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<math> \mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt
<math> \mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt
\\ =  \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]

Revision as of 17:36, 6 September 2011

Continuous-time Fourier transform computation (in terms of frequency f in hertz)

Compute the Continuous-time Fourier transform of the two following functions:

$ x(t)= \text{rect}(t) = \left\{ \begin{array}{ll} 1, & \text{ if } |t|<\frac{1}{2}\\ 0, & \text{ else} \end{array} \right. $

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} $

Justify your answer.

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Answer 1

Fourier Transform of rect(t):

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dx =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dx =\frac{e^{-j2\pi ft}}{-j2\pi f} $ from t=-1/2 to t=1/2

$ =\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f} =\frac{sin(\pi f)}{\pi f} $

Instructor's comments: Technically, you should look at the case f=0 separately, because your solution involves a division by f. -pm

Fourier Transform of $ \frac{sin(\pi t)}{\pi t} $:

Guess: $ X(f)=rect(t) $


$ x(t)=\int_{-\infty}^{\infty} X(f)e^{j2\pi ft} df =\int_{\frac{-1}{2}}^{\frac{1}{2}} e^{j2\pi ft} df =\frac{e^{j2\pi ft}}{j2\pi t} $ from f=-1/2 to f=1/2

$ =\frac{e^{j\pi t}-e^{j\pi t}}{j2\pi t} =\frac{sin(\pi t)}{\pi t} $

Instructor's comments: Guessing the answer and proving it using the inverse Fourier transform is a good trick. One could also obtain this Fourier transform using the duality property and your previous answer. -pm

Answer 2

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt =\int_{-\infty}^{\infty} rect(t)e^{-j2\pi ft} dt =\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

$ = -\frac{e^{-j2\pi ft}}{-j2\pi f} $ integrating from -0.5 to +0.5. $ = \frac{e^{-j\pi f} - e^{j\pi f}}{-j2\pi f} = \frac{sin(\pi f)}{\pi f} $

For y(t), we know that

$ y(t) = \int_{-\infty}^{\infty} Y(f)e^{j2\pi ft} df $

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t} $

For the above equation to be true,

$ Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t} $

Answer 3

$ X(f)=\int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt = \int_{-1/2}^{1/2} e^{-j2\pi ft} dt = \frac{e^-j2\pi ft}{j2 \pi f},where x from-1/2 to 1/2 =sinc(f) $ $ use duality, Y(f)=rect(x) $

Answer 4

Using Euler's equation, we know

$ y(t)= \frac{ \sin ( \pi t )}{\pi t} = \frac{e^{j\pi t} - e^{-j\pi t}}{j2\pi t} $

Do Fourier transform one component at a time,

$ \mathcal{F}[\text{rect}(t)] = \int_{-\infty}^{\infty} \text{rect}(t)e^{-j2\pi ft} dt \\ = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-j2\pi ft} dt $

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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