DESIGN WORKSHOP: Regional Multibeam Satellite System

Gsl (dB) = 20+20 – 10+10 – 0.1 – 3.1 – 1.1 – 0.1 – 0.2 – 5 – 25 – 55 = 110.4dB ... 'Posat : Goat L. OBO (9) es k. Bu. = AD OBO:6). 2 sat . U sat. (T) ES. Ap=.
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DESIGN WORKSHOP: Regional Multibeam Satellite System Pierre GRUYER, Simon PAILLARD, Max SUISSE DE SAINTE CLAIRE

pierre.gruyerenst-bretagne.fr, simon.paillardsupaero.fr, suissesupaero.fr 15 novembre 2005

1 Required





C N T

1.1 Number of arriers

There is one arrier per one-way link between station and zone. Given there are 3 zones, we get : Ncarriers = |{z} 3 × zones

8 |{z}

stations per zone

×

3 |{z}

= 72

1 carrier per one−way link

1.2 Carrier ara teristi s

There are 24 transponders available for 72 arriers : ⇒ 3 carriers / transponder

Given 10% of margin in ea h hannel, 3.6 Mhz are reserved as guard band per

hannel : ⇒ 10.8Mhz / carrier

1.3 Tra load

We assume the 5 million user are evenly reparted on the 24 earth stations. AEarth

Station

=

3 · 60} |24{z

·

mean call per day

1

5 · 106 | 72 {z }

users per carrier

= 145E

1.4 Required multiplex apa ity

A rule of thumb gives a required apa ity C = A+k·A0.5 for a blo king probability of p = 10−k . Sin e the blo king probability (here p = 1% thus k = 2) is relative to a given multiplexe pool of voi e hannels : required voi e hannels per arrier

⇒ Crequired = 170

A

ording to the standard multiplex apa ities, we will use what implies :   

Cper carrier = 192 fmin = 12kHz   fmax = 804kHz

1.5 Total network apa ity Ctotal = Cper

carrier



1.6 Required

· Ncarriers = 13824 voice channels = 6912 circuits

 C N0 T 

C N0



T

=

S N ∆Fr fm

2  

·

1 b

·p·w

Under lear sky onditions, Stest tone=1 mW and the 10000 pW Op noise is omposed of  7500 pW Op for all sour es of noise but external interferen e  2500 pW Op for external interferen es ⇒

−f Given ∆Fr = g·L with g = b = 3100 Hz , B = 10.8 Mhz : B 2



C N0

m



1.7 Required

T





1 mW S = ; N 7500 pW √

10

S N



and L = 10

= 51.3 dB

dB −1+4·log(192) 20

, p = 4 dB , w = 2.5 dB ,

(dB) = 51.3 + 3 + 34.9 − 4 − 2.5 = 82.7 dBHz  C N T 

C N



T required



C = N0

2



T

·

1 B



C N



T required

(dB) = 82.7 − 10 · log(10.8 · 106 ) = 12.3 dB

This value is higher than the demodulator threshold whi h is 7 dB . We should

ompare the values in order to know if the signal will be properly demodulated. 5.3 dB margin is quite few in lear sky onditions.

2



as a fon tion of the input ba k-o (IBO)



C N T

2.1 Uplink arrier over noise ratio 

C N



=

U

(P 1 ) Pi3 C = i sat · IBO = AU · IBO ⇒ AU = NU NU N 

To determine a numeri al value, we use : 

C N





Usaturation

CUsaturation k · TU · BN

=

Usaturation

 k : boltzman's onstant  TU : uplink system noise temperature  BN : bandwidth of one arrier of the earth station re eiver 2.1.1

Carrier level at saturation

P At the input of the re eiver, CU = G where GSL = Pi Gi − Pj Lj is the gain from the input of the redundant RX to the output of the TWTA. The

al ulation is done in the worst ase, in order to avoid saturation in all other ases. Thus, we hoose the more attenuated path. 1 Osat

saturation

dB

SL

GSL (dB) = 20 + 20 − 10 + 10 − 0.1 − 3.1 − 1.1 − 0.1 − 0.2 − 5 − 25 − 55 = 110.4dB

Given that PO1

sat

= 50 W ,

we get :

CUsaturation = 2.1.2

50 10+11.04

= −93.4dBW

Uplink system noise temperature

The uplink system noise temperature is given by : TAnt + TF TU = LF Rx



1 LF Rx



+ TRx

Considering the satellite at T0 = 290 K , any passive attenuator generates a noise gure F equal to its loss fa tor L. By using the Friss formula, we get FRx = 2.86 and then : TRx = (F − 1)T0 = 539.9K

3

The noise of earth seen by the antenna is : TAnt ≈ 290 K ≈ TF . Thus :  TU = 829.9 K AU = 3689   AU (dB) = 35.7dB  

2.2 Intermodulation noise 

C N

3



=

IM

Pout OBO · 1 PO sat OBO = = 1 PIM IM · PO sat IM

2.3 Downlink arrier over noise ratio 

Thus,

C N



D





1 G 1 · · LD T ES k · BN   G 1 1 1 · OBO · · = PO sat · Gsat · LD T ES k · BN   G = AD · OBO · T ES

=

3

PO · Gsat ·

1

AD =

(1) (2) (3)

PO sat · Gsat kLD BN

2.4 Satellite transmit antenna ree tor

Considering depointing values in yaw, pit h and roll axis, as well as in latitude, longitude and endly depointing of the boresight with respe t to the referen e axes of the satellite, the following table gives values in degrees () for the dierent depointing

omponents :

4

Zone ∆θSKx at node ∆θSKy at node ∆θSKx at vertex ∆θSKy at vertex ∆θACx ∆θACy ∆θx at node ∆θy at node ∆θx at vertex ∆θy at vertex ∆θm at node ∆θm at vertex ∆θm ∆θ θtotal

A

B

C

4.0410−3

4.7410−3

3.2310−3

1.1510−2 8.8810−3 7.7210−3 5.0010−2 4.2610−2 5.4010−2 5.4110−2 5.8910−2 5.0310−2 0.076 0.077 0.077 0.102 0.95

1.2710−2 9.1910−3 9.4110−3 5.0710−2 3.9810−2 5.5410−2 5.2410−2 5.9910−2 4.9210−2 0.076 0.077 0.077 0.102 0.95

1.3010−2 8.3610−3 1.0110−2 5.0010−2 3.5110−2 5.3210−2 4.8010−2 5.8410−2 4.5210−2 0.072 0.074 0.074 0.099 0.95

Global 3.9610−3 1.2710−2 8.9210−3 9.5010−3 5.0110−2 3.8310−2 5.4110−2 5.1010−2 5.9010−2 4.7810−2 0.074 0.076 0.076 0.101 2.98

As we know θ3dB , we will determine the diameter of the transmit antenna ree tor, in order to full requirements of all zones and every frequen ies. As : θ3dB =

70 · c 70 · c ⇔D= D·f f · θ3dB

Sin e for a given diameter D in the downlink, θ3dB orresponds to fmax so fmax is used in the al ulation. Thus, the overage of lower frequen ies signals will be better. min

DSL,T x = 1.88 m

2.5 Interferen e noise

The overall NC I has a downlink and a uplink ontribution. We do the following assumptions :  the free spa e loss for horizontally and verti aly polarised signals are identi al  the delta in the dierent re eiving levels in the primary lobe is negli table

ompared to the high relative gain of the primary lobe  all earth stations emit at the same level. Intermodulation noise are taken into a

ount in an other part. 



Two ases have to be studied for a given frequen y band in the downlink :  the both polarizations ome from the same zone, at a relative same level ;  the interfering polarization omes from a other zone than the wanted one, and is not re eived in the primary lobe of the antenna. 5

2.5.1

Uplink interferen es

The uplink frequen y plan is the downlink one with a xed frequen y shift. The satellite has an unique primary lobe on the re eiving side.  LI and LW represent spa e losses of respe tively the interfering signal and the wanted signal.  Xsat and XES are the ross-polarization isolation of respe ively satellite antennas and earth stations antennas, in dB .  GI and GW represent the antenna gain in the dire tion of intergering signal and wanted signal. We onsider a GG dB of −1dB in the worst ase. The interferen e noise is due to the interfering signal and the impa t of Xsat and XES on it. The ee t of both XES and Xsat is dire tly proportionnal to the path loss and the antenna gain, be ause there are from the same earth station and the same original emitted signal. Keeping the notation of the lesson, W I

                  

Finally :



w Cx = Px · G Lw Xsat Xsat − 10 i Ix = Cy · 10− 10 = Py · G · 10 Li es − X10 i i Jx = Qx · G = P · 10 ·G y Li Li as Px = Py Cx = GGwi · LLwi · − Xsat 1 − Xes Ix +Jx

10

C NI



U

=

10

+10

1 10

−Xsat 10

+ 10

·

−XES 10

10

LI GW · LW GI

interferring station A B C wanted station A 23.4 23.35 23.24 B 23.6 23.56 23.43 C 23.81 23.76 23.65 2.5.2

Downlink interferen es

We follow the same reasoning as in uplink, ex ept there are no terms taking into a ount free spa e loss, sin e both wanted and interfering signals follow the same path between the satellite and the earth station. Linterf ering lobe represents the absolute value of the dieren e of level between the rst lobe and the lobe of the interfering signal. We assume the satellite emits at the same power level whatever is the earth station. 

Linterf ering

lobe

C NI



D

=

1 10

−Xsat 10

+ 10

−XES 10

·

1 Linterf ering

an have one out of two values : 6

lobe





(Linterf ering

lobe )dB

= 0 dB

in the worst ase. Thus :

(Linterf ering

lobe )dB

if the both polarizations ome from the same zone 



= 24.6 dB D

≈ 17 dB = (20dB − Gmax ) − (3dB − Gmax ) if the interfering

polarization is not re eived in the primary lobe of the wanted zone, whi h is itself on the edge of the overage (at Gmax − 3 dB ). Thus : 

2.5.3

C NI

C NI



= 41.6 dB D

Overall arrier over interferen e ratio



C NI

−1

=



−1

C NI

+

U



C NI

−1 D

We minimize interferen es when there are no self beam interferen es in downlink, while we maximize when there are self beam interferen es in downlink. The previous formula provides these



C NI



dB

values :

interferring uplink station A B C wanted uplink station A 23.3482 23.2941 23.1821 B 23.5358 23.4817 23.3698 C 23.7444 23.6904 23.5785 Tab.

The worse



C NI



1



C NI



dB

minimizing downlink interferen es

without having self beam downlink interferen es is 23, 2dB .

interferring uplink station A B C wanted uplink station A 20.9503 20.9190 20.8538 B 21.0572 21.0266 20.9626 C 21.1739 21.1439 21.0813 Tab.

2



C NI



dB

maximizing downlink interferen es

7

The worse



C NI



with self beam downlink interferen es is 20, 9dB .

C/N)I from A towards any zone 24 C/N)I minimizing C/N)I,D C/N)I maximizing C/N)I,D 23.5

C/N)I from B towards any zone 24

23.5

C/N)I from C towards any zone 24

23.5

C/N)I min =23.6 dB→

C/N)I min =23.4 dB→

22.5

22.5

22

21.5

23

C/N)I in dB

23

C/N)I in dB

C/N)I in dB

C/N)I min =23.2 dB→ 23

22

21.5

21

22.5

C/N)I minimizing C/N)I,D C/N)I maximizing C/N)I,D

C/N)I minimizing C/N)I,D C/N)I maximizing C/N)I,D 22

21.5

21

C/N)I min =21 dB→

21

C/N)I min =21.1 dB→

C/N)I min =20.9 dB→

20.5 AA

AB AC uplink zones interfering

20.5 BA

BB BC uplink zones interfering

Fig.

1



C NI

20.5 CA

CB CC uplink zones interfering



dB

2.6 Frequen y plan & AD

Several riteria have to be taken into a

ount in the design of the frequen y plan  :  Firstly, the about 2 dB dieren e between self-zone and between-zones NC leeds us to avoid ata given frequen y the both polarization in the same zone. C  Moreover, sin e is a fun tion growing with f , the higher is the frequen y,  N D the better is NC D . Thus, we will prefer high frequen ies for further zones, like A zone. An other zone has to share the last third of frequen y band with zone A : here we will prefer higher frequen ies for zones with low elevation angle, i.e. zone B.  In order to avoid interferen e with side lobes in the same zone, the both downlinks of a given zone will not use the same polarization. I

All these onstrains leed to the following downlink frequen y plan : Frequen y band 11.2-11.36GHz 11.36-11.53 GHz 11.53-11.7 GHz Horizontal polarization B C A Verti al polarization C A B For the onsidered lear sky frequen y plan, we onsider the worst ase, i.d. the earth station is on the limit of downlink beam overage (GT = GT − 3dB ). We max

8

take into a

ount the losses due to S4/3, OMUX, Output Filter (HF) and TWT to antenna feeder loss. So, we get the following AD values : Zone AD (AD )dB

2.7 Minimum 

C N



A B C 11.33 11.63 12.1 10.5 10.7 10.8

 C N T

−1 T



C = N

−1 U



C + N

−1 D



C + N

−1 I



C + N

−1

IM

is not depending on station to zone ombination. As NC D is IM minimized by a low AD , we know that hoose     zone A as destination zone minimize C . As question 2.5 showed us that NC I is lowest with station from zone A as N D  C emetor, the station to zone ombination yielding the minimum N T for a given   G is : station of zone A to zone A. T ES +



 −1 C N

ZOOM of the left graph

C/N)T depending on sourceZone and destinationZone 15 14.7

A−>A A−>B A−>C B−>A B−>B B−>C C−>A C−>B C−>C

10 14.65 5 14.6 0

−5

−10

−15

14.55

A−>A A−>B A−>C B−>A B−>B B−>C C−>A C−>B C−>C

C/N)T in dB

−1 C N U

C/N)T in dB



14.5

14.45

14.4 −20 14.35 −25

Minimum extremum of C/N)T obtained when A−>A 14.3

−30

−35 −20

14.25 −15

−10 IBO in dB

−5

0

Fig.

−8.75

2

9



C N



TdB

−8.7

−8.65 IBO in dB

−8.6

−8.55



2.8



 C N T

as a fon tion of the input ba k-o (IBO) C/N)T depending on G/T)ES 30

20

C/N)T in dB

10

0

−10

−20 G/T=15dBK−1 − G/T=20dBK 1 G/T=25dBK−1 − G/T=30dBK 1 G/T=35dBK−1

−30

−40 −20

−18

Fig.

3

−16



−14



C N TdB

−12

−10 IBO in dB

−8

−6

−4

−2

0

fon tion of IBO for several GT

For lowvalues of IBO, the urve grows up be ause IBO is de reasing what  C implies, so N U is in reasing. And when IBO > −10, the urve goes down be ause   C is in reasing. This is reprensatative of the trade o between power delivered N IM by the TWT and intermodulation noise.

2.9 Minimum



values

 G T ES  

We ser h the minimum GT ES to obtain the required value of From the gure below, we dedu e :     G  T ESM =0dB  G T

ESM =2dB

10

= 9.5dB = 12dB



C N



T

.

minimum G/T)ES to have the C/N)T required without margin minimum G/T)ES to have the C/N)T required with 2dB margin 20 20 10

0

0

C/N)T in dB

C/N)T in dB

10

−10 −20

−10 −20

G/T=9.4dBK−1 − G/T=9.5dBK 1 C/N)T=12.3dB

−30 −40 −20

−15

−10 IBO in dB

G/T=11.9dBK−1 − G/T=12dBK 1 C/N)T=14.3dB

−30

−5

−40 −20

0

−15

14.5

12.4

14.4

12.3 12.2 G/T=9.4dBK−1 G/T=9.5dBK−1 C/N)T=12.3dB

12.1 12 −9

2.10 Largest

−8.5 IBO in dB



−5

0

ZOOM of the top graph

12.5

C/N)T in dB

C/N)T in dB

ZOOM of the top graph

−10 IBO in dB

14.3 14.2 G/T=11.9dBK−1 G/T=12dBK−1 C/N)T=14.3dB

14.1 14

−8

−9

−8.8

−8.6 −8.4 IBO in dB

−8.2

−8

 C N T maximum value of C/N)T

30

C/N)T max =20.8 dB 20

C/N)T in dB

10

0



−10

G/T=10dBK 1 G/T=20dBK−1 − G/T=30dBK 1 − G/T=40dBK 1 G/T=50dBK−1 − G/T=60dBK 1 G/T=70dBK−1 − G/T=80dBK 1 G/T=90dBK−1 G/T=100dBK−1

−20

−30

−40 −20

−18

−16

−14

−12

−10 IBO in dB

−8

−6

−4

−2

0

Even if GT in reases, there is still interferen es and intermodulation.  −1  −1  −1  −1      −1 C C C C G C As NC −1 = N + N + N + N , if is very high, is T N D T U D I   IM   C C negli table. So, there is a maximum value of N T whi h orresponds to N T = 20.8dB .  

max

11

3 Relationship between (EIRP )ES and 3.1

(EIRP )ES

(EIRP )ES =





G T ES

as a fun tion of IBO

P3 (P 1) (P 1 ) LF S LF S LF S · Pi3 =  G  · 1i · i sat =  G  · i sat ·IBO GRxSat (Pi )sat T T T

T

SL

|

SL

{z

AES

}

Hen e AES , whi h represents the value of EIRP when the satellite TWT is at saturation, is dened by : LF S AES =  G  · AU · BN k T

SL

Using the result of question 2.4, we dedu e the diameter of the earth station antenna, al ulated at fmax in order to always full the minimal θ3dB : DSLRx =

As AES must be minimized,

70 · c = 0.53m fuplinkmax · θ3dB

GSLant = ηT x  

G

πDfmax c 

ant G GSL = LFSL RXSL ⇒  T T USL = 829K

So we get AES = 79.4dBW .

12



!2

SL

= 34.7dB

= 4.4dB · K −1

3.2 Minimum values of (EIRP )ES C/N)T depending on G/T)ES 30

20 C/N)T required without margin

C/N)T in dB

10

0

−10



G/T=9.5dBK 1 − G/T=13.5dBK 1 G/T=17.5dBK−1 − G/T=21.5dBK 1 G/T=25.5dBK−1 G/T=29.5dBK−1 − G/T=33.5dBK 1 G/T=37.5dBK−1 C/N)T=12.3 dB

−20

−30

−40 −25

−20

−15

−10

−5

0

IBO in dB

The minimum IBO to the mission is the abs isse of the interse tion point     full C C between N required and N T . When we have the minimum IBO, we ould easily dedu t the minimum EIRP and ll the following array.  

G T ES

Tab.

in dB · K −1

9.5 13.5 17.5 21.5 25.5 29.5 33.5 37.5

IBOmin in dB

EIRPmin in dBW

-9 -15 -18.5 -20.8 -22.0 -22.6 -22.9 -23.0

70.4 64.4 60.9 58.6 57.4 56.8 56.5 56.4

3  Minimum values of (EIRP )ES with 0dB margin

13

 

G T ES

Tab.

3.3

in dB · K −1

IBOmin in dB

EIRPmin in dBW

-9.1 -14.7 -17.7 -19.3 -20.2 -20.5 -20.7 -20.8

70.3 64.7 61.7 60.1 59.2 58.9 58.7 58.6

12 16 20 24 28 32 36 40

4  Minimum values of (EIRP )ES with 2dB margin

(EIRP )ES

versus



 G T ES

minimum EIRP)ES versus G/T)ES to have the C/N)T required without margin minimum EIRP)ES in dBW

75

70

65

60

55

5

10

15

20

25

30

35

40



G/T)ES in dBK 1 minimum EIRP)ES versus G/T)ES to have the C/N)T required with 2dB margin minimum EIRP)ES in dBW

72 70 68 66 64 62 60 58 10

15

20

25

30

35

40

45



G/T)ES in dBK 1

4 Cost ee tive design of the earth station 4.1

GT

and tra king system

Three types of tra king systems are available : FMA, ATA step tra k or ATA monopulse. √  FMA : the parameter a = SKW 2 + SP U refers to the un ertainty of the satellite lo ation seen from the earth station, with SP U = 0.02 and SKW = 0.05 seen from the enter of the earth. So, a = 0.09 and b = 0.2 (θ3dB ). GTF M A = ηRx

πDf c

14

!2

2

· 10−1.2(b+ 70c ) aDf

Tra king system Diameter (m) 2 3 4 5 6 7 8

Fixed mount antenna GT x (dB ) Cost(k$ ) 44.2 2.4 47.2 7.0 49.1 14.7 50.4 26.3 51.3 42.2 51.9 63.0 52.2 89.1

Step tra k antenna GT x (dB ) Cost(k$) 44.9 21.1 48.4 27.0 50.9 35.7 52.9 48.1 54.4 64.7 55.8 86.1 56.9 112.8

Mono pulse antenna GT x (dB) Cost(k$) 45.4 61.1 48.9 67.0 51.4 75.7 53.3 88.1 54.9 104.7 56.2 126.1 57.4 152.8

 ATA step tra k : b = 0.2 (θ3dB ) GTAT Astep

track

= ηRx

πDf c

!2

· 10−1.2b

= ηRx

πDf c

!2

· 10−1.2b

2

 ATA monopulse : b = 0.1 (θ3dB ) GTAT Astep

track

2

4.2 Cost of the antenna as a fun tion of GT

From CAnt fun tion of DES and from the previous denitions of GT , we derive CAnt as a fon tion of GT , at fmin sin e the ost of the system is dened for fmin up to fmax and C(GT ) is maximum at fmin :  FMA : CF M A is not dedu able from GT (T ) whi h is not reversable. A interpolation based on a nite number of values onstitute the urve. F MA

 ATA step tra k : Cstep

track

c = 0.4· πf

!2.6

 ATA monopulse : Cmonopulse

c = 0.4· πf

!2.6

c +16.6· πf

!0.17

c +16.6· πf

!0.17



 0.048 1.3



 0.012 1.3

GT · 0.5 · 10

GT · 0.5 · 10

15



GT · 0.5 · 100.048



0.085

GT · 0.5 · 100.012

0.085

+40

4.3 Curves ost versus gain and type of tra king system Cost of the antenna in k$ depending on transmitting gain 160 fixed mounting step tracking monopulse

140

D=8m→

120

Cost of the antenna in k$

D=8m→ 100 D=8m→ 80

60 D=2m→

40

D=2m→ 20

D=2m→ 0 44

46

48

50

52

54

56

58

Gtx in dB

Fig.

4  Cost versus gain

For antenna diameter smaller than 5m, a xed mount antenna is the heaper tra king system. For higher antenna diameter, step tra king is better sin e it provides at a omparative pri e mu h better performan es. Hen e, we will hoose xed antenna when D < 5m and ATA ST when D > 5m.

4.4 ηrx



 G T ES

as a fun tion of diameter DES and type of tra king

Sin e the pointing of the antenna is done in re eive mode, we take are to use and to take LF Rx into a

ount. ES



G T



ES

= GT ·

1 LF RxES

·



1 T



ES

In lear sky onditions, earth station system noise is derived from : TES

        

1 TAnt + TF 1 − = LF RxES LF RxES

!

+ TRx

TAnt = TGround + TSky (elevation = 30) = 50K + 10K = 60K LF RxES = 0.2dB ⇒ TES = 190.3K TF = 290K TRx = 120K

16

Fixed mount antenna GR G/T (dBK −1 ) 44.1 21.1 47.3 24.3 49.3 26.4 50.8 27.8 51.8 28.8 52.6 29.6 53.1 30.1

Step tra k antenna GR G/T (dBK −1 ) 44.7 21.7 48.2 25.2 50.7 27.7 52.7 29.7 54.2 31.3 55.6 32.6 56.7 33.8

Mono pulse antenna GR G/T (dBK −1 ) 45.1 22.1 48.6 25.6 51.1 28.1 53.0 30.0 54.6 31.6 55.9 33.0 57.1 34.1

Grx/T)ES in dB at the input of LNA depending on D 35

Grx/T)ES in dB at the input of LNA

Tra king system Diameter (m) 2 3 4 5 6 7 8

30

25 fixed mounting step tracking monopulse

20

2

3

Fig.

4

5

5 D in meter

 

G T ES

6

7

versus diameter

17

8

4.5 Required power values per arrier minimum EIRP)ES in dBW

minimum EIRP)ES versus G/T)ES to have the C/N required without margin 59 ←D=2m FMA 58.5 58 ←D=3m FMA 57.5

←D=4m FMA ←D=5m FMA ←D=4m ATA ST

57 56.5 21

22

23

24

25

26

27

28

29

←D=5m ATA ST 30



G/T)ES in dBK 1 minimum EIRP)ES versus G/T)ES to have the C/N required with 2dB margin minimum EIRP)ES in dBW

61.5 ←D=2m FMA 61 60.5 ←D=3m FMA

60

←D=4m FMA

59.5 59 21

←D=5m FMA ←D=4m ATA ST 22

23

24

25

26

27

28

29

←D=5m ATA ST 30



G/T)ES in dBK 1

Fig.

6  Minimum (EIRP )ES versus

 

G T ES

4.6 Station RF equipments ar hite ture -8dB

L=1dB

HPA

HPA redundant

Fig.

7  Pre-ampli ation ase : Phpa = PT + 8 + 1 + 10 · log(3), Cost = 2.5 · Chpa

18

L=0.3dB L=1dB

HPA MUX

HPA

HPA

HPA redundant

Fig.

8  Post-ampli ation ase : Phpa = PT + 1.3, Cost = 4.75 · Chpa

4.7 Power and ost of the HPA D(m)

GT (dBi)

GR (dBi)

2m FMA 3m FMA 4m FMA 5m FMA 4m ATA | ST 5m ATA | ST

44.2 47.2 49.1 50.4 50.9 52.9

44.1 47.3 49.3 50.8 50.7 52.7

D(m)

Cant (k$)

2m FMA pre ampli 2m FMA post ampli

2.4

3m FMA

7

4m FMA

14.7

5m FMA

26.3

4m ATA | ST

35.7

5m ATA | ST

48.1

 

G T ES

Phpa (W )

688.8 39.0 266.1 15.1 155.3 8.8 108.9 6.2 97.3 5.5 58.4 3.3

19

(dB · K −1 )

EIRPES (dBW )

PT (dBW )

58.8 57.7 57.2 57.0 57.0 56.8

14.6 10.5 8.15 6.6 6.1 3.9

21.1 24.3 26.3 27.8 27.7 29.7

SSPA 135.9

81.9

92.9

56.0

74.9

45.1

65.0

39.1

62.1

37.4 50.6 30.5

Chpa (k$)

TWT Klystron 215.4 79.5 129.7 113.3 147.2 72.3 88.7 103.0 118.7 68.5 71.5

97.6

103.0

66.1

62.0

94.2

98.4

65.4

59.3

93.2

80.2

62.1

48.3

88.5

CF min (k$)

81.9 84.3 154.2 63.0 133.4 59.8

129.3 65.4 134.1 73.1 128.3 78.6

4.8 Main hara teristi s of the earth station Diameter(m) Type of tra king

Antenna

GT (dBi) Cant (k$) Phpa (W )

Te hnology Mode of RF oupling Chpa (k$)

HPA Total ost

Link budget

PT (dBW ) CF (k$) EIRP (dBW ) G/T (dBK −1 ) IBO (dB) OBO (dB) (C/N )U (dB) (C/N )D (dB) (C/N )IM (dB) (C/N )I (dB) (C/N )T (dB)

without margin 4.0 FMA 50.9 14.7 8.8 SSPA post ampli ation 45.1 8.1 59.8 57.2 26.3 -22.2 -16.7 13.5 20.1 34.0 23.2 12.3

with 2dB margin 4.0 FMA 50.9 14.7 14.7 SSPA post ampli ation 55.6 10.4 70.3 59.5 26.3 -19.9 -14.6 15.8 22.2 29.7 23.2 14.3

4.9 Chara teristi s of the earth station with 2dB margin

The additional ost of a 2dB margin is 10, 5k$ for one station. But we need to keep in mind that this extra ost must be multiplied by the number of stations. So we will have an extra ost of 24 · 10, 5k$ = 252k$ ! D(m)

GT (dBi)

GR (dBi)

2m FMA 3m FMA 4m FMA 5m FMA 4m ATA | ST 5m ATA | ST

44.2 47.2 49.1 50.4 50.9 52.9

44.1 47.3 49.3 50.8 50.7 52.7

 

G T ES

20

(dB · K −1 )

21.1 24.3 26.3 27.8 27.7 29.7

EIRPES (dBW )

PT (dBW )

61.2 60.0 59.5 59.3 59.3 59.0

17.0 12.8 10.4 8.9 8.4 6.1

D(m)

Cant (k$)

2m FMA pre ampli 2m FMA post ampli

2.4

3m FMA

7

4m FMA

14.7

5m FMA

26.3

4m ATA | ST

35.7

5m ATA | ST

48.1

Phpa (W )

1186.1 67.1 450.9 25.5 261.9 14.8 183.3 10.4 163.7 9.3 98.0 5.5

Chpa (k$)

SSPA

TWT Klystron 267.7 83.9 161.2 119.6 181.8 76.2 109.5 108.6 146.3 72.2 88.1 102.9 126.8 69.6

168.9 101.7 114.7

69.1

92.3

55.6

80.0

48.2

76.5

46.1

76.4

99.3

121.2

68.8

73.0

98.2

98.7

65.4

59.5

96.2

62.3

37.5

CF min (k$)

86.3 163.6 83.2 76.1 161.0 70.3

153.1 74.5 156.9 81.8 146.8 85.6

5 Performan e under rainy onditions Under rainy onditions, the path loss in reases with respe t to the polarization   G and frequen y of the signal. Moreover, the T ES is also modied, sin e rain ontributes to the noise temperature of the earth station antenna.

5.1 Loss in rainy onditions

is the attenuation due to rain ex eeded 0.3% of any month (hen e 0.075% of the year). We ompute A0.075 at the maximum frequen y of ea h third of the frequen y plan be ause attenuation due to rain in reases with frequen y, and in both polarizations in order to inverse polarization if it is relevant. Values for the

urrent frequen y plan are in bold. A0.075

Earth station

Zone A

Zone B

Zone C

f (GHz )

12.75

12.75

13.25

13.25

12.75

12.75

13.25

13.25

12.75

12.75

13.25

Polarization

V

H

V

H

V

H

V

H

V

H

V

13.25 H

A0.075dB f (GHz )

2.5

2.7

2.5

2.8

2.6

2.8

2.7

2.9

1.6

1.7

1.7

1.8

11.53

11.53

11.7

11.7

11.36

11.36

11.7

11.7

11.36

11.36

11.53

11.53

Polarization

V

H

V

H

V

H

V

H

V

H

V

H

A0.075dB

2.2

2.4

2.3

2.5

2.3

2.5

2.4

2.5

1.5

1.5

1.5

1.6

We noti e there is more attenuation in horizontal polarization than in verti al polarization. In uplink, the worst attenuation is 2.8dB while in downlink, it is 2.5dB .

21

5.2 Antenna noise in rainy onditions

In rainy onditions, antenna noise temperature is derived from : TAnt =





TSky 1 + TGround + Tm 1 − ARain ARain

The value used for ARain is the worst A0.075 in downlink (when the earth station re eives signal), i.d. 2.5dB . As Tm = 1.12Tamb − 50, we onsider Tamb = 290K thus Tm = 275K . Hen e, we get TAnt = 176K . By using the same formula as in 4.4 with the TAnt spe i to rainy onditions, we nd  TES = 301K , i.d. 2dB more noisy than in lear sky onditions. That implies a GT ES 2dB smaller with rain. dB

5.3 Oered



 C N T

in rainy onditions

Sin e the earth station emits at the same power level whatever are the weather

onditions, the loss due to rain translates in uplink as a lower IBO (−2.9dB ) and in downlink as a larger LD (+2.5dB ). EIRP (dBW ) G/T (dBK −1 ) IBO (dB) OBO (dB) (C/N )U (dB) (C/N )D (dB) (C/N )IM (dB) (C/N )I (dB) (C/N )T (dB)

5.4 Required



 C N T

without margin 57.2 24.3 -25.0 -19.3 10.7 13.0 39.4 23.2 8.5

with 2dB margin 59.5 24.3 -22.7 -17.2 13.0 15.1 35.0 23.2 10.7

in rainy onditions

In the same way we al ulated the required NC T in lear sky onditions in questions 1.6 and 1.7, we get : Stest tone=1 mW and the 50000 pW Op noise, in luded 2500 pW Op for other sour es of noise.  

S 1 mW = ; N 47500 pW 

C N0



T



S N



= 43.3 dB





dB

C (dB) = 74.7 dBHz ⇒ N

22

T

(dB) = 4.3 dB

Sin e the demodulator thresold is 7dB , this is the ee tive required NC T . Indeed, if the demodulator thresold was not rea hed, there would be no ommuni ations at all.  

The oered NC T in rainy onditions, either 8.5dB or 10.6dB depending on the margin, is higher than the demodulator thresold. The a tual design fulls requirements even in rainy onditions. Moroever, we noti e that by inverting the polarization of the frequen y plan, sin e the worst attenuation due to rain in uplink would hange from 2.8dB to 2.9dB , while the one in downlink would remain the same, i.d. 2.5dB .  

6 Design of the ommuni ations payload We onsider the lower frequen ies on the top of the gure and higher ones at the bottom.

23

7 Upgrade to digital transmission If we keep the same hara teristi s for the station as in question 4.8 we obtained with array knowing that the total network apa ity is  Rc = σ · B = 16.2Mbit/s with       R b =  ρ · Rc  Eb C  No req = No ρ · ρ · σ · B          C = Eb · ρ · σ N

B = 10, 8MHz

No ρ

req

·72 The total Network apa ity is Ct = 0,95·R . 64·10 b 3

ρ

1 7/8 3/4 1/2 1/3

Eb No

required (dB ) 10.5 7.3 6 5.2 4.7

C N

Tab.

ρ

1 7/8 3/4 1/2 1/3

Eb No

required (dB ) 8.4 6 4.7 3.8 3.4

required (dB ) Margin (dB ) Network apa ity (voi e hannels) 12.3 0.5 17314 8.5 3.3 15150 6.5 5.3 12985 4 7.8 8657 1.7 10.1 5771

5  Clear sky ondition BER = 10−6 C N

Tab.

required (dB ) Margin (dB ) Network apa ity (voi e hannels) 10.2 -2.2 17314 7.2 0.8 15150 5.2 2.8 12985 2.6 5.4 8657 0.4 7.6 5771

6  Rainy ondition BER = 10−4

Assuming that the NC T of the link budget will be de rease of 0, 5dB with the degradation due to the demodulator. We have a link budget :    In lear sky ondition without margin :  NC T =11.8dB  In rainy ondition without margin : NC T =8dB If we want to keep the hara teristi s of the question 4.8, we need to add oding to full the mission. We will hoose a ρ fa tor of 7/8 to full the mission with keeping a quite good network apa ity.  

24

Antenna diameter (m) 4 Transmitted power per arrier (W ) 8.1 Power of transmitted amplier (W ) 8.8 Cost fun tion CF (k$) 59.5 Coding rate 7/8 Margin in lear sky ondition (dB ) 3.3 Margin in rainy ondition (dB ) 0.8 Network Capa ity (voi e hannel) 15150

By using DCME (Digital Cir uit Multipli ation Equipment), a ompression fa tor of 5 : 1 applies on tra , and thus the network apa ity is improved by a fa tor of 5. Even if the impa t of digitalization on the ost of the network system and ustomers devi es has to be taken into a

ount, the network apa ity an be hugely improved, thanks to more ee tive modulation and digital ompression s hemes. Air interfa e Network apa ity (voi e hannels) FDM / FM / FDMA 13284 TDM / QPSK / FDMA 15150 TDM / QPSK / FDMA with DCME 75750

25