Calculating an image's fourier spectrum by hand?
Your particular case can be solved without a calculator. Fourier transforms are linear operations. You can decompose your input as:

$$3 \left( \begin{array}{cccc} 0 &amp; 0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; 1 \end{array} \right) + \left( \begin{array}{cccc} 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 \end{array} \right) $$

The first matrix is constant on the Y axis, so its 2D FFT is only the 1D-FFT of one row, times 4, and 0 everywhere else. The 1D FFT of $(0, 0, 1, 1)$ is $(2, -1+j, 0, -1-j)$, so this gives you, for the first term:

$$3 \times 4 \times \left( \begin{array}{cccc} 2 &amp; -1+j &amp; 0 &amp; -1-j \\
0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0 \end{array} \right)$$

As for the second term, there's only one value so there's only one component in the sum - it's just one of the basis functions, the checkerboard pattern:

$$\left( \begin{array}{cccc} 1 &amp; -j &amp; -1 &amp; j \\
-j &amp; -1 &amp; j &amp; 1 \\
-1 &amp; j &amp; 1 &amp; -j \\
j &amp; 1 &amp; -j &amp; -1 \end{array} \right)$$

Summing these two, you get:

$$\left( \begin{array}{cccc} 25 &amp; -12+11j &amp; -1 &amp; -12-11j \\
-j &amp; -1 &amp; j &amp; 1 \\
-1 &amp; j &amp; 1 &amp; -j \\
j &amp; 1 &amp; -j &amp; -1 \end{array} \right)$$

Note that this differs from Paul R's answer because of the normalization and sign conventions nobody agrees with... Divide by 4 and take the conjugate and you'll find the same result as his.