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, we can express it as a cosine function of time:
x(t) = M\cos(\omega t + \phi(t))Figure
Where \phi(t) is varying random phase with a uniform distribution: \phi(t) \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_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
Suppose the radio signal arrives at station b_1 is:
x_1(t) = M\cos(\omega t + \phi_1{\scriptstyle(t)})then signal x_2(t) at station b_2 , travels an extra length of d compared to b_1 , which infer a time delay \tau_g , so:
x_2(t)= x_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 the phase does not change after the extra time traveled, \tau_g , to reach b_2 :
\phi_1(t-\tau_g) \approx \phi_1(t)so we can simply x_2(t) to:
x_2(t)=M\cos\big( \omega (t-\tau_g) + \phi_1{\scriptstyle(t)}\big)1.1.1: Correlator Output
Next, signal x_2(t) and x_1(t) are passed to correlator to calculate the output v_{R_c} :
v_{R_c}(t,\tau_g)|_T =& \langle x_2(t) , x_1(t) \rangle_{\tau_g} \\ =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_2(\psi)x_1(\psi)} \, \color{green}d\psi\\ \color{black} =&{\color{green} \frac{1}{T} \int_{\psi = t-T/2}^{\psi=t+T/2}}{\color{blue} x_1(\psi-\tau_g)x_1(\psi)} \, \color{green}d\psi
Which is basically an autocorrelation function for x_1(t) , and T , is referred as Integration Time:
Let's break v_{R_c}(t,\tau_g)|_T down:
Step 1: Multiplication:
\color{blue}x_2(\psi)x_1(\psi) &= M^2\cos\big(\omega( \psi-\tau_g ) + \phi_1{\scriptstyle(\psi)} \big)\cos(\omega \psi + \phi_1{\scriptstyle(\psi)}) \\ &= \frac{1}{2} M^2 [ \cos({\scriptstyle \omega(\psi-\tau_g) + \phi_1{\scriptstyle(\psi)} + \phi_1{\scriptstyle(\psi)} +\omega \psi }) + \cos ( {\scriptstyle \omega(\psi-\tau_g) + \phi_1{\scriptstyle(\psi)} - \phi_1{\scriptstyle(\psi)} -\omega \psi }) ]\\ \\ &(\text{let } I=\frac{1}{2} M^2 , \text{ where $I$ is referred as Intensity and } I \propto M^2)\\ \\ &= I [ \cos ({\scriptstyle 2\omega \psi-\omega\tau_g + \phi_1{\scriptstyle(\psi)} + \phi_1{\scriptstyle(\psi)} }) + \cos({\scriptstyle-\omega\tau_g +\phi_1{\scriptstyle(\psi)}- \phi_1{\scriptstyle(\psi)}}) ] \\ &= I \cos ( 2\omega \psi-\omega\tau_g + 2 \phi_1(\psi) )+I\cos( -\omega\tau_g ) \\
Step 2: Average over integration time T :
(The first integral term = 0 is from Time average of cosine for fast varying random phase :\text{1-0-2} )
Also notice that the time instant t and integration time T are basically irrelevant compare to the geometric delay \tau_g , so we often time write the correlator as function of geometric delay \tau_g :
v_{R_c}(t,\tau_g)|_T \to v_{R_c}(\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_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}|} }