# Clarification on eigenfunction property from Oppenheim's Discrete-time signal processing, 3rd ed

Could anybody explain what is meant by the following statement on page 50 of Oppenheim and Schafer's book Discrete-Time Signal Processing (third edition)?

The eigenfunction property of complex exponentials depends on stability of the system, since at finite $$n$$, the transient response must have become zero, so that we only see the steady-state response $$H(e^{j\omega})e^{j\omega n}$$ for all finite $$n$$.

• It's not clear what you're not clear on. They're saying that if the system isn't stable it's not going to settle out at all, so it's meaningless to say what it'll settle out to. – TimWescott Dec 3 '20 at 6:46

This statement can only be understood given its context. Right before the sentence you quoted, we have this statement:

It is no surprise that the condition for existence of the frequency response is the same as the condition for dominance of the steady-state solution. Indeed, a complex exponential that exists for all $$n$$ can be thought of as one that is applied at $$n = -\infty$$.

Consequently, starting out at $$n=-\infty$$, if the system is stable, all transients must have died out at "finite $$n$$" because at finite $$n$$ the moment that the exponential was applied is infinitely long ago.

We also depend on stability because otherwise the eigenvalue $$H(e^{j\omega})$$ wouldn't exist, at least not in the classical sense.

• "It is no surprise that the condition for existence of the frequency response is the same as the condition for dominance of the steady-state solution". This statement is also not clear to me. Could you please prove this? – Lionel Iceberg Dec 3 '20 at 10:39
• @LionelIceberg: Isn't that proved in the equations right before that statement in the book? – Matt L. Dec 3 '20 at 11:33

In a stable system transient response is not necessarily zero at finite n, it becomes zero as n goes to infinity. Hence, the following statement is not clear to me:

"at finite n, the transient response must have become zero'

• You're right in general, but not if the input signal was applied at $n=-\infty$, i.e., has been existing forever. In that case the transients have died out at finite $n$. That's what the quoted sentence refers to. Also see my answer. – Matt L. Dec 3 '20 at 7:54