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=Prove of the CSFT of the signals= | =Prove of the CSFT of the signals= | ||
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| + | Yuanjun Wang | ||
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| + | 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. | ||
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| + | 1. <math>f(x,y)=\frac{ sin(\pi x)}{\pi x} \frac{ sin(\pi y)}{\pi y} </math> | ||
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| + | guess: <math>F(u,v) = rect(u) rect(v)</math> \\ | ||
| + | |||
| + | prove: | ||
| + | <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> | ||
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| + | <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> | ||
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| + | <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 05:39, 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)
