Fourier Transform of Delta Function: Solving Homework Statement

In summary: That is, if f(x) = delta(x), thenF(p) = integral from -infinity to infinity of delta(x)e^(-ipx)dx = e^(-ip*0) = 1.This follows from the sifting property of the delta function. Hence, if F(p) = 1, then f(x) = delta(x).In summary, the problem statement is to show that \int^{\infty}_{-\infty} e^{-ipt} dt = \delta(t), but this is not actually true. Instead, the Fourier transform of a constant function is a Dirac delta function, and we can interpret the given integral as the inverse Fourier transform of a constant, which leads us to the
  • #1
spaghetti3451
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Homework Statement



Show that [tex]\int^{\infty}_{-\infty} e^{-ipt} dt = \delta(t)[/tex].

Homework Equations





The Attempt at a Solution



I know that I must Fourier transform [tex]\delta(t)[/tex], but not sure how.
 
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  • #2
failexam said:

Homework Statement



Show that [tex]\int^{\infty}_{-\infty} e^{-ipt} dt = \delta(t)[/tex].
You can't. It's not true. The expression on the left is a function of p only, the expression on the right is a function of t only. Now, what is the problem really?

Homework Equations





The Attempt at a Solution



I know that I must Fourier transform [tex]\delta(t)[/tex], but not sure how.
 
  • #3
Ok. My mistake: the delta function should be a function of p, not t.
 
  • #4
failexam said:

Homework Statement



Show that [tex]\int^{\infty}_{-\infty} e^{-ipt} dt = \delta(t)[/tex].

The problem is that this isn't actually true. The integral on the left side doesn't exist.

What is true is that the constant function 1 has a Fourier transform (in the sense of distributions) equal to a Dirac delta function, possibly with a [itex]2 \pi[/itex] scale factor. However, you need quite a lot of mathematical machinery even to talk about what "in the sense of distributions" means.

If this is for an engineering course where these technicalities don't matter, then argue as follows:

1. The Fourier transform of a Dirac delta function is a constant
2. Therefore the inverse Fourier transform of a constant must be a Dirac delta function
3. Interpret the integral above as the inverse Fourier transform of a constant.

You should be able to prove 1. quite easily using the sampling/sifting property of the delta function. For 2 and 3 you will have to wave your hands a lot, but that's what engineering textbooks do in this case, too.
 
  • #5
Might I suggest simply placing the delta function in place of f(x) in the formula for the Fourier Transform?
 

1. What is the Fourier transform of a delta function?

The Fourier transform of a delta function, denoted as δ(x), is a mathematical operation that converts a time-domain signal into its equivalent frequency-domain representation. It is defined as the integral of the signal multiplied by the complex exponential function e^-iωt, where ω is the frequency variable.

2. How is the Fourier transform of a delta function related to the Dirac delta function?

The Fourier transform of a delta function is closely related to the Dirac delta function, which is a special type of function that has a value of zero everywhere except at one point, where it has an infinite value. The Dirac delta function can be represented as a limit of a sequence of delta functions, and its Fourier transform is a constant function equal to 1.

3. How is the Fourier transform of a delta function used in signal processing?

The Fourier transform of a delta function is commonly used in signal processing to analyze and manipulate signals in the frequency domain. It allows us to decompose a signal into its constituent frequencies, making it easier to identify important features and remove noise. It is also used in the convolution theorem, which relates the Fourier transforms of two signals to the Fourier transform of their convolution.

4. Can the Fourier transform of a delta function be calculated analytically?

Yes, the Fourier transform of a delta function can be calculated analytically using the formula δ(x) = 1/(2π) ∫ f(ω)e^iωx dω, where f(ω) is the Fourier transform of the original signal. However, for more complex signals, numerical methods may be used to approximate the Fourier transform.

5. How does the Fourier transform of a delta function relate to the sampling theorem?

The sampling theorem states that the sampling rate of a signal must be at least twice the highest frequency present in the signal in order to accurately reconstruct the original signal. The Fourier transform of a delta function can be used to prove this theorem, as it shows that a discrete signal in the time domain corresponds to a continuous spectrum in the frequency domain.

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