Delta function

  • Thread starter paalfis
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  • #1
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Can someone explain to me why in this equation (attached)
Captura de pantalla 2014-08-24 a la(s) 23.07.45.png


where ρ(t)=[itex]\sum[/itex]δ(t-ti) , dirac funtion.
in the left side we have the sum over h(t-ti) instead of the sum over h(ti) ?

It seems to me that the integral would work summing 1*h(t1)+1*h(t2)+...+1*h(ti) for all ti smaller than t.
 
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Answers and Replies

  • #2
Fredrik
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$$\int\mathrm d\tau\, h(\tau)\rho(t-\tau) =\int\mathrm d\tau\, h(\tau)\sum_i\delta((t-\tau)-t_i) =\sum_i\int\mathrm d\tau\, h(\tau)\delta(\tau-(t-t_i)) =\sum_i h(t-t_i)$$
 
  • #3
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Thanks, one more question, what does tau stands for in this type of integral. i.e. What is the meaning of (t-tau)?
 
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  • #4
Fredrik
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It's a dummy variable (i.e. it can be replaced by any other variable symbol), so it's not really significant. But since you plug it into the same functions where you plug in ##t## and ##t_i##, it can be thought of as representing the same sort of thing as those guys do. (For example, time or temperature, or whatever t stands for in this context). This is especially clear when you approximate an integral by a Riemann sum, ##\int f(x)\mathrm dx\approx\sum_i f(x_i)(x_{i+1}-x_i)##.
 

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