<p>The short answer is that the same formula applies, because the radiometer equation <a href="https://radio.astro.gla.ac.uk/ralectures/pw/radiometer.pdf" rel="nofollow noreferrer">is fundamentally just a statement about the noise that arises from sampling a signal</a>. If we have a telescope that experiences Gaussian-distributed variations in system temperature <span class="math-container">$T_{\mathrm{sys}}$</span>, we find that it experiences a noise from these fluctuations <span class="math-container">$\sqrt{2}T_{\mathrm{sys}}$</span>, and so an instantaneous single-sample observation of a source inducing antenna temperature <span class="math-container">$T_A$</span> would have a signal-to-noise ratio of
<span class="math-container">$$\mathrm{SNR}=\frac{\mathrm{signal}}{\mathrm{noise}}=\frac{T_A}{\sqrt{2}T_{\mathrm{sys}}}$$</span>
Observations aren't really instantaneous, though, and involve a whole bunch of data samples. Continuing our argument based on Gaussian statistics, if we have <span class="math-container">$\mathcal{N}$</span> samples, we need to multiply the above by a factor <span class="math-container">$\sqrt{\mathcal{N}}$</span>.</p>
<p>If we're talking about a single dish, the Nyquist sampling theorem says that <span class="math-container">$\mathcal{N}=2\tau\Delta\nu$</span>,<sup>1</sup> with <span class="math-container">$\tau$</span> the integration time and <span class="math-container">$\Delta\nu$</span> the bandwidth (see any detailed derivation of the theorem for more). We then get
<span class="math-container">$$\mathrm{SNR}=\frac{T_A}{\sqrt{2}T_{\mathrm{sys}}}\sqrt{\mathcal{N}}=\frac{T_A}{\sqrt{2}T_{\mathrm{sys}}}\sqrt{2\tau\Delta\nu}=\frac{T_A}{T_{\mathrm{sys}}}\sqrt{\tau\Delta\nu}$$</span>
This is the first equation you have,<sup>2</sup> albeit written differently. Your version has converted from antenna temperature to the power collected, which are related by
<span class="math-container">$$P=kT_A\Delta\nu$$</span>
and so the version in the question obscures how bandwidth first worked its way in.</p>
<p>Let's move to an interferometer. In this case, what are the data samples? Well, we still have the result from the Nyquist sampling theorem bringing integration time and bandwidth into the equation, but we also have to consider what our telescope is measuring: visibilities! At each moment in time, we get measurements from the correlation of signals from each pair of antennas. If we have <span class="math-container">$N$</span> antennas, there are <span class="math-container">$N(N-1)/2$</span> pairs. Because of the complex-valued nature of visibilities on a real-valued sky, the factor of 1/2 goes away and we end up with <span class="math-container">$N(N-1)$</span> data points (multiplied by the factor from Nyquist sampling), and so now our signal-to-noise ratio is
<span class="math-container">$$\mathrm{SNR}=\frac{T_A}{T_{\mathrm{sys}}}\sqrt{\tau\Delta\nu N(N-1)}$$</span>
which is your second equation.<sup>3</sup></p>
<p>The above results might well be familiar, so maybe this answer could be a bit shorter, but this particular derivation makes it explicit that the scaling with <span class="math-container">$\tau$</span>, <span class="math-container">$\Delta\nu$</span> and <span class="math-container">$N$</span> in the radiometer equation are just ways of representing the number of samples, and the equation itself is really one about the uncertainties that arise from Gaussian statistics. That scaling argument works for <em>any</em> radio telescope or interferometer.</p>
<p>Instrument-specific differences from telescope to telescope (or interferometer to interferometer) are still buried in the temperature ratio. In there you'll find factors representing the telescope gain, effective area, etc. -- all things the observatory should provide to anyone working out SNR estimates. But the general form of the radiometer equation -- how it scales with integration time, bandwidth and number of dishes -- is applicable to any antenna or collection of antennas, regardless of whether they come with dishes or not.</p>
<hr />
<p><sup>1</sup>There should arguably be a factor <span class="math-container">$n_p$</span> inside the square root describing the polarization measurements being made (<span class="math-container">$n_p=1$</span> for a lone dipole, <span class="math-container">$n_p=2$</span> for a crossed dipole), but it's fine to ignore that for this discussion.<br>
<sup>2</sup>Although mixing power and temperature there has led to a missing Boltzmann constant!<br>
<sup>3</sup>As noted in the comments, though, the scaling with <span class="math-container">$N$</span> is wrong! There needs to be that factor of <span class="math-container">$N(N-1)$</span> in a square root in the numerator, not the denominator, and for large <span class="math-container">$N$</span>, <span class="math-container">$\sqrt{N(N-1)}\approx N$</span>.</p>

CC BY-SA 4.0
2023-02-06T17:16:08.717Z
2023-02-06T17:16:08.717Z
HDE 226868
<p>It was wellknown that the SNR of single <strong>dish</strong> telescope reads
<span class="math-container">$$s/n=P_s/P_n=\frac{P_s}{T_n}\sqrt\frac{t}{B},$$</span>
where <span class="math-container">$P_s$</span>
is the collected power, <span class="math-container">$T_n$</span>
the noise temperature, <span class="math-container">$t$</span>
the measure time and <span class="math-container">$B$</span>
the bandwidth. When it comes to the array of multiple dish telescopes, like SKAO, that will change to
<span class="math-container">$$s/n=P_s/(\sqrt{N}P_{n,signal})$$</span>
where <span class="math-container">$N$</span>
indicates the number of telescopes.</p>
<p>Recently, I meet across with LOFAR, an telescope project with many stations that each consists of multiple <strong>dipole</strong> antennas. So, does it appropriate for me to treat the SNR of it like that of SKAO? (I think the difference between dish and diplole makes it impossible to use the equation above. )</p>
<p>But note that I'm not interested in sensitivity in webpage : <a href="https://science.astron.nl/telescopes/lofar/lofar-system-overview/observing-modes/lofar-imaging-capabilities-and-sensitivity/" rel="nofollow noreferrer">LOFAR Imaging capabilities and sensitivity</a>.</p>

CC BY-SA 4.0
2023-02-05T06:25:52.657Z
I would appreciate it if you can give me some guide or direct-related reference.
Hunter
CC BY-SA 4.0
2023-02-05T16:01:08.220Z
Can you provide a reference for where you saw the sqrt(N) dependance for the number of telescopes in the denominator?  I am expecting to see a dependance on N, not the square root of N.
GrapefruitIsAwesome
CC BY-SA 4.0
2023-02-05T20:30:47.677Z
Interesting question! Can you cite the source where the "well known" equations come from? *Thanks!*
uhoh
<observational-astronomy><radio-telescope>
What's the formula of SNR (Signal to Noise Ratio) to dipole antenna array (eg, LOFAR) look like?
Hunter
# What's the formula of SNR (Signal to Noise Ratio) to dipole antenna array (eg, LOFAR) look like?

(Asked by: username_0 on 2023-02-05)

It was wellknown that the SNR of single **dish** telescope reads
$$s/n=P\_s/P\_n=\frac{P\_s}{T\_n}\sqrt\frac{t}{B},$$
where $P\_s$
is the collected power, $T\_n$
the noise temperature, $t$
the measure time and $B$
the bandwidth. When it comes to the array of multiple dish telescopes, like SKAO, that will change to
$$s/n=P\_s/(\sqrt{N}P\_{n,signal})$$
where $N$
indicates the number of telescopes.

Recently, I meet across with LOFAR, an telescope project with many stations that each consists of multiple **dipole** antennas. So, does it appropriate for me to treat the SNR of it like that of SKAO? (I think the difference between dish and diplole makes it impossible to use the equation above. )

But note that I'm not interested in sensitivity in webpage : [LOFAR Imaging capabilities and sensitivity](https://science.astron.nl/telescopes/lofar/lofar-system-overview/observing-modes/lofar-imaging-capabilities-and-sensitivity/).

**Tags:** observational-astronomy, radio-telescope
> **Comments:**
> **username_0**: I would appreciate it if you can give me some guide or direct-related reference.  
> **username_1**: Can you provide a reference for where you saw the sqrt(N) dependance for the number of telescopes in the denominator?  I am expecting to see a dependance on N, not the square root of N.  
> **username_2**: Interesting question! Can you cite the source where the "well known" equations come from? *Thanks!*  

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## 1 answers

### Answer by username_3 on 2023-02-06. Score: 1

The short answer is that the same formula applies, because the radiometer equation [is fundamentally just a statement about the noise that arises from sampling a signal](https://radio.astro.gla.ac.uk/ralectures/pw/radiometer.pdf). If we have a telescope that experiences Gaussian-distributed variations in system temperature $T\_{\mathrm{sys}}$, we find that it experiences a noise from these fluctuations $\sqrt{2}T\_{\mathrm{sys}}$, and so an instantaneous single-sample observation of a source inducing antenna temperature $T\_A$ would have a signal-to-noise ratio of
$$\mathrm{SNR}=\frac{\mathrm{signal}}{\mathrm{noise}}=\frac{T\_A}{\sqrt{2}T\_{\mathrm{sys}}}$$
Observations aren't really instantaneous, though, and involve a whole bunch of data samples. Continuing our argument based on Gaussian statistics, if we have $\mathcal{N}$ samples, we need to multiply the above by a factor $\sqrt{\mathcal{N}}$.

If we're talking about a single dish, the Nyquist sampling theorem says that $\mathcal{N}=2\tau\Delta\nu$,1 with $\tau$ the integration time and $\Delta\nu$ the bandwidth (see any detailed derivation of the theorem for more). We then get
$$\mathrm{SNR}=\frac{T\_A}{\sqrt{2}T\_{\mathrm{sys}}}\sqrt{\mathcal{N}}=\frac{T\_A}{\sqrt{2}T\_{\mathrm{sys}}}\sqrt{2\tau\Delta\nu}=\frac{T\_A}{T\_{\mathrm{sys}}}\sqrt{\tau\Delta\nu}$$
This is the first equation you have,2 albeit written differently. Your version has converted from antenna temperature to the power collected, which are related by
$$P=kT\_A\Delta\nu$$
and so the version in the question obscures how bandwidth first worked its way in.

Let's move to an interferometer. In this case, what are the data samples? Well, we still have the result from the Nyquist sampling theorem bringing integration time and bandwidth into the equation, but we also have to consider what our telescope is measuring: visibilities! At each moment in time, we get measurements from the correlation of signals from each pair of antennas. If we have $N$ antennas, there are $N(N-1)/2$ pairs. Because of the complex-valued nature of visibilities on a real-valued sky, the factor of 1/2 goes away and we end up with $N(N-1)$ data points (multiplied by the factor from Nyquist sampling), and so now our signal-to-noise ratio is
$$\mathrm{SNR}=\frac{T\_A}{T\_{\mathrm{sys}}}\sqrt{\tau\Delta\nu N(N-1)}$$
which is your second equation.3

The above results might well be familiar, so maybe this answer could be a bit shorter, but this particular derivation makes it explicit that the scaling with $\tau$, $\Delta\nu$ and $N$ in the radiometer equation are just ways of representing the number of samples, and the equation itself is really one about the uncertainties that arise from Gaussian statistics. That scaling argument works for *any* radio telescope or interferometer.

Instrument-specific differences from telescope to telescope (or interferometer to interferometer) are still buried in the temperature ratio. In there you'll find factors representing the telescope gain, effective area, etc. -- all things the observatory should provide to anyone working out SNR estimates. But the general form of the radiometer equation -- how it scales with integration time, bandwidth and number of dishes -- is applicable to any antenna or collection of antennas, regardless of whether they come with dishes or not.

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1There should arguably be a factor $n\_p$ inside the square root describing the polarization measurements being made ($n\_p=1$ for a lone dipole, $n\_p=2$ for a crossed dipole), but it's fine to ignore that for this discussion.  
2Although mixing power and temperature there has led to a missing Boltzmann constant!  
3As noted in the comments, though, the scaling with $N$ is wrong! There needs to be that factor of $N(N-1)$ in a square root in the numerator, not the denominator, and for large $N$, $\sqrt{N(N-1)}\approx N$.


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