Assume that we have a causal, stable system that is rational in $s$. If we pass a complex exponential as input, we know that we get the frequency response of the system multiplied by the same exponential:

$$
x(t) = e^{i\omega t} \to \fbox{$H$} \to y(t) = H(\omega) e^{i\omega t}
$$

What is we apply the toneburst, $e^{i\omega t} u(t)$, a suddenly-applied sinusoid? The system is describable by a LCCDE, 

$$
\hat{H}(s) = \frac{ B(s) }{ A(s) } = \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n) } 
$$

The stability of the system implies that all of the poles are to the left of the $i\omega$-axis. In other words, the stability implies $\Real{p_\ell} \lt 0 $ $ \forall \ell$. We can now apply the toneburst to this system. We know that the transform of the toneburst is given by

$$
e^{i\omega_0 t} u(t) \ltp \frac{1}{s - i\omega_0} \quad \Real{s} \gt 0 
$$ 

We also know that convolution in the time domain corresponds to multiplication in the frequency domain, thus

\begin{align*}
\hat{Y}(s) &= \hat{H}(s) \hat{X}(s) \\
\hat{Y}(s) &= \frac{ B(s) }{ A(s) } = \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n)(s - i\omega_0) } \quad RoC \supseteq RoC(h) \cap RoC(x) \\
\end{align*}

The new region of convergence is $\Real{s} \gt 0$. Lets apply partial fraction decomposition to determine the time domain expression. For simplicity, assume that there are no duplicate poles.

\begin{align*}
\hat{Y}(s) &= \frac{R_1}{s - p_1} + \frac{R_2}{s - p_2} + \cdots + \frac{R_n}{s - p_n} + \frac{Q}{s - i\omega_0} \\
y(t) &= R_1 e^{p_1 t}u(t) + R_2 e^{p_2 t} u(t) + \cdots + R_n e^{R_n t} u(t) + Q e^{i\omega_0 t} u(t)  \\
\end{align*}

Since all poles $P_\ell$ were to the left of the $i\omega$-axis, and the pole from $Q e^{i\omega_0 t} u(t)$ is the $i\omega$-axis, we can see why the intersection of the regions of convergence is $\{ s \st \Real{s} \gt 0 \}$; the rightmost pole is zero. In this case, all of the poles that were to the left of the $i\omega$-axis are apart of the \emph{transient response}. This is because all $p_\ell \lt 0 \quad \forall \ell$, so the terms associated with these poles decay.

\begin{align*}
y_{tr}(t) &= R_1 e^{p_1 t}u(t) + R_2 e^{p_2 t} u(t) + \cdots + R_n e^{R_n t} u(t) \\
y_{ss}(t) &=  Q e^{i\omega_0 t} u(t) \\
\end{align*}

When we look at the transient response, we see that the terms emanate from the stable poles of the system. We can see that the \emph{steady state response} is due to the pole of the input. The steady state response's pole is to the right of the right-most pole of the transient state's. When you introduce an input with a pole to the right of the systems right-most pole it then becomes steady state response. \\
 
How can we find $Q$? 

\begin{align*}
\hat{Y}(s) &= \frac{R_1}{s - p_1} + \frac{R_2}{s - p_2} + \cdots + \frac{R_n}{s - p_n} + \frac{Q}{s - i\omega_0} \\
\hat{Y}(s) &= \sum \limits_{\ell = 1}^{n} \frac{R_\ell}{s - p_\ell} + \frac{Q}{s - i\omega_0} \\
(s - i\omega_0)\hat{Y}(s) &= \sum \limits_{\ell = 1}^{n} \frac{R_\ell(s-i\omega_0)}{s - p_\ell} + Q \\
\end{align*}

If we evaluate $\hat{Y}(s)(s - i\omega_0)$, we can get the coefficient $Q$:

\begin{align*}
Q &= \left. (s - i\omega_0)\hat{Y}(s) \right|_{s = i\omega_0} \\
Q &= \left. \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n) } \right|_{s = i\omega_0} \\
Q &= \left.\hat{H}(s) \right|_{s = i\omega_0} \\
Q &= H(\omega_0) \\
\end{align*}

Now we have our steady state response 
\begin{align*}
y_{ss}(t) &=  Q e^{i\omega_0 t} u(t) \\
y_{ss}(t) &=  H(\omega_0) e^{i\omega_0 t} u(t) \\
\end{align*}

If we apply a one-sided exponential to the system, $e^{i\omega_0 t} u(t)$, as $t$ becomes large enough, the system cannot distinguish between the one-sided exponential and the infinite duration exponential, $e^{i\omega_0 t}$. The coefficient of the steady state response should then be the frequency response of the system evaluated at the frequency of the input; this is what we knew from before when applying the infinite duration exponential as input to an LTI system:

\begin{align*}
e^{i\omega_0 t} \to \fbox{$H$} \to H(\omega_0) e^{i\omega_0 t}
\end{align*}


In other words, if $y(t) = y_{tr}(t) + y_{ss}(t)$, then as time goes on, we have

\begin{align*}
\lim_{t \to \infty} y(t) = y_{ss}(t) = e^{i\omega_0 t}u(t) = e^{i\omega_0 t}
\end{align*}

If $\omega_0 = 0$, then $x(t) = u(t)$, and then we have $y_{ss}(t) = H(0)u(t) = H(0)$. We find that we get the DC gain of the system. 


Steady State and Transient Responses of CT-LTI Systems