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===Answer 1===
 
===Answer 1===
Write it here.
+
Fourier Transform of rect(t):
 +
 
 +
<math> 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} </math> from t=-1/2 to t=1/2
 +
 
 +
<math> =\frac{e^{-j\pi f}-e^{j\pi f}}{-j2\pi f}
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      =\frac{sin(\pi f)}{\pi f} </math>
 +
 
 +
 
 +
Fourier Transform of <math>\frac{sin(\pi t)}{\pi t}</math>:
 +
 
 +
Guess: <math> X(f)=rect(t) </math>
 +
 
 +
Proof:
 +
 
 +
<math> 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} </math> from f=-1/2 to f=1/2
 +
 
 +
<math> =\frac{e^{j\pi t}-e^{j\pi t}}{j2\pi t}
 +
      =\frac{sin(\pi t)}{\pi t} </math>
 +
 
 
===Answer 2===
 
===Answer 2===
 
Write it here.
 
Write it here.

Revision as of 16:49, 3 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} $


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

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

Proof:

$ 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

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett