The ALMA (Atacama Large Millimetre/submillimetre Array) radio telescope's band-10 capabaility is now operational, per this answer. That's confirmed by NRAO's First Science with ALMA’s Highest-Frequency Capabilities.

According to this site ALMA's bands and frequencies are as follows:

band  wavelength   noise   frequency
        (mm)        (K)      (GHz)
 1    6.0 - 8.5     26     35 -  50
 2    3.3 - 4.5     47     65 -  90
 3    2.6 - 3.6     60     84 - 116
 4    1.8 - 2.4     82    125 - 163
 5    1.4 - 1.8    105    163 - 211
 6    1.1 - 1.4    136    211 - 275
 7    0.8 - 1.1    219    275 - 373
 8    0.6 - 0.8    292    385 - 500
 9    0.4 - 0.5    261    602 - 720
10    0.3 - 0.4    344    787 - 950

900 GHz (0.9 THz) is quite a high frequency for a radio receiver! Each ALMA dish down-converts received frequency to a baseband of a few GHz before they get digitized and sent to the correlator for digital interferometry, but you still need an ultra-stable local oscillator (LO) for downconversion, and all of the LO's of all of the dishes need to be mutually coherent. That's quite a feat considering they can be tens of kilometers apart!

Question: How does ALMA produce stable, mutually coherent ~THz local oscillators for all of their dishes?

As suggested in this partial answer to the question How does the Event Horizon Telescope implement the interferometry? hard drives collect digital data down-converted by local atomic clocks, probably using GPS as a secondary reference, then brought to a central location for post analysis.

I am guessing that they spend a significant amount of time trying to reconstruct coherence at the millimetre (picosecond) level, but that option is not available for ALMA as data is continuously recorded at some large fraction of 24/7, and the huge volume of data at a location where magnetic hard drives don't work is overwhelming. So they need to get it right the first time.


ALMA produces stable, mutually coherent ~THz LO (Local Oscillators) for all the antennas by...

Using a single central LO and piping it to every antenna via fiberoptic cable! The fiber expands and contracts due to temperature fluctuations so a laser system is used for calibration among the antenna. If you can believe it, they manually lengthen or shorten the fiber to each antenna to adjust for the contraction/expansion. The LO for the high frequency observations for ALMA requires femtosecond accuracy.

Check out this youtube for a more thorough explanation than I could provide:


  1. I think this is an excellent question, since solving the LO synchronization problem for high frequency observations was one of the most difficult technical and engineering problems associated with bringing the full capabilities of ALMA online.

  2. Building and designing interferometers decades ago, we always remembered that light travels about a foot per nanosecond. To get femtosecond accuracy for synchronization, they have to adjust fiber lengths by the micrometer. Picoseconds? Ain't nobody got time for that!

The video shows and explains details of how the LO signal included in the modulated laser is split and then distributed by optical fiber to each element in the array. From the explanation on the video's page:

Complex electronics accurately stitch ALMA’s individual wave detections together into one dataset. The first step in this process is to have exact measurements of where and when the antenna picked up its waves. On each antenna is a clock that timestamps the data using a kind of atomic metronome, or rhythm-keeping device, kept near the supercomputer. The timekeeping waves from this central oscillator beam out to each of ALMA’s antennas. Onboard the antennas, a local oscillator injects this timekeeping beat into a microscopic mixer with the waves coming through the receiver, and a mixed-down signal is digitized and sent back along the fiber into the supercomputer.

The timekeeping signals generated by the central local oscillator are sent through optical fiber, which presents an additional challenge. The length of the optical fiber can vary with temperature, but in order to achieve the incredible level of accuracy that ALMA requires, the fiber length must not change by more than one millionth of a millimeter over the entire 15-kilometer-distance to the farthest antenna. Stabilizing the length of the optical fiber is another job of the central local oscillator.

| improve this answer | |
  • 1
    $\begingroup$ Excellent! We only recently got the privilege of embedding YouTube videos in questions, and I see you've put it to excellent use! But we must always do as you've done and summarize a few key points in text, since links can rot/break over time and answers must retain value and utility. Il'll give it a watch soon. So this is a "centrally-located local oscillator" :-) $\endgroup$ – uhoh Nov 3 at 8:34
  • 1
    $\begingroup$ @uhoh I think they should just call it an oscillator. It isn't really a local oscillator if it's 15km away! $\endgroup$ – Connor Garcia Nov 3 at 17:23
  • $\begingroup$ If they didn't have single mode optical fiber, they'd have to use either optical or more radio telescopes to distribute it. Well I guess the term refers to the oscillator being here rather than in some other galaxy where the original oscillators (atoms or electrons) are. $\endgroup$ – uhoh Nov 3 at 23:04
  • 1
    $\begingroup$ Thanks for your ping. Two things; 1) I like to let answers sit for a period of time for other readers to review and comment and vote first. Accepting brings the question back into the active queue and if it happens later rather than immediately can bring fresh eyes to the post and more up votes for the answer's author. 2) It's always better to try to bring important information or key points from a linked source back into the answer itself. That way if/when the link breaks/rots the answer doesn't loose value to future readers. $\endgroup$ – uhoh Nov 13 at 0:49
  • 1
    $\begingroup$ @uhoh Your answer makes perfect sense. I am new so I am still trying to figure out how things work here! I will use your techniques when I work up the courage to ask a question of my own! $\endgroup$ – Connor Garcia Nov 13 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.