How Do You Apply Fourier Transform to sin(2t)/t?

[CE Trainee]

1. Homework Statement
f(t) = (sin(2t))/t

Homework Equations

3. The Attempt at a Solution
I know that sin(t)/t has the Fourier transform pi(w). I'm just not sure how to apply that fact to this problem. Knowing that sin(t)/t --> pi(w), I reasoned that sin(2t)/t --> 2pi(2w). I'm almost certain this is incorrect. I arrived at this conclusion by noticing that if i multiplied the top and bottom of sin(2t)/t by 2, I would have a sinc function 2sinc(2t). Then I could use the property sin(t)/t --> pi(w). Am I anywhere close?

 

[jbunniii]

1. Homework Statement
f(t) = (sin(2t))/t

Homework Equations

3. The Attempt at a Solution
I know that sin(t)/t has the Fourier transform pi(w). I'm just not sure how to apply that fact to this problem. Knowing that sin(t)/t --> pi(w), I reasoned that sin(2t)/t --> 2pi(2w). I'm almost certain this is incorrect. I arrived at this conclusion by noticing that if i multiplied the top and bottom of sin(2t)/t by 2, I would have a sinc function 2sinc(2t). Then I could use the property sin(t)/t --> pi(w). Am I anywhere close?

You have the right idea. Let's introduce the following notation:

$$h(t) = \frac{\sin(t)}{t}$$

and $H(\omega)$ is the Fourier transform of $h(t)$.

Notice that

$$h(2t) = \frac{\sin(2t)}{2t}$$

and so

$$f(t) = 2 h(2t) = \frac{\sin(2t)}{t}$$

Therefore you need to find the Fourier transform of [itex]2 h(2t)[/tex].<br /> <br /> Now suppose I write<br /> <br /> $$g(t) = h(2t)$$[/itex][tex] <br /> and let $G(\omega)$ denote the Fourier transform of $g(t)$. Do you know (in general, not necessarily for a specific function) how $G(\omega)$ can be expressed in terms of $H(\omega)$? If so, then you are essentially done, because<br /> <br /> $$\mathcal{F}(f(t)) = \mathcal{F}(2 h(2t)) = 2 \mathcal{F}(h(2t)) = 2 \mathcal{F}(g(t)) = 2 G(\omega)$$[/tex]