I need help with this problem. I can not get the answer in the back of the book.

I used the the Tables on pages 328 and 329 to get the following:

- $ \mathfrak{F} (sin(2\pi t))=\frac{\pi}{j}(\delta(\omega-2\pi)-\delta(\omega+2\pi)) $

By the Time Shift Property:

- $ \begin{align} \mathfrak{F} (sin(2\pi t+\frac{\pi}{4})) &= \mathfrak{F} (sin(2\pi(t+\frac{1}{8}))) \\ &=\frac{\pi}{j}e^{\frac{j\omega}{8}}(\delta(\omega-2\pi)-\delta(\omega+2\pi)) \end{align} $

However, the answer in the back of the book is:

- $ \frac{\pi}{j}(e^{\frac{j\pi}{4}}\delta(\omega-2\pi)-e^{-\frac{j\pi}{4}}\delta(\omega+2\pi)) $

They found this by finding the ak's and using the top Fourier Transform Pair in Table 4.2.

Based on the Tables, it seems as though both of these answers should be right, but how? They don't appear equivalent...

Is there some tricky manipulation that I don't see???

comments:

Hint --mireille.boutin.1, Fri, 05 Oct 2007 12:16:31 They are equivalent. The key is to observe that the unit impulses are zero all the time, except at one point. So you can take your answer and replace the $ \omega\ $ in each of the exponentials in front of the $ \delta\ $'s by a constant ($ 2\pi\ $ and $ -2\pi\ $ respectively). Does that make sense?

... --ross.a.howard.1, Fri, 05 Oct 2007 17:56:11 Thanks! Yes. that makes sense...I should have thought of that...