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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} $

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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} $

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} $

Answer 2

Write it here.

Answer 3

write it here.

Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang