Efficiently Compute the Fourier Transform of U(t) for Easy Homework Assignment

[Jamin2112]

Homework Statement

Computer the Fourier transform of U(t), where U(t) = 1 for |t| < 1, and U(t) = 0 for |t| > 1.

Homework Equations

Fourier Transform: F(w) = ∫U(t)e^-iwtdt (bounds: ∞, -∞)

The Attempt at a Solution

If |t| < 1, obviously F(w) = 0.

If |t| > 1,
F(w) = (-1/wt)*[cos(-wt) + i sin(-wt)] |^∞ - (-1/wt)*[cos(-wt) + i sin(-wt)] |^-∞.

How do I evaluate that? Obviously lim_t-->∞cos(-wt) and lim_t-->∞sin(-wt) don't exist. Or am I missing something important?

 

[jbunniii]

Homework Statement

Computer the Fourier transform of U(t), where U(t) = 1 for |t| < 1, and U(t) = 0 for |t| > 1.

Homework Equations

Fourier Transform: F(w) = ∫U(t)e^-iwtdt (bounds: ∞, -∞)

The Attempt at a Solution

If |t| < 1, obviously F(w) = 0.

If |t| > 1,
F(w) = (-1/wt)*[cos(-wt) + i sin(-wt)] |^∞ - (-1/wt)*[cos(-wt) + i sin(-wt)] |^-∞.

How do I evaluate that? Obviously lim_t-->∞cos(-wt) and lim_t-->∞sin(-wt) don't exist. Or am I missing something important?

I think you are missing something important.

Take this statement for example:

"If |t| < 1, obviously F(w) = 0."

This makes no sense. "F(w) = 0" is an equation containing no "t", so why would "|t| < 1" make it true?

Note that since U(t) = 0 unless -1 <= t <= 1, you can rewrite the integral as follows:

$$\int_{-\infty}^{\infty} U(t) e^{-i w t} dt = \int_{-1}^{1} e^{-i w t} dt$$

 

[lanedance]

Not quite, F(w) is dependent on the value of w, with the integral carried out over all t.

The effect of U(t) =0 for |t|>1 means you can change the interval of the integral to be [-1,1], as U(t) is zero outside this inetrval

 

[Jamin2112]

Ah, I see. I knew something was fishy. I guess that's what happens when it's 7 weeks into the quarter and I still haven't bought the textbook.