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Next: Impact on OTF observations Up: Generalization to a multi-pixel Previous: Impact on tracked observations

Imaging with HERA

HERA has a derotator, which ensures that the pixels do not rotate on the sky. The sky can thus be mapped by scanning along e.g. the right ascension or the declination axis in equatorial coordinates. We aim at obtaining a fully sampled map, implying a distance between the rows of $\ensuremath{\Delta}=
\ensuremath{\theta}/2.5$, where \ensuremath{\theta} is the beam full width at half maximum: At 1 mm, this corresponds typically to $4''$. However, the pixels are typically separated by $\ensuremath{\Delta}\simeq 2\ensuremath{\theta}$. We thus have to find the best scanning strategy which will fill the hole of the instantaneous footprint of the multi-pixel. To do this, we will use a property of the deroratator, i.e. it can be configured so that one of the main axes of the multi-pixel is rotated by an angle ( \ensuremath{\alpha}) from the scanning direction. Indeed, we can ask what is the value of \ensuremath{\alpha} needed so that the distance between the rows of two adjacent pixels is exactly \ensuremath{\Delta}. For a receiver of $\ensuremath{\sqrt{n_\ensuremath{\mathrm{pix}}}}\times\ensuremath{\sqrt{n_\ensuremath{\mathrm{pix}}}}$ pixels, we end up with \ensuremath{\sqrt{n_\ensuremath{\mathrm{pix}}}} groups of lines, the distance between two group of lines being noted \ensuremath{\delta'}. A bit of geometry gives

\begin{displaymath}
\ensuremath{\delta}= \ensuremath{\Delta}\,\sin\ensuremath{\...
...math{\delta'}= \ensuremath{\Delta}\,\cos\ensuremath{\alpha}.
\end{displaymath} (8)

If we now impose that
\begin{displaymath}
\ensuremath{\delta'}= \ensuremath{n_\ensuremath{\mathrm{sub...
...th{\sqrt{n_\ensuremath{\mathrm{pix}}}}\,\ensuremath{\delta},
\end{displaymath} (9)

we obtain
\begin{displaymath}
\tan\ensuremath{\alpha}= \frac{1}{\ensuremath{n_\ensuremath...
...{subscan}}}\,\ensuremath{\sqrt{n_\ensuremath{\mathrm{pix}}}}}.
\end{displaymath} (10)

We can fully sample without redundancy a given fraction of the sky in a single subscan ( $\ensuremath{n_\ensuremath{\mathrm{subscan}}}=1$) or in two parallel subscans (zigzag, $\ensuremath{n_\ensuremath{\mathrm{subscan}}}=2$).

For HERA, $\ensuremath{\sqrt{n_\ensuremath{\mathrm{pix}}}}= 3$ and the \ensuremath{\Delta} value is fixed to $4''$ by the observing wavelenght $\sim 1$ mm. $\ensuremath{n_\ensuremath{\mathrm{subscan}}}=1$ gives $\ensuremath{\alpha}=
18.4^\circ$, $\ensuremath{\Delta}\sim 12''$ and $\ensuremath{n_\ensuremath{\mathrm{subscan}}}=2$ gives $\ensuremath{\alpha}=
9.5^\circ$, $\ensuremath{\Delta}\sim 24''$. Current optical design implies a minimum distance between the pixels which is only compatible with the $\ensuremath{n_\ensuremath{\mathrm{subscan}}}=2$ solution.

In summary, by setting an angle of $9.5^\circ$ between one of the main axes of a $3\times3$ multi-pixels and the scanning direction, we can sweep in a fully sampled mode a given portion of the sky with two parallel scans separated by $3\ensuremath{\delta}= 12''$. The region of the sky fully sampled will then be rectangular: the length of the rectangular side perpendicular to the scanning direction is then $\ensuremath{d_\bot}= \ensuremath{n_\ensuremath{\mathrm{subscan}}}\ensuremath{n_\ensuremath{\mathrm{pix}}}\ensuremath{\delta}$, while the length of the rectangular size parallel to the scanning direction, \ensuremath{d_\Vert}, will depend on the observing strategy. However, there is an edge effect, due to the rotation of the array from the scanning direction. Indeed, the edges of the maps are not fully sampled: Thus must thus be considered as overheads. The area of the scanned sky must thus be larger than the targeted area, which must be fully sampled. Let's assume that the targeted area ( \ensuremath{A_\ensuremath{\mathrm{target}}}) is swept as a succession of \ensuremath{n_\bot} rectangles of size $\ensuremath{d_\bot}\times\ensuremath{d_\Vert}$. We get

\begin{displaymath}
\ensuremath{A_\ensuremath{\mathrm{target}}}= \ensuremath{n_\bot}\,\ensuremath{d_\bot}\,\ensuremath{d_\Vert}.
\end{displaymath} (11)

The area swept in the under-sampled edges ( \ensuremath{A_\ensuremath{\mathrm{edge}}}) is just the area of the rectangle whose side sizes are $\ensuremath{n_\bot}\,\ensuremath{d_\bot}$ and the scanning size of multi-pixel rotated by \ensuremath{\alpha}, i.e.
\begin{displaymath}
\ensuremath{d_\ensuremath{\mathrm{edge}}}= (\ensuremath{\sq...
...th{\Delta}\,(\cos\ensuremath{\alpha}+\sin\ensuremath{\alpha})
\end{displaymath} (12)

Indeed, the geometry of the edges show that half this area is covered on each size of the targeted area. Using Eqs. [*] and [*], we obtain
\begin{displaymath}
\ensuremath{d_\ensuremath{\mathrm{edge}}}= (\ensuremath{\sq...
...remath{\sqrt{n_\ensuremath{\mathrm{pix}}}})\ensuremath{\delta}
\end{displaymath} (13)

We now define the mapping efficiency \ensuremath{\eta_\ensuremath{\mathrm{edge}}} as

\begin{displaymath}
\ensuremath{\eta_\ensuremath{\mathrm{edge}}}= \frac{\ensure...
...suremath{d_\bot}\,\ensuremath{d_\ensuremath{\mathrm{edge}}}.
\end{displaymath} (14)

Replacing \ensuremath{A_\ensuremath{\mathrm{target}}} and \ensuremath{A_\ensuremath{\mathrm{edge}}} by their expressions [*] and [*], we derive
\begin{displaymath}
\ensuremath{\eta_\ensuremath{\mathrm{edge}}}= \frac{1}{1+\f...
...}}{\ensuremath{a}\,\ensuremath{n_\bot}\,\ensuremath{d_\bot}}}.
\end{displaymath} (15)

This expression indicates that the most efficient mapping strategy is to observe very wide scans. However, avoiding the edge overheads is only one aspect of wide-field mapping with a multi-pixels. In particular, we aim at having the most homogeneous map as possible. To achieve this, we need to scan as fast as possible so that the observing conditions are as comparable as possible on the whole map. We can then repeat the map as many time as possible so that the data affected by technical problems or bad weather happening during one coverage can just be discarded. In any case, at least two coverages obtained in perpendicular scanning direction is always advise to be able to use destriping algorithms (e.g. plait algorithms). Stripes happen because the system stability (weather, telescope, receiver and backend) evolves from one row to the other. Getting stripes is all the more probable than the time to scan a row is long. So this argues against making very wide scans, which are at the same time required to decrease the relative time spent in the edge overheads. A compromise is thus to map area chunks which are as close as possible to squares. A way to parametrize this is to introduce the map aspect ratio, defined as
\begin{displaymath}
\ensuremath{a}= \frac{\ensuremath{d_\Vert}}{\ensuremath{n_\...
... \quad \mbox{and} \quad
\ensuremath{n_\bot}\mbox{ integer.}
\end{displaymath} (16)

A given area \ensuremath{A_\ensuremath{\mathrm{map}}^\ensuremath{\mathrm{}}} will be mapped in chunks whose area ( \ensuremath{A_\ensuremath{\mathrm{chunk}}}) is defined by the linear scanning speed and the time of stability of the system ( \ensuremath{t_\ensuremath{\mathrm{chunk}}}). This gives

\begin{displaymath}
\ensuremath{n_\bot}\,\ensuremath{d_\bot}\,(\ensuremath{d_\V...
...nsuremath{d_\bot}\,\ensuremath{t_\ensuremath{\mathrm{chunk}}}.
\end{displaymath} (17)

Using [*] to replace \ensuremath{d_\Vert} by $\ensuremath{a}\,\ensuremath{n_\bot}\,\ensuremath{d_\bot}$, we yield
\begin{displaymath}
\ensuremath{n_\bot}^2 + \ensuremath{n_\bot}\,\frac{\ensurem...
...mathrm{chunk}}}}{\ensuremath{a}\,\ensuremath{d_\bot}^2} = 0.
\end{displaymath} (18)

This equation of the 2nd order has only one physical solution
\begin{displaymath}
\ensuremath{n_\bot}= \frac{1}{2}\,\frac{\ensuremath{d_\ensu...
...}}}{\ensuremath{d_\ensuremath{\mathrm{edge}}}^2}}-1 \right] }.
\end{displaymath} (19)

We note that this yields
\begin{displaymath}
\ensuremath{\eta_\ensuremath{\mathrm{edge}}}= \frac{1}{1+\f...
...rm{chunk}}}}{\ensuremath{d_\ensuremath{\mathrm{edge}}}^2}}-1}}
\end{displaymath} (20)


\begin{displaymath}
\quad \mbox{with} \quad
\frac{\ensuremath{a}\,\ensuremath{...
...ath{n_\ensuremath{\mathrm{subscan}}}}} \right) } \right] }^2}.
\end{displaymath} (21)

This expression can be used to understand how to get the highest mapping effiency ( \ensuremath{\eta_\ensuremath{\mathrm{edge}}}). This implies to get the largest value of the $(\ensuremath{a}\,\ensuremath{A_\ensuremath{\mathrm{chunk}}})/\ensuremath{d_\ensuremath{\mathrm{edge}}}^2$ ratio. We see that the larger the multi-pixel array, the smaller this ratio. Increasing the chunk area, either by increasing the linear velocity (i.e. increasing the dump rate, \ensuremath{f_\ensuremath{\mathrm{dump}}}) or by increasing the stability time ( \ensuremath{t_\ensuremath{\mathrm{chunk}}}) will increase the efficiency. The dump rate is fixed by the peak data rate, which gives typically $\ensuremath{f_\ensuremath{\mathrm{dump}}}= 2$ Hz. The stability time depends on the switching mode: It is the time between two off measurements in position switch (typically 1 or 2 minutes) and the time between two calibrations in frequency switch (typically 10 to 15 minutes).


Table: Mapping strategy to minimize edge effects.
\ensuremath{t_\ensuremath{\mathrm{chunk}}} \ensuremath{n_\bot} \ensuremath{a} \ensuremath{\eta_\ensuremath{\mathrm{edge}}}
min.      
1 1 3.7 0.83
2 2 1.9 0.83
5 4 1.2 0.86
10 6 1.1 0.90

Previous equations give the impression that the aspect ratio is a free parameter. This is not fully true because, \ensuremath{n_\bot} must be an integer. The following algorithm ensures that we get an integer value for \ensuremath{n_\bot} with the value of $\ensuremath{a}> 1$ and closest to 1. Starting with $\ensuremath{a}=
1$, Eq. [*] gives a value of \ensuremath{n_\bot}. We enforce the integer nature of \ensuremath{n_\bot} with

\begin{displaymath}
\ensuremath{n_\bot}= \ensuremath{\mathrm{floor}}(\ensuremath{n_\bot}),
\end{displaymath} (22)

and we recompute the associated aspect ratio with
\begin{displaymath}
\ensuremath{a}= \frac{\ensuremath{A_\ensuremath{\mathrm{chu...
...th{\mathrm{edge}}}}{\ensuremath{n_\bot}\,\ensuremath{d_\bot}}.
\end{displaymath} (23)

Table [*] gives the resulting values of \ensuremath{n_\bot}, \ensuremath{a} and \ensuremath{\eta_\ensuremath{\mathrm{edge}}} as a function of the stability time ( \ensuremath{t_\ensuremath{\mathrm{chunk}}}). We see that edge efficiencies are quite high. However, it is easier to have square chunks when the stability time is larger.

In summary, the time spent in edges is counted as overheads. It translates into a multiplicative efficiency ( \ensuremath{\eta_\ensuremath{\mathrm{edge}}}) because we enforce a mapping pattern through rectangular chunks. Although it is not intuitive (edge sizes are in general unrelated to area), this is not a big assumption because the use of a square multi-pixel anyway enforces mapping in rectangular chunks. We now summarize the algorithm to compute \ensuremath{\eta_\ensuremath{\mathrm{edge}}}:

Step #1: Computation of input quantities

\begin{displaymath}
\ensuremath{d_\bot}= \ensuremath{n_\ensuremath{\mathrm{subscan}}}\,\ensuremath{n_\ensuremath{\mathrm{pix}}}\,\delta,
\end{displaymath} (24)


\begin{displaymath}
\ensuremath{d_\ensuremath{\mathrm{edge}}}= (\ensuremath{\sq...
...math{\sqrt{n_\ensuremath{\mathrm{pix}}}})\ensuremath{\delta},
\end{displaymath} (25)


\begin{displaymath}
\ensuremath{t_\ensuremath{\mathrm{chunk}}}^\ensuremath{\mat...
...athrm{chunk}}}^\ensuremath{\mathrm{fsw}} = 10~\mbox{minutes}.
\end{displaymath} (26)


\begin{displaymath}
\ensuremath{A_\ensuremath{\mathrm{chunk}}}= \frac{\ensurema...
...thrm{subscan}}}}\,\ensuremath{t_\ensuremath{\mathrm{chunk}}}.
\end{displaymath} (27)

Step #2: Computation of \ensuremath{n_\bot} and \ensuremath{a}
Case $\ensuremath{A_\ensuremath{\mathrm{target}}}< \ensuremath{\eta_\ensuremath{\mathrm{edge}}^\ensuremath{\mathrm{min}}}\, \ensuremath{A_\ensuremath{\mathrm{chunk}}}$ with $\ensuremath{\eta_\ensuremath{\mathrm{edge}}^\ensuremath{\mathrm{min}}}= 0.8$

$\displaystyle 1.$   $\displaystyle \ensuremath{n_\bot}= \ensuremath{\mathrm{floor}}\ensuremath{\disp...
...t{\ensuremath{A_\ensuremath{\mathrm{target}}}}}{\ensuremath{d_\bot}} \right] },$ (28)
$\displaystyle 2.$   $\displaystyle \mbox{if $\ensuremath{n_\bot}= 0$, then send an error message: \lq\lq Area too small, use raster mapping.''},$ (29)
$\displaystyle 3.$   $\displaystyle \ensuremath{a}= \frac{\ensuremath{A_\ensuremath{\mathrm{target}}}}{(\ensuremath{n_\bot}\,\ensuremath{d_\bot})^2}.$ (30)

Case $\ensuremath{A_\ensuremath{\mathrm{target}}}\ge \ensuremath{\eta_\ensuremath{\ma...
...{edge}}^\ensuremath{\mathrm{min}}}\, \ensuremath{A_\ensuremath{\mathrm{chunk}}}$

$\displaystyle 1.$   $\displaystyle \ensuremath{n_\bot}= \ensuremath{\mathrm{floor}}\ensuremath{\disp...
...{chunk}}}}{\ensuremath{d_\ensuremath{\mathrm{edge}}}^2}}-1 \right] } \right\}},$ (31)
$\displaystyle 2.$   $\displaystyle \mbox{if $\ensuremath{n_\bot}= 0$, then send an error message: \lq\lq Area too small, use raster mapping.''},$ (32)
$\displaystyle 3.$   $\displaystyle \ensuremath{a}= \frac{\ensuremath{A_\ensuremath{\mathrm{chunk}}}}...
...emath{d_\ensuremath{\mathrm{edge}}}}{\ensuremath{n_\bot}\,\ensuremath{d_\bot}}.$ (33)

Step #3: Computation of \ensuremath{\eta_\ensuremath{\mathrm{edge}}}

\begin{displaymath}
\ensuremath{\eta_\ensuremath{\mathrm{edge}}}= \frac{1}{1+\f...
...}}{\ensuremath{a}\,\ensuremath{n_\bot}\,\ensuremath{d_\bot}}}.
\end{displaymath} (34)

Step #4: Recomputation of \ensuremath{A_\ensuremath{\mathrm{chunk}}} and \ensuremath{t_\ensuremath{\mathrm{chunk}}} when $\ensuremath{A_\ensuremath{\mathrm{target}}}< \ensuremath{\eta_\ensuremath{\mathrm{edge}}^\ensuremath{\mathrm{min}}}\, \ensuremath{A_\ensuremath{\mathrm{chunk}}}$

$\displaystyle 1.$   $\displaystyle \ensuremath{A_\ensuremath{\mathrm{chunk}}^\ensuremath{\mathrm{new...
...A_\ensuremath{\mathrm{target}}}}{\ensuremath{\eta_\ensuremath{\mathrm{edge}}}},$ (35)
$\displaystyle 2.$   $\displaystyle \ensuremath{t_\ensuremath{\mathrm{chunk}}^\ensuremath{\mathrm{new...
...hunk}}^\ensuremath{\mathrm{new}}}}{\ensuremath{A_\ensuremath{\mathrm{chunk}}}},$ (36)
$\displaystyle 3.$   $\displaystyle \ensuremath{A_\ensuremath{\mathrm{chunk}}}= \ensuremath{A_\ensure...
...chunk}}}= \ensuremath{t_\ensuremath{\mathrm{chunk}}^\ensuremath{\mathrm{new}}}.$ (37)

If $\ensuremath{t_\ensuremath{\mathrm{chunk}}}< 1$ minute, the targeted area is too small and the PI should use raster mapping instead of OTF mapping.


next up previous
Next: Impact on OTF observations Up: Generalization to a multi-pixel Previous: Impact on tracked observations
Gildas manager 2014-07-01