Question

Determine the Laplace transform of each of the following functions:

$\left(a\right)f_1\left(t\right)=\left\{\begin{array}{l}t,t=1,2,3,....,\\ e^t,t\ne 1,2,3,....\end{array}\right.$

$\left(b\right)f_2\left(t\right)=\left\{\begin{array}{l}{e}^t,t\ne 5,8,\\ 6,t=5,\\ 0,t=8,\end{array}\right.$

$\left(c\right)f_3\left(t\right)=e^t.$

Which of the preceding functions is the inverse Laplace transform of$\frac{1}{\left(s-1\right)}?$

Short answer

$$\begin{gathered} \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\frac{1}{s-1} \\ {\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}\left(t\right)=f_3\left(t\right)=e^t \end{gathered}$$

Step-by-step solution

Define Inverse Laplace transform

Given a function $F\left(s\right)$,if there is a function $f\left(t\right)$ that is continuous on

$[0,\infty )$and satisfies $\mathcal{L}\left\{f\right\}=F$,then we say that $f\left(t\right)$is the inverse Laplace transform of $F\left(s\right)$and employ the notation $f={\mathcal{L}}^{-1}\left\{F\right\}$

Non-repeated Linear Factors

If $Q\left(s\right)$can be factored into a product of distinct linear factors,

$Q\left(s\right)=\left(s-r_1\right)\left(s-r_2\right).....\left(s-r_n\right)$

where the $r_i$'s are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P\left(s\right)}{Q\left(s\right)}=\frac{A_1}{s-r_1}+\frac{A_2}{s-r_2}+.....+\frac{A_{n_{}}}{s-r_n}$

where the $A_i$'s are real numbers. There are various ways of determining the constants $A_1.....A_n$. In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If $s-r$is a factor of $Q\left(s\right)$and ${\left(s-r\right)}^m$is the highest power of $s-r$that divides $Q\left(s\right)$, then the portion of the partial fraction expansion of $P\left(s\right)/Q\left(s\right)$that corresponds to the term ${\left(s-r\right)}^m$is

$\frac{A_1}{s-r}+\frac{A_2}{{\left(s-r\right)}^2}+........+\frac{A_m}{{\left(s-r\right)}^m}$

where the $A_i$'s are real numbers.

Quadratic Factors

If ${\left(s-\alpha \right)}^2+{\beta }^2$is a quadratic factor of $Q\left(s\right)$ that cannot be reduced to linear factors with real coefficients and is the highest power of ${\left(s-\alpha \right)}^2+{\beta }^2$that divides $Q\left(s\right)$, then the portion of the partial fraction expansion that corresponds to ${\left(s-\alpha \right)}^2+{\beta }^2$is

$\frac{C_{1}s+D_1}{{\left(s-\alpha \right)}^2+{\beta }^2}+\frac{C_{2}s+D_2}{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}+......+\frac{C_{m}s+D_m}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^m}$

it is more convenient to express $C_{i}s+D_i$in the form $A_i\left(s-\alpha \right)+\beta B_i$when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_1\left(s-\alpha \right)+\beta B_1}{{\left(s-\alpha \right)}^2+{\beta }^2}+\frac{A_2\left(s-\alpha \right)+\beta B_2}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^2}+.......+\frac{A_m\left(s-\alpha \right)+\beta B_m}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^m}$

Find the factor of the denominator

a. b. c. Since the functions $f_1\left\{t\right\},f_2\left\{t\right\},$and $f_3\left\{t\right\}$differ at finite number of points and Laplace transform is a definite integral which doesn't depend on values of functions at finite number of points, we have that all three functions has the same Laplace transform.

Let's calculate it:

$$\begin{gathered} \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right) \\ =\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{e^t\right\}\left(s\right) \\ =\frac{1}{s-1} \\ Therefore, \\ \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\frac{1}{s-1} \end{gathered}$$

Since the inverse Laplace transform must be continuous function on $[0,\infty )$$f_1\left(t\right)$and $f_2\left(t\right)$ have discontinuities at some points we have that

${\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}\left(t\right)=f_3\left(t\right)=e^t$