Difference between revisions of "4.3 ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:06, 12 December 2008

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a

$ f(t) = sin(2 \pi t + \frac{\pi}{4}) $

This problem is done strictly from the properties in the tables 4.1 and 4.2.

First, find the time shift of the sine function by factoring out a $ 2\pi $

$ f(t) = sin\Big(2 \pi (t + \frac{1}{8})\Big) $

Then we have a new function:

$ g(t) = sin\big(2\pi t) $

From table 4.2

$ G(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi ) - \delta(\omega + 2\pi )\Big] $

From table 4.1, the time shift property

$ F(\omega) = e^{-j\omega t_0} G\big(\omega)\,\,\,\,,t_0 = -\frac{1}{8} $

Plug in -1/8 and the negatives cancel:

$ F(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{j\omega \frac{1}{8}} - \delta(\omega + 2\pi )e^{j\omega \frac{1}{8}}\Big] $

The final final answer comes when we realize the the delta functions multiplied by each of the exponentials are only valid when $ \omega $ is such that it is delta(0):

$ F(\omega) = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{2\pi \frac{1}{8}j} - \delta(\omega + 2\pi )e^{-2\pi \frac{1}{8}j}\Big] = \frac{\pi}{j}\Big[ \delta(\omega - 2\pi )e^{\frac{\pi}{4}j} - \delta(\omega + 2\pi )e^{-\frac{\pi}{4}j}\Big] $

b

$ f(t) = 1 + cos(6 \pi t + \frac{\pi}{8}) $

This is similar to the previous problem so I won't go through it in so many steps

$ F(w) = \mathcal{F}(1) + \pi\Big[ \delta(\omega - 6\pi )e^{\omega \frac{1}{48}j} - \delta(\omega + 6\pi )e^{\omega \frac{1}{48}j}\Big] $

$ = 2\pi\delta(\omega) + \pi\Big[ \delta(\omega - 6\pi )e^{\frac{\pi}{8}j} - \delta(\omega + 6\pi )e^{-\frac{\pi}{8}j}\Big] $