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All LTI systems possess the eigenfunction property for complex exponential inputs. That is (restricting our attention to periodic complex exponentials), if $e^{j\omega_k t}$ is an input to the LTI system, then the output is $H(j\omega_k) e^{j\omega_k t}$, where $H(j\omega_k)$ is the frequency response of the systsem evaluated at $\omega_k$. When a linear constant coefficient differential equation with zero initial conditions can represent a causal LTI system. Why is it that when we talk about systems, specifically systems with dynamics described by differential equations, we speak of a “steady-state” sinusoidal output? This seems to imply that transients are always present. But if the system is LTI to begin with, would their be transients in the output if a sinusoidal signal, or complex exponential, is applied? It can be shown mathematically that the output of an LTI system when a complex exponential is applied is just a scaled version of the input, with no transient terms. What’s going on here?

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  • $\begingroup$ Ideal complex exponentials and or sine waves do not actually exist. In order to truly be a sine wave it would have to go from minus infinity to plus infinity in time. Hence, all real world implementations have to deal with the "turn on" and "turn off" part $\endgroup$ – Hilmar Apr 9 '18 at 13:40
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It can be shown mathematically that the output of an LTI system when a complex exponential is applied is just a scaled version of the input, with no transient terms.

that's for an input that has been going on since forever ago (and will continue forever to the future):

if $$ x(t) = e^{j \omega_0 t} \qquad \forall t \in \mathbb{R} $$

then

$$ y(t) = H(j \omega_0) e^{j \omega_0 t} = H(j \omega_0) x(t) $$

What’s going on here?

problem is, sometimes we have to apply an input at some time we might call "$t=0$", and before the LTI settles down to the $y(t)$ shown above, there are transients. but if the LTI is stable (and not "marginally stable"), the transients will go away eventually.

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