Beat signals

  1. Hi

    On page 11 of this book, http://books.google.dk/books?id=Wv2OIC_SaDUC&printsec=frontcover&hl=da#v=onepage&q&f=false, it is stated that:

    "The phase modulation has an additional advantage: The first two sidebands at frequencies ω+Ω and ω-Ω have equal amplitudes, but opposite phases. A lock-in detector tuned to the modulation frequency Ω therefore recieves the superposition of the two beat signals between the carrier and the two sidebands, which cancel to zero if no absorption is present".

    I have been trying to show that the signal is 0 when there is no absorption. What I have done is to plot the two functions
    [tex]
    \cos((\omega - \Omega)t) + \cos(\omega t) \\
    -\cos((\omega + \Omega)t) + \cos(\omega t)
    [/tex]
    for some ω and Ω, but it is clear that these two functions when added do not cancel out. Have I misunderstood something here?

    Thanks for any help in advance.

    Best regards,
    Niles.
     
  2. jcsd
  3. f95toli

    f95toli 2,418
    Science Advisor
    Gold Member

    A lock-in will only detect terms oscillating at Ω so all other terms will drop out.
    Start by expanding cos (ω±Ω)t
     
  4. Hi

    Thanks for replying. OK, I get

    cos (ω±Ω)t = cos(ωt)cos(Ωt) - sin(ωt)sin(±Ωt).

    If I can only detect terms oscillating at Ω, then I get
    [tex]
    [\cos((\omega - \Omega)t)] - [\cos((\omega + \Omega)t) ]
    [/tex]
    which I by the above write as
    [tex]
    [\cos(\omega t)\cos(\Omega t)-\sin(\omega t)\sin(\Omega t)] - [\cos(\omega t)\cos(\Omega t)+\sin(\omega t)\sin(\Omega t)]
    [/tex]
    Shouldn't this yield 0?

    Best,
    Niles.
     
    Last edited: Mar 13, 2012
  5. If I lock-in on Ω, then I can do it with either cos(Ωt) or sin(Ωt). If I do it with the former, then I get 0 (after the low-pass filter), but if I use sin(Ωt), then it doesn't give 0. Is there something I have misunderstood here?

    Best,
    Niles.
     
    Last edited: Mar 13, 2012
  6. NascentOxygen

    Staff: Mentor

    Try cos(ωt) . cos(ω+Ω)t
    and keep the low frequency component.

    And cos(ωt) . cos(ω-Ω)t
    and keep the low frequency component.
     
  7. Hi

    Thanks. I guess we are interested in the power, since that is what we measure. So I get

    cos(ωt) . cos(ω+Ω)t = ½[cos(Ωt) + cos(2ω+Ω)] and
    cos(ωt) . cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)].

    Adding them and keeping terms at Ω gives cos(Ωt), so I get a nonzero DC-signal. Is the author wrong then?

    Best regards,
    Niles.
     
  8. NascentOxygen

    Staff: Mentor

    Not when you do it right. :smile:
     
  9. But cosine is an even function, so I thought that I could disregard the minus? But adding the signals with the minus still gives a nonzero DC-signal.

    Best regards,
    Niles.
     
  10. NascentOxygen

    Staff: Mentor

    EDIT
    [strike]cos (-A) = -cos(A)[/strike] [​IMG]

    Low frequency, not DC. :wink:
     
    Last edited: Mar 14, 2012
  11. NascentOxygen

    Staff: Mentor

    Oops! I was picturing 180-Ω, not -Ω! :redface:

    So, I got the result I was looking for, but method was wrong. I'll take another look at it.
     
  12. Thanks. Can a possible explanation be that the constant DC-signal can become 0 by an offset? Then from this any change in any of the amplitudes will result in a nonzero signal.
     
  13. NascentOxygen

    Staff: Mentor

    I intended looking at the technical publication you cited, but there's too much text to wade through searching for what I want. But I think you have provided the answer here: "The first two sidebands at frequencies ω+Ω and ω-Ω have equal amplitudes, but opposite phases."

    This is what I've been looking for. We agree that if they are in phase the two products add, but I see in the case cited you indicate one starts off with opposite phase. I don't know how that arises, but you can investigate. So the situation comes to:

    cos(ωt) . cos(ω+Ω)t = ½[cos(–Ωt) + cos(2ω+Ω)t] and
    cos(ωt) .cos(ω-Ω)t = ½[cos(Ωt) + cos(2ω-Ω)t]

    And the result is a low frequency component cos(Ωt) of zero amplitude, providing the sidebands are not attenuated unequally.
     
  14. NascentOxygen

    Staff: Mentor

    I'm a little uneasy with the concept of two sinusoids of different frequency being of opposite phase--wouldn't they be in-phase for exactly as much time as they'd be opposite-in-phase? But the maths looks good, and that's the main thing. :tongue:
     
  15. Ah, now I see it. OK, so the author was right after all, I have to give him that.

    One can show rigorously that when modulating the phase of an EM-wave of frequency ω with a small modulation amplitude of frequency Ω, two sidebands occur out-of-phase with frequency ω+Ω and ω-Ω, respectively. It is mentioned here (in the beginning), e.g.: http://en.wikipedia.org/wiki/Pound–Drever–Hall_technique#PDH_readout_function

    It most likely says so in the book too, but I haven't found it yet. I haven't read that many pages in it, because I was caught up with this. If I find it, I'll let you know on which page.

    Thanks for helping out with this.

    Best wishes,
    Niles.
     
  16. NascentOxygen

    Staff: Mentor

    Yes, I found it now, it's page 11 as you said. But I'm still trying to picture a graph of two sinewaves of different frequency being 180° out of phase. Won't they be constantly drifting through 0→360° relative phase? While they may start off in-phase, after a while they'll be precisely 180° out of phase.
     
  17. You are asking a very good question, because I thought about that very same issue. Intuitively I (still) don't really see why it cancels out, since the sidebands are not at the same frequency. This was the reason why I wanted to calculate it explicitly.

    I'm logging of now, but I'll check back tomorrow.

    Regards,
    Niles.
     
  18. NascentOxygen

    Staff: Mentor

    Certainly the sidebands themselves aren't of the same frequency, so additively they are never going to cancel. But when each is multiplied (aka mixed) with the carrier frequency, among the products of that process are two low frequency components each of frequency Ω. In this phase modulation, one of those low-frequency components is in anti-phase to the other.
     
  19. f95toli

    f95toli 2,418
    Science Advisor
    Gold Member

    It is because the sidebands are the results of a carrier being passed through a phase modulator where is is mixed with the IF, i.e. they are the products of the same process which means that they maintain a phase relationship when the signal is downmixed again in the lock-in.

    I use a variation of this technique in my experiment (although not for spectroscopy), and I can assure you that the math not only looks good but also works in practice:approve:
     
  20. NascentOxygen

    Staff: Mentor

    I'm coming to the conclusion that there is a bit of poetic licence here. When the author says that the upper sideband is out of phase with the lower sideband, I think he probably being conservative with words, and what he really means is that the modulation carried on that upper sideband, when demodulated returns a demodulated signal which is in antiphase with the demodulated signal recovered from the lower sideband.

    In which case, any mathematics is moot. [​IMG]

    If my hunch is right, this probably arises so often in communications engineering that it is a well-understood shorthand. Let's see if anyone can support me on this. :tongue2:
     
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