Inverse Laplace transform with partial fraction

Question

I have the transform below:

$$\frac{(7s+2)(2s-5)}{{s^2}(s-2)}$$

I think this is should be partial fraction to be solved. Can you please help me figure how to consider A, B, C and denominators?

Answer 1

Before doing partial function expansion, we need to make sure that the numerator has a degree smaller than the denominator. If not, we do long division in order to write the fraction as a polynomial and a fraction satisfies the degree condition.

Next, we factor the denominator. For the fraction in the question, both steps are unnecessary as the fraction already satisfies the degree condition and the denominator is already factored.

Now, we consider each term in the denominator:

• If it's of the form $(s-a)$, expect $\frac{A}{s-a}$ in the partial fraction expansion.
• If it's of the form $(s-a)^n$, expect $\frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$.
• $(s)$ can be treated as $(s - 0)$.
• There are other cases not needed here. See this answer for more information on them.

Given the above, we can write the fraction in the question as:

$$ \frac{(7s+2)(2s-5)}{{s^2}(s-2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-2} $$

What's left is plugging in values for $s$ to find $A$, $B$ and $C$. There is an quick way to do it.

• Multiply both sides by $(s-2)$ and plug in $s=2$ to get: $$ \frac{(14 + 2)(4 - 5)}{4} = 0 + 0 + C \Rightarrow C = -4 $$

• Multiply both sides by $s^2$ and plug in $s=0$ to get: $$ \frac{(0+2)(0-5)}{-2} = 0 + B + 0 \Rightarrow B = 5 $$

• Plug in $s=1$: $$ \frac{(7+2)(2-5)}{1\cdot(1-2)} = A + 5 + \frac{-4}{-1} \Rightarrow A = 18 $$

And the fraction becomes:

$$ \frac{(7s+2)(2s-5)}{{s^2}(s-2)} = \frac{18}{s} + \frac{5}{s^2} - \frac{4}{s-2} $$

Answer 2

Hint: write $\left( 7s+2\right) \left( 2s-5\right) /\left( s^{2}(s-2)\right) $ as

$$\frac{\left( 7s+2\right) \left( 2s-5\right) }{s^{2}(s-2)}=\frac{ 14s^{2}-31s-10}{s^{2}(s-2)}=\frac{A}{s^{2}}+\frac{B}{s}+\frac{C}{s-2}.\tag{1}$$

See e.g. this entry of the Wikipedia, where several methods are explained. The theory and the computation of each partial fraction are the same as when integrating a rational function by expansion into partial fractions.

Added: One of the methods is as follows. Since $$\begin{eqnarray*} \frac{A}{s^{2}}+\frac{B}{s}+\frac{C}{s-2} &=&\frac{A(s-2)+Bs(s-2)+Cs^{2}}{ s^{2}(s-2)} \\ &=&\frac{\left( B+C\right) s^{2}+\left( A-2B\right) s-2A}{s^{2}(s-2)},\tag{2} \end{eqnarray*}$$

then the numerators of $(1)$ and $(2)$ must be equal for all $s$ $$\begin{equation*} 14s^{2}-31s-10=\left( B+C\right) s^{2}+\left( A-2B\right) s-2A. \end{equation*}$$ So $$\begin{eqnarray*} -2A &=&-10\Rightarrow A=5 \\ A-2B &=&5-2B=-31\Rightarrow B=18 \\ B+C &=&18+C=14\Rightarrow C=-4,\tag{3} \end{eqnarray*}$$

which means that $$\begin{equation*} \frac{\left( 7s+2\right) \left( 2s-5\right) }{s^{2}(s-2)}=\frac{5}{s^{2}}+ \frac{18}{s}+\frac{-4}{s-2}. \end{equation*}\tag{4}$$

I've detailed in this answer the computation in a similar case.

Answer 3

$$\frac{(7s+2)(2s-5)}{s^2(s-2)} = \frac{5}{s^2} + \frac{18}{s} - \frac{4}{s-2}$$ Applying Inverse Laplace Transform, we get, $$ L^{-1} \left ( 5\frac{1}{s^2}\right) + 18 L^{-1} \left ( \frac{1}{s}\right) -4 L^{-1} \left ( \frac{1}{s-2}\right) = 5 t + 18- e^{2t}$$

Quick way to check.