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] $
