Dirac Delta using Fourier Transformation

In summary, the Dirac delta function is typically defined as having an infinite value at x=0 and a value of 0 everywhere else. However, using Fourier transformation, it can be shown that the Dirac delta function is actually the limit of the function sin(Ωx)/πx as Ω approaches infinity. This limit only exists when x=0, making it a more rigorous definition of the Dirac delta function.
  • #1
arpon
235
16
We know,
$$\delta(x) = \begin{cases}
\infty & \text{if } x = 0 \\
0 & \text{if } x \neq 0
\end{cases}$$
And, also,
$$\int_{-\infty}^{\infty}\delta(x)\,dx=1$$
Using Fourier Transformation, it can be shown that,
$$\delta(x)=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}$$
Let's look at the graph of ##y=\frac{\sin{(\Omega x)}}{\pi x}##, for large ##\Omega##.

Untitled.png


So, from this graph, it seems,
$$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}\neq 0 \text{ , for } x\neq 0 $$

But, this is actually the Fourier transformation of the function,
$$f(t)=1=\exp{(i\cdot 0\cdot t)}$$

So, intution tells me that the frequency spectrum of ##f(t)## should have infinite amplitude at ##x=0## and ##0## amplitude otherwise, just like the Dirac-delta function.

Now, how can I show that,
$$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x} = 0 \text{ when, } x\neq 0$$
 
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  • #2
The rigorous definition of dirac delta function is not the one you write at the beginning of your post. Even if we accept that definition (because intuitively that's what the dirac delta is) then I don't know how you come up with ##\delta(x)=\lim_{\Omega \rightarrow \infty} \frac{sin(\Omega x)}{\pi x}## cause that limit simply does NOT exist. (it does not exist for ##x\neq 0##, if we first take the limit x->0 then it exists and is infinite).
 
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  • #3
Delta² said:
The rigorous definition of dirac delta function is not the one you write at the beginning of your post. Even if we accept that definition (because intuitively that's what the dirac delta is) then I don't know how you come up with ##\delta(x)=\lim_{\Omega \rightarrow \infty} \frac{sin(\Omega x)}{\pi x}## cause that limit simply does NOT exist. (it does not exist for ##x\neq 0##, if we first take the limit x->0 then it exists and is infinite).

Using Fourier transformation, we have,
AS.PNG


Comparing with the equation,
$$f(t)=\int_{-\infty}^{\infty}\delta(t-u)f(u)\,du$$
we have,
ASS.PNG


Thus,
$$\begin{align}
\delta(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\,d\omega\\
&=\frac{1}{2\pi}\lim_{\Omega\rightarrow \infty}\left[\frac{e^{i\omega t}}{i\omega t}\right]^{\Omega}_{\omega=-\Omega}\\
&=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega t)}}{\pi t}
\end{align}$$

Reference:
Mathematical Methods for Physics and Engineering: A Comprehensive Guide, 3rd Edition by K. F. Riley, M. P. Hobson, S. J. Bence, Page/442
 
  • #4
The last two equalities would hold only if the limit ##\frac{1}{2\pi}\lim_{\Omega\rightarrow \infty}\left[\frac{e^{i\omega t}}{it}\right]^{\Omega}_{\omega=-\Omega}=
\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega t)}}{\pi t}

## existed but it does not exist, as you can easily prove for any x<>0 the limit ##\lim_{\Omega \rightarrow \infty}\sin{(\Omega x)}## does not exist.
 
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What is the Dirac Delta function?

The Dirac Delta function, denoted as δ(x), is a mathematical concept used in engineering and physics to represent a point mass or impulse at the origin. It has a value of infinity at x = 0 and is zero everywhere else.

How is the Dirac Delta function related to Fourier transformation?

The Fourier transformation is a mathematical operation used to decompose a function into its frequency components. The Dirac Delta function is used in the Fourier transformation to represent an impulse at the origin, making it a useful tool for analyzing signals with sharp changes at specific points.

What is the Fourier transform of the Dirac Delta function?

The Fourier transform of the Dirac Delta function is a constant value, with a value of 1 at all frequencies. This means that the Dirac Delta function contains equal amounts of all frequencies, making it a useful tool for analyzing signals with a wide frequency spectrum.

What are some applications of the Dirac Delta function and Fourier transformation?

The Dirac Delta function and Fourier transformation are commonly used in engineering and physics to analyze signals and systems with sharp changes at specific points. They are also used in signal processing, image processing, and quantum mechanics.

What are some properties of the Dirac Delta function and Fourier transformation?

The Dirac Delta function and Fourier transformation have several properties, including linearity, time and frequency shifting, and convolution. They also have a duality property, meaning that the Fourier transform of a function is equal to the inverse Fourier transform of its Fourier transform.

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