Sample Comsat Cost Calculation

8/14/2019 Sample Comsat Cost Calculation

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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 dBnote 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


(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


(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 (1,500Hz 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* 40

Spacecraft $15x3M * 45

Total System $105M

*(The system has two spacecraft in orbit plus one spare).The spacecraft carries twelve transponders renting for $C /year. The in-

orbit 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 + i)-n Rental Income Present Value

1 .89 5C 4.45C

2 .80 10C 8.00C

3 .71 15C 10.65C

4 .64 20C 12.80C

5 .57 24C 13.680C

6 .51 24C 12.24C

7 .45 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



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 comparewith 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 f Y l C



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 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 lesssalvage.

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 muchhigher rate of return.)

Sample Comsat Cost Calculation

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