Step 1

Consider the given function:

\(f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}\)</span>

Step 2

Now, by definition of Laplace Transform:

\(L\left\{f(t)\right\}=\int_0^\infty e^{-st}f(t)dt\)

\(=\int_0^1e^{-st}f(t)dt+\int_1^\infty e^{-st}f(t)dt\)

\(=\int_0^1e^{-st}(-1)dt+\int_1^\infty e^{-st}(1)dt\)

\(=\left[\frac{e^{-st}}{s}\right]_0^1+\left[\frac{e^{-st}}{-s}\right]_1^\infty\)

\(=\left(\frac{e^{-s}}{s}-\frac{1}{s}\right)+\left(0+\frac{e^{-s}}{s}\right)\)

\(L\left\{f(t)\right\}=\frac{2e^{-s}}{s}-\frac{1}{s}\)

Step 3

Now, Rewrite f(t) as the sum of step functions, \(u_c(t)\):

\(f(t)=f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}\)</span>

\(=(-1)(u_0,1(t))+(1)(u_1(t))\)

\(=(-1)(u_0(t)-u_1(t))+(1)(u_1(t))\)

\(=-u_0(t)+2u_1(t)\)

Step 4 Now, use the following formula for Laplace transform of step function:

\(L\left\{u_c(t)\right\}=\frac{e^{-cs}}{s}\)

Step 5

Thus, now take Laplace transform:

\(f(t)=-u_0(t)+2u_1(t)\)

\(L\left\{f(t)\right\}=L\left\{-u_0(t)\right\}+L\left\{2u_1(t)\right\}\)

\(=-\frac{e^{-(0)s}}{s}+2\frac{e^{-(1)s}}{s}\)

\(L\left\{f(t)\right\}=-\frac{1}{s}+2\frac{e^{-s}}{s}\)

Step 6

Thus, from equation (1) and (2), it can be seen that the Laplace transform of f(t) that is obtained from both the methods is same that is :

\(L\left\{f(t)\right\}=2\frac{e^{-s}}{s}-\frac{1}{s}\)

Consider the given function:

\(f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}\)</span>

Step 2

Now, by definition of Laplace Transform:

\(L\left\{f(t)\right\}=\int_0^\infty e^{-st}f(t)dt\)

\(=\int_0^1e^{-st}f(t)dt+\int_1^\infty e^{-st}f(t)dt\)

\(=\int_0^1e^{-st}(-1)dt+\int_1^\infty e^{-st}(1)dt\)

\(=\left[\frac{e^{-st}}{s}\right]_0^1+\left[\frac{e^{-st}}{-s}\right]_1^\infty\)

\(=\left(\frac{e^{-s}}{s}-\frac{1}{s}\right)+\left(0+\frac{e^{-s}}{s}\right)\)

\(L\left\{f(t)\right\}=\frac{2e^{-s}}{s}-\frac{1}{s}\)

Step 3

Now, Rewrite f(t) as the sum of step functions, \(u_c(t)\):

\(f(t)=f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}\)</span>

\(=(-1)(u_0,1(t))+(1)(u_1(t))\)

\(=(-1)(u_0(t)-u_1(t))+(1)(u_1(t))\)

\(=-u_0(t)+2u_1(t)\)

Step 4 Now, use the following formula for Laplace transform of step function:

\(L\left\{u_c(t)\right\}=\frac{e^{-cs}}{s}\)

Step 5

Thus, now take Laplace transform:

\(f(t)=-u_0(t)+2u_1(t)\)

\(L\left\{f(t)\right\}=L\left\{-u_0(t)\right\}+L\left\{2u_1(t)\right\}\)

\(=-\frac{e^{-(0)s}}{s}+2\frac{e^{-(1)s}}{s}\)

\(L\left\{f(t)\right\}=-\frac{1}{s}+2\frac{e^{-s}}{s}\)

Step 6

Thus, from equation (1) and (2), it can be seen that the Laplace transform of f(t) that is obtained from both the methods is same that is :

\(L\left\{f(t)\right\}=2\frac{e^{-s}}{s}-\frac{1}{s}\)