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One of the problems of my home assignment was this.

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.

Matt L.
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Francisco Fonseca
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  • 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. – Jazzmaniac Mar 06 '17 at 16:03

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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 $x[n]$.

Since this is a homework problem, instead of solving it for you I'll give you a few hints that should help you figure out the solution by yourself:

  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 $\omega=0$: $$\sum_{n=-\infty}^{\infty}x[n]=X(0)$$
Matt L.
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  • Alright , thank you. So the expression would be something like ∑j(dX(ω)/dω) / X(0) and from this i can simplify it and then apply it to the DTFT of a^nu[n]? – Francisco Fonseca Mar 06 '17 at 13:05
  • @LuceSkyWalker: You got the denominator right but not the numerator. Why should there be a sum? – Matt L. Mar 06 '17 at 13:09
  • Arent we applying the dtft only to what is inside the sum? Maybe this is real simple but i always hard problems with signal theory and all that – Francisco Fonseca Mar 06 '17 at 13:16
  • @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 $\omega=0$. – Matt L. Mar 06 '17 at 14:50
  • Btw, what´s the difference between using X(e^jw) or X(w)? – Francisco Fonseca Mar 06 '17 at 18:25
  • @LuceSkyWalker: Just different conventions. Take a look at this question. – Matt L. Mar 06 '17 at 19:07
  • But in this case when u say to evaluate the DTFT at w=0, is the result the same? with X(e^jw) at w=0, it would mean X(1) i think. – Francisco Fonseca Mar 06 '17 at 19:21
  • @LuceSkyWalker: Yes, but the function $X()$ of course depends on the argument that you choose: $X_1(\omega)=X_2(e^{j\omega})$, and clearly $X_1\neq X_2$. If you evaluate them at $\omega=0$ they must have the same value. – Matt L. Mar 06 '17 at 20:02