Sample Comsat Cost Calculation

Compare Three Approaches for the same customer characteristics. • 1,000 ground stations each with 1.2 kB/sec (1,500 Hz B and 15 dB C/N) • Each station can receive from the other • Duty cycle is 2%: I.e. the satellite transmitter average power is given by Per channel power x 1,000 / 50. • Cost of satellite $3 Million/watt (linear). • Cost of ground equipment from previous notes.

Two-way Data Station with Relay An 11-Meter Station is used to relay data from one 1.2-Meter (4-foot) station to another. Compare the cost of this relay with the optimum direct service of a 4.5 Meter service and a direct 1.2 Meter service. (a) Each station transmits 1,500 Hz Bandwidth with 15 dB C/N. (b) 4 GHz up-link 6 GHz down-link. (c) Transponder gain {32 Db EIRP/-84.5 dBw/M2 + 5 dB} (d) Satellite transmit gain = 32 dB, G/T = 3 dB (e) Cost of power “saturated” is $ 1 Million / watt (f) From earlier derivaion: X = - G’ TRS +4π R *2 + ( GRG/ GRS) – GRG/TRG) +20 log(fD/fU) G’TRS= 32 + 84.5+5 = 121.5 4π R *2= 163 (GRS/TRS) = 3 dB/deg K 20 log(fD/fU) = -3.5 dB

4.5 Meter Direct Service (a) For 4.5M and 150 oK (GRG/TRG) = 21.3 dB (b) X = -121.5 + 163 + 3 - 21 – 3.5 = 20 dB (c) (C/N)U = (C/N)T (1+X) = 10*1.5 ( 1 + 10*2.0) = 3,194 = 35 dB (d) (C/N)D= (C/N)T(1+1/X)= 10*1.5 ( 1+10*-2.0) = 31.9 = 15 dB (e) Up-link: “C/kT” = 31.7 dB Hz + 35 = 66.7 dB Hz !!! ERP = (“C/kT”) + k – G/T – Lfs6 = 66.7 - 228 – 3 + 200 = 35.7 (f) RP = ERP – G 4.5 @ 6GHz = 35.7 – 46.5 = 10.8 dBw => 0.10 watts (g) Down-link: “C/kT” = 31.9 + 15 = 46.9 G/T = 43-21.7 = 21.3 (h) ERP = 46.9 – 228 - 21.3 + 196.5 = -5.9 dBw (i) RP satellite = -5.9 - 25 = -30.9 dBw => 0.81 milli-watts

1.2 Meter (4-foot) Direct Service (a) For 1.2 M and 150 oK (GRG/TRG) = 9.6 dB/ oK (b) X = -121.5 + 163 + 3 – 9.6 – 3.5 = 31.4 dB (c) (C/N)U = (C/N)T (1+X) = 10*1.5 ( 1 + 10*3.14) => 46.4 dB (d) (C/N)D= (C/N)T(1+1/X)= 10*1.5 ( 1+10*-3.14) => 15 dB (e) Up-link: “C/kT” = 31.7 dB Hz + 46.4 = 78.1 dB Hz !!! ERP = (“C/kT”) + k – (G/T)SAT – Lfs6 = 78.1 - 228 – 3 + 200 = 47.1 (f) RP = ERP – G 1.2 @ 6GHz = 47.1 – 35 = 12.1 dBw => 16 watts (g) Down-link: “C/kT” = 31.7 + 15 = 46.7 GRG/TRG = 9.6 (h) ERP = 46.9 – 228 – 9.6 + 196.5 = +5.6 dBw (i) RP satellite = +5.6 - 25 = -19.4 dBw => 11.5 milli-watts

1.2 Meter (4-foot) To Hub Station (a) For 11 M and 150 oK (GRG/TRG) = 29.9 dB/ oK GRG= 51.7 dB (b) X = -121.5 + 163 + 3 – 29.9 – 3.5 = 11.1 dB (c) (C/N)U = (C/N)T (1+X) = 10*1.6 ( 1 + 10*1.11) => 27.1 dB (d) (C/N)D= (C/N)T(1+1/X)= 10*1.6 ( 1+10*-1.11) => 16.1 dB note we add 1 dB to C/N to account for two hops. (e) Up-link: “C/kT” = 31.7 dB Hz + 27.1 = 58.8 dB Hz ERP = (“C/kT”) + k – (G/T)SAT – Lfs6 = 58.8 - 228 – 3 + 200 = 27.8 (f) RP = ERP – G 1.2 @ 6GHz = 27.8 – 35 = -7.2 dBw => 0.19 watts (g) Down-link: “C/kT” = 31.7 + 16.1 = 48.8 GRG/TRG = 29.9 (h) ERP = 48.8 – 228 – 29.9 + 196.5 = -12.6 dBw (i) RP satellite = -12.6 - 25 = -37 6 dBw => 0.17 milli-watts

Hub Station to 1.2 Meter (4-foot) (a) For 1.2 M and 150 oK (GRG/TRG) = 9.6 dB/ oK GTG = 54.2 dB (b) X = -121.5 + 163 + 3 – 9.6 – 3.5 = 31.4 dB 1 dB higher C/N for two-hop (c) (C/N)U = (C/N)T (1+X) = 10*1.6 ( 1 + 10*3.14) => 47.4 dB (d) (C/N)D= (C/N)T(1+1/X)= 10*1.6 ( 1+10*-3.14) => 16 dB (e) Up-link: “C/kT” = 31.7 dB Hz + 47.4 = 79.1 dB Hz ERP = (“C/kT”) + k – (G/T)SAT – Lfs6 = 79.1 - 228 – 3 + 200 = 48.1 (f) RP = ERP – G 1.2 @ 6GHz = 48.1 – 54.2 = -6.1 dBw => 0.24 watts (g) Down-link: “C/kT” = 31.7 + 16 = 47.7 GRG/TRG = 9.6 (h) ERP = 46.9 – 228 – 9.6 + 196.5 = +7.6 dBw (i) RP satellite = +7.6 - 25 = -17.4 dBw => 18 milli-watts

Compare Three Approaches for the same customer characteristics. • 1,000 ground stations each with 1.2 kB/sec (1,500 Hz B and 15 dB C/N) • Each station can receive from the other • Duty cycle is 2%: I.e. the satellite transmitter average power is given by Per channel power x 1,000 / 50. • Cost of satellite $3 Million/watt (linear). • Cost of ground equipment from previous notes.

Comparison

Comparison (cont.)

COMPARISON OF SYSTEMS WITH DIFFERENT CAPITAL AND ANNUAL COSTS Interest on money: Put S .dollars in the bank In one year the money will be equal to: S ( 1+ I ) Where I is the bank’s interest rate If there is an inflation rate of R, one dollar in one year will be worth l / (l+R) in its buying power. .. The real value of the bank deposit after one year is

To avoid guessing the inflation rate we will use constant dollars, dollars that have a. value equal. to today's value. We will then use an interest rate “i” that represents a real increase in value. i=I-R

Present Value The value of a deposit is then after one year (in constant $) V1 = S ( l + I ) In two years: V2 = S (l + i) (l + i) In n years: Vn = S (l + i)** n

“Present Value” is the amount of money you commit now to have a given number of doller in year “n”. To have M dollars in year n:

Example I Launch a satellite with a 7-year life. What should be the charge for transponders? Assumed Budget: Development $20M 20 Launch $20x2M* Spacecraft $15x3M *

40 45

Total System $105M *(The system has two spacecraft in orbit plus one spare). The spacecraft carries twelve transponders renting for $C /year. The inorbit back-up carries twelve transponders, preemptible, renting for $1/2 C/year. ." Interest rate:(The expected earnings of the investors ) i = 12%

Total Present Value = $82.14 C

Initial Cost = $105 M = 82.14 C for desired earnings. C = $105 M / 82.14 = $1.28 Million per year

Example II Same cost of satellite but it has 24 transponders. They are rented out, a few at first and growing over the years, a more conservative business plan.

Year 1 2 3 4 5 6 7

(1 + i)-n .89 .80 .71 .64 .57 .51 .45

Rental Income Present Value 5C 4.45C 10C 8.00C 15C 10.65C 20C 12.80C 24C 13.680C 24C 12.24C 24C 10.80C Total Present Value =72.60 C For 12% interest (earnings) on $105 M investment: $ 105 M = 72.6 C C = $1.45 per year

. Example III – 14 years of service two launch sequences, replacement after seven years service

III. Cost Per transponder = $ 136.5 /100.9 = $1.35 / year

Present Value of constant yearly cost The present value of $C/year for n years is:

EXAMPLE I of Yearly Cost: Say a ground station lasts ten years. A satellite transponder rents for $1.35 M/year How many dollars should be used for the transponder to compare with the purchase price of the ground station? 10 years of satellite service can be purchased for: 10 years cost = PvlO x annual satellite transponder cost 10 years cost = 5.65 x $1.35M = $7.65 Million (If transponder has 8 watts, cost per watt = 7.65/8 = $.96M)

EXAMPLE II of Yearly Cost: . Ground station cost $20,000 to purchase. . It has an annual operation and maintenance cost of $2,000 per year. (10% of cost). What is the present value of ten years of operation? Pv = 20,000 + 2,000. PvlO = 20,000 + 5.65 x 2,000 Pv = $31.300

Example III of Yearly Cost: What if the ground station could be sold for $8,000 salvage at the end of the ten years? Then what is the “present value” cost of ten Years of use? Pv10 = 20,000+2000x Pv10 – 8000/(1+i)**10 Pv10 = 20,000 +11,300 -2,560 Pv10 = $28,740

EXAMPLE IV of Yearly Cost What annual rental fee would have to be charged for use of the station to give a 12% rate of return? 10 years of charges at $F/year PVincome = PV10 * F = $5.65 F This must equal the present value of costs less salvage. 5.65 F = $28,740 F =$5,086/year (Of Course the game is to borrow the money at a low interest rate and charge enough to earn a much higher rate of return.)