Fourier Transform Homework: Solving Integrals

In summary, Fourier Transform is a mathematical operation used to decompose a function or signal into its constituent frequencies, providing insights into the underlying components of a complex signal. It can also be used to solve integrals and has various applications in fields such as signal processing and physics. However, it has limitations such as only being applicable to square-integrable functions and may not accurately capture highly complex signals with rapid changes.
  • #1
Nusc
760
2

Homework Statement


[tex]
\begin{subequations}
\begin{eqnarray}
\dot{\hat{{\cal E}}}(t) &=& -\kappa \hat{{\cal E}}(t) + i g\int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta)\, \Bigg\{
e^{-(\gamma + i\Delta)(t-t_{0})}\hat{\sigma}_{ge}(t_{0},\Delta)+ ig\int_{t_{0}}^{t} d t' \hat{{\cal E}}(t')e^{-(\gamma + i\Delta)(t-t')} \Bigg\}
\nonumber\\
& & + \sqrt{2\kappa}\, \hat{{\cal E}}_{in},
\\
&=& -\kappa \hat{{\cal E}}(t) + i g\int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta)\, \hat{\sigma}_{ge,0}(t_{0},\Delta)e^{-(\gamma +i\Delta)(t-t_{0})}
\nonumber\\
& & -g^{2} \int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta)\, \int_{t_{0}}^{t} d t' \hat{{\cal E}}(t')e^{-(\gamma + i\Delta)(t-t')} + \sqrt{2\kappa}\, \hat{{\cal E}}_{in},
\\
\nonumber
\end{eqnarray}
\end{subequations}
[/tex]

I need to find the Fourier Transform of these integrals.

Homework Equations

When looking at this expression, the integrals on the right are evaluated first then proceed to the left.

The Attempt at a Solution


[tex]
After applying the Fourier transform to the integral, we obtain:
\begin{subequations}
\begin{eqnarray}
i\omega \tilde{\hat{{\cal E}}}(\omega)
&=& -\kappa \tilde{\hat{{\cal E}}}(\omega) + i g\int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta) \int_{-\infty}^{\infty} dt e^{-i\omega t} \hat{\sigma}_{ge,0}(t_{0},\Delta)e^{-(\gamma +i\Delta)(t-t_{0})}
\nonumber\\
& & -\frac{g^{2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta) \int_{-\infty}^{\infty} dt e^{-i\omega t} \int_{t_{0}}^{t} d t' \hat{{\cal E}}(t')e^{-(\gamma + i\Delta)(t-t')}
\nonumber\\
& & + \sqrt{2\kappa}\, \tilde{\hat{{\cal E}}}_{in}(\omega),
\\
\nonumber
&=& -\kappa \tilde{\hat{{\cal E}}}(\omega) + i g\int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta) \int_{-\infty}^{\infty} dt e^{-i\omega t} \hat{\sigma}_{ge,0}(t_{0},\Delta)e^{-(\gamma +i\Delta)(t-t_{0})}
\nonumber\\
& & -\frac{g^{2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta) \int_{-\infty}^{\infty} dt e^{-i\omega t} \int_{t_{0}}^{t} d t' \hat{{\cal E}}(t')e^{-(\gamma + i\Delta)(t-t')}
\nonumber\\
& & + \sqrt{2\kappa}\, \tilde{\hat{{\cal E}}}_{in}(\omega),
\\
\nonumber
\end{eqnarray}
\end{subequations}
[/tex]
 
Physics news on Phys.org
  • #2
We can now use the following identity to simplify the integrals:\begin{equation}\int_{-\infty}^\infty e^{-i\omega t}e^{-(\gamma + i\Delta)(t-t_0)}dt= \frac{1}{\gamma + i (\Delta - \omega)}e^{-(\gamma + i\Delta)(t - t_0)}.\end{equation}The resulting expression is now:\begin{subequations}\begin{eqnarray}i\omega \tilde{\hat{{\cal E}}}(\omega) &=& -\kappa \tilde{\hat{{\cal E}}}(\omega) + i g\int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta)\frac{1}{\gamma + i (\Delta - \omega)}\, \hat{\sigma}_{ge,0}(t_{0},\Delta)e^{-(\gamma +i\Delta)(t-t_{0})} \nonumber\\ & & -\frac{g^{2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} d \Delta\; {\cal \rho}(\Delta) \frac{1}{\gamma + i (\Delta - \omega)} \int_{-\infty}^{\infty} dt e^{-i\omega t} \int_{t_{0}}^{t} d t' \hat{{\cal E}}(t')e^{-(\gamma + i\Delta)(t-t')} \nonumber\\& & + \sqrt{2\kappa}\, \tilde{\hat{{\cal E}}}_{in}(\omega), \\\nonumber \end{eqnarray}\end{subequations}The Fourier transform of the integrals is now in a much simpler form.
 

Related to Fourier Transform Homework: Solving Integrals

1. What is a Fourier Transform?

A Fourier Transform is a mathematical operation that decomposes a function or signal into its constituent frequencies. It converts a time-domain signal into a frequency-domain representation, which can be useful for understanding the underlying components of a complex signal.

2. How is a Fourier Transform used in solving integrals?

A Fourier Transform can be used to solve integrals by converting the integral into a simpler form in the frequency domain. This can make the integral easier to solve and can also provide additional insights into the underlying components of the function being integrated.

3. What is the difference between a Fourier Transform and a Laplace Transform?

While both Fourier Transform and Laplace Transform are used to analyze functions in the frequency domain, they differ in the types of functions they can handle. Fourier Transform is used for periodic and non-periodic signals, while Laplace Transform is used for more general functions that are not necessarily periodic. Additionally, Laplace Transform takes into account the initial conditions of a system, while Fourier Transform does not.

4. What are some common applications of Fourier Transform?

Fourier Transform has a wide range of applications in various fields such as signal processing, image processing, acoustics, and physics. It is commonly used in audio and video compression techniques, noise reduction, spectral analysis, and solving differential equations.

5. Are there any limitations or drawbacks to using Fourier Transform?

While Fourier Transform is a powerful mathematical tool, it does have some limitations. It can only be applied to functions that are square-integrable, meaning that they have a finite energy. Additionally, it may not be suitable for highly complex signals with rapid changes, as it may not accurately capture all the details of the signal. In such cases, other techniques such as Wavelet Transform may be more suitable.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
900
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
889
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
31
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Replies
4
Views
623
Back
Top