Fourier Transform in the Form of Dirac-Delta Function

In summary, the Fourier transform of x(t)=8cos(70\pi t)+4sin(132\pi t)+8cos(24\pi t) is X(f)=4\pi (\delta (\omega +70\pi)+\delta (\omega -70\pi))+\frac{2\pi}{j} (\delta (\omega +132\pi)-\delta (\omega -132\pi))+4\pi (\delta (\omega +24\pi)+\delta (\omega -24\pi)).
  • #1
Captain1024
45
2

Homework Statement


Given [itex]x(t)=8cos(70\pi t)+4sin(132\pi t)+8cos(24\pi t)[/itex], find the Fourier transform [itex]X(f)[/itex] in the form of [itex]\delta[/itex] function.

Homework Equations


[itex]X(f)=\int ^{\infty}_{-\infty}x(t)e^{-j\omega _0t}dt[/itex]
[itex]cos(\omega t)=\frac{e^{j\omega t}+e^{-j\omega t}}{2}[/itex]
[itex]sin(\omega t)=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}[/itex]
[itex]\int ^{\infty}_{-\infty}cos(\omega _0t)e^{-j\omega t}dt=\frac{\pi}{2}(\delta (\omega +\omega _0)+\delta (\omega -\omega _0))[/itex]
[itex]\int ^{\infty}_{-\infty}sin(\omega _0t)e^{-j\omega t}dt=\frac{\pi}{j2}(\delta (\omega +\omega _0)-\delta (\omega -\omega _0))[/itex]

The Attempt at a Solution


[itex]X(f)=\frac{8\pi}{2}(\delta (\omega +70\pi)+\delta (\omega -70\pi))+\frac{4\pi}{j2}(\delta (\omega +132\pi)-\delta (\omega -132\pi))+\frac{8\pi}{2}(\delta (\omega +24\pi)+\delta (\omega -24\pi))[/itex]

Simplifying: [itex]X(f)=4\pi (\delta (\omega +70\pi)+\delta (\omega -70\pi))+\frac{2\pi}{j} (\delta (\omega +132\pi)-\delta (\omega -132\pi))+4\pi (\delta (\omega +24\pi)+\delta (\omega -24\pi))[/itex]
 
Physics news on Phys.org
  • #2
All clear, Captain!
(You could changew +1/2j to -j/2 but that would be quibbling!)
Nice work.
 

Related to Fourier Transform in the Form of Dirac-Delta Function

1. What is a Fourier Transform in the Form of Dirac-Delta Function?

A Fourier Transform in the form of Dirac-Delta function is a mathematical tool used to analyze the frequency components of a signal or function. It represents the signal in terms of a combination of sinusoidal functions with different frequencies, amplitudes, and phases. The Dirac-Delta function is a special mathematical function that is used to describe impulses or sudden changes in a signal.

2. How is the Fourier Transform in the Form of Dirac-Delta Function different from a regular Fourier Transform?

The Fourier Transform in the form of Dirac-Delta function is a generalized version of the regular Fourier Transform. It allows us to analyze signals with discontinuities or impulses, which cannot be analyzed using the regular Fourier Transform. The Dirac-Delta function acts as a weight to the different frequency components, giving more importance to the frequencies close to the impulse or discontinuity.

3. What are the applications of the Fourier Transform in the Form of Dirac-Delta Function?

The Fourier Transform in the form of Dirac-Delta function has various applications in signal processing, image and audio compression, and data analysis. It is used to analyze signals with sudden changes or impulses, such as in communication systems and medical imaging. It is also used in solving differential equations and in the study of quantum mechanics.

4. How is the Fourier Transform in the Form of Dirac-Delta Function calculated?

The Fourier Transform in the form of Dirac-Delta function is calculated using the formula F(f(t)) = ∫f(t)e^(-j2πft)dt, where f(t) is the original signal and F(f(t)) is the Fourier Transform. The Dirac-Delta function is represented as δ(t) in the formula, and it acts as a weighting function to the different frequency components of the signal.

5. Are there any limitations or drawbacks of using the Fourier Transform in the Form of Dirac-Delta Function?

One limitation of using the Fourier Transform in the form of Dirac-Delta function is that it cannot be used for signals that are not time-limited, meaning signals that extend to infinity. Also, if the signal contains multiple impulses or discontinuities, the Fourier Transform may become complicated and difficult to interpret. Additionally, the use of the Dirac-Delta function may introduce numerical errors in the calculation of the Fourier Transform.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
350
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Replies
4
Views
640
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
8K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
912
Replies
1
Views
1K
  • Topology and Analysis
Replies
7
Views
2K
Back
Top