Requirements: Solving for area of solar array (m2) and number of cells.
Solution
Power that must be produced by the solar cells P sa = power requirement = 1.30 kW
For multi-junction cells, power output P0 = 1300W/ (0.03*0.06) m2 = 722.22 W/m2
Beginning of life (BOL) power production capability, P BOL
= P0*I d*Cos Ɵ
= 722.22*0.82*Cos 6 = 588.98 kW/m2
End of life (EOL) power production capability, P EOL = L d* P BOL
L d = (1-degradation/year) design life = (1-0.03)8.5 = 0.77
P EOL = 0.77*588.98 kW/m2 = 453.51 W/m2
Solar array size to meet the requirement = Psa/ P
EOL = 1300/453.51 = 2.87 m2
Number of cells required = 2.87/ (0.03*0.06), approximately 1594 cells.
Question 2:
Given:
Batteries used –Nickel-Hydrogen batteries
Number of cycles around the Earth = 1×105
Requirement: Maximum DOD that we can use
Solution:
A Nickel-hydrogen battery used in 15000 cycles has 40% Depth of Discharge (DOD)
Therefore, the maximum DOD that we can use,
= 2.667 %.
The calculated value of DOD is expected to be less than 40% since a conservative depth of discharge is required for the larger number of cycles and since the spacecraft will take more time in space.
Question 3
Given:
Received power in the satellite, PR = 0.5 Pw,
Bit rate of the BPSK module Rb = 1.544 Mb/s and
Noise power density N0 = 0.5×10-19 J
PR = 0.5e-12 W
Rb = 1.544e6 b/s
N0 = 0.5e-19 J
Solution
To determine the bit error rate I used the equation for the ratio of the energy per bit to the noise power density.
= [0.5e-12/1.544e6]/ 0.5e-19
1.7601
The ratio was used to determine the corresponding bit error rate using the table of BER against Eb/ N0 for BPSK/QPSK module.
a)
This is the ratio of the energy per bit to the noise power density
b) Eb/ N0 = [0.5e-12/0.5e-19] = 1096.6332
c) Using fig. 16-16, the approximate BER was found to be 2×10-1
