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

[tex] x(t) = [/tex] a triangle with points (-2,0), (0,1), and (2,0).

I am supposed to compute the Fourier transform ([itex] \hat x (\omega) [/itex]) of [itex] x(t) [/itex].

"Solution"

[tex] m = \frac{y_2-y_1}{x_2-x_1}=\frac{1}{2} [/tex]

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

Let [tex] k(t) = \frac{dx(t)}{dt} [/tex]

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

Let [tex] x'(t) = [/tex] a pulse from -1 to 1 with height 1

Then, [tex] \frac{1}{2} x'(t) \leftrightarrow \frac{\sin \omega}{\omega} [/tex]

and, [tex] k(t) = \frac{1}{2}x'(t+1) - \frac{1}{2}x'(t-1) [/tex]

Thus, [tex] \hat k(\omega) = e^{j\omega}\frac{\sin \omega}{\omega} - e^{-j\omega}\frac{\sin \omega}{\omega} [/tex]

[tex] \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)) [/tex]

note: [tex] \sin_c(x) = \frac{sin x}{x} [/tex] (I don't know how to write the sinc function)

Does this look right?

Thanks

[tex] x(t) = [/tex] a triangle with points (-2,0), (0,1), and (2,0).

I am supposed to compute the Fourier transform ([itex] \hat x (\omega) [/itex]) of [itex] x(t) [/itex].

"Solution"

[tex] m = \frac{y_2-y_1}{x_2-x_1}=\frac{1}{2} [/tex]

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

Let [tex] k(t) = \frac{dx(t)}{dt} [/tex]

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

Let [tex] x'(t) = [/tex] a pulse from -1 to 1 with height 1

Then, [tex] \frac{1}{2} x'(t) \leftrightarrow \frac{\sin \omega}{\omega} [/tex]

and, [tex] k(t) = \frac{1}{2}x'(t+1) - \frac{1}{2}x'(t-1) [/tex]

Thus, [tex] \hat k(\omega) = e^{j\omega}\frac{\sin \omega}{\omega} - e^{-j\omega}\frac{\sin \omega}{\omega} [/tex]

[tex] \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)) [/tex]

note: [tex] \sin_c(x) = \frac{sin x}{x} [/tex] (I don't know how to write the sinc function)

Does this look right?

Thanks

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