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I read that the New Horizon's probe will send data to back to Earth at 3000 bits per second by the time it reaches Pluto.

I don't understand why a spacecraft has to have a lower data tranfer rate the further away it gets from Earth. The only reason I can think of is that the further away from Earth, the less remaining power for the spacecraft to send data.

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Look up the inverse square law. With the same power available at the spacecraft, the signal strength received at Earth diminishes as the square of the distance.

inverse square law

If the vehicle is solar powered, then it is possible that it will also reduce the transmitted power as it goes farther from the Sun. However it is more likely to reduce the transmission duration by drawing on the batteries, since you'd like the best signal to noise ratio possible. Either way, the effective result is that if all the power is going to the radio, then the average data rate integrated over the time spent transmitting and not transmitting will go down as the fourth power of distance (assuming that the distance from the Sun and the distance from the Earth are about the same, which it is for a probe at Pluto).

New Horizons is not solar powered. Now you know why.

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  • $\begingroup$ Nice illustration why signal strength falls with inverse square. I don't understand the last sentence though. Fourth power? $\endgroup$ – HopDavid Jan 12 '15 at 16:30
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    $\begingroup$ If the power available for the transmitter from the solar arrays goes as $1/r^2$, and the received signal strength at Earth goes as $1/r^2$, then the received power at Earth is the product of those, which then goes as $1/r^4$. (The two $r$'s are really $r_{Earth}$ and $r_{Sun}$, but if you're far enough away, they're about equal.) $\endgroup$ – Mark Adler Jan 12 '15 at 16:58
  • $\begingroup$ Okay, I think I understand. A solar powered probe falls with inverse fourth but a nuke powered probe's signal strength would still fall with inverse square? $\endgroup$ – HopDavid Jan 12 '15 at 19:17
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    $\begingroup$ It's a bit worse than inverse square degradation even with that RTG. An RTG, being based on radioactive decay, loses power at an exponential decay rate. The half life of plutonium 238 half life is about 87.7 years, which means more than a 10% reduction in power between launch and the Pluto flyby. $\endgroup$ – David Hammen Jan 12 '15 at 21:42
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    $\begingroup$ Yes, but that's true at any distance. $\endgroup$ – Mark Adler Jan 12 '15 at 21:52
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New Horizons also uses very strong error correction coding. Specifically, rate 1/6 turbo coding and BPSK (binary phase shift keying). This expands the signal bandwidth by a factor of six, but it still reduces the required energy per user data bit. This allows a higher data rate for a given transmitter power. The minimum Eb/N0 for this code is about 0 decibels (dB), about the lowest that's ever been achieved. Shannon's famous channel capacity theorem says it's impossible to achieve a low error rate with an Eb/N0 less than ln(2) = -1.6 dB even with infinite bandwidth, so we're pretty close to the theoretical limit. Turbo coding wasn't even discovered until the early 1990s.

What is Eb/N0? It's the ratio of the signal energy per bit to the noise power spectral density. Bit energy is measured in joules and power spectral density is in watts/Hz, which is equivalent to joules so the ratio Eb/N0 is dimensionless. Eb/N0 is a fundamental figure of merit in all digital modulation and coding schemes.

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    $\begingroup$ Huh, that's pretty specific info. Can you tell us how you know? As in, are you this Phil Karn? en.wikipedia.org/wiki/Phil_Karn $\endgroup$ – kim holder Jul 15 '15 at 1:13
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    $\begingroup$ Yes, that's me, KA9Q. I'm referencing several published papers on the New Horizons communication system. They seem to have been written back when the project was in development so they don't have actual, measured performance figures. Most other deep space NASA missions have very detailed volumes on their telecom systems in the DESCANSO publication series (see descanso.jpl.nasa.gov/) but I don't see one yet for New Horizons. The turbo code is described in CCSDS 131.0.B-2. $\endgroup$ – Phil Karn Jul 15 '15 at 19:46
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    $\begingroup$ Well, it's a pleasure to meet you, KA9Q. $\endgroup$ – kim holder Jul 16 '15 at 0:10
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TL;DR: Data rate is directly related to effective noise floor, which is related to bandwidth and received power (and lack of interference). Lower received power means increasing transmit power or reducing bandwidth (speed). APL had to choose the latter due to constraints. Read on for a much more detailed and long-winded explanation (sorry if too long winded!).

There are a bunch of inter-related reasons for this that boils down to Link Budget, which is basically how much signal you have above the noise floor to work with in your receiver. Thanks to the Inverse-Square Law mentioned above, the incredible distance to Pluto (at 32 Astronomical Units distance--1 AU is the Sun-Earth distance) means a very, very weak signal to work with regardless of transmit power.

New Horizons transmits TWELVE (12) MEASLY WATTS from each of its TWTA amps, which can be used primary-secondary or in a cross-polarized mode for 2 separate data streams. The full-size 2.1m/7ft high-gain dish antenna has a 42dBi gain, about the same as an SNG satellite truck's--only, instead of having to shoot to GEO at 35,000km or 22k miles, you're trying to shoot SOOOOOOO much further at around a tenth (-10dB) of the transmit power. On the plus side, they will be using the largest Deep Space Network dishes to receive and talk to it, which are 70 meters in diameter, but that only makes up so much for all this.

The lack of power is due to using an RTG (Radioisotope Thermal Generator) which uses the heat from decaying Plutonium-238 (one neutron less than the Pu-239 used in nuclear fission) to supply a thermocouple setup. The documents state just under 4kW thermal output and around 245 watts electric at launch, dropping to 200-220 watts at the Pluto encounter. That little bit of electricity has to be able to supply EVERYTHING on the spacecraft, and it decreases with time, which is why the Voyagers can no longer send images or much else that requires significant power; NH will end up the same way eventually.

To come full circle, the massive distance and low power means that there won't be much signal left by the time it arrives at either destination. The effective noise floor is directly affected by the bandwidth of the signal, which is why automated transmission systems--like the modems in your cell phones, land lines, etc--have both fast rates for good-quality links and slower rates for weaker links: this is especially true with LTE. Ham radio operators (which I'm one) have use CW and Morse Code for talking using moon or aurora bounce because the distortion is too bad to use any kind of voice mode, and now there are new data modes that optimize for VERY SLOOOOOOW data rates (one bit per second or less) that exceed CW-by-ear for reliability by sacrificing bandwidth and data rate, but this obviously would be useless for this type of science mission other than health telemetry.

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  • $\begingroup$ Bandwidth to a mobile phone is also worse in areas with poor signal coverage. I've heard this described in terms of packet loss; essentially, you have to repeat yourself a lot to get a given message through, like two people talking in an noisy room, or yelling back and forth from far away. Maybe that's another way of describing the "signal to noise" problem. $\endgroup$ – Nathan Long Jul 13 '15 at 13:31
  • $\begingroup$ Is the limitation to twelve watts a consequence of transmitter weight? I would think that transmitting 24 watts, half the time, even with a short duty cycle, would improve SNR greatly [if one uses a half-second on, half-second off, the amount of energy received from the transmitter each second would be the same either way, but the amount of noise received each second would be cut in half using higher power intermittently]. Or is duty cycle already factored into the "12 watts" figure? $\endgroup$ – supercat Jul 13 '15 at 17:48
  • $\begingroup$ No, transmitting twice the power for half the time would not improve things much. In theory, not at all since the required energy per bit is set by the rest of the system and that wouldn't change. In practice, higher data rates are somewhat more efficient in deep space communications because a lower fraction of the total power can be spent on carrier synchronization but that may have changed in recent years; the trend is toward suppressed carrier modulation and extremely stable onboard oscillators. $\endgroup$ – Phil Karn Jul 15 '15 at 20:05
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    $\begingroup$ Also, on New Horizons your peak power is limited by whatever the RTG can produce in real time. Some missions, like the Curiosity rover, have rechargeable batteries to cover demand peaks but I don't think New Horizons has one. Batteries are too short-lived and unreliable for such a lengthy mission. $\endgroup$ – Phil Karn Jul 15 '15 at 20:11
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The probability of correct detection of a data bit is a function of the energy (not the power) in the received pulse. When the received power diminishes due to increasing distance, we can still meet the required energy by making the bit last longer (E = P x T). Longer pulses mean reduced data rate.

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In short

The capacity of the downlink between New Horizons and a DSN station, decreases when the distance between them increases, according approximately to C=log2(1/distance). The actual bit rate is lower than this limit, depending on the type of modulation, and the error correction code. The decrease of this maximal bit rate is explained with two elements:

  1. The strength of a radio signal decreases with the distance from the emitter.

  2. The Shannon–Hartley theorem:
    enter image description here
    says that:

    The maximum capacity of a communication channel is limited by a value, known as the signal-to-noise ratio (SNR), which is the ratio between the strength of the signal at the receiving end of the channel and the amount of noise introduced by the receiver and the rest of the link.

In July 2015, New Horizons is now far from us, by about 40 times the distance Earth-Sun. The radio signal is very faint while the noise is still the same. With a small SNR, the maximum capacity of the radio channel is very limited.


Details

This formula links 4 quantities:

  • C, which is the channel maximal capacity. It's expressed in bits * second. This is what you ask for in your question.

    Note that this is what non-engineers may call the bandwidth (this is inaccurate, and it especially matters in this formula): "My Internet ADSL connection has a 10 Mbps bandwidth", capacity should have been used.

  • B is the bandwidth used for the communication.

    The bandwidth is the space taken up by the channel in the frequency spectrum. A communication usually happens around a carrier frequency, and the information create sidebands around this frequency, by using some more or less clever principle (New Horizons uses binary PSK). The larger the amount of information to transmit per unit of time, the larger the frequency slot required. But there are two problems in extending the bandwidth: The peak power decreases, and the noise increases (see below).

    Note: The bandwidth is expressed in Hz, and mustn't be confused with C as explained earlier.

  • S is the strength of the signal at the receiver side.

    S depends mostly on the emitter power, the gain of both antennas, and the distance between the emitter and the receiver. For a given power, and given antennas, S is essentially dependent on the distance. Each time the distance is doubled, S is divided by 4, this is a case of inverse square law.

  • N is the channel noise, that is random false signals collected or created by the receiver. The most important is created by the receiver, due to the limits of electronic components (thermal noise). Low noise receiver are difficult to build. N is quite constant for a given system (antenna and amplifier) at a given temperature.

    S and R are usually expressed together as S/N, the signal-to-noise ratio (or SNR), and the dimensionless value is expressed in dB, e.g. a 18 dB SNR.

With all approximations made earlier: For a given receiving system (antennas and amplifiers), and when the carrier frequency and the bandwidth are constant, then C varies according to log2(1/distance).

To deal with that and to regain the previous S/N ratio, we need to either increase S or decrease N. This means larger antennas (larger gain), more accurate antenna alignment, and lower noise amplifiers. None of these possibilities is really possible for New Horizons downlink.

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The transmitter has only so much power to work with, no matter what the mission phase. Of course the power budget would probably drop over time. Probably more significantly, the greater the distance, the less signal strength at the receiving station according to the inverse square law, so the less it stands out against background noise. Reducing the data rate allows the receiver to "integrate" the incoming signal (filter out the noise) so that the original signal can be recovered.

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  • $\begingroup$ That's the most simple and accurate answer, in spite of its low vote rate (1) today. $\endgroup$ – mins Jul 20 '15 at 5:58

protected by Deer Hunter Jul 25 '15 at 22:17

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