I am confused with determining transfer function, with Laplace transformation.
First of all, I would like to make some intro. It is know that transfer function is:
$G(s) = \frac{Y(s)}{X(s)}$, where $X(s)$ is Laplace transformation of $x(t)$, where x(t) is input variable.
Can someone explain me these Laplace transformations:
$\mathcal{L}$ {$\dot{x}(t-t_0)$, where $x(t)=u(t)$.
$\mathcal{L}$ {$\dot{x}(t-t_0)$, where $x(t)=u(t-t_1)$.
$u(t)$ is step function.
I assume that first Laplace transformation is this (maybe I am wrong...):
$\mathcal{L}$ {$\dot{x}(t-t_0)\}$ = $e^{-t_0s} \cdot (s \cdot U(s) - u(t_0))$ , where $x(t)=u(t)$.
First, is this equation above correct? If so, $u(t_0) = 1$ ? Is this correct, since step function has value of 1 in $t=t_0$ (we will assume it is 1, not 0.5, due to first-order discontinuity).
Now lets move to the second Laplace transformation: $\mathcal{L}$ {$\dot{x}(t-t_0)$, where $x(t)=u(t-t_0)$.
So, because I am determining Transfer function and I need only X(s) (as Laplace transformation of x(t) - input), do I need to care about what is x(t) at all?). Will it be the same result ?
$\mathcal{L}$ {$\dot{x}(t-t_0)\}$ = $e^{-t_0s} \cdot (s \cdot U(s) - u(t_0))$ , where $x(t)=u(t-t_1)$
or maybe we should consider $t_1$ somehow? Does this delay $t_1$ has any role here?
I hope there will be someone who will answer my question.
Note: This is not homework, I am practicing for my exam