A simple delta function properties, sifting property

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hojoon yang
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I don't know why this is possible

To use delta function properties( sifting property)

integral range have to (-inf ,inf)

or at least variable s should be included in [t_0,t_0+T]

but there is no conditions at all (i.e. t_0 < s < t_0+T)

am I wrong?
 
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It is the sameness of the integration limits that eliminates the need for them to be +- infinity. Consider the integral over ##t## only
$$
\int_{t_0}^{t_0+T} \delta(t-s) dt
$$.
The integrand is a delta function centered at ##t=s##, but the range of ##s## is the same as the range of ##t##. This means the integrand ##\delta(t-s)## is guaranteed to always be located inside the integration limit of ##t##. So,
$$
\int_{t_0}^{t_0+T} \delta(t-s) dt = 1
$$.
 
u mean delta function property does not care about integral range?
 
blue_leaf77 said:
$$
\int_{t_0}^{t_0+T} \delta(t-s) dt = 1
$$.
This is only true under the situtaion being considered, namely there is a second integral with respect to ##s## present with the same limits as the first integral.

EDIT: It might have been more justifiable if I had written
$$
\int_{t_0}^{t_0+T} \delta(t-s) dt = 1 \hspace{1.2cm} \text{for} \hspace{1.22cm} s=[t_0,t_0+T]
$$
 
Last edited:
hojoon yang said:
Taking one integral at a time: ∫ δ(t-s)dt from t0 to t0 + T = 0 unless s is within range of t0 to t0 + T, in which case it = 1. So you have to assume s is within range of t0 to t0 + T, and I agree that should have been specified.

Then the second integral is obviously T so the whole thing is N0/2 times 1 times T = N0T/2.
 
hojoon yang said:
View attachment 88419

I don't know why this is possible

To use delta function properties( sifting property)

integral range have to (-inf ,inf)

or at least variable s should be included in [t_0,t_0+T]

but there is no conditions at all (i.e. t_0 < s < t_0+T)

am I wrong?

Do you mean "shifting" or "sifting"? These are both relevant aspects of transforms such as the delta function, but they refer to vastly different properties.