Link Budget
The best case scenario will be
outlined:
- stream = 2 bits/sec (coded)
- 2 bits per symbol for QPSK modulation
- k = 0.5 (roll off of raised cosine pulse)
Ts = 2 bits per symbol / 2 bits per second = 1 symbol per second
Bandwidth = 1/Ts*(1+k) = 1.5Hz
Thus, the calculation for noise floor is as follows:
Pn=kTB=(1.38*10-23)*(3+28)*(1.5)=6.8342*10-22= -211.93dB
To
minimize projected bit error rates, the design will assume the received
signals are at least a few times more powerful than the noise power of
space. If the satellite transmits a signal too powerful, the power
supply of the satellite will be exhausted before all the bits can be
sent. Thus, the signal to noise ratio (SNR) will be three decibels.
What is the maximum distance of feasible propagation?
The link budget for this system is shown below:
10log(Pr) = 10log(PtGt) - 10log(Gr)-20log(4πR/λ)
- Pr is power received
- Pt is power transmitted
- Gr is the gain of the receiver antenna
- Gt is the gain of the transmission antenna
- λ is the wavelength of the propagating wave
- R is the distance between two antennas
We
suspect that our signal will not transmit safely all the way back to
Earth. For the time being, we assume the transmitting and receiving
antennas are the same in case we need to relay the signal across space,
between similar satellites.
λ=(3*108)/(60*109)=0.005m
Gt=Gr=η*(4π2(d/2)2)/λ = 0.75*(4*π2(25/2)2)/0.0052= 1.8531*108= 83.65dB
Pr=Pn+2= -209.93dB
10log(Pr) = 10log(PtGt) - 10log(Gr)-20log(4πR/λ)
Substituting terms
-209.93 = 102.67 - 83.65-20log(4πR/0.005); R = 3.058*1016m
Note
that one light year is 9.4608*1015m, and the distance to be covered is
10.5 light years. Also note that NASA’s largest dish in its Deep Space
Network array is 70m. Completely neglecting the benefits of the array of
satellites, some additional distance may be accounted by this one
satellite from NASA:
- d = 70
- η = 0.94
- λ = 0.005
Gr=η*(4π2(d/2)2)/λ = 0.94*(4*π2(70/2)2)/0.0052= 1.8209*109= 92.6dB
Pr=Pn+2= -209.93dB
10log(Pr) = 10log(PtGt) - 10log(Gr)-20log(4πR/λ)
Substituting terms
-209.93 = 102.67 - 92.6-20log(4πR/0.005); R = 7.2358*1016m
Knowing all of this information, we can see that our satellite must propagate as far as is shown below:
(10.5*9.4608*1015)-(6.2547*1016) = 3.6791*1016m
Because
this total distance to propagate is greater than that which can safely propagate via one
satellite, we will launch a second relay satellite approximately 40000
years after the original launch. This second satellite functions to
receive and improve the signal for its final transmission to Earth.
Provided that our satellites can sustain a 100W transmission for a
relatively long time, this is the best means of sending pictures back
from the exoplanet.
And the enormous magnetic field around Epsilon Eridani?
With
absolute certainty, the pictures to be sent back from the original
satellite will be corrupted by the observed magnetic field radiating
from the star Epsilon Eridani. This magnetic field is projected to
affect all space about one tenth of a light year away from the star.
Therefore, to ensure the transmission of at least some decent pictures of the exoplanet revolving about Epsilon Eridani, a
telescope is projected to be fitted onto the camera of the satellite.
This telescope will be front-mounted and auto-focused to record
pictures. At least, the light radiating from Epsilon Eridani that
bounces off of the exoplanet has a chance to be captured by the
satellite. To protect the lens of the telescope, blast caps can be
installed to the lens-cover which will be blasted off after 40000 years
of travel. After that time, the first photos of this star system and/or
exoplanet will be propagated.
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