Basic Interferometry

Short Topical Videos

• Basic Radio Interferometry I (by Karol Sanchez)

Basic Interferometry

Interferometry is the practice of using a two-or-more-element radio telescope array to observe astronomical sources. The array itself, along with the electronics used to synthesise the signals detected by the telescopes, are what we call the interferometer.

The Two-Element Interferometer

A two-element interferometer consists of two telescopes separated by a vector $\mathbf {b}$, called the baseline. Both antennas receive electromagnetic radiation from an astronomical source in the sky, which is in the direction of the unit vector $\mathbf {\hat {s}}$. Because astronomical sources are far away, the radiation received by the antennas is in the form of plane waves. As the plane waves reach the antennas, one will actually receive the signal first and the second will not receive the signal until a certain amount of time has passed. The time that needs to pass is called the geometric delay, and is denoted by the variable $\tau$.

In order to find the value of $\tau$, we need to know the extra distance that the plane waves had to travel to reach the second telescope. This distance equal to the baseline vector $\mathbf {b}$ dotted with the unit vector $\mathbf {\hat {s}}$:

${\begin{aligned}\mathbf {b} \cdot \mathbf {\hat {s}} \ .\end{aligned}}\,\!$

Since $\mathbf {\hat {s}}$ is a unit vector, this distance is actually the projection of the baseline vector, $\mathbf {b}$ onto the vector $\mathbf {\hat {s}}$ and is given by:

$\mathbf {b} \cdot \mathbf {\hat {s}} =\mid \mathbf {b} \mid \cos \theta ,\,\!$

where $\theta$ is the angle between the baseline and direction vectors.

Knowing the distance, we just divide by the velocity of the plane waves, which is the speed of light, c:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. TeX parse error: Undefined control sequence \emph"): {\displaystyle \tau ={\frac {\mathbf {b} \cdot \mathbf {\hat {s}} }{\emph {c}}}\ .\,\!}

Suppose there were another source in the sky, in a different direction from the first. This second source would also cause a geometric delay between the two antennas; however, it would be different from the geometric delay of the first source because they would have different directional unit vectors:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. TeX parse error: Undefined control sequence \emph"): {\displaystyle \tau _{1}={\frac {\mathbf {b} \cdot \mathbf {\hat {s_{1}}} }{\emph {c}}}\ ,\tau _{2}={\frac {\mathbf {b} \cdot \mathbf {\hat {s_{2}}} }{\emph {c}}}\ .\,\!}

Now each antenna is receiving a signal from each source, and what they detect is actually the total signal of the two sources. In order to separate the signals from each source at each antenna, we need to correlate these total signals with each other.

Correlation

First, we define the total signal detected by each antenna as :

$E_{i}(t)=e_{1}(t)+e_{2}(t)\ ,\,\!$ $E_{j}(t)=e_{1}(t-\tau _{1})+e_{2}(t-\tau _{2})\ ,\,\!$

where $e_{1}$ and $e_{2}$ are the signals given by each source at times $t$ and $t$ minus either $\tau$1 or $\tau _{2}$.

We will use the correlation equation

$(f\star g)(\tau )=\int {f(t)g^{*}(t-\tau )}dt\,\!$

and substitute

$f=E_{i}\ ,g=E_{j}.\,\!$

The substitution and the expansion are

$(f\star g)(\tau )=\int {[e_{1}(t)+e_{2}(t)][e_{1}(t-\tau _{1}-\tau )+e_{2}(t-\tau _{2}-\tau )]}\ dt\,\!$ $(f\star g)(\tau )=\int {e_{1}(t)e_{1}(t-\tau _{1}-\tau )+e1(t)e_{2}(t-\tau _{2}-\tau )+e_{2}(t)e_{1}(t-\tau _{1}-\tau )+e_{2}(t)e_{2}(t-\tau _{2}-\tau )}\ dt\ .\,\!$

Notice that we are working in the real-valued time domain, which is why $g^{*}=g$. Also, because $e_{1}$ and $e_{2}$ are different and independent of each other, the 2nd and 3rd term integrate away by averaging to zero as random noise. We are then left with

$(f\star g)(\tau )=\int {e_{1}(t)e_{1}(t-\tau _{1}-\tau )+e_{2}(t)e_{2}(t-\tau _{2}-\tau )}\ dt\ .\,\!$

In order for us to extract any meaningful signal information out of the above integral, they must not both average to zero as well. This is possible when $\tau$ is equal either to $-\tau _{1}$ or $-\tau _{2}$.

Taking the case when $\tau =\tau _{1}$, the above integral becomes

$(f\star g)(\tau )=\int {e_{1}(t)e_{1}(t-\tau _{1}-(-\tau _{1}))\ dt\ =\int {e_{1}(t)e_{1}(t)\ dt\ =<e^{2}>.}}\,\!$