Let's start with a simplify universe. In this universe, there is only 1 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 . By the time the radio wave reaches a point on Earth, assuming no noise, we can express the signal as a cosine function of time:
x(t) = M\cos(\omega t )Figure
\text{ base stations}:
b_1 and b_2 , both stations' antennas have perfect Antenna Gain, meaning they can received signal from all directions with no phase distortion nor magnitude attenuation. Assuming the antennas are capable of observing just a single frequency.
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.
Figure
Say the signal arrives at station b_1 is:
x_1(t) = M\cos(\omega t )The same signal need to travel an extra distance of d to reach station b_2 . , which infer a time delay, \tau_g , so the signal at station b_2 is:
x_2(t)= x_1(t-\tau_g) = M\cos\big( \omega (t-\tau_g) \big)1.1.1: Correlator Output
Next, signal x_2(t) and x_1(t) are passed to correlator to calculate their correlation v_{R_c} :
\large v_{R_c}(\bar t,\tau_g)|_T =&\large \frac{1}{2T} \int_{\bar t-T}^{\bar t +T} x_1(t)x_2^*(t) dt \; \;, \text{ where $2T$ is referred as Integration Time}\\ =&\large \frac{1}{2T} \int_{\bar t-T}^{\bar t +T} x_1(t)x_1^*(t-\tau_g) dt\\ =&\large \frac{1}{2T} \mathcal{Re} \left\{ \int_{\bar t-T}^{\bar t +T} Me^{j\omega t } Me^{-j\omega (t-\tau_g)} dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ \int_{\bar t-T}^{\bar t +T} e^{j\omega \tau_g} dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ e^{j\omega \tau_g} \int_{\bar t-T}^{\bar t +T} 1 dt \right\} \\ =&\large \frac{M^2}{2T} \mathcal{Re} \left\{ e^{j\omega \tau_g} 2T \right\} \\ =&\large M^2 \cos(\omega \tau_g) \\
As you can see, T, \bar t is irrelevant, so we can rewrite the correlation function result to:
\large v_{R_c}(\tau_g) = M^2 \cos(\omega \tau_g)And with the relation between Intensity(I ) and Magnitude(M ) being:
I \propto |M|^2We will express the final correlation function result to:
\large v_{R_c}(\tau_g) = I \cos(\omega \tau_g)1.1.2: Time delay geometric vectors
Next we want to replace the time delay \tau_g with geometric variables from 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_2 and b_1
\vec b : the baseline vector that points from station b_2 to station b_1
See the figure below:
Figure
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 v_{R_c} can be rewritten and simplified as:
v_{R_c}(\tau_g) &= 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) \\Also note a useful fact:
\nu \tau_g = \frac{\vec{b}\cdot \hat s}{\lambda}1.1.3: Put Geometric Vectors on Cartesian Grid
Let's further simplify the v_{R_c}(\tau_g) 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:
Figure
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 :
Figure
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 function v_{R_c}(\tau_g) :
v_{R_c}(\tau_g) &= I \cos\left(\omega \tau_g\right) \\ &= I \cos\left(2\pi\frac{\vec{b} \cdot \hat{s}}{ \lambda} \right) \\ &= I \cos(2\pi ul ) \\ &= v_{R_c}(u,l) \\ &\text{or}\\ &= v_{R_c}(u,\theta) \\ \\ \\ &(\omega \tau_g = 2\pi u l )u:
is determined by baseline, i.e., locations of the stations.
l:
is determined by source, i.e., location of the source.
In short, when people say fringe function, it is a function of geometric delay \tau_g , which is equivalent to function of baseline u and angle \theta .
1.1.5: Fringe
For a single u value, we can plot out the v_{R_c} function for \frac{-\pi}{2} \le \theta \le \frac{\pi}{2} , then you will see a sinusoidal pattern. We refer them as Fringe Pattern as shown in the figure below.
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 v_{R_c} .
Figure :
\cos Fringe for monochromatic Wave: draggable {\color{blue}\text{source } s}, {\color{brown}\text{baseline }u}
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.)
Figure :
Angular Resolution demonstration with 2 sources: draggable {\color{green}\text{source }s_1}, {\color{blue}\text{source }s}, {\color{brown}\text{baseline }u}
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 } v_{R_c}(u,\theta_{01} ) &= I \cos(2\pi {\color{brown} u \sin\theta_{01}} ) \\ &( \because 1^{st}\text{ bright fringe is at } I \cos(2\pi \ast {\color{brown}1}) ) \\ \to & \; {\color{brown} u \sin(\theta_{01})} = 1\\ &(\text{use small angle approximation : } \sin(\theta_{01}) \approx \theta_{01}) \\ \ \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}|} }