Revision as of 18:01, 2 March 2011

Homework 5 Solutions, ECE301 Spring 2011 Prof. Boutin

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Question 1

$ \begin{align} \mathcal{X}(\omega)&= \int_{-\infty}^{\infty} e^{-3|t|}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{3t}e^{-j\omega t} dt + \int_0^{\infty} e^{-3t}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{(-j\omega +3)t} dt + \int_0^{\infty} e^{-(j\omega +3)t} dt \\ &= \frac{1}{-j\omega +3} \left[e^{(-j\omega +3)t}\right]_{-\infty}^0 - \frac{1}{j\omega +3} \left[e^{-(j\omega +3)t}\right]^{\infty}_0 \\ &= \frac{1}{-j\omega +3} \left[1-0 \right] - \frac{1}{j\omega +3} \left[0-1\right] \\ &= \frac{1}{-j\omega +3} + \frac{1}{j\omega +3} \\ &= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\ &= \frac{6}{\omega^2+9} \end{align} $

To verify our answer using the table, we first write:

$ x(t)=e^{-3|t|}=e^{3t}u(-t)+e^{-3t}u(t) $.

From the table:

$ \mathfrak{F}\left\{e^{-3t}u(t)\right\}=\frac{1}{j\omega+3} $.

Using time-reversal property from the table:

$ \mathfrak{F}\left\{e^{3t}u(-t)\right\}=\frac{1}{-j\omega+3} $.

Now, since the Fourier transform (FT) is linear, we have that:

$ \begin{align} \mathcal{X}(\omega)&=\mathfrak{F}\left\{e^{-3t}u(t)\right\}+\mathfrak{F}\left\{e^{3t}u(-t)\right\} \\ &=\frac{1}{j\omega+3}+\frac{1}{-j\omega+3} \\ &=\frac{6}{\omega^2+9} \end{align} $.

Question 2

$ \begin{align} \mathcal{X}(\omega)&= \int_{-\infty}^{\infty} \sin^2\left(\pi t +\frac{\pi}{12}\right)e^{-j\omega t}dt \\ &= \int_{-\infty}^{\infty} \left[\frac{1}{2} - \frac{1}{2}\cos\left(2\pi t + \frac{\pi}{6}\right) e^{-j\omega t} \right]dt \\ &= \frac{1}{2} \int_{-\infty}^{\infty} e^{-j\omega t} dt - \frac{e^{j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty} e^{j2\pi t} e^{-j\omega t} dt -\frac{e^{-j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty}e^{-j2\pi t}e^{-j\omega t} dt \\ &= \frac{1}{2} \int_{-\infty}^{\infty} e^{-j\omega t} dt - \frac{e^{j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty} e^{j(2\pi -\omega)t} dt -\frac{e^{-j\frac{\pi}{6}}}{4} \int_{-\infty}^{\infty}e^{-j(2\pi + \omega)t} dt \\ &=\pi \delta(\omega) - \frac{\pi e^{j\frac{\pi}{6}}}{2} \delta(\omega - 2\pi) - \frac{\pi e^{-j\frac{\pi}{6}}}{2} \delta(\omega + 2\pi), \end{align} $

where we have used the following property:

$ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-j\omega t} dt = \delta(\omega) $

Now to verify our answer we write:

$ \begin{align} x(t)&=\sin^2\left(\pi t - \frac{\pi}{12}\right) \\ &= \frac{1}{2}-\frac{1}{2}\cos\left(2\pi t + \frac{\pi}{6}\right) \\ &= \frac{1}{2}-\frac{1}{2}\cos\left[2\pi\left( t + \frac{1}{12}\right)\right] \end{align} $

Now using (from the table) the time-shift property, FT of a constant, and FT of a cosine we get:

$ \begin{align} \mathcal{X}(\omega) &= \pi\delta(\omega) - \frac{\pi e^{j\frac{\omega}{12}}}{2} \delta(\omega - 2\pi)- \frac{\pi e^{j\frac{\omega}{12}}}{2} \delta(\omega + 2\pi) \\ &= \pi\delta(\omega) - \frac{\pi e^{j\frac{\pi}{6}}}{2} \delta(\omega - 2\pi)- \frac{\pi e^{-j\frac{\pi}{6}}}{2} \delta(\omega + 2\pi) \end{align} $

where we have used the fact that $ f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0) $.

HW5

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