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Evaluate the signal expression

  1. Jun 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate (simplify) the following expression:


    y(t) = ∫0 e2(t + 2τ)δ(t + τ)dτ =



    2. Relevant equations

    Shortcut signal expression techniques, integration of an exponential signal and impulse signal, other techniques that are probably unknown to me

    3. The attempt at a solution

    (t + τ) = a;

    τ = -t + a; dτ = 0;

    =1

    but this can't be right at all. I know you're supposed to integrate something but don't really know what (or how it works in this case). Any hints would be greatly appreciated, thank you!
     
    Last edited: Jun 27, 2014
  2. jcsd
  3. Jun 27, 2014 #2

    rude man

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    How about we rewrite this as

    y(T) = ∫0 exp{2(T + 2t)}δ(t + T) dt

    that has to be the same thing except the answer is of course y(T), not y(t).
    So at which point in t is δ(t+T) non-zero? Then, look at the limits of integration.

    (I rewrote the xpression solely because it's easier, for me at least, to visualize t as the independent variable). Comments/refutations welcome!
     
  4. Jun 27, 2014 #3
    Sorry for sounding dumb, but t was given that way, as y(t), but by changing it, you mean it does mean the same thing anyway, correct? (τ is Tau and I couldn't find a more proper text symbol to write it down here).


    Anyway, I didn't even know you could rewrite e^(anything) that way, but I guess that does make it easier.

    I do not know how to see which point where δ(t+T) is not zero. I have it down though that, since you are integrating the δ function it "cancels out," so would it still make a difference?
     
  5. Jun 27, 2014 #4

    rude man

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    Sure. I only did it because I and most people think of t as the independent variable and T as the delay time.

    You mean write exp(x) instead of ex? Sure, that way you don't need awkward exponents and everything is aligned up. Very common I hope, at least it was in my day! :-)

    Look at the definition of δ(x). It's
    δ(x) = 0, x ≠ 0
    δ(x) = ∞, x = 0.

    BUT with the proviso that ∫δ(x)dx = 1 if the limits of integration include the point x = 0. Note that the dimension of δ(x) is always x-1. So actually the δ function is defined only in terms of an integral expression:

    ∫δ(x)dx = 0, integration limits outside of x=0
    = 1, integration limits surrounding x=0.

    The most interesting thing about the δ function is its sampling property:
    f(T) = ∫f(t)δ(t-T)dt if the limits of integration include t=T

    So with the definition I have given you for δ(x) and looking hard at your given limits of integration it should be apparent what f(T) is.
     
  6. Jun 28, 2014 #5
    Are you saying, T = -t? Seems there is a shortcut you used, where you can say like if δ(x - anything), then x - anything = 0. I take it I would be changing the exp part from T = -t as well.

    Something else bad to tell you; I have never seen those definitions either, but those are way better than what I was given before (though I'm sure they actually mean the same thing), thank you.
     
    Last edited: Jun 28, 2014
  7. Jun 28, 2014 #6

    rude man

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    Instead of saying T = -t I prefer t = -T since as I said I consider t the independent variable.

    What you said about δ(x-anythig) is correct. Whatever is inside the parentheses, the δ function is zero except when the parenthetical expression is zero.

    Look at your limits of integration and you won't be worrying about the exponential expression. We can talk about that once you figured out the answer.
     
  8. Jun 30, 2014 #7
    I see, and then this is based on the definition you gave, right? ("δ(x) = 0, x ≠ 0"?)



    From what you gave then, the exp part would be

    =exp{2[T + 2(-T)]}

    =exp{2[T - 2T]}

    =exp{2[-T]}

    =exp(-2T)

    What happens to the exp part of the expression now? Don't you still integrate this? I see from the limits of integration and the other definition you gave the δ(x) will just become 1.
     
  9. Jun 30, 2014 #8

    rude man

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    Look at where δ(t+T) is non-zero and compare that to your limits of integration!
     
  10. Jun 30, 2014 #9

    rude man

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    Right.


    From what you gave then, the exp part would be

    =exp{2[T + 2(-T)]}

    =exp{2[T - 2T]}

    =exp{2[-T]}

    =exp(-2T)

    What happens to the exp part of the expression now? Don't you still integrate this? I see from the limits of integration and the other definition you gave the δ(x) will just become 1.[/QUOTE]

    See my post # 8.
     
  11. Jun 30, 2014 #10
    Sorry, can't figure out what you mean by this. Isn't it already 1? I can't really see what is going on with the rest of the function.

    δ(x) = 0 (when x is not 0), but the integral limits are from 0 to ∞ ?
     
  12. Jul 1, 2014 #11

    rude man

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    For what value of t is your delta function not zero? Does that value of t fall within your limits of integration?
     
  13. Jul 2, 2014 #12
    Let's see

    "δ(x) = 0, x ≠ 0
    δ(x) = ∞, x = 0."

    "What you said about δ(x-anythig) is correct. Whatever is inside the parentheses, the δ function is zero except when the parenthetical expression is zero."

    (t+T = 0) => δ(x) ≠ 0.

    Really bad at this, but I'm going to say (disregarding T), that t can be 0 which is on the limit of integration.
     
  14. Jul 3, 2014 #13

    rude man

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    Why are you disregarding T?
    For what value of t is δ(t+T) non-zero?
    What are the limits of integration on t?
    Those two answers give you your final answer.
     
  15. Jul 4, 2014 #14
    ok then t = -T, and the limits of integration are from 0 to ∞. It is probably that I am rusty with integration, but you changed it to y(T), do the integration limits still apply?
     
  16. Jul 4, 2014 #15

    rude man

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    abf(t)dt = F(b) - F(a) where f(t) = dF(t)/dt. Fundamental theorem of calculus.

    In your case a = 0 and b = ∞. The limits are on t. When you perform the definite integral, t disappears.

    If f(t) = 0 for all a <= t <= b then that interal is zero.

    You need to review basic calculus.
     
  17. Jul 6, 2014 #16
    Then there's still more to know about this. You were supposed to take the derivative of f(t) (and basically the entire integral) all along?

    The reason I kept asking is because, the integral would just go to ∞ if you plug it back in for t (assuming all you do is take the antiderivative of e-2t). Was/is there another property regarding this? It does make a lot of sense now though.


    It's still just e-2t then, right?
     
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