Fourier Transform- is this right?

[FrogPad]

If someone wouldn't mind checking if this answer is correct, that would be awesome.

$$x(t) =$$ a triangle with points (-2,0), (0,1), and (2,0).

I am supposed to compute the Fourier transform ($\hat x (\omega)$) of $x(t)$.

"Solution"
$$m = \frac{y_2-y_1}{x_2-x_1}=\frac{1}{2}$$

Thus, $$\frac{dx(t)}{dt} =$$ a pulse with height [itex] \frac{1}{2} [/tex] from (-2 to 0), and a pulse with height $\frac{-1}{2}$ from (0 to 2).

Let $$k(t) = \frac{dx(t)}{dt}$$

Since, $$x(t) = \int_{-\infty}^x k(t)dt \leftrightarrow \frac{1}{j\omega} \hat k(\omega) + \hat k(0) \delta(\omega)$$

Let $$x'(t) =$$ a pulse from -1 to 1 with height 1

Then, $$\frac{1}{2} x'(t) \leftrightarrow \frac{\sin \omega}{\omega}$$

and, $$k(t) = \frac{1}{2}x'(t+1) - \frac{1}{2}x'(t-1)$$

Thus, $$\hat k(\omega) = e^{j\omega}\frac{\sin \omega}{\omega} - e^{-j\omega}\frac{\sin \omega}{\omega}$$

$$\hat x (\omega) = \frac{2 \sin_c \omega}{\omega} \left( \frac{e^{j\omega}-e^{-j\omega}}{2j} \left) + 2 \sin_c(0)\cos(0)\delta(\omega) = 2 ( \sin_c^2 \omega + \delta(\omega))$$note: $$\sin_c(x) = \frac{sin x}{x}$$ (I don't know how to write the sinc function)
Does this look right?

Thanks

 

[mighty2000]

for your post! Your solution looks correct to me. However, I would suggest using the notation \mathrm{sinc}(x) = \frac{\sin x}{x} for the sinc function. It is a common notation and will make your solution clearer for others to read. Additionally, you may want to include some explanations or steps in your solution to make it easier for others to follow. Overall, great job!