The problem arises from frequency response method (control systems ) and most books try to explain why we replace $'s'$ by $'j\omega'$ in frequency response. So, let me get to the math part; Our system is$$ y(t)=ae^{-j\omega t} +be^{j\omega t}$$ Which is the steady state response.
The input is sinusoid $x(t)=A\sin(wt)$; and thus the output is also a sinusoid. In the Laplace domain, the relation between input($x(s)$) and output ($y(s)$) is $$y(s)=G(s)x(s)$$ where $G(s)$ is the transfer function.now here is my problem, when finding $a$ and $b$ here is what they did. $$a=G(s)\frac{(\omega A)}{(s^{2}+\omega^{2})}(s+j\omega)$$ and $$b=G(s)\frac{(\omega A)}{(s^{2}+\omega^{2})}(s-j\omega)$$ ( I know how this is derived but I don't know the next step which is:-)
When evaluating a and b at $-j\omega$ and $j\omega$ respectively we get; $$a=-A\frac{G(-j\omega)}{2j}$$ $$b=A\frac{G(j\omega)}{2j}$$ So, my question is why did we pick the point $j\omega$ in place of $'s'$ For $b$ and $-j\omega$ For $a$?
$TRIAL$
May be initial conditions....
