| Line 22: | Line 22: | ||
<math> F^{-1}(u,v) = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{j2\pi ux} du e^{j2\pi vy} dy </math> | <math> F^{-1}(u,v) = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{j2\pi ux} du e^{j2\pi vy} dy </math> | ||
| − | <math> = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \frac{e^{j\pi x} - e^{-j\pi x}}{j\pi x} e^{j2\pi vy} dy | + | <math> = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \frac{e^{j\pi x} - e^{-j\pi x}}{j\pi x} e^{j2\pi vy} dy </math> |
| + | |||
| + | <math> = \frac{ sin(\pi x)}{\pi x} \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) e^{j2\pi vy} dy </math> | ||
Revision as of 07:17, 13 December 2013
Prove of the CSFT of the signals
Yuanjun Wang
Below are CSFT of six signals. The general way we solve CSFT questions is to guess its Fourier Transform, then prove it by taking the inverse F.T. of the signals.
1. $ f(x,y)=\frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} $
guess: $ F(u,v) = rect(u) rect(v) $ \\
prove: $ F^{-1}(u,v) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} rect(u) rect(v) e^{j2\pi (ux+vy)} dx dy $
because we know that $ rect(u) = \left\{ \begin{array}{ll} 1, & \text{ if } |t|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $
$ F^{-1}(u,v) = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{j2\pi ux} du e^{j2\pi vy} dy $
$ = \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) \frac{e^{j\pi x} - e^{-j\pi x}}{j\pi x} e^{j2\pi vy} dy $
$ = \frac{ sin(\pi x)}{\pi x} \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) e^{j2\pi vy} dy $
