Find, using Table 2.1, the Fourier transform of $$ f(t)=\left\\{\begin{array}{lc} 1 & -1 \leq t \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$

The given function is a rectangular function defined as follows: $$ f(t)=\left\\{\begin{array}{lc} 1 & -1 \leq t \leq 1 \\\ 0 & \text { otherwise } \end{array}\right. $$

The function f(t) we determined in Step 1 can be rewritten as a product of the unit step function, as follows: $$ f(t) = u(t+1) - u(t-1) $$ With u(t) being the unit step function. Now, we can find the corresponding Fourier transform in Table 2.1.

Since f(t) = u(t+1) - u(t-1), the linearity property of the Fourier transform allows us to calculate the Fourier transform of f(t) by subtracting the transforms of u(t-1) and u(t+1). Let's denote the Fourier Transform of f(t) as F(ω). The Fourier transform of the unit step function is given in Table 2.1 as: $$ \mathcal{F}\left\\{u(t-a)\\right\\} = \frac{1}{j\omega} e^{-j\omega a} $$ For a=1, we have the Fourier transform of u(t-1) and u(t+1): $$ \mathcal{F}\left\\{u(t-1)\\right\\} = \frac{1}{j\omega} e^{-j\omega} $$ $$ \mathcal{F}\left\\{u(t+1)\\right\\} = \frac{1}{j\omega} e^{j\omega} $$ Now, we can find the Fourier transform of f(t): $$ F(ω) = \mathcal{F}\left\\{f(t)\\right\\} = \mathcal{F}\left\\{u(t+1) - u(t-1)\\right\\} $$ Applying the linearity property, we get: $$ F(ω) = \frac{1}{j\omega} e^{j\omega} - \frac{1}{j\omega} e^{-j\omega} $$ Finally, we can simplify the expression to obtain the Fourier transform of f(t): $$ F(ω) = \frac{2}{ω} \sin(\omega) $$ Thus, the Fourier transform of the given function f(t) is: $$ F(ω) = \frac{2}{ω} \sin(\omega) $$