Fourier transform of signum function*exponential

In summary, the conversation discusses the calculation of the Fourier transform of a function using properties and theorems related to Fourier transforms. The attempt to reduce the boundaries of the integral using the signum function is not successful due to the exponential term in the function. The suggestion is to use the convolution theorem or to work through the calculation step by step.
  • #1
quebecois22
23
0

Homework Statement



Using properties of the Fourier transform, calculate the Fourier transform of: sgn(x)*e^(-a*abs(x-2))

Homework Equations



FT(f(x))= integral from -∞ to +∞ of f(x)*e^(-iwx) dx

The Attempt at a Solution



I've realized that with the signum function, the boundaries of the integral can be reduced to: integral from 0 to 4 of e^(-a*abs(x-2))*e^(-iwx) dx from 0 to 4. However I'm guessing that I just can't use the properties and theorems related to Fourier transforms as the integral does not have the same boundaries as the original... Maybe I just souldn't have changed the integral...Any help?

Sorry for not using latex I really don't understand it =/
 

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  • #2
What was your reasoning that you could change the limits to 0 and 4?
 
  • #3
The function from -∞ to 0 is the opposite of the function from 4 to ∞ so they cancel out. Can I do that?
 
  • #4
That's true for f(x) but not for f(x)e-iωx. The exponential messes things up.
 
  • #5
Ahh indeed :(... How should I approach it then?
 
  • #6
I suggest using the convolution theorem. Or just grind it out — it doesn't look like there's anything tricky going on.
 

FAQ: Fourier transform of signum function*exponential

What is the Fourier transform of the signum function?

The Fourier transform of the signum function is not defined in the traditional sense, as the signum function has infinite discontinuities and does not meet the requirements for integration in the Fourier transform formula. However, it can be defined in the distributional sense, where it is equal to 2/pi divided by the frequency.

How does the Fourier transform of the signum function differ from other functions?

The Fourier transform of the signum function is unique in that it is not defined in the traditional sense and must be defined using distribution theory. This is due to the infinite discontinuities of the signum function, which do not meet the requirements for integration in the Fourier transform formula.

What is the role of the exponential in the Fourier transform of the signum function?

The exponential in the Fourier transform of the signum function serves as a smoothing function, allowing for the Fourier transform to be defined in the distributional sense. It helps to "smooth out" the discontinuities of the signum function and make it integrable in the Fourier transform formula.

Can the Fourier transform of the signum function be visualized graphically?

Yes, the Fourier transform of the signum function can be visualized graphically. The resulting graph will show a sinc function with a peak at the origin and zeros at all other frequencies. However, it is important to note that this graph is only a distributional representation of the Fourier transform, and does not correspond to a traditional Fourier transform in the mathematical sense.

In what applications is the Fourier transform of the signum function useful?

The Fourier transform of the signum function is useful in signal processing and communications, where it can be used to filter out high-frequency components of a signal. It is also used in image processing, particularly in edge detection and image enhancement algorithms. Additionally, it has applications in control systems and circuit analysis.

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