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Signals and math confusion.

  1. Jan 31, 2012 #1
    I am a bit confused.

    http://pokit.org/get/bedc0ac7e1d17e01d7d58b021b81663c.jpg [Broken]

    The function is y=x(2t)

    and the point of it is to show the property of time-invariance.(which we should fail in this example, because it isn't time invariant.)

    Input signal is x1(t)

    Output signal is shown for that signal y1(t)

    But when you shift it, x1(t-2) it doesn't give y1(t-2).

    I am both ok and not ok with that.

    First lets go over the math.

    If I shift my signal x1(t) by -2

    I get [itex] y_2(t)=x(2(t-2)) [/itex]

    Is that correct?

    But that is [itex] y_2(t)=x(2(t-2))=x(2t-4) [/itex] and that is original signal, shifted by 4, and then scaled by 2, which is not what it should be.

    y2(0)=x(-4) which is 0, not 1.

    I am probably stuck on something simple, but nevertheless I am stuck.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 1, 2012 #2
  4. Feb 1, 2012 #3

    jim hardy

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    i dont know anything about that notation.

    is it analog or digital?
  5. Feb 1, 2012 #4
    I believe its analog.
  6. Feb 1, 2012 #5
    This might help.

    Its a quote from Oppenheim, signals and systems.

  7. Feb 1, 2012 #6
    it looks like the time axis is compressed towards the origin between a) and b), and between c) and d). Doing that means that every point on the time axis is mapped to a new point, except the origin, which remains unchanged. This makes t=0 special. I think the idea of a time invariant system is that no time t is special, and you are free to choose your origin wherever you like.

    What are these images supposed to represent? Input and output signals of some system? I think the output y requires knowledge of the future of x which violates causality
  8. Feb 1, 2012 #7

    jim hardy

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    i'm a plodder and haven't ruled out that the paradox might stem from 2 X 2 = 2 + 2

    is operator y half of operator x?
    and offset also 2?
    That's too many twos for me.

    what if operator y were one-third of operator x ?
    or 1/e ?

    please excuse if that's a dumb question but my simple mind has to rule out the dumb things first. And i do just plod.

    thanks for your tolerance.

    old jim
  9. Feb 2, 2012 #8
    I am stuck just like you guys. I think I will have just skip this one, and accept it.

    This represents a system with function y=x(2t).
  10. Feb 13, 2012 #9
    This is wrong:

    If I shift my signal x1(t) by -2

    I get y2(t)=x(2(t−2))

    The function is H{x(t)}=x(2t)
    What it means is take any occurrence of t in your input, and replace it with 2t.

    So we look at x2(t)=x1(t-2)
    And we want to get its output: we replace any t with 2t:
    so we get y2(t)=x2(2t)=x1(2t-2)=x1(2(t-1))

    Which agrees with the plots in your post

    Think hard about it, and do more examples, I remember these kind of things were really a pain in the *** in my signals & systems course
  11. Feb 13, 2012 #10
    I will. This was just a brainer. I didn't have a lot of time to tackle this. I did in the end, get a very good mark on this course 9/10.

    Thank you, I will give it some deep thought tomorrow. Gotta study Theory of information now :D
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