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I have the following piecewise function:

$$ x(t) = \begin{cases} 1 & |t| \le T_0, \\[6pt] 0 & |t| > T_0. \end{cases} $$

I apologize for the formatting.

I need to compute the Fourier transform of $x(t)$ when $T_0$ is equal to $1$. Then equal for two and I am supposed to see some pattern and be able to discuss it. The issue is I have absolutely no idea where to start. We briefly touched on these and never touched how to solve them. I have some equations but nothing clear enough for me to really make any progress alone. I'd like to solve this alone so I can make sure I grasp it but I just desperately need a push in the right direction.

Thank you in advance, I appreciate any advice.

EDIT: Thank you Michael and Rob. You really helped me out here. It all makes so much sense now.

Joe
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2 Answers2

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$$ \hat x(f) = \int_{-\infty}^\infty e^{-ift} x(t) \, dt = \int_{-1}^1 e^{-ift}\cdot1 \, dt. $$

If I understand correctly, you need to do the same with $\displaystyle\int_{-2}^2$.

Michael Hardy
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  • I really appreciate it. I suppose I just have to calculate it with all the constants still in there correct? – Joe Sep 26 '13 at 22:08
  • I'm not sure what your question in the comment means. You need to find $\displaystyle\int_{-1}^1 e^{-ift},dt$. – Michael Hardy Sep 26 '13 at 22:10
  • I guess I was referring to the fact that I don't know $i$ or $f$. I must be missing something. – Joe Sep 26 '13 at 22:43
  • The number $i$ is the imaginary square root of $-1$. If you don't know about that, you're not going to understand Fourier transforms. The number $f$ is the argument to the function that is the Fourier transform. It could be any number. The answer should be a function of $f$. I used the letter $f$ on the theory that $t$ could stand for "time" and thus $f$ would stand for "frequency". Multiplying time by frequency gives a dimensionless number. – Michael Hardy Sep 26 '13 at 23:03
  • Thank you. I was just so worked up over these few questions that I had no direction on that I did not actually look and realize what was right in front of me. – Joe Sep 26 '13 at 23:29
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Hopefully, this is not too much of a hint: $$ \begin{align} \hat{x}_1(\tau)&=\int_{-\infty}^\infty x_1(t)\,e^{-2\pi it\tau}\,\mathrm{d}t\\ &=\int_{-1}^1e^{-2\pi it\tau}\,\mathrm{d}t\\ &=\left.\frac1{-2\pi i\tau}e^{-2\pi it\tau}\,\right]_{-1}^{\hphantom{+}1} \end{align} $$ A useful formula: $$ \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i} $$

robjohn
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  • Note that my answer and Michael Hardy's differ in the normalization factor. I feel that the formulas work out nicer using $e^{-2\pi it\tau}$ as the character for the Fourier Transform, but almost any non-zero, real constant will work in place of $2\pi$. – robjohn Sep 26 '13 at 22:31
  • I am still somewhat confused. There are a number of constants such as $i$ that I am unsure how to calculate. Should I forge ahead with this and just leave the constants or am I missing something? – Joe Sep 26 '13 at 22:37
  • @Joe: $i$ is such that $i^2=-1$. In this computation, it can be treated simply as a constant. – robjohn Sep 26 '13 at 22:44
  • Alright, that makes sense. I'm sorry to be nit picking through everything like someone who has never done an integral but I want to make sure I understand what everything is and where it is coming from.

    So that leads me to this, what is $τ$?

     – Joe Sep 26 '13 at 22:49
  • @Joe: what are you using as the definition of the Fourier Transform? The Fourier Transform takes a function to another function. The destination function has an argument, and I am calling that $\tau$. I have now written out the Fourier Transform of $x_1(t)$ which is $\hat{x}_1(\tau)$ (my $\tau$ plays the same role as Michael Hardy's $f$). – robjohn Sep 26 '13 at 22:52
  • I guess I really wasn't using anything as concrete definition. For whatever reason we were assigned this problem and the definition we have of the Fourier Transform is this formula where I was not sure what anything was. 
      Oh! I think I understand, since you still need to keep the function as function you need $τ$ in order to actually use it once you transform it.
    – Joe Sep 26 '13 at 23:03