To: File	DRAFT

From: Jon Hagen

Date: April 26, 2004

Subject: VLBI Principles

1. Introduction

This memo is intended to provide an explanation of VLBI for maintenance engineers. It deals only with 2-element interferometry; the simplest VLBI technique. However, for multiple baseline VLBI observations, the equipment and procedures at the individual stations remains the same. Keywords: baseline, fringes, coherence., spectrometry, .....(outline of memo)

2. Two-element interferometer

A single radiotelescope using a dish of diameter D has a beam angle of lambda/D (radians). This angle is the "resolving power" of the dish; all sources of angular size smaller than lambda/D will produce the same drift scan pattern on a chart record, as if they were point sources. However, when two such antennas, separated by a very long distance, L (the baseline), are pointed in the same direction, it becomes possible to resolve sources whose angular sizes as small as /(bL), where b is the fractional bandwidth of the receiver.

The signals from the two antennas are brought to a common point, where they are multiplied. The product, suitably low-pass filtered, is the the interferometer's output. For interferometry to work, the time taken for the signal to propagate from the source to the multiplier must be equal for the path through Antenna 1 and the path through Antenna 2.\

Before seeing how the interferomenter works, let us clarify the requirement for equal propagation times. Sources are always extended regions. A more exact statement is that the time taken for signal to propagate from any arbitrary point in the source to the multiplier via the Antenna 1 path be equal to the time taken via the Antenna 2 path - within a tolerance of 1/B, where B is the bandwidth of each receiver. Thus, for a bandwidth of 100MHz, the times must be equal within 10nsec. This condition requires that a suitable delay element be placed in the otherwise shorter path. Note that the longest delay required for two antennas located on the surface of the earth would be 22msec (the time for light to propagate half an earth radius). In VLBI experiments, the signals are recorded at each antenna. These recordings are later played back simultaneously at a central processing facility (the "Correlator"), so it is finally the tape player outputs that are multiplied. A delay of 22msec would simply require that one tape player be started 22msec later than the other. (Fine adjustments are made with a variable-length electronic delay line). Note that the the propagation times from the source to the multipler via the paths through Antenna 1 and Antenna 2 might be extended equally by several weeks, depending on how fast the tapes are shipped and then analyzed. But hey, those signals already took years to arrive from the source to the antennas.

Before going on to see how the interferometer works, let us show that the 'almost equal propagation time' requirement can, indeed, be met by extended, astronomically large sources. Since we are allowed to use a delay element (e.g. the tape position), it is enough to show that, for every point on the source, the difference between the distances to Antenna 1 and Antenna 2 is almost the same, to be specific, that this difference must not vary by more than c/B, where B is the bandwidth and c is the speed of light.

In the figure, L is the baseline, i.e. a vector

connecting the two antennas. R is a fixed vector to

an arbitrary point in the source, while r roams to

select any point in the source. R1 and R2 are the

distances from the point in the source to the two

antennas. Note that |R| is much greater than

|L| and |r|. We can evaluate |R1| - |R2| as follows:

|R1| - |R2| = ( |R1|2 - |R2|2 ) / (|R1| + |R2| )

( |R1|2 - |R2|2 ) / (2|R|)

=( |R+r|2 - ( |R-L+r|2) / (2|R|)

= (2R L - |L|2 )/(2|R|) - (r L)/ (|R|)

The first term is a constant, which can be compensated by adding a constant delay element. The second term, then, must remain less than c/B, i.e. (r L)/ (|R|) < c/B. Since |r|/|R| is the angular size of the source, , we have < c/(BL). Writing B as c/(b ), where b is the fractionalbandwidth, B/f, we find that < /(b L).

Example: if =0.2m , B=100MHz, and L = 3000 km,

then c/BL = 3 108/(100 106 3000 103 ) = 10-6 radian = 0.2 arcsec.

How it works

Each antenna uses a standard receiver front-end, consisting of a low-noise amplifier followed by a filter to define the bandpass, for example, 1400 MHz to 1500 MHz. It is convenient to regard the power spectrum of the signal as a comb of discrete sine waves, at uniformly spaced fixed frequencies throughout the band. White noise corresponds to a comb of a great many sine wave components with equal amplitudes.

Following the front end, a series of one or more frequency translators ('mixers') shift the bandpass channel down to base band. Our Mark IV terminal uses a 'single-sideband converter' which shifts the band edge all the way down to dc; the example band from 1400 MHz to 1500 MHz becomes a band from 0MHz to 100MHz. Frequency translators work by multiplying an input signal voltage by a sinewave voltage from a local oscillator (LO). This produces outputs at the sum and difference frequencies. The difference frequency is used when the signal is to be shifted downward. When the input is a comb of sinewaves, each component in the comb is shifted downward by the same amount (the LO frequency). If the phase of the LO is shifted, every component in the output comb will have its phase shifted by this amount. This equal phase shift property is essential to the operation of the interferometer.

Some figues of mixer, usb, lsb, comb, etc.......

Suppose that the two antennas are aimed at a compact source and that there is no other power in the passband. If the two antenna/receiver systems are identical and the path lenghs to the multiplier are equal, the multiplier is fed by identical multiplicands and is equivalent to a squarer. It will produce a positive dc output voltage, just as if we had square-law detected the signal from one of the antennas. But now let us change the phase of one of the local oscillators in one of receiver systems by 180 degrees. At the multiplier, the phase of each tooth in the sinewave comb will change by 180 degrees, which is equivalent to reversing the polarity of each tooth in the comb and hence of the multiplicand. The output from the multiplier will therefore become negative. If we had changed the phase by 90 degrees, the output of the multiplier would have fallen to zero, since at each of the frequencies in the comb, the multiplicands would be 90 degrees out of phase, and the average of sin (omega t) x cos (omega t) is zero. (The products of sinwave components at different frequencies always average to zero).

What if the source were not compact? Considering an arbitrary tooth in the combs, the phase the signal from Antenna 1 will not be the same as the phase as the signal from Antenna 2. Moreover, the phase difference from one comb tooth to anotherwill have a random value. The multiplier output, sum over i of sin( i t + 1i ) sin( i t + 2i) will average to zero, and will remain zero for any LO phase shift.

The signal at the output of the receiver is a superposition of several such combs; one of which is from a presumably compact source that is the target of the VLBI observation. Call this comb the 'signal'. Other combs, usually very much stronger, are from other sources that may be in the beam, background sky noise, and receiver noise. Call these combs, added together, the 'noise'. At each of the frequencies in the hypothetical comb there is a sine wave that is the sum of the noise sine wave and the signal sine wave. While their frequencies are the same, their phases are uncorrelated.