{"id":341,"date":"2010-09-22T17:30:14","date_gmt":"2010-09-22T21:30:14","guid":{"rendered":"https:\/\/blogs.mathworks.com\/steve\/2010\/09\/22\/fourier-transforms-vertical-lines-and-horizontal-lines\/"},"modified":"2019-10-29T13:34:57","modified_gmt":"2019-10-29T17:34:57","slug":"fourier-transforms-vertical-lines-and-horizontal-lines","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/steve\/2010\/09\/22\/fourier-transforms-vertical-lines-and-horizontal-lines\/","title":{"rendered":"Fourier transforms, vertical lines, and horizontal lines"},"content":{"rendered":"
\r\n

A reader asked in a blog comment recently why a vertical line (or edge) shows up in the Fourier transform of an image as a\r\n horizontal line. I thought I would try to explain this using the simplest example I could think of.\r\n <\/p>\r\n

I'll start with an image that is a constant black except for a single vertical line through the middle of it.<\/p>

x = zeros(200, 200);\r\nx(:, 100) = 1;\r\nimshow(x)<\/pre> 
Next I'll compute and display the log magnitude of the 2-D Fourier transform.<\/p>
X = fft2(x);\r\nimshow(fftshift(log(abs(X) + 1)), [])<\/pre> 
There's the question in black and white, so to speak: Why is there a horizontal line in the 2-D Fourier transform?<\/p>\r\n   
Although I'm going to avoid equations and other complicated mumbo-jumbo in my answer, I do have to explain a couple of things\r\n      about 1-D and 2-D Fourier transforms.\r\n   <\/p>\r\n   
First, the magnitude of the 1-D Fourier transform an \"impulse sequence\" is a constant. An \"impulse sequence\" is a sequence\r\n      that is nonzero only at one place, like this one:\r\n   <\/p>
x1 = [0 0 1 0 0]<\/pre>
\r\nx1 =\r\n\r\n     0     0     1     0     0\r\n\r\n<\/pre>
The magnitude of the 1-D Fourier transform of x<\/tt> is constant:\r\n   <\/p>
abs(fft(x1))<\/pre>
\r\nans =\r\n\r\n    1.0000    1.0000    1.0000    1.0000    1.0000\r\n\r\n<\/pre>
Second, the magnitude of the 1-D Fourier transform of a constant sequence is an impulse. That is, the Fourier transform is\r\n      nonzero only at one place.\r\n   <\/p>
x2 = [1 1 1 1 1]<\/pre>
\r\nx2 =\r\n\r\n     1     1     1     1     1\r\n\r\n<\/pre>
abs(fft(x2))<\/pre>
\r\nans =\r\n\r\n     5     0     0     0     0\r\n\r\n<\/pre>
Finally, computing the 2-D Fourier transform is mathematically equivalent to computing the 1-D Fourier transform of all the\r\n      rows and then computing the 1-D Fourier transforms of all the columns of the result. The order (row-first or column-first)\r\n      doesn't actually matter.\r\n   <\/p>\r\n   
Now let's go back to x<\/tt>, our 200-by-200 matrix with a vertical line down the middle of it.  Each row of x<\/tt> is an impulse sequence:\r\n   <\/p>
plot(x(50,:))<\/pre> 
So if we compute the Fourier transform of all the rows, they'll all be constant (in magnitude):<\/p>
X_rows = fft(x, [], 2);\r\nplot(abs(X_rows(50, :)))\r\nylim([0 2])<\/pre> 
The next step in computing the 2-D Fourier transform is to compute the 1-D Fourier transforms of the columns of X_rows<\/tt>. But those columns are constant. Here's the 100th column of X_rows<\/tt>:\r\n   <\/p>
plot(abs(X_rows(:, 100)))\r\nylim([0 2])<\/pre> 
As I said above, the Fourier transform of a constant sequence is an impulse. If you stack up all the resulting columns containing\r\n      impulse sequences, the result looks like a horizontal line.\r\n   <\/p>
X = fft(X_rows, [], 1);\r\nimshow(fftshift(log(abs(X) + 1)), [])<\/pre> 
That's a step-by-step computational explanation. Let me leave you with more of a conceptual (that is, hand-wavy) explanation.\r\n      Let's look at the input image and the 2-D Fourier transform side by side.\r\n   <\/p>
subplot(1,2,1)\r\nimshow(x)\r\ntitle('x'<\/span>)\r\nsubplot(1,2,2)\r\nimshow(fftshift(log(abs(X) + 1)), [])\r\ntitle('Log magnitude of fft2(x)'<\/span>)<\/pre> 
You can think in terms of horizontal and vertical cross-sections. Each horizontal cross-section of x<\/tt> is an impulse sequence. The Fourier transform of an impulse sequence is constant, so horizontal cross-sections of the Fourier\r\n      transform are constant.\r\n   <\/p>\r\n   
Similarly, each vertical cross-section of x<\/tt> is a constant sequence. The Fourier transform of a constant sequence is an impulse sequence, so the vertical cross-sections\r\n      of the Fourier transform are impulses.  The impulses all line up each other, resulting in the appearance of a horizontal line.\r\n   <\/p>\r\n   
Do these explanations work for you?  Do you know of another way to think about it that makes a better explanation?  Post your\r\n      thoughts as comments below.\r\n   <\/p>