A simple delta function properties, sifting property

In summary: Shifting means changing the location of the limit point, while sifting means removing all but a certain number of elements.
  • #1
hojoon yang
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0
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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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  • #2
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,
$$
\int_{t_0}^{t_0+T} \delta(t-s) dt = 1
$$.
 
  • #3
u mean delta function property does not care about integral range?
 
  • #4
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:
  • #5
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.
 
  • #6
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.
 

Related to A simple delta function properties, sifting property

1. What is a delta function?

A delta function is a mathematical concept used in the field of calculus and analysis. It is often denoted as δ(x) and is defined as a function that is zero everywhere except at x=0, where it is infinitely large and has an area of 1. It is also known as a unit impulse function or Dirac delta function.

2. What is the sifting property of a delta function?

The sifting property of a delta function states that when the delta function is integrated over an interval containing its point of singularity (x=0), the result is 1. Mathematically, this property can be written as: ∫f(x)δ(x)dx = f(0).

3. How is the sifting property useful in mathematics?

The sifting property is useful in simplifying mathematical expressions and solving equations. It allows us to evaluate integrals involving delta functions easily, and it is also used in solving differential equations and Fourier transforms.

4. Can the sifting property be extended to higher dimensions?

Yes, the sifting property can be extended to higher dimensions. In multiple dimensions, the delta function is defined as a function that is zero everywhere except at a single point, where it has an infinite value. The sifting property in higher dimensions states that when the delta function is integrated over a region containing its point of singularity, the result is 1.

5. Are there any limitations to using the sifting property?

While the sifting property is a powerful tool in mathematics, it is important to note that it is not a function in the traditional sense and cannot be graphed. It is also defined only at a single point, which may limit its use in certain applications. Additionally, care must be taken when dealing with delta functions in mathematical operations to avoid potential errors.

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