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In the related Astronomy SE questions below and their answers, it has become apparent that radio telescope arrays down convert their highest frequency bands before the first stage of amplification.

In the online course at nrao.edu Essential Radio Astronomy the Introduction argues that 1 THz is the absolute upper boundary to far infrared in astronomy (there's semantic overlap between millimeter waves and far infrared that context).

In Yuh-Jing Hwang's 2018 ASIAA Summer Student Lecture Series 1 slides Engineering Aspects for Radio Astronomy the crossover between amplify-first and down convert-first is circa 100 GHz, as shown in the two screenshots below, and that might correspond to the following, from here:

Just for example ALMA has ten different frequency bands listed at ESO's ALMA Receiver Bands.

The table says that the lowest two bands (35–50 and 65–90 GHz) use High-electron-mobility transistor receiver technology, while the top eight use "SIS".

Note that according to the linked presentation's slide #33-35 SIS (also ) is a barrier Superconducting-Insulator-Superconductor barrier tunneling detector.

Question: Why can't quantum phase noise in radio astronomy amplifiers just be filtered out after down conversion, instead of down-converting first?

I'm wondering what it is about quantum phase noise induced by the amplifier that allows it to survive down-conversion to a much lower frequency. The spatial resolution of the interferometer will be linked to the primary submillimeter wavelength (~1 THz), not the 2 GHz baseband after down-conversion. A 1 radian $\Delta \phi$ radian at 1 THz would correspond to 2 milli-radians at 2 GHz, and yet this is more easily achieved.

The reason phase noise is so critical is that these receivers are used in large arrays of dish antennas for interferometric imaging. Also note that the signal is then digitized with a very few bit ADC, then transmitted digitally to a central computer called a correlator where interferometry takes place computationally in real time.

Also related in Astronomy SE:


From the linked presentation:

Heterodyne receiver:

– for low frequency, the signal is amplified and then divided into different frequency bands and then digitized, correlated and generated the digital spectral information.

– For millimeter-wave frequency, the signal is amplified and then down-convert the frequency (by mixer) into intermediate frequency (IF), amplified and then divided into different frequency bands and then digitized, correlated and generated the digital spectral information.

For submillimeter-wave frequency, no low-noise amplifier with quantum-limited performance available, signal is down-convert the frequency (by mixer) into intermediate frequency (IF), amplified and then divided into different frequency bands and then digitized, correlated and generated the digital spectral information.

Suitable for fRF < 116 GHz

  • Receiver Feed
  • RF Amplifier
  • Mixer and Local Oscillator
  • IF Amplifier
  • IF Low-pass filter
  • Spectrometer or Correlator

Suitable for fRF > 84GHz

  • Receiver Feed
  • RF Amplifier (omitted)
  • Mixer and Local Oscillator
  • IF Amplifier
  • IF Low-pass filter
  • Spectrometer or Correlator

click for full size:

screenshot from Yuh-Jing Hwang's '2018 ASIAA Summer Student Lecture Series 1' slides "Engineering Aspects for Radio Astronomy" https://events.asiaa.sinica.edu.tw/ssp/viewFile.php?i=e0ac09c7b674289316abbcb7ed062f65&t=L screenshot from Yuh-Jing Hwang's '2018 ASIAA Summer Student Lecture Series 1' slides "Engineering Aspects for Radio Astronomy" https://events.asiaa.sinica.edu.tw/ssp/viewFile.php?i=e0ac09c7b674289316abbcb7ed062f65&t=L

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  • $\begingroup$ Note that I don't argue that one would want to down-convert first, but only ask why this would not work. $\endgroup$
    – uhoh
    Mar 19, 2022 at 5:42
  • $\begingroup$ "Why can't quantum phase noise in radio astronomy amplifiers just be filtered out after down conversion, instead of down-converting first?" This seems to be circular. Doesn't down-converting first imply filtering after down conversion? $\endgroup$
    – TimWescott
    Mar 19, 2022 at 6:14
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    $\begingroup$ Yes I agree, don't change your question as it is now aligned with the answer given. I had assumed you had a "phase filtering" technique in mind that wasn't in fact a frequency selective filter, or assumed they would be different filters. That may in fact lead to another and more concise question from you-- I welcome that vs ongoing discussion in the comments. $\endgroup$ Mar 19, 2022 at 10:02
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    $\begingroup$ Do you already have a good understanding of phase noise in a communication system and the related considerations for receivers? If not, that may be the good place to start-- there are some existing links here already I can point you to to start with that may fill in some of the gaps. $\endgroup$ Mar 19, 2022 at 10:13
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Mar 19, 2022 at 10:24

1 Answer 1

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I'm wondering what it is about quantum phase noise induced by the amplifier that allows it to survive down-conversion to a much lower frequency.

In general, the down-conversion process conserves phase shifts, not timing shifts. The basic heterodyne equation is just trigonometry (assuming $\omega_2> \omega_1$ and after the low pass filter): $$\cos \omega_1 t \cdot \cos (\omega_2 t + \phi) = \frac{1}{2} \cos \left((\omega_2 - \omega_1)t + \phi \right ) \tag 1$$

So a timing error at $1\mathrm{THz}$ that leads to a 1 milliradian phase shift will result in a 1 milliradian phase shift once the signal is converted down to $100 \mathrm{GHz}$.

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  • $\begingroup$ You posted an answer before seeing my reply to your comment $\endgroup$
    – uhoh
    Mar 19, 2022 at 6:22
  • $\begingroup$ ...and your equation is both unsourced and at first glance possibly incorrect. I tried $\omega_1, \omega_2 = 10, 1$ and $t, \phi = 4.2, e$. The left side is -0.322 and the right is 0.203 and 0.500 Adding back the missing $t$ doesn't seem to help the equality. I understand "Well, I'm pretty smart" but I think this answer attempts to oversimplify the problem. $\endgroup$
    – uhoh
    Mar 19, 2022 at 6:43
  • $\begingroup$ I believe Tim's equation is now correct, not just due to him being pretty smart (which he really is!) but also simple trigonometry and understanding the down-conversion process is a product of the LO at freq $\omega_1$ with the received signal at freq $\omega_2$ . If you need a source: sosmath.com/trig/prodform/prodform.html Importantly, the main point of his answer is correct and gets to the root of your question (I believe). $\endgroup$ Mar 19, 2022 at 9:29
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    $\begingroup$ @S.H.W Okay! Yes that now works. I think the original equation and its edit were simply shorthand, i.e. "you know what I mean" notation. $\endgroup$
    – uhoh
    Mar 19, 2022 at 9:55
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    $\begingroup$ I'm not touching the "quantum phase noise" because this is a DSP group, and "quantum" is about physics. So if the "quantum" part is retained, through demodulation, somehow (i.e., if quantum states stay entangled, and how you demodulate it either retains the entanglement or destroys it), then that's physics, not signal processing. $\endgroup$
    – TimWescott
    Mar 19, 2022 at 23:43

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