I've read the following passage in the answer to the Quora question How can Voyager send a signal strong enough for us to receive, in spite of its enormous distance from us? And how can it have the power to do so more than 20 years after its launch?.

For example:

  • TODAY, when Voyager (I) sends a signal at 22W (13.42dBW) power:

  • We can receive the signal power as 7.22e-19Watt (-181.4dBW).

  • Very weak signal because path losses are too big (316.77 dB).

  • Voyager I’s data bit rate was 21.6 kbps at the beginning, now it is decreased to 160 bit per second (so slow)

  • Received energy per bit is 4.5e-21 Joule (-203.4 dBJ)

Is there a simple way to understand mathematically the relationship between these numbers? I understand that these were reported values by the Deep Space Network system, but is there a way ton at least estimate the resulting low received power and data rates from known information like Voyager's power and the receiving station's capabilities?


2 Answers 2


OK let's first understand units. The decibel (dB) is a base-10 log scale without units and dBm is a similar decibel scale for power referenced to 1 milliwatt. They also include a factor of 10, so for example 10 dB is a ratio of 10^1, 20 dB is a ratio of 10^2, etc, while 10 and 20 dBm would be 10 mW and 100 mW.

But in the block quote, they use dBW instead of mW, so $log_{10}(22)$ = 1.342 and it's shown as 13.42 dbW. While dBm is more common, let's stick with Watts here.

The standard way to calculate the received power on Earth is to use a link budget calculation. This is one way to calculate the received power in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other. Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.

$$ P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX} $$

  • $P_{RX}$: received power on Earth
  • $P_{TX}$: transmitted power by Voyager
  • $G_{TX}$: Gain of Voyagers transmitting antenna (compared to isotropic)
  • $L_{FS}$: Free space Loss, what we usually call $1/r^2$
  • $G_{RX}$: Gain of Earth's receiving antenna (compared to isotropic)

We know that $P_{TX}$ is 13.4 dBW already, and on page 17 of the DESCANSO Design and Performance Summary Series Article 4: Voyager Telecommunications we can see that Voyager's high gain (X-band, around 8.4 GHz) antenna has a $G_{TX}$ gain of 48 dBi, where the "i" means relative to a theoretical isotropic radiator.

The gain of the receiving dish antenna $G_{RX}$ can be caluladed (from here) as

$$G_{Dish} \sim \left( \frac{\pi d}{\lambda} \right)^2 e_A$$

where $d$ is the diameter of the dish, $\lambda$ is the wavelength, which is the speed of light of 3E+08 m/s divided by the frequency of 8.4E+09 Hz or about 0.036 meter (3.6 centimeters), and $e_A$ is some aperture efficiency term between 0 and 1 for a realistic dish, which we'll set to 1 to make things simple. For the Deep Space Network's largest diameter dish antenna of 70 meters, this becomes about 1.9E+07 which after applying $10 \times \log_{10}$ becomes about 73 dB.

The Free Space path loss is calculated by calculating the fraction of an expanding spherical wave (from an isotropic radiator) that would be received by an area similar to one square wavelength. The exact equation in dB is:

$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$

The reason the fraction flipped, but a minus sign did not appear outside is because by convention, loss is expressed in positive dB, and then subtracted by the minus sign in the "master equation". Currently Voyager 1 is about 2.1E+13 meters (yes, 21 billion kilometers!) away, so $L_{FS}$ is about 7.3E+16 or 317 dB.

$$ P_{RX} \ dBW = 13.4 \ dBW + 48 \ dB - 317 \ dB + 73 \ dB = -182.6 \ dBW$$

which is darn close to the -181.4 dBW shown in the question!

When receiving the signal, the limit to the data rate is the ratio of received signal power to the total noise power (received plus system). We calculate both for a fixed range in frequency, which should be roughly the bandwidth that Voyager is using.

For an effective receiver temperature of say 20 Kelvin, the noise equivalent power will be about $k_B T \times \Delta f$ where $k_B$ is the Boltzmann constant.

I'll do some handwaving${}^†$ here and just estimate the bandwidth used by Voyager's spread-spectrum transmission to be about 1 kHz, a few times larger than the quoted bit rate of 160 bits/second would require. That makes the noise effective power about 1.3E-20 Watts or -199 dBW, and that gives a signal to noise ratio (S/N) of -182.6 dBW minus -199 dBW of 16.4 dB, which is more than ample for good reception!

update: Thanks to @TomSpilker's careful review: That makes the noise effective power about 2.7E-19 or -182.6 dBW minus -185.6 dBW = 3 dB, which is sufficient when used with some combination of redundancy and error correction.

edit: @Hobbes' comment points out that I don't really know if Voyager uses spread-spectrum for data communications or not, since I've recently asked Have deep-space spacecraft always used some form of spread-spectrum for data downlink?. I had assumed that it would have been used to improve the S/N ratio, but that was a groundless assumption. Stay tuned for more updates!

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  • 1
    $\begingroup$ Your explanation is amazing! The only thing that still remains in doubt is the calculation of the data rate, having the snr we can use the shannon theorem? R = H log2 (1 + S / N). I tried to use your data and this formula and in the end I get a data rate of 5.48 kbps. is it right to use this formula or should I use another one? $\endgroup$
    – Tommytii
    Jan 8, 2018 at 11:04
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    $\begingroup$ @Tommytii Right. I'm somewhat familliar with it (see Am I using Shannon-Hartley Theorem and thermal noise correctly here? where the actual rate also turns out to be lower than theoretical as well) and so I agree that it first appears that a higher data rate that 160 bits/sec theoretically possible. But just because it is possible doesn't mean that Voyager must use the highest possible data rate. There could be other issues related to the spacecraft or other measurements (e.g. doppler), I don't know. $\endgroup$
    – uhoh
    Jan 8, 2018 at 12:05
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    $\begingroup$ I thank you for your help, which helped me to understand some concepts that I will need to do a university exam in 2 weeks. Really, many thanks for your explanation! $\endgroup$
    – Tommytii
    Jan 8, 2018 at 13:03
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    $\begingroup$ Are you sure Voyager uses spread-spectrum transmissions? Your sentence "I'll do some handwaving here and just estimate the bandwidth used by Voyager's spread-spectrum transmission to be about 1 kHz, " - indicates Voyager uses spread spectrum. If that was not your intent, you'll have to rephrase that sentence $\endgroup$
    – Hobbes
    Mar 12, 2018 at 15:55
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    $\begingroup$ @uhoh , excellent analysis! Done without getting mired in details of Eb/No, coding schemes, etc. But I was going through the calculations and I'm getting a different result for the noise power, = kTB (where bandwidth B is the same as your ∆f), = (1.38 x 10^-23 J/K) * (20 K) * (10^3 s^-1) = 2.76 x 10^-19 J/s (J/s = W), larger than your result by a factor of 20, which is T. I was with the Voyager Radio Science team 1981-1989; the DSN ran the downlink bandwidth not a whole lot larger than the Nyquist criterion, so maybe not spread-spectrum, and you get to decrease the noise power maybe ~3 dB. $\endgroup$ May 6, 2018 at 5:34

The receiver will likely use a filter matched to the 160 cps bit rate. This will reduce the transmit power in 1KHz BW by about 7.95 dB. The filter will however have a 1 to 2 dB mismatch loss. If this mismatch loss is 1 dB then the SNR =3dB+7.95dB-1dB=9.95 dB based on the initial SNR derivation.

  • 1
    $\begingroup$ Voyager's communication system is well documented in this pdf linked in my answer. I'm not sure such a filter exists (there are practical reasons) but if it does, you'll be able to support your answer with the appropriate section of the documentation. $\endgroup$
    – uhoh
    Dec 13, 2018 at 0:52

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