- #36
Color_of_Cyan
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rude man said:check δ(t - 2).
Ok so it was actually this then?: (t - T - 2):
-∞∫∞[(1/3)[-exp(-4t) + exp(2t)][(t - T)2δ(t - T - 2)]dT ??
rude man said:Fact: δ(x) is an even function: δ(-x) = δ(x).
Sampling function: ∫f(x)δ(x-a)dx = f(a). The integration limits can be anything so long as they include x=a. So they can for example be + and - infinity. Notice you don't need to know how to integrate f(x)!
I can't seem to understand or see what is f(a) or f(x) in this case. But, if everything above is correct so far now, then according to only the 'tricks' I already know with this then:
t = T + 2 (because it is inside the 'δ')
Which is then substituted above into this:
(1/3)[-exp(-4t) + exp(2t)][(t - T)2
and so it becomes this:
(1/3)[-exp(-4(T+2)) + exp(2(T+2))][(2)2
= (4/3)[-exp(-4(T+2)) + exp(2(T+2))]
or was it supposed to be
T = t + 2 or something?
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