The DFT of an image: phase vs. magnitude


Posted by Diego Assencio on 2017.09.01 under Computer science (Digital signal processing)

Let $X$ be an $M \times N$ matrix (a two-dimensional array) of complex numbers $x_{mn}$ for $m = 0, 1, \ldots, M - 1$ and $n = 0, 1, \ldots, N - 1$. The discrete Fourier transform of $X$ is another $M \times N$ matrix $\tilde{X}$ of complex numbers $\tilde{x}_{pq}$ such that: $$ \tilde{x}_{pq} = \sum_{m = 0}^{M - 1} \sum_{n = 0}^{N - 1} x_{mn} e^{-i2\pi pm/M} e^{-i2\pi qn/N} \label{post_81059bc883bd6156c2e83f83e2e38c2b_dft_2d} $$ for $p = 0, 1, \ldots, M - 1$ and $q = 0, 1, \ldots, N - 1$. The original matrix $X$ can be recovered from $\tilde{X}$ through its inverse discrete Fourier transform: $$ x_{mn} = \displaystyle\frac{1}{MN} \sum_{p = 0}^{M - 1} \sum_{q = 0}^{N - 1} \tilde{x}_{pq} e^{i2\pi pm/M} e^{i2\pi qn/N} $$ Since a grayscale image can be represented as a matrix whose elements are 8-bit integers representing pixel values ranging from $0$ (black) to $255$ (white), we can compute its DFT using equation \eqref{post_81059bc883bd6156c2e83f83e2e38c2b_dft_2d}. Interestingly, the magnitude of the DFT of such an image does not contain much information, but its phase does. To clarify what this means, let us define two $M \times N$ matrices $\tilde{X}_{\textrm{mag}}$ and $\tilde{X}_{\textrm{phase}}$ whose elements are given by: $$ \begin{eqnarray} (\tilde{X}_{\textrm{mag}})_{pq} &=& \left|\tilde{x}_{pq}\right| \label{post_81059bc883bd6156c2e83f83e2e38c2b_M}\\[5pt] (\tilde{X}_{\textrm{phase}})_{pq} &=& \tilde{x}_{pq} / \left|\tilde{x}_{pq}\right| \label{post_81059bc883bd6156c2e83f83e2e38c2b_P} \end{eqnarray} $$ Being a complex number, $\tilde{x}_{pq} = \rho_{pq} e^{i\phi_{pq}}$, where $\rho_{pq}$ is its magnitude and $\phi_{pq}$ is its phase. From equations \eqref{post_81059bc883bd6156c2e83f83e2e38c2b_M} and \eqref{post_81059bc883bd6156c2e83f83e2e38c2b_P}, we have that $(\tilde{X}_{\textrm{mag}})_{pq} = \rho_{pq}$ and $(\tilde{X}_{\textrm{phase}})_{pq} = e^{i\phi_{pq}}$, so the matrices $\tilde{X}_{\textrm{mag}}$ and $\tilde{X}_{\textrm{phase}}$ contain only magnitude and phase information of the elements of $\tilde{X}$ respectively.

If we compute the inverse DFTs of $\tilde{X}_{\textrm{mag}}$ and $\tilde{X}_{\textrm{phase}}$, what type of "images" will we recover? Our original image had only real values, but these inverse DFTs will in general not, so we will consider only their real components. Additionally, the pixel values of the resulting images will not necessarily fall within the closed interval $[0, 255]$, but we can clip the negative values (set them to zero) and rescale the nonnegative values to have them fall within this interval. An example of what the resulting images look like is shown on figure 1 (all figures in this post were generated using the script available here).

Fig. 1: An original image (left) and the images produced by computing the inverse DFTs of the magnitude (center) and phase (right) components of the original image's DFT.

From figure 1, we can see that the inverse DFT of the magnitude matrix $\tilde{X}_{\textrm{mag}}$ produces a nearly black image, but the inverse DFT of the phase matrix $\tilde{X}_{\textrm{phase}}$ shows well-defined contours from the original image (if you cannot see them, try increasing the brightness of your screen or click on the figure to see a larger version of it). This indicates that the phase component of the image's DFT carries more information than the magnitude component. To better see this, we can apply histogram equalization to the images produced from the inverse DFTs. The results, shown in figure 2, serve as further confirmation of our hypothesis.

Fig. 2: An original image (left) and the images produced (after histogram equalization) by computing the inverse DFTs of the magnitude (center) and phase (right) components of the original image's DFT.

Comments

Franziska on Sep 01, 2017:
cute dog

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