# Gravity center of a discrete signal

If i use the definition to solve this(I'm not seeing other way) i have an expression like $\sum\sum nx[n]e^{-j\omega n}$ on the numerator and $\sum\sum x[n]e^{-j\omega n}$ on denominator. How do i simplify those expressions? The double sum is making me confuse.

Thanks.

• Your definition of the "center of gravity" of a signal is not a very good one. The denominator of your expression may evaluate to 0 in the general case. A better and more common definition uses either the magnitude of the signal samples or the squared magnitude. Mar 6 '17 at 16:03

Your double sums do not make sense. The numerator and the denominator of the expression for the center of gravity are numbers, and it's pointless to take the discrete-time Fourier transform (DTFT) of these numbers. What you are supposed to do is express the center of gravity in terms of the DTFT of $x[n]$.
1. The discrete-time Fourier transform (DTFT) of the sequence $nx[n]$ is given by $$\text{DTFT}\{nx[n]\}=j\frac{dX(\omega)}{d\omega}$$
2. The sum over a sequence (if it exists) is equal to the DTFT of that sequence evaluated at $\omega=0$: $$\sum_{n=-\infty}^{\infty}x[n]=X(0)$$
• @LuceSkyWalker: It's the same as with the denominator. You have the sum over a signal, and as soon as you have the DTFT of that signal (which is $nx[n]$), you can compute the sum by evaluating the DTFT at $\omega=0$. Mar 6 '17 at 14:50