# Fourier transform of the components of a vector

• I
• redtree
In summary, the question asks whether given the Fourier conjugates ##\vec{r}## and ##\vec{k}##, where ##\vec{r} = [r_1,r_2,r_3]## and ##\vec{k} = [k_1,k_2,k_3]##, ##r_1##/##k_1##, ##r_2##/##k_2##, ##r_3##/##k_3## are also Fourier conjugates. The answer is yes, as the coordinates in a multi-dimensional Fourier transform are treated independently. Additionally, for the Fourier transform theory to work, there must be a linear response of the output vector field to the
redtree
Given the Fourier conjugates ##\vec{r}## and ##\vec{k}## where ##\vec{r} = [r_1,r_2,r_3]## and ##\vec{k} = [k_1,k_2,k_3]## , are ##r_1## /##k_1##, ##r_2##/##k_2##, ##r_3##/##k_3## also Fourier conjugates, such that:
##
\begin{split}
f(\vec{r})&=[f_1(r_1),f_2(r_2),f_3(r_3)]
\\
F[f(\vec{r})]&=F[f_1(r_1),f_2(r_2),f_3(r_3)]
\\
F[f_1(r_1),f_2(r_2),f_3(r_3)]&=[\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)]
\\
[\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)]&=\hat{f}(\vec{k})
\end{split}
##

redtree said:
Given the Fourier conjugates ##\vec{r}## and ##\vec{k}## where ##\vec{r} = [r_1,r_2,r_3]## and ##\vec{k} = [k_1,k_2,k_3]## , are ##r_1## /##k_1##, ##r_2##/##k_2##, ##r_3##/##k_3## also Fourier conjugates, such that:
##
\begin{split}
f(\vec{r})&=[f_1(r_1),f_2(r_2),f_3(r_3)]
\\
F[f(\vec{r})]&=F[f_1(r_1),f_2(r_2),f_3(r_3)]
\\
F[f_1(r_1),f_2(r_2),f_3(r_3)]&=[\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)]
\\
[\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)]&=\hat{f}(\vec{k})
\end{split}
##
I believe the answer is yes, that e.g. ## k_x ## and ## x ##, etc. are Fourier conjugates, by themselves, just as ## \omega ## and ## t ## are Fourier conjugates. In the multi-dimensional Fourier transform, the 3 coordinates are treated independently, and the time/frequency is also treated independently for a 4-dimensional F.T. ## \\ ## editing... One additional comment I could add is that for the Fourier transform theory and equations such as the convolution equation to work in F.T. space, there must be a linear response (editing) of the output vector field to the input vector field. e.g. in linear circuit theory, if you write ## \tilde{V}_{out}(\omega)=\tilde{m}(\omega) \tilde{V}_{in}(\omega) ## you must have ## V_{out}(t)=\int m(t-t')V_{in}(t') \, dt' ##. The last integral equation is indicative of a linear temporal response.

Last edited:

## 1. What is Fourier transform of the components of a vector?

The Fourier transform of the components of a vector is a mathematical operation that decomposes a vector into its individual sinusoidal components. It is commonly used in signal processing and image analysis to analyze the frequency content of a signal or image.

## 2. How does Fourier transform of the components of a vector work?

The Fourier transform of the components of a vector involves converting a vector from its time or space domain into its frequency domain. This is done by decomposing the vector into its individual sinusoidal components, each with a specific amplitude and phase. The resulting frequency spectrum provides information about the different frequencies present in the original vector.

## 3. What is the relationship between Fourier transform and vectors?

The Fourier transform of the components of a vector is a mathematical operation that can be applied to any type of vector, including complex vectors. It is used to analyze the frequency content of a vector, which can be beneficial in understanding the behavior and characteristics of the vector.

## 4. What are the applications of Fourier transform of the components of a vector?

The Fourier transform of the components of a vector has various applications in fields such as signal processing, image analysis, and data compression. It is used to analyze and filter signals, identify patterns in images, and extract features from data.

## 5. Are there any limitations to Fourier transform of the components of a vector?

The Fourier transform of the components of a vector has limitations, such as the inability to accurately represent signals with infinite or discontinuous components. It also requires the signal to be sampled at a specific rate, known as the Nyquist rate, in order to avoid aliasing. Additionally, it may not be suitable for analyzing non-stationary signals that change over time.

• General Math
Replies
7
Views
1K
• Calculus
Replies
4
Views
4K
• General Math
Replies
42
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
632
• Differential Geometry
Replies
10
Views
5K
Replies
2
Views
956
Replies
2
Views
1K
Replies
5
Views
2K
• Topology and Analysis
Replies
4
Views
731
• Classical Physics
Replies
3
Views
1K