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As we discussed earlier, an aperture synthesis telescope can be regarded as a collection of two element interferometers. Hence, for understanding the sensitivity of such a telescope, it is easier to first start with the case of a two element interferometer. Consider such an interferometer composed of two antennas $i,j$, (of identical gains, but possibly different system temperatures), looking at a point source of flux density S. We assume that the point source is at the phase center5.1and hence that in the absence of noise the visibility phase is zero. Let the individual antenna gains5.2be G and system temperatures be T$_{s_i}$ and T$_{s_j}$. If $n_i(t)$ and $n_j(t)$ are the noise voltages of antennas $i$ and $j$ respectively,then $\sigma^2_i = \left<n^2_i(t)\right> = \rm {T}_{s_i}$, and $\sigma^2_j = \left<n^2_j(t)\right> = \rm {T}_{s_j}$. Similarly if $v_i(t)$ and $v_j(t)$ are the voltages induced by the incoming radiation from the point source, $\left<v^2_i(t)\right> = \left<v^2_j(t)\right> = {\rm GS}$. The instantaneous correlator5.3 output is given by:

r_{ij}(t) = \left(v_i(t) + n_i(t)\right) \left(v_j(t) + n_j(t)\right)

The mean5.4 of the correlator output is hence:

$\displaystyle \left<r_{ij}(t)\right>$ $\textstyle =$ $\displaystyle \left<
\left(v_i(t) + n_i(t)\right) \left(v_j(t) + n_j(t)\right)
  $\textstyle =$ $\displaystyle \left<v_i(t) v_j(t)\right>$  
  $\textstyle =$ $\displaystyle {\rm GS}$ (5.1.1)

where we have assumed that the noise voltages of the two antennas are not correlated, and also of course that the signal voltages are not correlated with the noise voltages. $r_{ij}(t)$ is hence an unbiased estimator of the true visibility.

To determine the noise in the correlator output, we would need to compute the rms of $r_{ij}(t)$ for which we need to be able to work out:

\left<r_{ij}(t)r_{ij}(t) \right> =\left<(v_i + n_i)(v_j + n_j)(v_i + n_i)(v_j + n_j)\right>

where for ease of notation we have stopped explicitly specifying that all voltages are functions of time. This quantity is not trivial to work out in general. However, if we assume that all the random processes involved are Gaussian processes5.5 the complexity is considerably reduced because for Gaussian random variables the fourth moment can then be expressed in terms of products of the second moment. In particular5.6, if $x_1,\ x_2,\ x_3,\ \&\ x_4$ have a joint gaussian distribution then:

$\displaystyle \left<x_1 x_2 x_3 x_4\right>$ $\textstyle =$ $\displaystyle \left<x_1 x_2\right>\left<x_3 x_4\right> + \left<x_1 x_3\right>\left<x_2 x_4\right> +$  
    $\displaystyle \left<x_1 x_4\right>\left<x_2 x_3\right>$ (5.1.2)

Rather than directly computing $\left<r_{ij}(t)r_{ij}(t)\right>$, it is instructive first to consider the more general quantity

\left<r_{ij}(t)r_{kl}(t) \right> =
\left<(v_i + n_i)(v_j + n_j)(v_k + n_k)(v_l + n_l)\right>

viz. the cross-correlation between the outputs of interferometers $(ij)$ and $(kl)$. We have:

$\displaystyle \left<r_{ij}(t)r_{kl}(t)\right>$ $\textstyle =$ $\displaystyle \left<(v_i + n_i)(v_j + n_j)\right>\left<(v_k + n_k)(v_l + n_l)\right> +$  
    $\displaystyle \left<(v_i + n_i)(v_k + n_k)\right>\left<(v_j + n_j)(v_l + n_l)\right> +$  
    $\displaystyle \left<(v_i + n_i)(v_l + n_l)\right>\left<(v_k + n_k)(v_j + n_j)\right>$  
  $\textstyle =$ $\displaystyle (\left<v_i v_j\right>+\left<n^2_i\right>\delta_{ij})(\left<v_k v_l\right> + \left<n^2_k\right>\delta_{kl}) +$  
    $\displaystyle (\left<v_i v_k\right>+\left<n^2_i\right>\delta_{ik})(\left<v_j v_l\right> + \left<n^2_j\right>\delta_{jl}) +$  
    $\displaystyle (\left<v_i v_l\right>+\left<n^2_i\right>\delta_{il})(\left<v_k v_j\right> + \left<n^2_k\right>\delta_{kj})$  
  $\textstyle =$ $\displaystyle {\rm (GS)}^2 + {\rm GS}(\sigma^2_i \delta_{ij} + \sigma^2_k \delta_{kl})
+\sigma^2_i\delta_{ij} \sigma^2_k\delta_{kl} +$  
    $\displaystyle {\rm (GS)}^2 + {\rm GS}(\sigma^2_i \delta_{ik} + \sigma^2_j \delta_{jl})
+\sigma^2_i\delta_{ik} \sigma^2_j\delta_{jl} +$  
    $\displaystyle {\rm (GS)}^2 + {\rm GS}(\sigma^2_i \delta_{il} + \sigma^2_k \delta_{kj})
+ \sigma^2_i\delta_{il} \sigma^2_k\delta_{kj}$ (5.1.3)

The case we are currently interested in is $\left<r_{ij}(t)r_{ij}(t)\right>$, which from eqn(5.1.3) is:
$\displaystyle \left<r_{ij}(t)r_{ij}(t)\right>$ $\textstyle =$ $\displaystyle 3{\rm (GS)}^2 + (\sigma^2_i + \sigma^2_j){\rm GS}
+ \sigma^2_i \sigma^2_j$  
  $\textstyle =$ $\displaystyle 2{\rm (GS)}^2 + ({\rm GS} + {\rm T}_{s_i})({\rm GS} + {\rm T}_{s_j})$ (5.1.4)

To get the variance of $r_{ij}(t)$ we need to subtract the square of the mean of $r_{ij}(t)$ from the expression in eqn(5.1.4). Substituting for $\left<r_{ij}(t)\right>^2$ from eqn(5.1.1) we have:
\sigma^2_{ij} = {\rm (GS)}^2 + ({\rm GS} + {\rm T}_{s_i})
({\rm GS} + {\rm T}_{s_j})
\end{displaymath} (5.1.5)

Note that the angular brackets denote ensemble averaging. In real life of course one cannot do an ensemble average. Instead one does an average over time, i.e. we work in terms of a time averaged correlator output $\bar{r}_{ij}(t)$, defined as

\bar{r}_{ij}(t) = {1 \over {\rm T}}\int^{t+{\rm T}/2}_{t -{\rm T}/2}
r_{ij}(t^{'}) dt^{'}

As can easily be verified, $\left<\bar{r}_{ij}\right> = \left<r_{ij}\right>$. However, computing the second moment, viz., $\bar{\sigma}^2_{ij} = \left<\bar{r}_{ij}\bar{r}_{ij}\right> -
\left<\bar{r}_{ij}\right>^2$ is slightly more tricky. It can be shown5.7 that if $x(t)$ is a zero mean stationary process and that $\bar{x}(t)$ is the time average of $x(t)$ over the interval $(t-T/2, t+T/2)$, then
\bar{\sigma}^2_x = {1 \over T} \int^{{\rm T}/2}_{-{\rm T}/2}...
...1 - {\vert\tau\vert \over {\rm T}}\right)
R_{xx}(\tau)\ d\tau
\end{displaymath} (5.1.6)

where $R_{xx}(\tau)$ is the auto-correlation function of $x(t)$, and $\bar{\sigma}$ is the variance of ${x(t)}$. Now, if $x(t)$ is a quasi-sinusoidal process with bandwidth $\Delta \nu$, then the integral of $R_{xx}(\tau)$ will be negligible outside the coherence time $1/\Delta\nu$. Further, if $T >> 1/\Delta\nu$, then the factor in parenthesis in eqn(5.1.6) can be taken to be $\sim 1$ for $\tau < 1/\Delta\nu$. Hence we have:
$\displaystyle \bar{\sigma}^2_x$ $\textstyle \simeq$ $\displaystyle {1 \over T} \int^{{\rm T}/2}_{-{\rm T}/2}
R_{xx}(\tau)\ d\tau
~~~\simeq {1\over T} \int^{\infty}_{-\infty}
R_{xx}(\tau)\ d\tau$  
  $\textstyle =$ $\displaystyle { 1\over T}{\rm S}_{xx}(0)
~~~\hskip 0.5in = { 1\over T}{\sigma^2_x \over 2\Delta\nu}$ (5.1.7)

where ${\rm S}_{xx}(\nu) = \sigma^2_x /2\Delta \nu$ is the power spectrum5.8 of $x(t)$. From eqn(5.1.7) and eqn(5.1.5) we hence have
\bar{\sigma}^2_{ij} = { 1 \over 2{\rm T}\Delta\nu}\left(
...({\rm GS} + {\rm T}_{s_i})
({\rm GS} + {\rm T}_{s_j}) \right)
\end{displaymath} (5.1.8)

Putting all this together we get that the signal to noise ratio of a two element interferometer is given by:
{\rm snr} = {(\sqrt{2{\rm T}\Delta\nu}{\rm GS} ) \over
...}^2 + ({\rm GS} + {\rm T}_{s_i})
({\rm GS} + {\rm T}_{s_j})}}
\end{displaymath} (5.1.9)

There are two special cases which often arise in practice. The first is when the source is weak, i.e. ${\rm GS} \ll {\rm T}_s$. In this case the snr becomes
{\rm snr} = {(\sqrt{2{\rm T}\Delta\nu}{\rm GS} ) \over
\sqrt{ {\rm T}_{s_i}{\rm T}_{s_j}}}
\end{displaymath} (5.1.10)

For a single dish with the collecting area equal to the sum of the collecting areas of antennas $i$ and $j$ (i.e. with gain $2$G), and with system temperature ${\rm T}_s= \sqrt{\rm {T}_{s_i}\rm {T}_{s_j}}$ the signal to noise would have been a factor of $\sqrt{2}$ better5.9. The loss of signal to noise in the two element interferometer is because one does not measure the auto-correlations of antennas $i$ and $j$. Only their cross-correlation has been measured. In a sigle dish one would have effectively measured the cross-correlation as well as the auto-correlations.

The other special case of interest is when the source is extremely bright, i.e. ${\rm GS} \gg {\rm T}_s$. In this case, the signal to noise ratio is:

{\rm snr} = {(\sqrt{2{\rm T}\Delta\nu}{\rm GS} ) \over
\sqrt{ 2 {\rm (GS)^2}}} = \sqrt{{\rm T}\Delta\nu}
\end{displaymath} (5.1.11)

This is as expected, because for very bright sources, one is limited by the Poisson fluctuations of the source brightness, and hence one would expect the signal to noise ratio to go as the square root of the number of independent measurements. Since one gets an independent measurement every $1/\Delta\nu$ seconds, the total number of independent measurements in a time T is just T$\Delta \nu$.

Having derived the signal to noise ratio for a two element interferometer, let us now consider the case of an N element interferometer. This can be considered as ${}^N{\rm }C_2$ two element interferometers. Let us take the case where the source is weak. Then from eqn(5.1.3) the correlation between $r_{12}(t)$ and $r_{13}(t)$ is given by

$\displaystyle \left<r_{12}(t)r_{13}(t)\right>$ $\textstyle =$ $\displaystyle \sigma^2_1\delta_{12} \sigma^2_1\delta_{13} +
\sigma^2_1\delta_{13} \sigma^2_1\delta_{21} +
\sigma^2_1\delta_{11} \sigma^2_2\delta_{23}$  
  $\textstyle =$ $\displaystyle 0$ (5.1.12)

The outputs are uncorrelated, even though these two interferometers have one antenna in common5.10. Similarly, one can show that (as expected) the outputs of two two-element interferometers with no antenna in common are uncorrelated. Since the $r_{ij}$'s are all uncorrelated with one another, the rms noise can simply be added in quadrature. In particular, for an N element array, where all the antennas are identical and have the same system temperature, the signal to noise ratio while looking at a weak source is:
{\rm snr} = \sqrt{{\rm N(N-1) T}\Delta\nu} \ {\rm GS} \over
{\rm T_s}
\end{displaymath} (5.1.13)

This is the fundamental equation5.11 that is used to estimate the integration time required for a given observation. The signal to noise ratio for an N element interferometer is less than what would have been expected for a single dish telescope with area N times that of a single element of the interferometer, but only by the factor N/ $\sqrt{\rm N(N-1)}$. The lower sensitivity is again because the N auto-correlations have not been measured. For large N however, this loss of information is negligible. For the GMRT, N $= 30$ and N/ $\sqrt{\rm N(N-1)} = 1.02$, hence the snr is essentially the same as that of a single dish with 30 times the collecting area of a single GMRT dish.

For a complex correlator5.12, the analysis that we have just done holds separately for the cosine and sine channels of the correlator. If we call the outputs of such a correlator $r^c_{ij}$ and $r^s_{ij}$ then it can be shown that the noise in $r^c_{ij}$ and $r^s_{ij}$ is uncorrelated. Further since the time averaging can be regarded as the adding together of a large number of independent samples ( $\sim \sqrt{T\Delta\nu}$), from the central limit theorem, the statistics of the noise in $\bar{r}^c_{ij}$ and $\bar{r}^s_{ij}$ are well approximated as Gaussian. It is then possible to derive the statistics of functions of $\bar{r}^c_{ij}$ and $\bar{r}^s_{ij}$, such as the visibility amplitude ( $\sqrt{\bar{r}^c_{ij}+\bar{r}^s_{ij}}$) and the visibility phase ( $\tan^{-1}{\bar{r}^s_{ij}/\bar{r}^c_{ij}}$). For example, it can be shown that the visibility amplitude has a Rice distribution5.13

For an extended source, the entire analysis that we have done continues to hold, with the exception that S should be treated as the correlated part of the source flux density. For example, at low frequencies, the Galactic background is often much larger than the receiver noise and one would imagine that the limiting case of large source flux density (i.e. eqn(5.1.11) is applicable. However, since this background is largely resolved out at even modest spacings, its only effect is an increase in the system temperature.

Finally we look at the noise in the image plane, i.e. after Fourier transformation of the visibilities. Since most of the astronomical analysis and interpretation will be based on the image, it is the statistics in the image plane that is usually of interest. The intensity at some point $(l,m)$ in the image plane is given by:

I(l,m) = {1 \over {\rm M}}\sum_p w_p\mathcal{V}_p e^{-i2\pi(lu_p + mv_p)}

where $w_p$ is the weight5.14 given to the $p$th visibility measurement $\mathcal{V}_p$, and there are a total of M independent measurements. The cross-correlation function in the image plane, $\left<I(l,m)I(l^{'},m^{'})\right>$ is hence:

\left<I(l,m)I(l^{'},m^{'})\right> = {1 \over {\rm M}^2}\sum_...
e^{-i2\pi(lu_p + mv_p)}e^{i2\pi(l^{'}u_q + m^{'}v_q)}

In the absence of any sources, the visibilities are uncorrelated with one another, and hence, we have

\left<I(l,m)I(l^{'},m^{'})\right> = {1 \over {\rm M}^2}\sum_m w^2_p\sigma^2_p
e^{-i2\pi((l-l^{'})u_p + (m-m^{'})v_p)}

Hence in the case that all the noise on each measurement is the same, and that the weights given to each visibility point is also the same, (i.e. uniform tapering), the correlation in the map plane has exactly the same shape as the dirty beam. Further the variance in image plane would then be $\sigma^2_\mathcal{V}/{\rm M}$, where $\sigma^2_\mathcal{V}$ is the noise on a single visibility measurement. This is equivalent to eqn(5.1.13), as indeed it should be.

Because the noise in the image plane has a correlation function shaped like the dirty beam, one can roughly take that the noise in each resolution element is uncorrelated. The expected statistics after simple image plane operations (like smoothing) can hence be worked out. However, after more complicated operations, like the various possible deconvolution operations, the statistics in the image plane are not easy to derive.


... center5.1
See Chapter 4.
... gains5.2
Here the gain is taken to be in units of Kelvin per Jansky of flux in the matched polarization
... correlator5.3
Here we are dealing with an ordinary correlator, not the complex correlator introduced in the chapter on two element interferometers.
... mean5.4
Note that the average being taken over here is ensemble average, and not an average over time.
... processes5.5
Recall from the discussion of sensitivity of a single dish telescope that the central limit theorem ensures that the signal and noise statistics will be well approximated by a Gaussian. This of course does not include `systematics', like eg. interference, or correlator offsets because of bit getting stuck in the on or off mode etc.
... particular5.6
The derivation of this expression is particularly straightforward if one works with the moment generating function; see also the derivation sketched in Chapter 1.
... shown5.7
Papoulis, `Probability, Random Variables & Stochastic Processes', Third Edition, Chapter 10
... spectrum5.8
Where we have made the additional assumption that $x(t)$ is a white noise process, i.e. that its spectrum is flat. The power spectrum for such processes is easily derived from noting that $\int_{-\infty}^\infty {\rm S}_{xx}(\nu) d\nu = \sigma^2_x$, and that for a quasi-sinosoidal proccess of bandwidth $\Delta \nu$, the integrand is non zero only over an interval $2\Delta\nu$ (including the negative frequencies).
... better5.9
As you can easily derive from eqns 5.1.1 and 5.1.3 by putting $i=j=k=l$. Note that in this case eqn 5.1.1 becomes $\left<r_{ii}(t)\right> = \left(v_i(t) + n_i(t)\right) \left(v_i(t)
+ n_i(t)\right) = 2{\rm GS} + {\rm T}_s$
... common5.10
This may seem counter intuitive, but note that the outputs are only uncorrelated, they are not independent.
... equation5.11
In some references, an efficiency factor $\eta$ is introduced to account for degradation of signal to noise ratio because of the noise introduced by finite precision digital correlation etc. This factor has been ignored here, or equivalently one can assume that it has been absorbed into the system temperature.
... correlator5.12
See the chapter on two element interferometers
... distribution5.13
Papoulis, `Probability, Random Variables & Stochastic Processes', Third Edition, Chapter 6.
... weight5.14
As discussed in Chapter 11, this weight is in general a combination of weights chosen from signal to noise ratio considerations and from synthesized beam shaping considerations.

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