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Periodic functions

  1. Aug 1, 2015 #1
    Hey.

    Assume you have a signal ##f## with period ##T_f## and a signal ##g## with period ##T_g##. Then the signal ##h= f+g## is periodic iff ##T_f/T_g \in \mathbb{Q}##.

    So if ##T_f/T_g## is an irrational number, the signal ##h## will not be periodic. Why is this actually the case?
     
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  3. Aug 1, 2015 #2

    Orodruin

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    Have you tried workin it out? What does it mean to say that the sum is periodic and how must the period of the sum be related to the periods of the individual functions?
     
  4. Aug 3, 2015 #3
    Hey! Sorry for late reply, things have been hectic lately.

    Well, I've thought about it.

    Assume the scale of the periods is time. Let's simultaneously release the two signals ##f## and ##g## with a phase difference ##\delta## and with periods ##T_f## and ##T_g##. If the two signals have different periods, then the length of ##h=f+g##'s period is dependent on how many "iterations" it takes for both signals to again have a phase difference ##\delta##. So if ##f## has a period ##T##, and ##g## has a period of ##1.1T##, then ##h## will have a period of ##10T##. This means that the ratio ##T_f/T_g= 1/1.1 = 10/11 ## represents the lengths of time periods needed (10 for ##f## & 11 for ##g##) it takes for the signals to match up.

    Now, I am tempted to say that this ratio must be rational in general because a whole number of ##f## and ##g## periods must transpire in time for the two signals to get back to where they started. But I'm a bit unsure still.
     
  5. Aug 3, 2015 #4
    It must be ##f## needs ##11## and ##g## needs ##10,## for that they both spend ##11T.##
    That's an intuitive opinion, and is correct though.
     
  6. Aug 3, 2015 #5

    Orodruin

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    As was stated, your intuition is correct. However, to convince a mathematician, you will need to prove this more rigorously. The "if" is easily shown as you have done, but more generally:
    Assume f(t+Tf) = f(t) and g(t+Tg) = g(t) for all t. If Tf/Tg = n/m, then define T = n Tg = m Tf. We now have f(t+T) + g(t+T) = f(t+m Tf) + g(t + n Tg) = f(t) + g(t), showin that f+g is periodic (note that n and m need to be coprime for T to be the actual period).

    The more involved part is showing the only if, i.e, to show that f+g is not periodic if the ratio of their periods is irrational. Can you think of a way?
     
  7. Aug 3, 2015 #6
    Well, let's assume two signals ##f## and ##g## are unleashed with some phase difference ##\delta##. Then if ##n/m## is an irrational number there never will be a whole number of periods ##nT_g## and ##mT_f## that are equal to each other => the signal ##h=f+g## will not repeat itself.

    I guess this isn't a proof though. I'd have to somehow prove that if ##n/m## is irrational then ##h## will have a period of infinity or something, but I don't even know how to derive the relationship between ##h##'s period and that of ##f## and ##g##.

    By the way, it's kind of interesting how the non-repeating decimals trait of irrational ##n/m## is carried over to ##h## never repeating itself either.

    Yeah you're completely right of course, it was a slip of the mind. And also ##h## will have a period of ##11T##.
     
  8. Aug 3, 2015 #7

    olivermsun

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    It's almost a complete proof. What does irrational mean, after all?
     
  9. Aug 4, 2015 #8
    What needs to be done to make it complete?
     
  10. Aug 4, 2015 #9

    olivermsun

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    I don't think you're missing anything except a more formal construction. One way might be to suppose that the ratio of periods is irrational, but that the signal is periodic (with finite period T). Then show that a contradiction results.
     
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