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If the Fourier transform of the function \(f(x, y)\) is \(F(m, n)\), then the Fourier transform of the function \(f(2x, 2y)\) is :

  • A.\(\frac{1}{4}F\left(\frac{m}{2}, \frac{n}{2}\right)\)
  • B.\(\frac{1}{4}F(2m, 2n)\)
  • C.\(\frac{1}{4}F(m, n)\)
  • D.\(\frac{1}{4}F\left(\frac{m}{4}, \frac{n}{4}\right)\)

Answer: A

For the two dimensional Fourier transform, scaling the function in space scales its transform in the opposite way: if f(x, y) has the transform F(m, n), then f(ax, by) has the transform

\[ \frac{1}{|ab|} F\left(\frac{m}{a}, \frac{n}{b}\right) \]

With a = b = 2 this gives \(\frac{1}{4}F(m/2, n/2)\). The option with F(2m, 2n) has scaled the wrong way, the one with F(m, n) has no scaling, and the one with m/4, n/4 has used a factor of 4 for the arguments. The keyed answer is the first option.

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