Radio interferometer
Assumptions for now- source is infinitesimal, waves are plane, receivers have very narrow bandpass filters allowing almost only a single frequency, antenna spacing is east-west.
Phase switching interferometer: output voltages are multiplied and then time-averaged, that has the effect of filtering out high frequencies. Geometric delay between the two antennas,
<math>\tau_g = \frac{\mathbf{D}.\mathbf{s}}{c} = \frac{D}{c}\sin\theta</math>
Output of the multiplier is proportional to,
<math>F = 2\sin(2\pi\nu t)\sin2\pi\nu(t-\tau_g)</math>
<math>F = \cos 2\pi\nu\tau_g - \cos(4\pi\nu t)\cos(2\pi\nu\tau_g) - \sin(4\pi\nu t)\sin(2\pi\nu\tau_g)</math>
Rate of variation of theta equals to rotational velocity of Earth and there is a limit to the baselines on Earth. Thus variation of <math>\nu\tau</math> is at least six orders of magnitude lower than that of <math>\nu t</math>. More rapidly varying terms are filtered out giving the fringe function,
<math>F = \cos 2\pi\nu\tau_g = \cos\left(\frac{2\pi D\sin\theta}{\lambda}\right) = \cos\left(\frac{2\pi Dl}{\lambda}\right)</math>
where <math>l=\sin\theta</math>
Earth rotates, theta varies and quasi-sinusoidal fringes are created at the correlator as shown in the right figure. It represents the directional power reception pattern of the interferometer ignoring the difference in the responses of individual antennas.
Because of the rotation of the Earth two antennas have different components of velocity in the direction of the source, thus the signals reaching the antennas suffer different Doppler shifts. When these two signals with different frequencies are multiplied in the correlator the sinusoidal output arises due to their interference or in other words due to the beats between them.
Effect of bandwidth
In practice signal has a continuum of frequencies. For signal at two different frequencies the fringe functions will be have different coefficients and thus fringe periods will be different. Due to interference between these two Fourier components the fringe function will be modulated, in case of a signal with uniform spectral density by a sinc function envelope,
<math>F(l) = \frac{1}{\Delta\nu} \int_{\nu_0-\Delta\nu}^{\nu_0+\Delta\nu} \cos\left(\frac{2\pi Dl\nu}{c}\right)d\nu = \cos\left(\frac{2\pi Dl\nu_0}{c}\right) \frac{\sin(\pi Dl\Delta\nu/c)}{\pi Dl\Delta\nu/c}</math>
It implies, envelope of the fringe pattern is the Fourier transform of the instrumental frequency response.
Output from the correlator after including the instrumental time delay,
<math>r = \frac{1}{2T} \int_{-T}^T V(t)V(t-\tau)\ dt</math>
Assuming integration time 2T very small compared to the inverse of bandwidth we get the unnormalized autocorrelation function,
<math>r = \lim_{T\rightarrow \infty} \frac{1}{2T} \int_{-T}^T V(t)V(t-\tau)\ dt</math>
Squared amplitude of a frequency spectrum of a signal is called the power spectrum which is the Fourier transform of the autocorrelation of that signal (Wiener–Khinchin theorem). If H is the amplitude response of the signal than power spectrum is,
<math>|H(\nu)|^2 = \int_{-\infty}^\infty r(\tau) e^{-j2\pi\nu\tau}\ d\tau</math>
and oppositely,
<math>r(\tau) = \int_{-\infty}^\infty |H(\nu)|^2 e^{j2\pi\nu\tau}\ d\nu</math>
As we assumed that the cosmic signal has a spectrum of constant amplitude the spectrum H is determined only by the passband characteristics of the amplifiers. Now consider a Gaussian passband centred at <math>nu_0</math>,
<math>|H(\nu)|^2=\frac{1}{2\sigma\sqrt{2\pi}}\left\{exp\left[-\frac{(\nu-\nu_0)^2}{2\sigma^2}\right]+exp\left[-\frac{(\nu+\nu_0)^2}{2\sigma^2}\right]\right\}</math>
Where sigma is the bandwidth factor, <math>FBHM=\sigma\sqrt{8\ln2}</math> Than the interferometer response is,
<math>r(\tau) = e^{-2\pi^2\tau^2\sigma^2}\cos(2\pi\nu_0\tau)</math>
This is a cosinusoidal function multiplied by an envelope function (shown on the right), in this case a Gaussian. The envelope function is called bandwidth pattern, delay pattern or fringe washing function.
Setting <math>\tau_i=0</math> and expressing in terms of geometric delay,
<math>r(\tau_g) = exp\left[-2\left(\frac{\pi D\sigma}{c}\sin\theta\right)\right]\cos\left(\frac{2\pi\nu_0 D}{c}\sin\theta\right)</math>
Period of the fringes varies with the term inside cosine, so it varies as <math>D/ \lambda</math>, but it does not depend on the bandwidth parameter, sigma. The width of the bandwidth pattern depends both on D and sigma. Wide bandwidth and long baseline results in narrow fringe envelope. Now, instead of Gaussian if we use a rectangular passband response than the envelope of the bandwidth pattern will be a sinc function.
Except in VLBI, fringe envelope is usually considered a nuisance. It is desirable to observe fringes in the vicinity of the maximum of the pattern where fringe amplitude is greatest. To do this we can change the instrumental delay continuously so as to keep the total delay very small. If tau is smallest its cosine will be the largest. If <math>\tau_i</math> is adjusted in steps of the reciprocal of the central frequency of the band than the response remains cosinusoidal with <math>\tau_g</math> and we can get rid of the envelope.
As bandwidth approaches the central frequency envelope function becomes so narrow that only the central fringe remains which in case of optics is called white light fringe.
One dimensional source synthesis
Angles of the antenna beam and other variables are defined w.r.t. the phase reference position or phase tracking centre. Instrumental delay is constantly adjusted to equal the geometric delay at this position.
<math>\tau = \tau_g |_{\theta=\theta_0}-\tau_i = 0 = \frac{D}{c}\sin\theta_0 - \tau_i</math>
Now for radiation from a direction theta_0-delta_theta the fringe response term is,
<math>\cos(2\pi\nu_0\tau) = \cos\left\{2\pi\nu_0\left[\frac{D}{c}\sin(\theta_0-\Delta\theta)-\tau_i\right]\right\} \simeq \cos\left[2\pi\nu_0\frac{D}{c}\sin\Delta\theta \cos\theta_0\right]</math>
Angular resolution determined by the baseline projected onto a plane normal to the direction of the source,
<math>u=\frac{D\cos\theta_0}{\lambda} = \frac{\nu_0 D\cos\theta_0}{c}</math>
As delta_theta is small, the response from this direction is,
<math>F(l) = \cos(2\pi\nu_0\tau) = \cos(2\pi ul),\ \text{where } l=\sin\Delta\theta</math>
The spatial variable l can be expressed in radians and thus the spatial frequency u can be expressed in cycles per radian.
Interferometer response as a convolution
Total output power from an antenna with power reception pattern A for a source with 1D brightness <math>I_1</math>,
<math>\int_{source} A(\theta'-\theta)I_1(\theta')d\theta' = \int_{source} \mathcal{A}(\theta-\theta')I_1(\theta')d\theta' \text{, where } \mathcal{A}(\theta)=A(-\theta)</math>
We can say, output power of a single antenna is the convolution of source with the power reception pattern of that antenna. Mirror image reception pattern is the response of the antenna to a point source. In case of interferometer just replace the antenna reception pattern with the total power reception pattern of the interferometer.
Response of an interferometer is determined by 3 functions,
- Reception patterns of the antennas, <math>A(l)</math>
- Fringe pattern, <math>F(l)</math> that is proportional to power
- Bandwidth pattern, <math>F_B(l)</math>, can be a sinc function or Gaussian or anything
Source intensity is modified by antenna and bandwidth patterns. So the interferometer response can be written as the convolution of the fringe pattern and the modified intensity,
<math>R(l) = \int_{source} \cos[2\pi u(l-l')] A(l')F_B(l')I_1(l')dl' = \cos(2\pi ul) \ast [A(l)F_B(l)I_1(l)]</math>
Neglecting A and <math>F_B</math> as they can be deducted we can write the simplified response,
<math>R(l) = \cos(2\pi ul) \ast I_1(l)</math>
Now introduce convolution theorem,
<math>f \ast g \rightleftharpoons \mathcal{F}\ \mathcal{G}</math>
Now write the three terms of simplified response in light of this response. Fourier transform of fringe term for a particular value of <math>u=u_0</math>,
<math>\cos(2\pi u_0l) = \frac{1}{2} [\delta(u+u_0)+\delta(u-u_0)]</math>
Fourier transforms of <math>R(l)</math> and <math>I_1(l)</math> are <math>r(u)</math> and <math>V(u)</math> respectively where V is visibility. Thus in Fourier space,
<math>r(u) = \frac{1}{2} [\delta(u+u_0)+\delta(u-u_0)] V(u) = \frac{1}{2} [V(-u_0)(u+u_0)+V(u_0)(u-u_0)]</math>
See that, the interferometer response consists of 2 delta functions situated at <math>+u_0</math> and <math>-u_0</math>. <math>V(u)</math> represents the amplitudes and phases of the sinusoidal component of the intensity profile, I. So interferometer acts as a filter that responds only to the spatial frequencies <math>\pm u_0</math>.
Fringe visibility was defined by Michelson,
<math>V_M = \frac{\text{intensity of maxima}-\text{intensity of minima}}{\text{intensity of maxima}+\text{intensity of minima}}</math>
All images are from the book Interferometry and Synthesis in Radio Astronomy written by Thompson, Moran and Swenson Jr. The text is also a summary of some of the chapters of that book.
