1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physically what is phase of an Image?

  1. Apr 17, 2015 #1
    Hello everyone,I want to ask very basic question related to
    multidimentional signals like an
    image or a video signal.

    Physically what is phase of an Image? Also what is its physical significance of phase compared to magnitude of an image?
  2. jcsd
  3. Apr 17, 2015 #2


    User Avatar
    Gold Member

    In what connection?
    The Fourier-transform of an image?
  4. Apr 18, 2015 #3
    yes you are right.
  5. Apr 18, 2015 #4


    User Avatar
    Gold Member

    Say you have an image which simply repesents a dot exactly in the center of the image ( black background, white dot). You at first fourier-transform this image row by row (horizontally). All the rows, but the centerrow, will be black ( the coefficients to all harmonics will be zero ). But the centerrow will contain a pulse, which will cause that the absolute value of all the (complex) coefficients to the harmonics in this row will be same. As such you know it's a pulse (dot), but you don't know its location.

    Now you fourier-transform this first "horizontal" fourier-transform vertically (column by column), and all these vertical fourier-transforms will "see the horizontal line" having the same absolute coefficients. Again, by this vertikal transform, a pulse is seen in every column, which will cause that the absolute value of all the (complex) coefficients to the harmonics in all columns will be same. As such you know it's a pulse (line), but you don't know where it is located.

    If you move this original dot in the image, the absolute values ( amplitude/power ) of all the coefficients in the whole image will be the same, but the phases of the complex coefficients to the harmonics will change, so you know the image represents a dot, and by inspection of the phases of these complex coefficients, you know where this dot is located in the image, and you can reconstruct the original image by inverse fourier-transform.

    Rough: As for a dot, the absolute amplitudes tells you what it is ( how it looks ), the phases tells you where it is located. As for more complicated images, you can look at them as a sum of dots.
    Last edited: Apr 18, 2015
  6. Apr 19, 2015 #5
    @Hesch Thank very much .But it's very hard to understand the above concept without any image. so if possible can you add few images so that it can be very easy to understand it.
  7. Apr 19, 2015 #6
    also it's very hard to understand concepts like local phase and global phase,calculation of local phase at each frequency of an image
  8. Apr 19, 2015 #7


    User Avatar
    Gold Member

    You cannot make an "image" of the fourier-transform of a spatial image, but you can make a plot, showing the power of (absolute value of the coefficients) in the transform. Below two images of "Guffy" along with a plot of their transforms. Near the center of the plot are shown the power of lower harmonics, and near the edges are shown the power of higher harmonics. Bright dots means high power, dark dots means low power. The left image is sharp, and in the left plot you see some high power near the edges. Now someone has dampened (by computer) the powers in the right plot (has dampened the higher harmonics/frequencies in the image). The right image is a result of an inverse transform of the right plot, and is blurred due to the lack of high frequencies. It's as listening to music with no treble.
    Likewise, if you have a blurred photo, you can sharpen it by enhancement of higher frequencies (more treble).
    Other purposes: Dynamic contrast enhancement of images and recognition of patterns.


    Another example:


    More ingenious use of fourier-transform: Say you have a photo (A) of a car passing by. As you have used a long shutter-time the photo has been blurred, so you cannot read the registry-number. But maybe you can spot a stretched reflection on the car, and thereby determine the exact length and direction of the blur. You now draw a picture (B) of a dot, and draw picture (C) of a line (blurred dot) with the same length and direction as the blurred reflection. Fourier-transform A, B, C into Af, Bf, Cf. Then you calculate: Df = Af*Bf/Cf, and calculate D = inverse fourier-transform of Df. You can now read the registration-number in D.

    I don't know what a local phase is.
    Last edited: Apr 19, 2015
  9. Apr 21, 2015 #8
    Thanks to all. But again i am disturbing you. If i want to find phase at each frequency or coordinate of image , mathematically what should i do ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook