Can someone help me in a step-by-step derivation for the Fourier Transform of a line ? It appears to be simple but still cannot figure out. I know what is the end result but I am unable to figure out the intermediate steps. I tried to use the Shifting and the Similarity theorems but I am not getting the final result.

We have a line $Ax+By+C=0$ which can be called a line impulse $\delta(Ax+By+C)$ where $\delta$ is the delta function.

We would like to derive the Fourier transform $\mathcal{F}$ of this line impulse which is mentioned in $[1]$ to be:

$$\mathcal{F}(\delta(Ax+By+C))(u,v) = e^{i2 \pi C (Au+Bv)} \delta(-Bu+Av)$$

where $u,v$ are the frequency components in two dimensions.

$[1]$: Line Segment Sampling with Blue-Noise Properties By Xin Sun et al 2013


2 Answers 2


The result seems to be only valid for $A^2 + B^2= 1$.

first approach (brute force):

Let us simply start with the definition $$\mathcal{F}(\delta(Ax+By+C))(u,v) = \int dx\! \int dy\, \delta (A x + B y +C) e^{-2\pi i (x u + y v)}. \tag{F}$$

In a first step, we perform the integration over $y$. Note that $$\delta (A x + By +C) = \delta ( B (y + C/B +A x /B)) = \frac{1}{|B|} \delta(y + C/B +A x /B)$$ (as $\delta(\alpha x) = |\alpha|^{-1} \delta(x)$) and therefore $$\begin{align}\int dy\,\delta (A x + B y +C) e^{-2\pi i (x u + y v)} &= \frac1{|B|} \exp[-2\pi i ( x u -(C/B +A x /B) v)]\\ &= \frac1{|B|} \exp[ -2\pi i x (u - A v/B) + 2\pi i C v/B].\end{align} $$

We are left with the integral over $x$. Using the standard result $$\int\! dx\, e^{-2\pi i x \alpha} = \delta(\alpha),$$ we obtain $$\mathcal{F}(\delta(Ax+By+C))(u,v) = \frac1{|B|} e^{2\pi i C v/B} \delta(u - A v/B).$$ We have $|B|^{-1} \delta(u - A v/B) = \delta(B u - A v)$. Because of the $\delta$-function, we can use $ B u = Av$ in the prefactor. Thus, we can "symmetries" ($g$ is an arbitrary function) $$\frac{1}{|B|} g(v/B) \delta(u- Av /B) = g(v/B) \delta(B u -A v)=g[(v/B)(A^2 +B^2) /(A^2 + B^2)]\delta(B u -A v)=g[(A (Av/B) + B v) /(A^2 + B^2)]\delta(B u -A v) = g[(A u + B v) /(A^2 + B^2)] \delta(B u -A v). $$

We obtain the final result $$ \mathcal{F}(\delta(Ax+By+C))(u,v) = \exp[2\pi i C (A u + B v) /(A^2 + B^2)] \delta(B u - A v). \tag{res}$$

second approach:

To make maximal use of the $\delta$-function, we would like to introduce a new variable $s= Ax + By + C$ in (F). We want to use a second variable whose coordinate axis are perpendicular, so we choose $t= ( Ay -B x)/(A^2+B^2)$. The factor is chosen such as to make the Jacobian for the variable change from $(x,y)$ to $(s,t)$ unity.

A simple calculation shows that $$ x u + y v =(s -C) (A u + B v)/(A^2+B^2) + t (-Bu + Av) . $$ So we have $$\mathcal{F}(\delta(Ax+By+C))(u,v) = \int \!dt \int\! ds\, \delta(s) \exp\{-2 \pi i[(s -C) (A u + B v)/(A^2+B^2) + t (-Bu + Av)] \}.$$

The integral over $s$ is readily performed and we have $$\mathcal{F}(\delta(Ax+By+C))(u,v) = \int \!dt \exp\{2 \pi i C (A u + B v) /(A^2+B^2) - 2\pi i t (-Bu + Av) ] \}.$$

The integral over $t$ leads to a $\delta$-function $$\int\!dt \exp\{-2 \pi i t (-Bu + Av) \} = \delta(-Bu +Av).$$ And we end up with the result (res) quoted above.

  • $\begingroup$ Thanks a lot @Fabian for providing a step by step derivation. I have just one question in the burte-force appraoch. You mentioned that we can use Bu=Av in the prefactor, I don't really understand this step. Can you explain how you symmetrize here ? $\endgroup$
    – sinner
    Jan 12, 2016 at 2:14
  • $\begingroup$ Because of the $\delta$-function, we can replace $Bu$ with $Av$ everywhere in the factor multiplying the $\delta$ function. I added a few steps in the derivation for clarification. $\endgroup$
    – Fabian
    Jan 12, 2016 at 21:19
  • $\begingroup$ Is this result a line ? $\endgroup$
    – haleh
    Apr 17, 2017 at 13:24

Here I'll give an answer for arbitrary dimension.

For $\vec{x},\vec{w}_1\in \mathbb{R}^n$, what is the Fourier transform along the line $\vec{w}_1\cdot\vec{x} + c = 0$? Assume $\vec{w}_1$ is normalized.

Using the dirac delta to define the line in $\mathbb{R}^n$ we therefore must calculate,

\begin{align} \mathcal{F}[\delta(\vec{w}_1\cdot\vec{x} + c)](\vec{\nu}) = \int d\vec{x} \, \delta(\vec{w}_1\cdot\vec{x} + c) \exp{[-2\pi i \vec{x}\cdot\vec{\nu}]},\tag{1} \end{align} where $\vec{\nu}$ are the fourier space variables.

Let $\{\vec{w}_2,...,\vec{w}_n\}$ be the orthonormal basis that spans $\mathrm{Null}[\vec{w}_1]$. You can determine this for any dimension with the Wolfram command NullSpace, e.g. for 3D.

Define a change of variables by, \begin{align} s_1 =& \vec{w}_1\cdot\vec{x} + c\\ s_2 =& \vec{w}_2\cdot\vec{x}\\ &...\\ s_n =& \vec{w}_n\cdot\vec{x}. \end{align}

Written more precisely, \begin{align} \vec{s} =& W \vec{x} + \hat{s}_1 c\\ I =& W W^T. \tag{2} \end{align} The second line comes from the orthonmality of the basis of choosen to span the null space of $\vec{w}_1$. Note: The determinant of an orthogonal matrix satisfying (2) is $1$, and represents a rotation matrix, and therefore the Jacobian of the transform is $1$.

With these new variables equation (1) becomes, \begin{align} \mathcal{F}[\delta(s_1)](\vec{\nu}) = \int d\vec{s} \, \delta(s_1) \exp{[-2\pi i W^T(\vec{s}-\hat{s}_1 c)\cdot\vec{\nu}]}. \end{align} Now we carry out the integral over $s_1$ by using $\vec{s} = \sum_{i=1}^n s_i \hat{s}_i$, \begin{align} \mathcal{F}[\delta(s_1)](\vec{\nu}) =& \int ds_n ... \int ds_1 \, \delta(s_1)\exp{[-2\pi i W^T(\sum_{i=1}^n s_i \hat{s}_i-\hat{s}_1 c)\cdot\vec{\nu}]}\\ =& \exp{[2\pi i c(W^T\hat{s}_1)\cdot\vec{\nu}]} \int ds_n ... \int ds_2 \, \exp{[-2\pi i \sum_{i=2}^n s_i (W^T\hat{s}_i)\cdot\vec{\nu}]}. \end{align} Performing the remaining integrations using the dirac delta definition $\delta(y) = \int dp\, \exp[-2\pi i p y]$, we get, \begin{align} \mathcal{F}[\delta(s_1)](\vec{\nu}) =& \exp{[2\pi i c(W^T\hat{s}_1)\cdot\vec{\nu}]} \prod_{i=2}^n \delta(W^T\hat{s}_i\cdot\vec{\nu}). \end{align}

In order to evaluate $W^T\hat{s}_i$, note the basis vectors in the rotated space are related to the original basis vectors via the same rotation matrix, i.e. $[\hat{s}_1, ..., \hat{s}_n] = W [\hat{x}_1, ..., \hat{x}_n]$. Therefore $W^T\hat{s}_i = \hat{x}_i = \hat{x}_i (\vec{w}_1\wedge\vec{w}_1 + ... + \vec{w}_n\wedge\vec{w}_n)$.


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