# Measurement Equation

## 1 Measurement Equation

Let’s begin by restricting our discussion of interferometers 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”.

## 2 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.