Gil VK1GH attended a talk at the Australian Academy Of Science in Canberra, ACT, on the development work for the Square Kilometre Array (SKA).  The receiving sensitivity of the SKA is measured in Jansky.  Given my background as a somewhat-lapsed amateur astronomer, Gil emailed me to ask if I had heard of a Jansky, and what would be the equivalent expressed in terms of dBm.  Gil forwarded my email response to the VK1-reflect mailing list on 06 Dec 2014 - part of my email is provided below.

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I have heard of Janskys as a unit, but not since I used to attend Canberra Astronomical Society meetings (which is a few years ago now). I visited Google and found this page about Jansky: http://en.wikipedia.org/wiki/Jansky.

They have already figured out the conversion factor, viz.
1 Jansky = -260 dBW/m^2/Hz

P [dBW/m^2/Hz] = 10 * log (P [Jy]) - 260
P [Jy] is power spectral flux density (PSFD) expressed in Janskys.

So, a receiving sensitivity of 600 micro Jansky is 600 x 10E-6 Jy, or 6 x 10E-4 Jy.

And (while not using a calculator, because I'm too lazy to find where it is buried under the pile of papers on this desk, and my wife's laptop computer which I'm using at the moment doesn't have the scientific calculator function loaded);
  10*log (6 x 10E-4)
= 10*log (2) + 10*log (3) + 10*log (1 x 10E-4)

=         3  +        4.8 +         -40
= 7.8 - 40
= -32.2 dB

We can check this result by looking at the multiplicative inverse of 6 x 10E-4, which is 1 / (6 x 10E-4) = 10000 / 6 = 1666 or 1600 approximately. 10 * log (1600) = 12 + 20 = 32. Take the multiplicative inverse by subtracting from zero which gives -32 dB, close to what was calculated above (but less accurate because I rounded 1666 down to 1600).

Anyway, back to Janskys: so 600 micro Janskys = -32.2 - 260 = -292.2 dBW/m^2/Hz. Adding 30 dB to convert from dBW to dBm yields a power spectral flux density of -262.2 dBm/m^2/Hz.

So what does this mean in terms that a radio communications engineer could relate to?

Let's use an example where hopefully I can remember the numbers off the top of my head: a 14m diameter parabolic antenna operating at 12.5 GHz (which was the middle of the AUSSAT Ku-band downlink) exhibiting a gain of 63 dBi (including efficiency) and a G/T of 36 dB/K (it had relatively noisy LNAs compared to what one can get now). Let's say this antenna is receiving a 1 Mbps QPSK carrier using 1/2 rate Forward Error Correction (FEC), so it occupies roughly 1 MHz bandwidth, and we'll say that the digital demodulator needs an input Energy per bit / Noise power spectral density (Eb/No) of 6 dB to recover the baseband data at a Bit Error Rate (BER) of, say, roughly 1 x 10E-6, or 1 bit error per million bits received. The aim is to compute the downlink power spectral density (PSD) then the PSFD at the antenna. BTW, if I remember correctly, the isotropic area at 12.5 GHz is roughly -43.6 dBm^2.

Eb/No = PFD + Ai + G/T - k - R
Eb/No is Energy per bit / Noise power spectral density = 6 dB
PFD is power flux density in dBW/m^2, which is unknown,
Ai is isotropic area = -43.6 dBm^2,
G/T is gain-to-temperature ratio = 36 dB/K,
k is Boltzmann's Constant = -228.6 dBW/K/Hz, and
R is bit rate = 60 dBHz (which is 10 * log (1 Mbps) ).

Re-arranging the equation gives
PFD = Eb/No - Ai - G/T + k + R
    = 6 + 43.6 - 36 + -228.6 + 60
    = 13.6 - 168.6
    = -155 dBW/m^2

Check if this is right: -155 dBW/m^2 hits the face of the dish, add -43.6 dBm^2 to convert to the power received by an isotropic antenna gives -155 - 43.6 = -198.6 dBW. Add the gain to this while taking away the noise temperature, by adding G/T of 36 dB/K gives -198.6 + 36 = -162.6 dBW/K. Take away Boltzmann's constant to convert temperature to noise PSD, so - 162.6 - -228.6 = 228.6 - 162.6 = 66 dBHz (the C/No). Take away the data rate of 60 dBHz to arrive at Eb/No, ie 66 - 60 = 6 dB, which is what was specified above. So this all seems to stack up OK.

So we know that -155 dBW/m^2 is the PFD at the antenna. This signal occupies 1 MHz of spectrum, which can be expressed as 60 dBHz (ie 10 * log (1 x 10E6). So the PSFD can be calculated as follows:

PSFD = -155 - 60 = -215 dBW/m^2/Hz

Adding 30 dB converts this to -185 dBm/m^2/Hz.

Even bearing in mind that this figure includes allowance for 6 dB Eb/No input to the detector in the demodulator, so if the bit energy equalled that of the noise it would be -191 dBm/m^2/Hz; the SKA sensitivity is -262.2 dBm/m^2/Hz, which is substantially less PSFD, or substantially greater sensitivity!

What is the SATCOM antenna's sensitivity in Janskys? Let's use the -191 dBm/m^2/Hz figure: this is -221 dBW/m^2/Hz. Using the equation near the beginning of this email means adding 260 dB which gives -221 + 260 = 39 dB, which is 8000 Janskys. That's quite deaf in comparison to the SKA! :

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This page was created by Mike Dower VK2IG: 24 Jan 2015.  Material may be copied for personal or non-profit use only.