# Compute the given Fourier transform by using the given tables

• s3a
In summary, the conversation is about trying to solve a problem involving Fourier transforms using tables, but the person is unsure of how to use the tables and why certain steps are necessary. They mention the possibility of using the Scaling Theorem in 1D and 2D, and ask for clarification on their understanding. They also mention using (i) and (iv) from the tables and replacing variables in the given signal, but are unsure if they are on the right track. Ultimately, they are looking for help in fully understanding the process and reasoning behind it.
s3a
Homework Statement
"Compute the Fourier transform G_B(u,v) of the signal g_B(x,y) = 0.5 circ(x/B,y/B) - 0.5 rect(2x/B,2y/B) + 0.8 circ(4x/B,4y/B) - 0.8 circ(8x/B,8y/B), where B is a constant, using these tables ( https://docdro.id/dPaHriL ) (which is a two-page document)."
Relevant Equations
(i), (iv) and possibly (v) or (vi) from these tables ( https://docdro.id/dPaHriL (which is a two-page document) ).
Hello, everyone. :)

I'm trying to do a certain problem regarding Fourier transforms (but one that's supposedly easy, because of just using tables, rather than fully computing stuff), and I know how to do it, but I don't know why it works. Here's the problem statement.:
"Compute the Fourier transform G_B(u,v) of the signal g_B(x,y) = 0.5 circ(x/B,y/B) - 0.5 rect(2x/B,2y/B) + 0.8 circ(4x/B,4y/B) - 0.8 circ(8x/B,8y/B), where B is a constant, using these tables ( https://docdro.id/dPaHriL ) (which is a two-page document)."

Here ( https://docdro.id/oWMtf7i ) is the answer.

For both the rect and circ (which should probably be ellipse, right?) functions, I think I'm supposed to use (i) and (iv) (from the second page of the tables pdf document), and I get the solution by replacing all occurrences of u and v in each of the four subparts of g_B(x,y), 0.5circ(x/B,y/B), -0.5rect(2x/B,2y/B), 0.8circ(4x/B,4y/B) and -0.8circ(8x/B,8y/B), by u multiplied by [1/(B/denominator)]^(-T) (-T means inverse transpose) = B/denominator and v multiplied by [1/(B/denominator)]^(-T) = B/denominator, respectively, but by multiplying the whole frequency-domain function by 1/det [1/(B/denominator)] = B/denominator twice (as in squared, not times two), instead of once. To try to justify this, I'm thinking one has to perhaps think of the frequency-domain functions as a product of two functions and use (v) or (vi), but if I'm even on the right track, the details of that are not clicking in my mind.

If someone could help me fully understand what's going on, I would GREATLY appreciate it!

s3a said:
but I don't know why it works
Can be read in two ways:
1. you don't know that FTs are linear
2. you don't know why these tables are as they are
To make an inroad other than: 'which is it ?' my question to you is: what do you know about 1D FTs ?

FactChecker
BvU said:
what do you know about 1D FTs ?
@s3a , Specifically, are you familiar with the Scaling Theorem in 1 D? The 2 D version just uses matrices instead of 1-dimensional scaling.

Thank you both for your responses. :)

Before getting the epiphany I just got (in large part because I was thinking of scaling in two dimensions because of focusing more on that aspect because of seeing "scaling theorem" and "1-dimensional scaling" (which made me think of two-dimensional scaling)), I had written what I wrote below (which I'm not removing, just in case), but it appears that the solution to my problem was that I was setting A = [1/(B/denominator)] = 1/(B/denominator), instead of A = [1/(B/denominator),0;0,1/(B/denominator)]. Is that the correct solution to my problem, though, or is it just a fluke?

What I was going to post before my epiphany:
BvU said:
Can be read in two ways:
1. you don't know that FTs are linear
2. you don't know why these tables are as they are
To make an inroad other than: 'which is it ?' my question to you is: what do you know about 1D FTs ?
Well, (i) makes it seem to me that Fourier transforms are linear.

And, I guess I don't know how these tables are as they are because I didn't compute any of them (because our class doesn't focus too much on those mathematical details since we're mostly just expected to know when to call them and their inverse functions in computer code, in addition to how to compute them by hand using the tables I posted). Having said that, I vaguely suspect that it's probably (Eq.1) from here ( https://en.wikipedia.org/wiki/Fourier_transform#Definition ), followed by replacing the part with the natural exponential that involves an imaginary number with a trigonometric equivalent that only involves real numbers by using a certain identity.

FactChecker said:
@s3a , Specifically, are you familiar with the Scaling Theorem in 1 D? The 2 D version just uses matrices instead of 1-dimensional scaling.
I didn't know about that theorem, however if I think about scaling each dimension, then I can see why the the 1/det A is multiplied twice (as in squared, not times two) (since each of the dimensions is scaled by the same amount). Having said that, I guess the real issue is that I'm misunderstanding how to use (iv) somehow, but what is wrong with my thinking?

## 1. What is a Fourier transform?

A Fourier transform is a mathematical tool used to transform a signal from the time domain to the frequency domain. This allows us to analyze the different frequency components present in a signal.

## 2. How is a Fourier transform computed?

A Fourier transform is computed by integrating the signal over all possible frequencies and then expressing the result as a complex number with a magnitude and phase. This complex number is known as a Fourier coefficient.

## 3. Why do we use tables to compute Fourier transforms?

Using tables to compute Fourier transforms allows us to quickly and accurately calculate the Fourier coefficients for a given signal without having to perform the integration manually. These tables contain pre-calculated values for common signals and are based on the properties of the Fourier transform.

## 4. What are the benefits of using a Fourier transform?

The Fourier transform allows us to analyze signals in the frequency domain, which can provide valuable insights into the underlying components and patterns of a signal. It is also a powerful tool for solving differential equations and has numerous applications in engineering, physics, and other fields.

## 5. Are there any limitations to using Fourier transforms?

While the Fourier transform is a versatile and useful tool, it does have limitations. For example, it assumes that the signal is periodic and can only be used on signals that are finite and have a well-defined frequency spectrum. Additionally, it is sensitive to noise and can lead to inaccurate results if the signal is not properly processed beforehand.

• Calculus and Beyond Homework Help
Replies
1
Views
938
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
5K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• MATLAB, Maple, Mathematica, LaTeX
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus
Replies
2
Views
2K