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A demodulator problem

  1. Dec 8, 2011 #1
    Hi all.

    Consider the following demodulator:

    2b5kdh.jpg

    [itex]s(t)=A_c\cos[2\pi f_ct+\beta \sin(2\pi f_mt)][/itex] , [itex]\beta<1[/itex].

    The delay line causes π/2 radians delay in phase. The problem is to find the output signal of the demodulator under the following assumptions:

    [itex]\cos(2\pi f_mT)\approx 1[/itex]
    [itex]\sin(2\pi f_mT)\approx 2\pi f_mT [/itex]

    T is the delay in time caused by the delay line which leads to π/2 radians delay in phase.

    It's not a homework, although once it was! What is the simplest way to solve this?

    Thanks in advance.
     
  2. jcsd
  3. Dec 8, 2011 #2
    I don't understand why

    [tex]2\pi f_mT→0[/tex]

    [tex]\hbox { Let n=1 }\;\Rightarrow \;2\pi f_mT = \pi f_m.\;\hbox { and }\; \cos (\pi f_m)≠1[/tex]

    I never saw demodulation this way. Demodulation usually mix a modulated signal with an unmodulated frequency to get the phase change ( or frequency change.). The way you show, both side have modulation and I don't see how it would work. If this work, we don't need PLL anymore!!!
     
  4. Dec 8, 2011 #3
    I think maybe this means [itex]f_m/f_c\ll 1[/itex].

    Let's just find the output signal and forget about the demodulator.

    Thanks.
     
  5. Dec 13, 2011 #4
    This a viable detector. I've seen it used, though not recently.
    Consider that the delay line is 1/4 wave, hence delays by 90 deg at fo.

    The governing equation is similar to:
    sin(x)cos(x) = 1/2 sin(2x). Thus, the baseband result is 0.

    As the frequency varies from fo, the resulting phase shift either increases or decreases resulting in a +/- baseband signal as the vector sum of the shifted signal contains either +a sin(x) or -a sin(x) depending on the deflection, a, and relative frequency shift, + or -.
     
  6. Dec 13, 2011 #5
    I still don't understand how this work as demodulator. Just work with me as I am not the expert.

    If you use a quarter wave line for the carrier frequency, so you basically mixing the [itex]\sin(\omega t +\phi_m) \;\hbox { and }\;\cos(\omega t +\phi_m+ Δ\phi)[/itex] where [itex]Δ\phi\;[/itex] is the change of phase within the quarter wave delay time:

    [tex] \sin(\omega t +\phi_m)\cos(\omega t +\phi_m+Δ \phi)= \frac 1 2 \sin(2\omega +2\phi_m+ Δ \phi)+ \frac 1 2 \sin (Δ\phi)[/tex]

    But the modulating frequency is much slower than the carrier frequency which mean [itex]Δ\phi≈0[/itex].

    If this work, we don't need PLL in cell phone as they operate in one single frequency and all you need is a about 1/2 inch copper trace for demodulation!!!!
     
    Last edited: Dec 13, 2011
  7. Dec 13, 2011 #6

    NascentOxygen

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    What if we make it a quarter wave at the modulating frequency?
     
  8. Dec 13, 2011 #7
    Your modulation frequency is not constant!!!! It is a spread of frequencies.
     
  9. Dec 13, 2011 #8

    NascentOxygen

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    A quarter wave at the centre of a narrow modulation band.
     
  10. Dec 13, 2011 #9
    Base band start at DC!!! If the BW is 1MHz, you have to do a quarter wave at 500KHz. That is an awfully long line!!!! I think even if that can be done, the error is too big because at lower end, there is nothing coming out!!!
     
  11. Dec 14, 2011 #10

    NascentOxygen

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    A quarter wave at the centre of the subcarrier frequency.
     
  12. Dec 14, 2011 #11
    What do you mean? You have the carrier frequency, then you have the base band which is a spread from DC to some frequency.
     
  13. Dec 14, 2011 #12
    Thanks for the replies.
    I think the assumptions say that T is small enough so that we can take s(t)-s(t-T)≈s'(t)T.
    So the input of the envelope detector is:
    [tex]-TA_c \left (2\pi f_c+\frac{\beta}{2\pi f_m}\cos(2\pi f_mt) \right )\sin(2\pi f_ct+\beta\sin(2\pi f_mt) )[/tex]
    And the output is:
    [tex]-TA_c \left (2\pi f_c+\frac{\beta}{2\pi f_m}\cos(2\pi f_mt) \right )[/tex]
     
  14. Dec 14, 2011 #13
    I spent like 15minutes expanding s(t-T) and tried to expand s(t)s(t-T), it just get longer and longer using Euler expansion. I give up. It'll take hours to expand and particularly double and triple check the steps to make out. I can't comment any further unless I can work out the formulas as I am skeptical about the validity of the demodulation scheme where you can replace the whole PLL with just a short copper trace of less than an inch in the cell phone, all the 802.11 wireless LAN and Blue Tooth. I always very skeptical about those college professors that never really have real life experience!!! Sorry!!!

    Good luck in working on this. I would be willing to spend lot more time if someone can point me to a real life circuit in production using this. At least I would be convinced that I can learn something if I spend the time.
     
  15. Dec 15, 2011 #14

    NascentOxygen

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    Does this scheme have a name? You are citing assumptions, so I think you should be able to cite the method's name. Is it from a text book? Author? Title? If it's only in class notes, can you photocopy the relevant half-page explanation and post it so that those who may be interested can take a look at it?
     
  16. Dec 15, 2011 #15

    sophiecentaur

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    It looks like a type of 'slope detector', to me and is the easiest way to detect fm with a simple AM receiver - just off tune it so that the received carrier frequency is half way down one of the skirts of the IF filter. Crude but effective and, of course, will only give a linear response for relatively low deviation FM.
     
  17. Dec 15, 2011 #16
    I was going to post back about the slope detector because for quarter wave delay as soon as I get up............Which is 9am!!!:rofl::rofl:. All you get is the comparison of the two wave of time T apart. All you can see is the relative increase or decrease in phase between the two. You have no absolute phase information with this setup. Even with this in mind:

    [tex] \sin(\omega t +\phi_m)\cos(\omega t +\phi_m+Δ \phi)= \frac 1 2 \sin(2\omega +2\phi_m+ Δ \phi)+ \frac 1 2 \sin (Δ\phi)[/tex]

    Where

    [tex] \frac 1 2 \sin (Δ\phi)≈\frac {Δ\phi} 2 [/tex]

    The problem is this term is so so small compare to [itex]\frac 1 2 \sin(2\omega +2\phi_m+ Δ \phi)\;[/itex] where it swing from -0.5 to +0.5 that it would be very very hard to filter out to retrieve that very low amplitude. It might work if the modulating frequency is much higher in order to make the
    [itex] \frac 1 2 \sin (Δ\phi)\; [/itex] larger so it is easier to detect. But that would violate every rule of real life modulation that I can think of regarding side band interference and all!!!
     
  18. Dec 15, 2011 #17

    sophiecentaur

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    The problem with using a simple temporal filter, as suggested, is that its response is periodic for ever. A good old lumped component resonator works only over a limited frequency range and you can get a 'fast slope', which will discriminate a small frequency deviation.

    The quarter wave delay must, surely, refer to the wavelength of the carrier and not to the modulation. It's always much easier to think of what goes on when you have a high deviation (wrt the modulating frequency) because it is possible just to consider the carrier 'sweeping up and down' over the discriminator slope and rely on the fact that the modulation sidebands (although infinite, in theory) all lie comfortably within the span of the filter. If the modulating frequency is not low compared with the carrier (small fractional bandwidth) then it becomes a bit of a nightmare due to the folding back of the lower sidebands which could fall below Zero Hz.
     
  19. Dec 23, 2011 #18
    This was onced used in exceptional quality FM tuners. IF frequency was knocked down to a few MHz, and the delay was a long, thin coaxial cable that was spireled up in a 4" diameter x 1" high sheild can. The output was beat against the original signal in a fairly common mixer chip and the output was the demodulated signal.
    Using a glass or quartz delay line, one should be able to fit this technology in just a few square inches.
     
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