In an LTI system, when the input is the characteristic function (i.e. an everlasting exponential function say \$e^{st}\$ ), the ZSR which is obtained by the convolution integral with \$h(t)\$ is \$H(s)e^{st}\$, where \$h(t)\$ is the impulse response of the system. Clearly it does not contain any characteristic mode terms. How do these terms get vanished?
Now if I go for the classical approach I get the particular solution for the input \$e^{st}\$ to be \$ce^{st}\$ . The constant c comes out to be \$H(s)\$ and hence I get \$H(s)e^{st}\$ as the particular solution which is same as ZSR. What does it mean, did I commit some error or are they actually equal? With the help of Laplace Transform I can represent most of the signals in terms of everlasting exponential components. And hence as it follows from the above query, I should get ZSR and PI to be the same for most of the input signals. This can't be true. I'm sure I've gone wrong somewhere. Where have I gone wrong?
Best Answer
I do not believe there is any problem with the notation used by the author. He just wanted to express the following (\$x\$ input, \$y\$ output - the response to a eigenfunction \$e^{st}\$):
$$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)d\tau $$ $$ = \int_{-\infty}^\infty h(\tau)e^{s(t-\tau)}d\tau $$ $$ = e^{st} \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$
Thus, the response to \$e^{st}\$ is of the form
$$y(t) = H(s)e^{st}$$
where \$H(s)\$ is a complex constant whose value depends on \$s\$ and wich is related to the impulse response by
$$ H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$
There is no conflict time x frequency here.
Also, note:
If \$\lambda\$ is a pole of \$H(s)\$, the mode \$e^{{\lambda}t}\$ can be generated at the output by an initial state without the application of any input. If \$\lambda\$ is not a pole of \$H(s)\$, then this is not possible. In this case, the only way to generate \$e^{{\lambda}t}\$ at the output is to apply \$e^{{\lambda}t}\$ at the input. In other hand, if the input \$x(t)\$ is of the form \$e^{{\lambda}t}\$, where \$\lambda\$ is not a pole of \$H(s)\$, then the output due to input \$x(t) = e^{{\lambda}t}\$ is equal to \$y(t) = H(\lambda)e^{{\lambda}t}\$.