MHB Laplace Transform with two steps

[Dustinsfl]

Given the transfer function
\[
H(s) = \frac{Y(s)}{U(s)} = \frac{1}{s + 1}
\]
and
\[
u(t) = 1(t) + 1(t - 1).
\]
How do I find U(s)? I know I take the Laplace transform of u(t) but with the two step functions how can this be done?

The Laplace transform of the step function is \(\frac{1}{s}\) for \(t > a\) where \(a\) is the shift. If I take the Laplace of \(u(t)\), do just get
\[
\frac{2}{s}\mbox{?}
\]
That seems strange though since \(\frac{1}{s}\) is for \(t > 0\) and the other \(\frac{1}{s}\) is for \(t > 1\).

The end goal is to find \(y(t)\).

 

[chisigma]

Given the transfer function
\[
H(s) = \frac{Y(s)}{U(s)} = \frac{1}{s + 1}
\]
and
\[
u(t) = 1(t) + 1(t - 1).
\]
How do I find U(s)? I know I take the Laplace transform of u(t) but with the two step functions how can this be done?

The Laplace transform of the step function is \(\frac{1}{s}\) for \(t > a\) where \(a\) is the shift. If I take the Laplace of \(u(t)\), do just get
\[
\frac{2}{s}\mbox{?}
\]
That seems strange though since \(\frac{1}{s}\) is for \(t > 0\) and the other \(\frac{1}{s}\) is for \(t > 1\).

The end goal is to find \(y(t)\).

If $\displaystyle g(t) = f(t - a)\ \mathcal{U}\ (t-a)$ and $\displaystyle \mathcal{L}\ \{f(t)\} = F(s)$ then $\displaystyle \mathcal {L}\ \{ g(t)\} = e^{- a\ s}\ F(s)$...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

If $\displaystyle g(t) = f(t - a)\ \mathcal{U}\ (t-a)$ and $\displaystyle \mathcal{L}\ \{f(t)\} = F(s)$ then $\displaystyle \mathcal {L}\ \{ g(t)\} = e^{- a\ s}\ F(s)$...

Kind regards

$\chi$ $\sigma$

So \(U(s) = \frac{1}{s} + e^{-s}\frac{1}{s}\)?

I don't think U(s) is correct. I then would have
\[
Y(s) = \frac{s}{(s + 1)(1 + e^{-s})}.
\]