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This is the statement from the book:enter image description here

Eq. (129) only ensures that $y_t[n]$ will be bounded for all values of $n$. For example, this may include a case where $y_t[n]$ oscillates and hence never becomes small. Hence, it is not clear how one can conclude that the transient response must become increasingly smaller as $n\to\infty$ when Eq. (129) is satisfied.

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It's true that Eq. $(129)$ just presents an upper bound, from which it can't directly be concluded that the transient response $y_t[n]$ decays to zero. However, stability requires that

$$\sum_{k=0}^{\infty}|h[k]|<\infty\tag{1}$$

which can only be true if $\lim_{n\to\infty}h[n]=0$, i.e., if the impulse response decays to zero. If this is the case, then the transient response

$$y_t[n]=-\sum_{k=n+1}^{\infty}h[k]e^{j\omega (n-k)}$$

must also decay to zero for $n\to\infty$.

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  • $\begingroup$ You should write a new signal processing book! Oppenheim-Schafer is a good book but the "wordings" are really problematic sometimes. $\endgroup$ – DSPinfinity Dec 3 '20 at 12:27
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In the context of discrete-time LTI systems, with (causal) impulse response $h[n]$, the analysis of suddenly applied complex-exponential; $x[n] = e^{j\omega_0 n}~ u[n]$, yields the following result:

$$ \begin{align} y[n] &= \sum_{k=0}^{\infty} h[k] ~ e^{j \omega_0 (n-k) } ~ u[n-k] ~~~ ,~~~n \ge 0\\ \\ & = e^{j \omega_0 n } ~\sum_{k=0}^{n} h[k] ~e^{-j \omega_0 k } \\ \\ & = e^{j \omega_0 n } \left( \sum_{k=0}^{\infty} h[k] e^{-j \omega_0 k } - \sum_{k=n+1}^{\infty} h[k] e^{-j \omega_0 k }\right) \\ \\ &= y_{ss}[n] - y_{t}[n] ~~~ ,~~~n \ge 0\\ \\ \end{align} $$

where $~y_{ss}[n] = e^{j \omega_0 n } ~ \sum_{k=0}^{\infty} h[k] e^{-j \omega_0 k } = e^{j \omega_0 n } H(\omega_0)~ $ is the ideal steady-state response , and $y_{t}[n] = e^{j \omega_0 n } ~ \sum_{k=n+1}^{\infty} h[k] e^{-j \omega_0 k }$ is the transient-response.

Eq.(129) in the quotation, states that the magnitude of $y_t[n]$ is bounded by the sum of absolute values of impulse response $h[n]$. And it follows that if the LTI system is stable; i.e. its impulse response is absolutely summable,

$$ \sum_{k=0}^{\infty} |h[k]| < \infty \tag{1} $$

then the transient response will not only be bounded for all $n$, but will also be decaying to zero as $n$ goes to infinity. To see this latter condition note that Eq. (1) implies that $$ \lim_{k \to \infty} |h[k]| = 0 \tag{2}$$ Which is a necessary condition for the sers in Eq. (1) to converge. Which is also a necessary condition for any LTI system impulse response to be stable. Then you can see that the partial sum

$$ \lim_{n \to \infty} \sum_{n+1}^{\infty} |h[k]| = 0 \tag{3}$$

When you object to this condition with an oscillating transient response assumption, you violate the fact that the impulse response be absolutely summable; $h[n]$ be stable. Because for an oscillating transient response, you need an oscillating impulse response which cannot meet the condition in Eq. (2), which is necessary for $h[n]$ to be stable.

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  • $\begingroup$ perfect answer also. the system did not allow to accept two answers! thank you. $\endgroup$ – DSPinfinity Dec 3 '20 at 12:32

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