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A question in DSB modulation

  1. Mar 13, 2012 #1
    A signal,
    [itex]x(t)=4sin(0.5\pi t), [/itex]​
    is transmitted by a DSB modulator. What is the frequency range of the carrier wave corresponding to the given message (data) signal?

    My basic misunderstanding here is if the given signal is the message signal that I should demodulate from a DSB modulated signal, or is it maybe the modulated signal itself?
    Now, given
    [itex]x_{DSB}(t)=V_{m} cos(\omega_{c}t) cos(\omega_{m}t)[/itex]​

    is the form of a DSB modulated signal, the question is how could I actually decide the value range for [itex]f_{c}[/itex]? I tried also to use some trigonometric identities to form a cosine multiplication as a sum of two cosines, but I just cannot figure out what is [itex]\omega_{m}[/itex] and what is [itex]\omega_{c}[/itex] here.

    Does anyone have an idea how to handle that task?
    Thanks. :redface:
     
  2. jcsd
  3. Mar 13, 2012 #2

    sophiecentaur

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    If you know the formula for Cos(a)Cos(b) then a is one ωt and b is the other.
    You should get just two terms in your expansion (the sidebands). This is because your formula shows the result of multiplying the two signals together - giving double sideband with 'suppressed carrier' (DSBSC). For the more familiar DSBAM, the formula has a constant term in it which also produces a component at the carrier frequency.
    Wiki it - there's plenty to read there.
    You should be able to find that info without too much help from me, initially.

    There is no limitation to the values of the frequencies involved but it would be practical to ensure that the modulation frequency is less than the carrier frequency.
     
  4. Mar 14, 2012 #3
    My problem is how to continue from here

    tadmc5ojhk2m.jpg

    and to use the given x(t)=4sin(0.5πt) to find what I need to. I just cannot see it yet..
     
  5. Mar 14, 2012 #4

    sophiecentaur

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    0.5π corresponds to ωm
    the factor 4 is just a constant multiplier of the amplitude when you multiply the two signals together (modulate).
    so the two terms you have arrived at correspond to signals at frequencies (carrier ω +ωm) and (carrier ω - ωm). There is no product at the carrier frequency.
    See it now? I'm sure you are thinking this is harder than it realy is.
     
  6. Mar 14, 2012 #5
    You say [itex]0.5\pi[/itex] corresponds to [itex]\omega_{m}[/itex] because the given [itex]x(t)[/itex] is the message signal and therefore the [itex]\omega_{m}[/itex]?
    How can you actually know there's no product in the carrier frequency?

    Another question is if I can be sure the frequency range of [itex]\omega_{c}[/itex] is [itex]-\omega_{m}<\omega_{c}<\omega_{m}[/itex]?
     
  7. Mar 15, 2012 #6

    sophiecentaur

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    Let's go back to the beginning.
    What do you understand that your original formula
    xDSB(t)=Vmcos(ωct)cos(ωmt)
    represents and do you 'believe' your sums?

    The above formula does not describe conventional Double sideband Amplitude modulation (AM radio); it is straight 'multiplication', which is Double Sideband Suppressed Carrier Modulation. Not surprisingly, the result has no trace of the original carrier. The result that you got (handwritten) shows this.

    If you want to represent normal AM then you need a different formula:
    xDSB(t)=Vmcos(ωct)(1+Bcos(ωmt))

    Where -1<B<1 and affects the depth of modulation.

    I think you mean the 'frequency range of the final signal' because ωc is fixed.
    If the modulator is 'perfect' then your calculation shows the result. You need to accept the consequences of what the Maths tell you - same as when you do mechanics or dynamics calculations.
     
  8. Mar 15, 2012 #7
    I understand that the original formula represents a multiplication of two signals, [itex]x(t)[/itex] (better called [itex]x_{m}(t)[/itex] , just for a better distinction) and [itex]x_{c}(t)[/itex] . I also agree that using a trigonometric relation could transform that multiplication and represent it as a sum of two signals. Applying the fourier transform on each of them results in two dirac's delta functions in the frequency domain. so I do accept the matter of double sideband here.

    What I don't understand now is how can you actually tell that the result has no trace of the original carrier? I can see [itex]\omega_{c}[/itex] there, so the frequency carrier does appear (and therefore the misunderstanding). How (and where) I can see that the original signal's carrier is suppressed?

    Yes, considering that range is between the two delta functions I mentioned above.

    So I should finally conclude that if [itex]\omega_{m}[/itex] is [itex]0.5\pi[/itex] then [itex]f_{c}=0.25 Hz[/itex]? Is that correct?
     
    Last edited: Mar 15, 2012
  9. Mar 15, 2012 #8

    sophiecentaur

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    If you start with a mathematical model then the result is the result. Forget the Fourier transform - your output function just consists of two frequencies - where could there be any others?

    If you are talking of a practical modulator with real signals then that is a different matter and you can expect finite levels of other modulation products but I don't think that's what you are talking about, is it?

    This just looks like a matter of confidence in relating maths to reality.
    Would you have the same problem is you were studying problem in dynamics, about an experiment involving a low speed collision on a high speed jet. If the flight were straight and level then you would accept that maths would give you the right result, despite the fact that, to another observer, all the velocity measurements might need +150m/s added to them. You wouldn't expect +150m/s to come into the prediction of a simple two-ball on-board collision situation would you?

    I could also ask you how the upper sideband could 'know' that its frequency is made up of the sum of two others. The frequency is only a number, however you generate it.

    Is that enough to convince you? I can try again - next time with a base ball bat. :biggrin:
     
  10. Mar 15, 2012 #9
    :tongue:
    thanks.
     
  11. Mar 15, 2012 #10

    sophiecentaur

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    You're welcome.
    Things like this just need sorting out and it's always worth seeing them through.
     
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