude of99.5° and a latitude of29.5° north. The satellite of interest is Satcom 2. Determine the look angles for the earth sta-tion antenna.

Problems 299

._-~ ., ",- !I:' ~",":i"''',''''"''''" ""'" -..- "'-- - ._-""" , - ".",,,"",.....

7-2. A satellite system operates at l4-GHz uplink and ll-GHz downlink and has a projected P(e)of 10-7. The modulation scheme is 8-PSK, and the system will carry 120Mbps. The equiva-lent noise temperature of the receiver is 400 K, and the receiver noise bandwidth is equal tothe minimum Nyquist frequency. Determine the following parameters: minimum theoreticalC/N ratio, minimum theoretical Ei/No ratio, noise density, total receiver input noise, mini-mum receive carrier power, and the minimum energy per bit at the receiver input.

7-3. A satellite system operates at 6-GHz uplink and 4-GHz downlink and has a projected P(e) of10-6. The modulation scheme is QPSK and the system will carry 100 Mbps. The equivalentreceiver noise temperature is 290 K, and the receiver noise bandwidth is equal to the mini-mum Nyquist frequency. Determine the C/N ratio that would be measured at a point in thereceiver prior to the BPF where the bandwidth is equal to (a) It times the minimum Nyquistfrequency, and (b) 3 times the minimum Nyquist frequency.

7-4. Which systlm has the best projected BER?(a) 8-QAM, C/N = 15dB, B = 2fN'fb = 60 Mbps.(b) QPSK, C/N = 16 dB, B = fN,fb = 40 Mbps.

7-5. An earth station satellite transmitter has an HPA with a rated saturated output power of10,000 W. The back-off ratio is 6 dB, the branching loss is 2 dB, the feeder loss is 4 dB, andthe antenna gain is 40 dB. Determine the actual radiated power and the EIRP.

7-6. Determine the total noise power for a receiver with an input bandwidth of 20 MHz and anequivalent noise temperature of 600 K.

7-7. Determine the noise density for Problem 7-6.

7-8. Determine the minimum C/N ratio required to achieve a P(e) of 10-5 for an 8-PSK receiverwith a bandwidth equal tofN.

7-9. Determine the energy per bit-to-noise density ratio when the receiver input carrier power is-100 dBW, the receiver input noise temperature is 290 K, and a 60-Mbps transmission rateis used.

7-10. Determine the carrier-to-noise density ratio for a receiver with a -70-dBW input carrierpower, an equivalent noise temperature of 180 K, and a bandwidth of 20 MHz.

7-11. Determine the minimum C/N ratio for an 8-PSK system when the transmission rate is60 Mbps, the minimum energy of bit-to-noise density ratio is 15 dB, and the receiver band-width is equal to the minimum Nyquist frequency.

7-12. For an earth station receiver with an equivalent input temperature of 200 K, a noise band-width of 20 MHz, a receive antenna gain of 50 dB, and a carrier frequency of 12 GHz, deter-mine the following: G/Te,No, and N.

7-13. For a satellite with an uplink Ei/No of 14 dB and a downlink Ei/No of 18 dB, determine theoverall Ei/No ratio.

7-14. Complete the following link budget:

Uplink Parameters

1. Earth station transmitter output power at saturation, 1 kW

2. Earth station back-off loss, 3 dB

3. Earth station total branching and feeder losses, 3 dB

4. Earth station transmit antenna gain for a lO-m parabolic dish at 14GHz5. Free-space path loss for 14GHz

6. Additional uplink losses due to the earth's atmosphere, 0.8 dB

t: sateIilte transponc!ercTl1e, -4.0 dBK-~

8. Transmission bit rate, 90 Mbps, 8-PSK


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Downlink Parameters

1. Satellite transmitter output power at saturation, lOW

2. Satellite transmit antenna gain for a 0.5-m parabolic dish at 12GHz3. Satellite modulation back-off loss, 0.8 dB

4. Free-space path loss for 12 GHz

5. Additional downlink losses due to earth's atmosphere, 0.6 dB

6. Earth station receive antenna gainJor a lO-m parabolic dish at 12GHz

7. Earth station equivalent noise temperature, 200 K

8. Earth station branching and feeder losses, 0 dB

9. Transmission bit rate, 90 Mbps, 8-PSK~

Problems 301

J---

w. tomasi - pp

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