Types of Convergence of the DTFT & Relation to Summability of x[n]

[CAVision]

Given a discrete time signal x[n] that has a DTFT (which exists in the mean square convergence or in the uniform convergence sense), how can we tell if the signal x[n] converges absolutely?

I know the following:

x[n] is absolutely summable <=> $X(e^{j \omega})$converges uniformly (i.e. the ROC of the Z-transform includes the unit circle)

x[n] is square summable <=>$X(e^{j\omega})$ converges in the mean-square sense (i.e. the ROC of the Z-transform does not include the unit circle)

Specifically, given x[n] with DTFT
$X(e^{j\omega}) = \frac{1 + 0.55e^{-j\omega} -0.2e^{-j2\omega} }{(1 + 0.8665e^{-j\omega} + 0.5625e^{-j2\omega})(1+2e^{-j\omega})}$
converges (uniformly or in mean-square). Is x[n] absolutely summable?

Thanks.

 

[NateD]

Yes, x[n] is absolutely summable. This can be seen from the fact that the region of convergence (ROC) of its Z-transform includes the unit circle. The ROC is given by the set of all points $\omega$ in the complex plane such that the denominator of $X(e^{j\omega})$ is zero. Since the denominator has two zeros at $\omega=0$ and $\omega=\pi$, it follows that the ROC contains the entire unit circle. Therefore, the DTFT of x[n] converges uniformly, meaning that x[n] is absolutely summable.