| 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=== | ||
| − | + | <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 14:31, 6 September 2011
Contents
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.
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.
