# Question about a transient response statement (Oppenheim-Schafer"s discrete-time signal processing book, 3rd ed)

This is the statement from the book: 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.

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 $$n\to\infty$$.

• You should write a new signal processing book! Oppenheim-Schafer is a good book but the "wordings" are really problematic sometimes. Dec 3 '20 at 12:27

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.

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