# Basic Interferometry II

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# Interferometry II

## 1 Recap of Interferometry I

You’ll recall from the Interferometry I lecture that for two antennas $i,j$ , we call the vector separating them ${\vec {b}}_{ij}$ the baseline. If a source in the sky is in direction ${\hat {s}}$ , then we derived the time delay between the two antennas from the source’s point of view was $\tau _{ij}={\frac {{\vec {b}}_{ij}\cdot {\hat {s}}}{c}}$ . We then showed how the fact that these delays as a function of baseline could in principle be used to reverse-engineer where sources were on the sky. In this lecture, we will begin there and develop a bit more formalism about how this works, and then walk through how imaging works, to first order. The basic two-element interferometer.

## 2 The Visibility Equation

Let’s begin by restricting our discussion to a single frequency $\nu$ , with corresponding wavelength $\lambda$ . Then just as ${\frac {{\vec {b}}_{ij}\cdot {\hat {s}}}{c}}$ was the time delay between antennas $i,j$ , we can also describe the number of wavelengths this is:

$\tau _{ij}\nu ={\frac {{\vec {b}}\cdot {\hat {s}}}{\lambda }}\,\!$ Knowing the number of wavelengths between the two antennas, we can now say that for a signal of a particular frequency emanating from a source in direction ${\hat {s}}$ , the complex phase difference between that signal measured at antenna $i$ and the signal at antenna $j$ will be $e^{-i\theta }$ , where $\theta$ is the angle swept out by the wave as it propagates from $i$ to $j$ . Using that $\theta$ is just $2\pi$ times the number of wavelengths, we know the phase difference $\Delta \phi$ is:

$\Delta \phi =e^{-2\pi i\tau _{ij}\nu }=e^{-2\pi i{\frac {{\vec {b}}\cdot {\hat {s}}}{\lambda }}}\,\!$ The phase difference $\Delta \phi$ is, of course, frequency dependent. And at a given frequency, it also varies with position on the sky. The pattern that this complex phase traces on the sky (see below for a graph of just the real component) is called the “fringe pattern” of an interferometer.

A graph of the real component of $e^{-2\pi i{\frac {{\vec {b}}\cdot {\hat {s}}}{\lambda }}}$ at a fixed frequency, as a function of direction on the sky. The complex response of a baseline along the sky is called the “fringe pattern”, and it is suspiciously close to a sine wave.

Now we will define a few variables that will help us extrapolate from a single baseline in a single direction to a picture of how a whole array might respond to the whole sky that falls within the primary beam of the correlated antennas. First, we will define coordinates representing the length of a baseline in units of wavelength:

${\frac {\vec {b}}{\lambda }}\equiv (u,v,w),\,\!$ where $u$ is the east-west component of the baseline, $v$ is the north-south component, and $w$ is the vertical (up-down) component. We will also split the source direction vector ${\hat {s}}$ into its components:

${\hat {s}}\equiv (l,m,{\sqrt {1-l^{2}-m^{2}}}),\,\!$ where $l$ is the east-west direction on the sky, $m$ is the north-south direction, and the third component comes from the fact that we restrict ${\hat {s}}$ to have unit length (it’s a direction vector).

Using these components, we can now write down the response of a baseline (called the “visibility” $V$ ) as a function of the $u,v,w$ separation of the antennas, integrating over all the source intensity $I$ on the sky as a function of $l,m$ :

$V(u,v)=\int \!\!\int {A(l,m)\cdot I(l,m)\cdot e^{-2\pi i(ul+vm+w{\sqrt {1-l^{2}-m^{2}}})}dl\ dm}.\,\!$ The equation above is the full form of the “visibility equation”, otherwise known as the “measurement equation” of an interferometer. The only variable that we haven’t yet defined is $A$ , which is the response of the primary beams of the antennas as a function of direction on the sky. In general, $A$ and $I$ are always grouped together, because the sky is always seen through the filter of the primary beam. The product $A\cdot I$ is sometimes called the “perceived intensity”.

## 3 Understanding the Visibility Equation as a Fourier Transform

The equation we derived above can be much easier to understand if we make a simplifying assumption, known as the “flat-sky” approximation. This approximation is either that $w=0$ , or alternately, that the primary beam $A(l,m)$ is sufficiently small that $l,m\ll 1$ , making ${\sqrt {1-l^{2}-m^{2}}}\approx 1$ . In either case, we are asking that the response of a baseline not need to account for the fact that the sky is a curved surface of a sphere. Under this assumption, the term $e^{-2\pi iw{\sqrt {1}}}$ is no longer a function of $l,m$ , and can be removed from the integral to give us:

$V(u,v)=e^{-2\pi iw}\int \!\!\int {A(l,m)\cdot I(l,m)\cdot e^{-2\pi i(ul+vm)}dl\ dm}.\,\!$ This formulation of the Visibility Equation is much more illuminating. It says that when phased to a “phase center” via a choice of a corresponding $e^{-2\pi iw}$ , with $w$ being the baseline component along the direction toward the phase center in wavelengths, the visibility $V(u,v)$ is just the Fourier Transform of the perceived sky.

So in addition to thinking about the fringe-pattern of a baseline on the sky, we can equivalently think of the following process. We take an image of the sky in $l,m$ coordinates and Fourier transform it: The true image of the sky (left) and the true UV plane (right).

The result is called the “uv-plane”, and its coordinates are inverse angles. An inverse angle is the same thing as a wavelength, so the uv-plane has coordinates (not surprisingly) of $u,v$ .

Next, this uv-plane is sampled at particular $u,v$ -coordinates by various baselines in an antenna array. The sampling pattern can be computed from the antenna configuration by choosing all of the antenna-to-antenna spacings. (Interestingly, this sampling pattern is the convolution of the antenna placement pattern with itself): The array sampling pattern in the uv-plane (right), and the associated synthesized beam pattern or “dirty beam” (left).

Note that for each pair of antennas you get two samples: one at $u,v$ , and one at $-u,-v$ . Because the sky is real-valued (no complex fluxes), these two Fourier components are related by a complex conjugate. That is, if you measure $V(u,v)$ at $u,v$ , you will measure $V^{*}(u,v)$ at $-u,-v$ .

Now, the sampling of the uv-plane is simply multiplying the true uv-plane by the sampling pattern you just computed. This is what you would get if you took visibilities recorded from an interferometer, and then placed each measured visibility $V(u,v)$ at the corresponding $u,v$ (and $-u,-v$ ) coordinates of a matrix. Finally, if you take the inverse Fourier Transform of this sampled uv-plane, you get an image: The “dirty image” (left) and the associated sampled uv-plane.

As you may notice, this image is somewhat degraded from our original. In fact, it is usually called a “dirty image”. Why is it dirty? Because we lost information when we sampled the uv-plane. We multiplied the true uv-plane by our sampling function. This is equivalent to convolving the true sky by the Fourier Transform of our sampling function:

The Fourier transform of our sampling function is often called the “dirty beam”. The dirty beam is what convolved the true sky to yield the dirty image. It is possible to recover something resembling the true sky by attempting to deconvolve the dirty image by the dirty beam. This is a complex process that will be described in detail in another lecture. Broadly, deconvolving attempts to compensate for the information that was lost by only sampling part of the uv-plane by injecting prior information about the sky. This information might be along the lines of “I know the sky is just point-sources” or “I want the smoothest sky that fits the data to the level of noise”. Either way, deconvolution is not a well-posed problem until you decide exactly what prior information you have about the sky. A “cleaned” dirty image. Note that point sources are recovered, but the extended square in the middle is a bit worse for the wear as a result of using a point-source model in the deconvolution process.