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Yuanjun Wang | Yuanjun Wang | ||
| − | Below are CSFT of | + | Below are CSFT of 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. <math>f(x,y)=\frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} </math> | 1. <math>f(x,y)=\frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} </math> | ||
| Line 11: | Line 11: | ||
prove: | prove: | ||
<math> F^{-1}(u,v) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} rect(u) rect(v) e^{j2\pi (ux+vy)} dx dy </math> | <math> F^{-1}(u,v) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} rect(u) rect(v) e^{j2\pi (ux+vy)} dx dy </math> | ||
| + | |||
| + | because we know that <math> rect(u) = \left\{ | ||
| + | \begin{array}{ll} | ||
| + | 1, & \text{ if } |t|<\frac{1}{2}\\ | ||
| + | 0, & \text{ else} | ||
| + | \end{array} | ||
| + | \right. | ||
| + | </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> | ||
| + | |||
| + | <math> = \frac{ sin(\pi x)}{\pi x} \int_{-\frac{1}{2}}^{\frac{1}{2}} rect(v) e^{j2\pi vy} dy </math> | ||
| + | |||
| + | <math> = \frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} </math> | ||
| + | |||
| + | so <math> f(x,y) = \frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} </math> | ||
| + | |||
| + | so CSFT (f(x,y)) = rect(u) rect(v) | ||
Latest revision as of 04:39, 14 December 2013
Prove of the CSFT of the signals
Yuanjun Wang
Below are CSFT of 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 $
$ = \frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} $
so $ f(x,y) = \frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} $
so CSFT (f(x,y)) = rect(u) rect(v)
