Line 47: Line 47:
 
:<span style="color:green">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 </span>
 
:<span style="color:green">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 </span>
 
===Answer 2===
 
===Answer 2===
Write it here.
+
<math> 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
 +
</math>
 +
 
 +
<math> = -\frac{e^{-j2\pi ft}}{-j2\pi f}
 +
</math>
 +
integrating from -0.5 to +0.5.
 +
<math> = \frac{e^{-j\pi f} - e^{j\pi f}}{-j2\pi f}
 +
      = \frac{sin(\pi f)}{\pi f}
 +
</math>
 +
 
 +
 
 +
For y(t), we know that
 +
 
 +
<math> y(t) = \int_{-\infty}^{\infty} Y(f)e^{j2\pi ft} df
 +
</math>
 +
 +
<math> y(t)= \frac{ \sin ( \pi t )}{\pi t}
 +
      = \frac{e^{-j\pi t} - e^{j\pi t}}{\pi t}  </math>
 +
 
 +
For the above equation to be true,
 +
 
 +
<math> Y(f) = \frac{\delta(f - \frac{1}{2})}{\pi t} - \frac{\delta(f + \frac{1}{2})}{\pi t}
 +
</math>
 +
 
 
===Answer 3===
 
===Answer 3===
 
write it here.
 
write it here.
 
----
 
----
 
[[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 15:31, 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.


Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


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) $

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} $

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

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