A simple delta function properties, sifting property

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SUMMARY

The discussion centers on the properties of the delta function, specifically its sifting property. It is established that the integral of the delta function, $$\int_{t_0}^{t_0+T} \delta(t-s) dt$$, equals 1 if the variable s is within the limits of integration, specifically within the range [t_0, t_0+T]. The necessity for the limits of integration to be the same for both t and s is emphasized, as it guarantees the delta function is evaluated correctly. The conversation also clarifies the distinction between "shifting" and "sifting" properties of the delta function, which are crucial for understanding its application in transforms.

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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.
 

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