Solution to Q1 of Week 14 Quiz Pool
Using the definition of the CSFT,
$ \begin{align} F(u,v) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-j2\pi (ux+vy)}dxdy \\ F(u,0) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{-j2\pi (ux)}dxdy \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty}f(x,y) dy \right) e^{-j2\pi ux}dx \\ &= \int_{-\infty}^{\infty} p(x) e^{-j2\pi ux}dx \\ &= P(u) \\ \end{align} $
so F(u,0) is the same as P(u) which is the CTFT of the function p(x).
Credit: Prof. Bouman
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