# [SOLVED]Laplace Transform with two steps

#### dwsmith

##### Well-known member
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

##### Well-known member
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

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
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
$Y(s) = \frac{s}{(s + 1)(1 + e^{-s})}.$

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