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According to [1, page 11], the unit step sequence is given by $$ u[n] = \begin{cases} 1, & n \geq 0, \\ 0, & n < 0. \end{cases} \tag{1} $$ The discrete-time Fourier transform of $u[n]$ is [1, Table 2.3] $$ U(e^{j \omega}) = \frac{1}{1 - e^{-j \omega}} + \sum_{k = -\infty}^\infty \pi \delta(\omega + 2 \pi k). \tag{2} $$ One of my students, Mr. Chang, wants to derive (2). The following is his derivation: $$ U(e^{j \omega}) = \sum_{n = -\infty}^\infty u[n] e^{-j \omega n} = \sum_{n = 0}^\infty e^{-j \omega n} = \frac{1}{1 - e^{-j \omega}}. \tag{3} $$ $$ u[n] = \sum_{k = 0}^\infty \delta[n - k] \leftrightarrow U(e^{j \omega}) = \sum_{k = 0}^\infty e^{-j \omega k} = \frac{1}{1 - e^{-j \omega}}. \tag{4} $$

The question is: Why do (3) and (4) not have the term $$ \sum_{k = -\infty}^\infty \pi \delta(\omega + 2 \pi k)? \tag{5} $$

Thank you very much in advance.

Reference:

[1] A. V. Oppenheim and R. W. Schafer, Discrete-Time Signal Processing, 2nd ed., Prentice Hall, Inc., 1999.

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  • $\begingroup$ The formula $$\sum_{n=0}^{\infty}e^{-i\omega n} = \frac{1}{1 - e^{-i\omega}}$$ is false. In general, the formula $$\sum_{n=0}^{\infty}z^n = \frac{1}{1-z}$$ holds only if $|z| < 1$. In particular, if $z = e^{-i \omega}$, then $|z| = 1$, so the formula does not hold. $\endgroup$ – Bungo Sep 5 '16 at 2:53
  • $\begingroup$ Moreover, the series $\sum_{n=0}^{\infty}e^{-i\omega n}$ does not converge at all, because the terms do not converge to $0$. (The absolute value of each term is $1$.) So the claimed identity holds only in the sense of distributions. $\endgroup$ – Bungo Sep 5 '16 at 2:55
  • $\begingroup$ @Bungo Thank you for your comments, Bungo. :-) $\endgroup$ – Wei-Cheng Liu Sep 5 '16 at 7:12

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