Fourier Transform of rect and time shifting.

May 2010
1
0
The transform is simple, but I am sure I am missing something in the end.

f(t) =
0 for [-inf --> 2]
8 for [2 --> 6]
0 for [6 --> +inf]

If i just put a standart Fourier transform equation into it and put limits for integral as 2 (for lower) and 6 (for higher) I come up with two e^-j terms in the end and I cant use any of the euler's equations 'Relationship to trigonometry' (wiki reference) quations to simplify it.

I was thinking to shift rect function in time so i would have -2 and 2 as for integral limits, that would help, or if i could make one of the terms disappear by putting integral limits as 0 and 4.
 

Ackbeet

MHF Hall of Honor
Jun 2010
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Before or After...

I'm guessing you're an electrical engineering major, what with using \(\displaystyle j=\sqrt{-1}\). I'll use the same, to avoid confusion. You might be able to simplify the expression either before or after the integral. Certainly, if you shift the limits of the integral, that ends up as multiplying by an exponential. But, going through with the integral as you've stated it (here I'll ignore any multipliers out front, since there are loads of different conventions - I prefer the symmetric factor myself), then you get

\(\displaystyle \hat{f}(x)=\int_{2}^{6}8\,e^{j\omega x}\,dx=\frac{8}{j\omega}(e^{j\omega x})|_{2}^{6}=\frac{8}{j\omega}(e^{6j\omega}-e^{2j\omega})\).

It sounds like this is as far as you got. There is a little trick I learned for my thesis: create symmetry where there isn't any before. You'd like the exponentials to be equal in magnitude, but opposite in sign. Then you could use some nice formulas for converting to trig functions. So, multiply top and bottom by what you need:

\(\displaystyle \frac{8}{j\omega}(e^{6j\omega}-e^{2j\omega})
=\frac{8e^{-4j\omega}}{j\omega e^{-4j\omega}}(e^{6j\omega}-e^{2j\omega})
=\frac{8}{j\omega e^{-4j\omega}}(e^{2j\omega}-e^{-2j\omega})\).
In turn, this equals
\(\displaystyle \frac{8e^{4j\omega}}{j\omega}(e^{2j\omega}-e^{-2j\omega})\).

Now use the identities to simplify further. As I hinted earlier, this procedure is entirely analogous to time-shifting before taking the transform. The result is the same. I would not, however, put one of the limits at 0, because exponentials go to 1 there; they do not disappear.