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HomeUncategorizedHow many photons are received per bit transmitted from Voyager 1?

# How many photons are received per bit transmitted from Voyager 1?

For an exact calculation we need to address a few choices: (you can change them, the answer will not be tremendously affected)

1. What is the receiver? Let’s assume a 70 m dish, like this one [CDSCC] in the Deep Space Network.
2. [Voyager 1] can transmit at \$2.3 {rm GHz}\$ or \$8.4 {rm GHz}\$. Let’s assume \$8.4 {rm GHz}\$, for better beam forming (but probably it can only use the lowest frequency at the highest power, so this could be too optimistic).
3. Does “received” mean all photons hitting the antenna dish, or only those entering the electronic circuit of the first LNA? A similar question can be asked for the transmitter in the space craft. We’ll ignore this here since losses related to illuminators or Cassegrain construction will not even be one order of magnitude, insignificant compared with the rest.

A) Voyager sends \$160\$ bits/second with \$23{rm W}\$. Using \$8.3 {rm GHz}\$ this is \$4 cdot 10^{24}\$ photons per second, or \$2.6 cdot 10^{22}\$ per bit, because for frequency \$f\$ the energy per photon is only
\$\$E_phi=hbar, omega=2pihbar f=5.5cdot10^{-24}{rm J}
text{or} 5.5 text{yJ (yoctojoule)}.\$\$

B) The beam forming by Voyager’s \$d=3.7{rm m}\$ dish will direct them predominantly to Earth, with \$(pi d/lambda)^2\$ antenna gain, but still, at the current distance of \$R=23.5\$ billion kilometers, this only results in \$3.4cdot10^{-22}\$ Watt per square meter reaching Earth, so a receiver with a \$D=70{rm m}\$ dish will collect only \$1.3\$ attowatt (\$1.3cdot 10^{-18}{rm W}\$), summarized by:
\$\$
P_{rm received} = P_{rm transmit} Big(frac{pi d}{lambda}Big)^2 frac1{4pi R^2} frac{pi D^2}4
\$\$

Dividing by \$E_phi\$ we see that this power then still corresponds to c. \$240000\$ photons per second, or \$1500\$ photons per bit. If we assume \$f=2.3{rm GHz}\$ this becomes \$415\$ photons per bit. And if we introduce some realistic losses here and there perhaps only half of that.

C) (Although not asked in the question) how many photons per bit are needed? The [Shannon limit] \$ C=B , log_2(1+{largefrac{S}{N}})\$, relates bandwidth \$B\$, and \$S/N\$ ratio to maximum channel capacity. It follows that with only thermal noise \$N=k, T_{rm noise},B\$, the required energy per bit is:
\$\$ E_{rm bit} = frac S C = k, T_{rm noise} frac{2^{,C/B}-1}{C/B}
Rightarrow lim_{Cll B} E_{rm bit} = k, T_{rm noise} log 2,
\$\$

where \$Cll B\$ is the so-called “ultimate” Shannon limit. With only the CMB \$(T_{rm noise}!=!3{rm K})\$ we would then need \$41 {rm yJ}\$, or \$41 cdot 10^{-24},{rm J}\$, per bit.
That’s only \$7.5\$ photons at \$8.3 {rm GHz}\$. But additional atmospheric noise and circuit noise, even with a good cryogenic receiver, could easily raise \$T_{rm noise}\$ to about \$10{rm K}\$ and then we need \$25\$ photons per bit at \$8.3 {rm GHz}\$, and even \$91\$ at \$2.3 {rm GHz}\$. So clearly there is not much margin.

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