7
$\begingroup$

It has always been said that $F(0)$ is the "DC component" in fourier transform. However, I don't get what it means to say that $F(0)$ is "DC" in the context of image processing.

The zero in this case just meant zero frequency, and hence no changes isn't it? Then in the context of image processing, how should I imagine the picture to be like at $F(0)$? What exactly is that "DC" component?

$\endgroup$
3
  • 3
    $\begingroup$ In the context of an image the "DC" component is just the average (arithmetic mean) of the entire image. $\endgroup$
    – Paul R
    Commented Mar 12, 2013 at 13:58
  • 3
    $\begingroup$ The term "DC" comes from the field of electrical engineering, where "DC" refers to direct current, or current flow that doesn't cycle periodically like alternating current (AC). Thus, DC is stated to have zero frequency, hence the association with the Fourier transform component at $f=0$. $\endgroup$
    – Jason R
    Commented Mar 12, 2013 at 14:17
  • 1
    $\begingroup$ An image that is entirely black has all frequency components 0, including DC. An entirely white image has value 1.0 at DC, and 0 for all other frequencies. A completely gray image has 0.5 at DC, and 0 at all other frequencies. $\endgroup$
    – endolith
    Commented Mar 12, 2013 at 16:26

2 Answers 2

8
$\begingroup$

The zero frequency of an image DCT is the mean gray value of the pixels of the input image (for graylevel images). $F(0)$ is not an image: it is a single coefficient.

$\endgroup$
2
  • $\begingroup$ Isn't that only true for gray-scale images? For color images shouldn't there be three coefficients- one for blue, one for red, and one for yellow? $\endgroup$
    – Jim Clay
    Commented Mar 13, 2013 at 13:13
  • $\begingroup$ Yes, as I wrote in the answer it is valid for grey level images. In the general case of vectorial images, it usually makes sense to take the 2D FFT for each channel (3 in your case), which yields 1 DC coefficient per channel. Vectorial variants may be used, but you have to make sure that they do really have a signification before using them. $\endgroup$
    – sansuiso
    Commented Mar 13, 2013 at 13:20
6
$\begingroup$

In the Fourier series, $x(t) = \sum\limits_{-\infty}^{\infty} C_n e^{jnwt}$ where $C_n =\frac{1}{T} \int\limits_0^T x(\lambda)e^{-jnwt} d\lambda$. $T$ is the period, $w$ the frequency, $j = \sqrt{-1}$.

So if we plug in $n=0$, we get the following: $x(t) = \sum\limits_{-\infty}^{\infty} C_n e^{0}$, with $C_n =\frac{1}{T} \int\limits_0^T x(\lambda)e^{0} d\lambda$. Any number raised to the $0$th power is $1$, so $C_n =\frac{1}{T} \int\limits_0^T x(\lambda) d\lambda$. This is exactly how we define the average value. So in image processing, $F(0)$ corresponds to the average value of all the pixels.

And yes, this is the term that looks at a frequency of 0. It's simply an offset to be added. It's not the whole image, it's just a coefficient. So yeah, I guess it could be the whole image, just a horrible approximation. It would have every pixel equal to the average.

$\endgroup$
1
  • $\begingroup$ Best answer in my opinion. Nice job. $\endgroup$
    – user2718
    Commented Mar 13, 2013 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.