Stuck on inverse fourier transform pair

In summary: Why shouldn't a function of time have an imaginary part? It all depends on the application. Sometimes you're calculating something that, because of the physics it represents, must be a real number, and in those cases the imaginary parts must indeed cancel out. But you say that you know what the phase means for your application, which (correct me if I misunderstand) means that for your purposes, complex numbers do have a physical meaning. This is not uncommon, actually.
  • #1
DragonPetter
830
1
I have been trying to solve the inverse Fourier transform:

[itex]\int_{-\infty}^{\infty}\left[e^{-j2\pi ft_0}e^{j\theta}\right]e^{j2\pi ft}df[/itex]

I know that the Fourier transform pair says

[itex]e^{-j2\pi ft_0}e^{j\theta} \leftrightarrow \delta(t-t_0)[/itex]

but the extra phase term [itex]e^{j\theta}[/itex] makes it so I can't use this pair. Can I just consider it a constant? If so, then it gives me a weird time based function of an imaginary number.
 
Physics news on Phys.org
  • #2
Yes, [itex]e^{j\theta}[/itex] is independent of f (unless there's something you're not telling us), so you can just pull it out of the integral. I don't understand why you say it gives you "a weird time based function of an imaginary number." It gives you [itex]e^{j\theta}\delta\left(t-t_0\right)[/itex], which is just the delta function multiplied by a phase factor (assuming theta is real), with maybe a factor of [itex]\frac{1}{\sqrt{2\pi}}[/itex] or something like that.
 
  • #3
pmsrw3 said:
Yes, [itex]e^{j\theta}[/itex] is independent of f (unless there's something you're not telling us), so you can just pull it out of the integral. I don't understand why you say it gives you "a weird time based function of an imaginary number." It gives you [itex]e^{j\theta}\delta\left(t-t_0\right)[/itex], which is just the delta function multiplied by a phase factor (assuming theta is real), with maybe a factor of [itex]\frac{1}{\sqrt{2\pi}}[/itex] or something like that.

Yes, but this phase factor has an imaginary component for any angle that is not a whole multiple of 0 or pi. I know what this phase means for my application, but I don't understand how a time function can have an imaginary component. In Euler's formula, there are imaginary components, but they always cancel out to a real value, and in this case they don't.
 
  • #4
DragonPetter said:
Yes, but this phase factor has an imaginary component for any angle that is not a whole multiple of 0 or pi. I know what this phase means for my application, but I don't understand how a time function can have an imaginary component. In Euler's formula, there are imaginary components, but they always cancel out to a real value, and in this case they don't.
Why shouldn't a function of time have an imaginary part? It all depends on the application. Sometimes you're calculating something that, because of the physics it represents, must be a real number, and in those cases the imaginary parts must indeed cancel out. But you say that you know what the phase means for your application, which (correct me if I misunderstand) means that for your purposes, complex numbers do have a physical meaning. This is not uncommon, actually.
 
  • #5


I would suggest checking the properties of the Fourier transform to see if there is a way to handle the extra phase term. It is possible that there is a property or formula that can help simplify the problem. If not, then considering the extra phase term as a constant could be a valid approach, but it may lead to a complex-valued function as you mentioned. In this case, it may be helpful to plot the function to gain a better understanding of its behavior. Additionally, consulting with a colleague or referencing a textbook or online resource may provide further insight and assistance in solving the inverse Fourier transform.
 

Related to Stuck on inverse fourier transform pair

1. What is an inverse Fourier transform pair?

An inverse Fourier transform pair refers to a mathematical relationship between a signal and its frequency spectrum. It allows us to convert a signal from the time domain to the frequency domain and vice versa.

2. How do I perform an inverse Fourier transform?

To perform an inverse Fourier transform, you need to take the complex conjugate of the frequency spectrum of the signal and then apply the Fourier transform formula. This will result in the signal being transformed back to the time domain.

3. What is the significance of an inverse Fourier transform pair?

An inverse Fourier transform pair is significant because it allows us to analyze signals in both the time and frequency domains. This is essential in many scientific fields, such as signal processing, telecommunications, and image processing.

4. Can an inverse Fourier transform pair be applied to any signal?

Yes, an inverse Fourier transform pair can be applied to any signal as long as the signal is finite, and its frequency spectrum is well-defined. This means that the signal must have a definite start and end time, and it cannot contain infinite or undefined values.

5. Is there a difference between an inverse Fourier transform and a regular Fourier transform?

Yes, there is a significant difference between an inverse Fourier transform and a regular Fourier transform. While an inverse Fourier transform converts a signal from the frequency domain to the time domain, a regular Fourier transform does the opposite. Additionally, the formula used for each transform is different.

Similar threads

  • Differential Equations
Replies
4
Views
2K
  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
4
Views
2K
Replies
4
Views
601
  • Differential Equations
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
310
  • Calculus and Beyond Homework Help
Replies
5
Views
616
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top