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===Answer 1=== | ===Answer 1=== | ||
| − | + | 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} | ||
| + | =\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 15:49, 3 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} $
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.
