- #1
EvLer
- 458
- 0
hello again,
i have an integral to solve and not sure how to approach this:
[tex]\int f(q+T)\delta (t-q)dq [/tex]
and the boundaries of integral are -inf +inf couldn't figure it out with latex.
what I know about this is that if delta function is integrated like this, it would be just the value of the function f(q) at some point. What bothers me is that f(q) is shifted to the right and I am not sure where the dirac delta function samples f(q). actually, I think I'm sort of confused...with all the variables in there q and t??
Any help is very much appreciated.
edit: my best estimation of the solution to this is f(t+T)?
not sure if that's correct, but [tex]\delta (-t) = \delta (t) [/tex], so [tex] \delta (t-q) = \delta (q-t) [/tex] which means that the integral = f(q + T) evaluated at q = t, i.e. value of f(t + T)?
i have an integral to solve and not sure how to approach this:
[tex]\int f(q+T)\delta (t-q)dq [/tex]
and the boundaries of integral are -inf +inf couldn't figure it out with latex.
what I know about this is that if delta function is integrated like this, it would be just the value of the function f(q) at some point. What bothers me is that f(q) is shifted to the right and I am not sure where the dirac delta function samples f(q). actually, I think I'm sort of confused...with all the variables in there q and t??
Any help is very much appreciated.
edit: my best estimation of the solution to this is f(t+T)?
not sure if that's correct, but [tex]\delta (-t) = \delta (t) [/tex], so [tex] \delta (t-q) = \delta (q-t) [/tex] which means that the integral = f(q + T) evaluated at q = t, i.e. value of f(t + T)?
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