Summation of Sinc function approximated by sines and cosines Suppose we can express the summation of a Sinc function as shown below:
$$\sum_{n=-\infty}^{\infty}Sa(3\pi(t-n)) = a+bcos(\omega t)+csin(\omega t)$$
Then what is the relationship between those constants, namely $$c<a<b<\omega$$ or $$c<b<a<\omega$$ or it could be any other not listed here.
I know that the Sinc function comes from the Fourier transform of the rect function, but it doesn't seem helpful here. 
 A: Applying the Fourier Transform over the time variable $t$, you have that $$\mathcal{F}(Sa(3\pi(t-n)))=\frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} e^{i\omega n}$$ assuming that $Sa(3\pi(t-n))=\frac{\sin(3\pi(t-n))}{3\pi(t-n)}$ in your notation (there are two different $sinc$ functions in the literature, usually for Mathematics and Signal Processing). This means that we can express your equation with the Inverse Fourier Transform as:
\begin{align}\sum_{n=-\infty}^\infty Sa(3\pi(t-n)))&=\sum_{n=-\infty}^\infty  \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} e^{i\omega n} e^{i\omega t}d\omega \\&=  \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} \left[ \sum_{n=-\infty}^\infty e^{i\omega n} \right]e^{i\omega t} d\omega\\&\stackrel{*}{=} \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} 2\pi \sum_{m=-\infty}^\infty \delta(\omega - 2\pi m) e^{i\omega t} d\omega \\&= \frac{2\pi}{3}\int_{-\infty}^\infty \left[\delta(\omega+2\pi) + \delta(\omega) + \delta(\omega - 2\pi) \right]e^{i\omega t} d\omega \\&= \frac{2\pi}{3}\left[ 1 + 2\cos(2\pi t) \right] \end{align}
where (*) follows from the inverse Fourier Series of signal of $1$ with period $2\pi$, and the last step from the inverse Fourier Transform.
