Step 1

Since f(t)=0 for \(2\leq t \text{ and } t<0\)</span> so create step function: \(u(t)-u(t-2)\) which will give those zero values.

Thus, \(f(t)=e^{-t}[u(t)-u(t-2)] \text{ or } f(t)=e^{-t}u(t)-e^{-t}u(t-2)\)

Step 2

Calculate Laplace transform:

\(F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt\)

\(=\int_{-\infty}^\infty(e^{-t}u(t)-e^{-t}u(t-2))e^{-st}dt\)

\(=\int_{-\infty}^\infty(e^{-t}u(t))e^{-st}dt - \int_{-\infty}^\infty(e^{-t}u(t-2))e^{-st}dt\)

\(=\int_0^\infty e^{-(s+1)t}dt - \int_2^\infty e^{-(s+1)t}dt\)

\(=\frac{1}{(s+1)}- \frac{e^{-2(s+1)}}{(s+1)}\)

\(=\frac{1-e^{-2(s+1)}}{(s+1)}\)

Step 3

Thus, Laplace transform is \(F(s)=\frac{1-e^{-2(s+1)}}{(s+1)}\)

Since f(t)=0 for \(2\leq t \text{ and } t<0\)</span> so create step function: \(u(t)-u(t-2)\) which will give those zero values.

Thus, \(f(t)=e^{-t}[u(t)-u(t-2)] \text{ or } f(t)=e^{-t}u(t)-e^{-t}u(t-2)\)

Step 2

Calculate Laplace transform:

\(F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt\)

\(=\int_{-\infty}^\infty(e^{-t}u(t)-e^{-t}u(t-2))e^{-st}dt\)

\(=\int_{-\infty}^\infty(e^{-t}u(t))e^{-st}dt - \int_{-\infty}^\infty(e^{-t}u(t-2))e^{-st}dt\)

\(=\int_0^\infty e^{-(s+1)t}dt - \int_2^\infty e^{-(s+1)t}dt\)

\(=\frac{1}{(s+1)}- \frac{e^{-2(s+1)}}{(s+1)}\)

\(=\frac{1-e^{-2(s+1)}}{(s+1)}\)

Step 3

Thus, Laplace transform is \(F(s)=\frac{1-e^{-2(s+1)}}{(s+1)}\)