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Problem 4.4

a

$ X_1\big(j\omega) = 2\pi \delta(\omega) + \pi \delta(\omega - 4\pi) + \pi \delta(\omega + 4 \pi) $

Apply the inverse fourier transform integral:

$ x_1(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\Big(2\pi \delta(\omega)e^{-j\omega t}\Big)\,d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\Big(\pi \delta(\omega - 4\pi)e^{-j\omega t} + \pi \delta(\omega + 4 \pi)e^{-j\omega t}\Big)\,d\omega $

Cancel the pi:

$ x_1(t) = \int_{-\infty}^{\infty}\Big(\delta(\omega)e^{-j\omega t}\Big)\,d\omega + \frac{1}{2}\int_{-\infty}^{\infty}\Big( \delta(\omega - 4\pi)e^{-j\omega t} + \delta(\omega + 4 \pi)e^{-j\omega t}\Big)\,d\omega $

Apply the sifting property:

$ x_1(t) = e^{0} + \frac{1}{2}\Big( e^{-4\pi j t} + e^{4\pi j t}\Big) $

Simplify using euler's formula

$ x_1(t) = 1 + cos\big(4\pi t) $

b

$ X_2\big(j\omega) = \begin{cases} 2, \,\,\,\,\,\,\,\, 0 \le \omega \le 2 \\ -2,\,\, -2 \le \omega < 0 \\ 0,\,\,\,\,\,\, |\omega| > 2 \end{cases} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva