Causal LTI system having exponential input

Question

I know that for an LTI system having complex exponential input,
i.e, $x(t)=\exp(j w_o t)$ & $h(t) \to $ LTI System ;
then, its output { $y(t) \} =M \exp(j w_o t + \phi)$ , where $M= |H(j w)|_{|w= w_o}$ & $\phi = \angle \{ H(j w)\} _{|w= w_o}$

But for a causal LTI system having exponential input,
i.e, $x(t)=\exp(a t)$ & $h(t) \to $ Causal LTI System ;
Is it true to say its output $\{ y(t)\}$ is given by:
$y(t) =K \exp(a t )$ , where $K= H(s)_{|s=a} \quad$ ?

Answer 1

Any LTI system with transfer function $H(s)$ reacts to an input signal $x(t)=e^{s_0t}$ with an output signal $y(t)=H(s_0)e^{s_0t}$, if $s_0$ is inside the region of convergence (ROC) of $H(s)$. This is equivalent to saying that the functions $e^{s_0t}$ are eigenfunctions of LTI systems, and $H(s_0)$ is the corresponding eigenvalue.

Note that $s_0$ is an arbitrary complex constant inside the ROC of $H(s)$, so the relation mentioned above is also valid for $s_0\in\mathbb{R}$. Also note that causality is not required for the above relation to hold.

However, if your question is about input signals of the form $x(t)=e^{at}u(t)$, where $u(t)$ is the unit step function, i.e., for input signals that are zero for $t<0$, then the output is generally not equal to $H(a)e^{at}u(t)$, because the functions $e^{at}u(t)$ are no eigenfunctions of LTI systems.