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Question concerning convolution

  1. Jul 17, 2013 #1
    Hi,
    I have a question concerning convolution. I don't quite understand the boundaries of the integration in the attachment. The way I see it, the areas do not overlap at all for 1<t<3. You might say that they do overlap at 0, but that is usually not considered an overlap. Even if they were overlapping, why would the upper boundary be t-1 for that range?
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2013 #2
    I'm confused by this. The convolution will be a three part convolution the first section being an increasing linear, and then it will be a constant slope, and then it will become a decreasing linear line.

    The piecewise functions will be something like:
    0 -> -inf<t<0
    Ax -> 0<t<2
    B -> 2<t<3
    -Cx -> 3<t<5

    unless you are looking at the tau as negative shift then it's all backwards and I don't understand your question
     
  4. Jul 17, 2013 #3
    First, let me clarify that I was merely referring to the first range. That is, this is not the complete answer. Please see attachment for the rest. I couldn't even understand the integration boundaries for the first range, which is why I only attached that part of the solution. The solution was provided by the professor, so it is not likely that it is wrong.
     

    Attached Files:

  5. Jul 17, 2013 #4
    Would you please explain this to me?
     
  6. Jul 17, 2013 #5
    I would agree completely with that solution.

    Are you asking about why the solution is such on one of the ranges? you are asking why the first range is the way it is? Are you confused by the slope or the range?



    Keep in mind that the x-values for the solution DO NOT correspond to the x-values of the functions. The x-values of the convolution refer to the AMOUNT of times the function has been shifted.
     
  7. Jul 17, 2013 #6
    I don't quite understand the boundaries of the integration. Let's look at the first range (1<t<3). The way I see it, the areas do not overlap at all for 1<t<3. You might say that they do overlap at 0, but that is usually not considered an overlap. Even if they were overlapping, why would the upper boundary be t-1 for that range?
     
  8. Jul 17, 2013 #7
    It's hard to discuss a visual convolution over text, but I'll try.

    So in the picture where are you do is reflect the wider rectangle across the Y-axis is t = 0 (TAU not to be mixed up with T: time)

    These images are confusing for a first time convolver. They barely differentiate between time and tau (shift amount)

    so when you move the blue function to the right one unit, that is tau = 1. Between tau = 0 and tau = 1 the functions are overlapping. This means that exactly after tau = 0 there will be an increasing function in the convolution.

    Now increase the tau again to tau = 2.


    wait.

    I think I see what you are saying. You are saying that the convolution should have an increasing value on the interval from tau = [0,1]? because now that I really study that solution, I don't think it's correct.
     
  9. Jul 17, 2013 #8
    All I was saying was that I didn't quite understand why there would be any overlapping between 1 and 3. By the way the rectangles are drawn in the attachment, they do not seem to overlap at all. If you could explain to me why the integration for that range would be between 0 and t-1, perhaps I could pick it up from there on my own.
     
  10. Jul 18, 2013 #9
    you are asking a vague question, are you asking is there an overlap between T=[1,3] or Tau=[1,3]?

    Oh wow I was misunderstanding the question, that answer is correct okay now I see it. I didn't realize that the first frame of the second thumbnail was an already shifted function.
     
  11. Jul 18, 2013 #10
    when the function is not shifted, there is no overlap. This is why the function solution is 0 from tau=[0,1]. then once you shift the blue rectangle right one unit, there is initial overlap at tau=1. so the function solution begins to increase after tau=1. as tau (shifting) keeps increasing there is more and more overlap. This happens until tau = 3 then the overlap remains the same for tau=4. I'm trying to grasp your question here it's a bit hard to get what you're asking.
     
  12. Jul 18, 2013 #11
    keep in mind that you are comparing always the reflected blue function- NOT the blue function itself
     
  13. Jul 18, 2013 #12
    I am sorry, but I still don't quite understand why based on the diagram there would be any overlap between the two rectangles for 1<t<3. Doesn't the diagram show that they intersect only at 0?
     
  14. Jul 18, 2013 #13
    you are mistaking t and tau. There is definitely no overlap from 1<t<3 there is definitely overlap from 1<tau<3
     
  15. Jul 18, 2013 #14
    do you understand the difference I am trying to make? the solution isn't the overlap at any time, t. The solution is the amount of area contained after shifting any distance, Tau.
     
  16. Jul 18, 2013 #15
    Those diagrams are absolutely terrible with notation, I understand your confusion.
     
  17. Jul 18, 2013 #16
    And what is the relation between t and tau?
     
  18. Jul 18, 2013 #17
    the teacher differentiates between T and Tau by using T and T' which is not a very clear distinction. T is simply the function value at any time. T is in your standard time domain, that we are all so comfortable using. now TAU is the magical convolution x-axis. tau is a measure of how far you have shifted a function. f(t-TAU) is a shift in TIME by the amount TAU. does this make sense? so the convolution is in "tau space"- it isn't a time representation of anything, it is simply a representation of the overlap of the functions (in time) at different shifting amounts.
     
  19. Jul 18, 2013 #18
    it's a way to get some weird conceptual value of the commonality of two functions at every point in both functions at every shifting value.
     
  20. Jul 18, 2013 #19
    I didn't understand convolution that well until I saw (cross)correlation concepts.
    http://en.wikipedia.org/wiki/Cross_correlation

    the only difference is your shifting goes in the opposite direction in correlation functions.

    Basically if you have two strings of numbers (maybe different slightly) that are phase shifted, when you correlate them, and they have a huge peak in the correlation, the strings of numbers are strongly correlated. Because at some point, there is a lot of overlap in the functions. This is a fairly common communications concept.
     
  21. Jul 18, 2013 #20
    please don't yell at me for leaving out the integration of tau and being solely in time space for the solution. When you are doing this graphical convolution kinda stuff, you are considering shifts by tau units.
     
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