Fourier transform of convolution of sinusoidal signals, or product of distributions (generalized functions)

Course Queries Syllabus Queries 2 years ago

0 2 0 0 0 tuteeHUB earn credit +10 pts

5 Star Rating 1 Rating

Posted on 16 Aug 2022, this text provides information on Syllabus Queries related to Course Queries. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

Take Quiz To Earn Credits!

Turn Your Knowledge into Earnings.

tuteehub_quiz

Answers (2)

Post Answer
profilepic.png
manpreet Tuteehub forum best answer Best Answer 2 years ago

 

I will unashamedly say that this was at least spurred by homework. However I have gone far beyond the syllabus of the course and still can't find an authoritative answer. And it seems an interesting question to me that I doubt the professor will answer (if I wasn't ashamed to ask).

I am asked to calculate the Fourier transform of the convolution of two signals, for generality:

F{sin3(at+b)cos3(ct+d)}F{sin3⁡(at+b)∗cos3⁡(ct+d)}.

I have tried two approaches.

First, take the product of the Fourier transforms of the sinusoids. This leads to an expression that contains terms of the form δ(ωa)δ(ωb)δ(ω−a)δ(ω−b). According to [1] and unless I missed it, the product of two distributions, unlike other operations, is not defined.

Secondly calculate the convolution directly. This leads me to an integral of the form:

sin3(aτ+b)cos3(c(τt)+d)dτ∫−∞∞sin3⁡(aτ+b)cos3⁡(c(τ−t)+d)dτ

This also I think is non-convergent.

So am I right to think that this convolution and its Fourier transform are not defined?

[1] Zemanian: Distribution Theory and Transform Analysis

profilepic.png
manpreet 2 years ago

It is not difficult to find a test function φφ so that

(sincos)(φ)=sinxcostφ(tx)dtdx(sin∗cos)(φ)=∫sin⁡x∫cos⁡tφ(t−x)dtdx
does not exist. Hence sincossin∗cos is not a (tempered) distribution, so you cannot expect the existence of its Fourier Transform. The same is true for the product of its separate Fourier Transforms.

 

There are specific situations in which you can try to extend either the convolution or the multiplication, but for your case I don't expect anything useful.


0 views   0 shares

No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.

tuteehub community

Join Our Community Today

Ready to take your education and career to the next level? Register today and join our growing community of learners and professionals.

tuteehub community