Let's start with a simplify universe. In this universe, there is only one source and 2 base stations:
\text{source } s:
A monochromatic point source that emits a radio wave with wavelength \lambda , frequency \omega = 2\pi \nu , and magnitude M , which can be express as a cosine function of time:
M(t) = M\cos(\omega t + \phi(t))Where \phi is a phase shift function which is a continuous uniform random variable: \phi \sim \mathcal{U}([0,2\pi])
And \phi(t) has a coherence time = \tau_c , meaning for \tau \ll \tau_c : \phi(t-\tau) \approx \phi(t)
\text{ base stations}:
b_0 and b_1 , both stations' antennas have perfect Antenna Gain, meaning they can received signal from all directions with no phase distortion nor magnitude attenuation.
Now the source s sits somewhere in the sky far away from the stations, and the two antennas pointing directly up are receiving wave signals from source s , and the signals are passed to the correlator for processing, as shown in the figure below.
Suppose the radio signal arrives at station b_1 is:
M_1(t) = M\cos(\omega t + \phi_1{\scriptstyle(t)})then signal M_0(t) at station b_0 , travels an extra length of d compared to b_1 , which infer a time delay \tau_g , so:
M_0(t)= M_1(t-\tau_g) = M\cos\big( \omega (t-\tau_g) + \phi_1{\scriptstyle(t-\tau_g)}\big)We assume that \tau_g \ll \tau_c , the coherence time, meaning \Delta \phi_1 = 0 during the extra time traveled, \tau_g , to reach b_0 :
\phi_1(t-\tau_g) \approx \phi_1(t)so we can simply M_0(t) to:
M_0(t)=M\cos\big( \omega (t-\tau_g) + \phi_1{\scriptstyle(t)}\big)1.1.1: Correlator - Multiple & Average
Next, signal M_0(t) and M_1(t) are passed to correlator:
In the correlator, the two signals are multiplied and then average:
1. Multiplication:
M_0(t)M_1(t) &= M^2\cos\big(\omega( t-\tau_g ) + \phi_1{\scriptstyle(t)} \big)\cos(\omega t + \phi_1{\scriptstyle(t)}) \\ &= \frac{1}{2} M^2 [ \cos({\scriptstyle \omega(t-\tau_g) + \phi_1{\scriptstyle(t)} + \phi_1{\scriptstyle(t)} +\omega t }) + \cos ( {\scriptstyle \omega(t-\tau_g) + \phi_1{\scriptstyle(t)} - \phi_1{\scriptstyle(t)} -\omega t }) ]\\ &= \frac{1}{2} M^2 [ \cos ({\scriptstyle 2\omega t-\omega\tau_g + \phi_1{\scriptstyle(t)} + \phi_1{\scriptstyle(t)} }) + \cos({\scriptstyle-\omega\tau_g +\phi_1{\scriptstyle(t)}- \phi_1{\scriptstyle(t)}}) ] \\ &= \frac{1}{2} M^2 \cos ( {\scriptstyle 2\omega t-\omega\tau_g + \phi_1{\scriptstyle(t)} + \phi_1{\scriptstyle(t)}} )+ \frac{1}{2} M^2 \cos( {\scriptstyle-\omega\tau_g +\phi_1{\scriptstyle(t)} - \phi_1{\scriptstyle(t)} }) \\
2. Average:
E[M_0(t)M_1(t)] &= E \left[ \underbrace{ \bcancel{\frac{1}{2} M^2 \cos (2\omega t-\omega\tau_g +\phi_1{\scriptstyle(t)} + \phi_1{\scriptstyle(t)})} }_{\text{average}=0 } + \frac{1}{2} M^2 \cos( \underbrace{ \omega\tau_g }_{\text{constant}} ) \right]\\ &= E \left[ \frac{1}{2} M^2 \cos(\omega\tau_g ) \right] \\ &= \tfrac{1}{2} M^2 E \left[ \cos(\omega\tau_g ) \right] \\ &(\text{let } I=\frac{1}{2} M^2 , \text{ where $I$ is referred as Intensity, since } I \propto M^2)\\ &= I \cos(\omega\tau_g) \\
So the correlator output R_c is:
R_c &=E[M_0(t)M_1(t)] \\ &= I \cos(\omega\tau_g) \\Which is a function of the time delay \tau_g .
1.1.2: Time delay geometric vectors
Next we want to replace the time delay \tau_g with geometric variables of the stations and source.
We introduce 2 new vectors to the picture:
\hat s : a unit vector that points at source s from the stations, because source s is far far away, we can treat the wave paths to be parallel for station b_0 and b_1
\vec b : the baseline vector that points from station b_0 to station b_1
See the figure below:
In the above figure, we can solve for d :
d &=|\vec{b} | \cos(\alpha) \\ &=|\vec{b}| |\hat{s}| \cos(\alpha) \\ &=\vec{b} \cdot \hat{s} \\Now we can express \tau_g with d :
\tau_g &= \frac{d}{c} \\ &(\text{$c=$ speed of light} = \nu \lambda)\\ &= \frac{\vec{b} \cdot \hat{s}}{\nu \lambda} \\So the correlator output R_c can be rewritten and simplified as:
R_c &= I \cos(\omega\tau_g) \\ &= I \cos\left( (2\pi\nu)\frac{\vec{b} \cdot \hat{s}}{\nu \lambda} \right) \\ &= I \cos\left(2\pi\frac{\vec{b} \cdot \hat{s}}{ \lambda} \right) \\1.1.3: Put Geometric Vectors on Cartesian Grid
Let's further simplify the R_c formula, let's see how to do it:
We put the stations and source on a cartesian (x,y) grid, with \hat x as unit vector:
From the figure above, we can rework \frac{\vec b \cdot \hat s}{\lambda} :
\frac{\vec b \cdot \hat s}{\lambda} &= \frac{(|\vec b| \hat x) \cdot \hat s}{\lambda} \\Next, we introduce a new variable u :
u \coloneqq \frac{|\vec b|}{\lambda}u is a scalar that lies on the same \bold{x}- axis, it is just x but scaled by \frac{1}{\lambda} .
Basically, u is the baseline length in units of wavelength.
So with u , we can simplify further:
\frac{\vec b \cdot \hat s}{\lambda} &= \frac{(|\vec b| \hat x) \cdot \hat s}{\lambda} \\ &= \frac{|\vec b|( \hat x \cdot \hat s)}{\lambda} \\ &= u (\hat x \cdot \hat s) \\Next let's see how to simplify (\hat x \cdot \hat s) . Let's focus on the unit vectors \hat x and \hat s :
From the figure above, we can express (\hat x \cdot \hat s) as:
\hat x \cdot \hat s &= |\hat s | | \hat x | \cos(\frac{\pi}{2} - \theta) \\ &= |\hat s | | \hat x | \sin(\theta) \\ &= \sin(\theta) \\ &= lSo back to \frac{\vec b \cdot \hat s}{\lambda} = u (\hat x \cdot \hat s) :
\frac{\vec b \cdot \hat s}{\lambda} &= u (\hat x \cdot \hat s) \\ &= ul1.1.4: Putting It All Together
Finally we plug this result back to the correlator output:
R_c &= I \cos\left(2\pi\frac{\vec{b} \cdot \hat{s}}{ \lambda} \right) \\ R_c(u,l) &= I \cos(2\pi ul )u:
is determined by baseline, i.e., locations of the stations.
l:
is determined by source, i.e., location of the source.
1.1.5: Cosine Fringe
If we plot out the R_c function, you will see a sinusoidal pattern. We refer them as Fringe Pattern.
In the interactive figure below, feel free to drag around the blue triangle source \color{blue} s and the brown circle \color{brown} u , and observe that:
Varying \color{brown} u changes the fringe pattern function. The smaller the \color{brown}u values, the lower the frequency in the fringe pattern.
Varying \color{blue} s changes value of R_c .
1.1.6: Angular Resolution
Now we can show you that to distinguish 2 point sources, the minimum angular separation between them is:
\theta_{\min} = \frac{\lambda}{|\vec b|}\theta_{\min} is also referred as the angular resolution.
To prove the above equation, let's first look at the interactive figure below that shows fringe patterns for 2 sources (\color{blue}s and \color{green}s_1 ) that have same brightness I .
(Feel free to drag around the triangles for source \color{blue} s & \color{green}s_1 , and observe their respective correlator outputs and the angular distance between them.)
Suppose the angular distance between k^{th} and (k+1)^{th} peak (bright fringe) is:
\theta_{k(k+1)} \;\;, \;\;\; k \in \mathbb{Z}Looking at the figure above, notice that:
\min(\theta_{k (k+1)}) = \theta_{01} =\theta_{(-1)0 }\;\; , \;\;\; k \in \mathbb{Z}Meaning that the angular distance between 0^{th} and 1^{th} bright fringe is minimum compare to other bright fringe pairs.
To be able to distinguish the 2 sources, it is required for those 2 sources to be at least \theta_{01} angular distance apart because if the angular distance between the two sources is anything smaller, then the two sources will look like a single bright spot.
Now let's we solve for \theta_{01} :
\text{given } R_c(u,\theta ) &= I \cos(2\pi u \sin\theta ) \\ & \because 1^{st}\text{ bright fringe is at } I \cos(2\pi \ast 1) \\ \to & \; u \theta_{01} = 1\\ \to & \; \frac{|\vec{b}|}{\lambda}\theta_{01} = 1\\ \to & \; \theta_{01} = \frac{\lambda}{|\vec{b}|} \\There we have it, the minimum angular separation (angular resolution) is:
\boxed{ \theta_{\min} = \frac{\lambda}{|\vec{b}|} }