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Convolution of delta function

  1. Oct 5, 2007 #1

    zai

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    for linear time invariant system,
    y(t)=h(t)*x(t) where y(t) is the output , x(t) is the input and h(t) is the impulse response.(* is the convolution)

    The definition of convolution is
    y(t)=integration from -infinity to +infinity (h(tau)x(t-tau)d(tau)

    p/s: i don't know how to use mathematical equation inhere. just joining the group

    i know that if h(t) =delta(t) then y(t)=x(t).

    now here comes my confusion, if h(t)=f(t)delta(t), then what is y(t)?
    can anybody give me any ideas?

    thanks for the help.

    zai
     
  2. jcsd
  3. Oct 10, 2007 #2
  4. Oct 11, 2007 #3

    zai

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    Thank you for the reply. For this particular property i knew it already. my question is if
    f(t)=[x(t).y(t)]*z(t) and y(t) is a delta function. How do i solve this problem?

    Zai
     
  5. Oct 11, 2007 #4


    What happens when you multiply a function by a delta function?

    What is [tex] x(t)\delta(t) [/tex] equal to? Thinking about this graphically may help.
     
  6. Oct 11, 2007 #5
    if delta(t) = 1 for t = 0,1,2,3 and 0 for else
    then,
    you get x(t) for t = 0,1,2,3 which is also a delta(t) function.
     
  7. Oct 16, 2007 #6

    zai

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    Dear Frogpad,
    thank you for the reply. As far as i know if [tex] x(t)\delta(t) [/tex], the answer is x(t) at t=0, i.e x(0). isn't it? So, do we need to consider x(0) as a constant convolve with z(t) then? it doesn't seem right to me.

    zai
     
  8. Oct 19, 2007 #7
    Hi Zai,

    I'm a bit late with my answer.

    "i know that if h(t) =delta(t) then y(t)=x(t)."
    Actually it is x(0) because of one of the delta function properties:

    integration from -infinity to +infinity (delta(t-tau)f(t)dt) = f(tau).

    Convolution of a function f(t) with a delta distribution moves the function to the point (in time) where the delta impulse takes place.

    I guess your hole integration becomes:
    y(t)=integration from -infinity to +infinity (f(tau)delta(tau)x(t-tau)d(tau)

    You could use the product rule of integration and use the convention that the integral over the delta distribution is the Heaviside Step function Theta.

    Now you have:
    Theta(tau) f(tau) x(t-tau) - int(Theta(tau) [f(tau)x(t-tau)]' d tau
    where ' is the derivative. Usually the derivative of f(tau)x(t-tau) should be easily to obtain by product rule again.

    You could also use the Laplace transform and your convolution becomes a multiplication. There are tables with often used formulas and possibly you can find one with f(t) delta(t). Afterward you have to transform it back in the time domain again.
     
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