Partial fraction - calculate $k_1$ $k_2$ and $k_3$

Question

I have the following problem:

Calculate $k_1$, $k_2$ and $k_3$ in this equation:

$$\frac{s+13}{(s+2)(s+11)(s+8)}=\frac{k_1}{s+2}+\frac{k_2}{s+11}+\frac{k_3}{s+8}$$

I've searched on the internet and read in the book my instructor but it is very poorly explained and I don't know how to solve this problem.

Can someone help me with this?

Answer 1

The goal is to find the unknowns by making true the identity you wrote above. For this you have to take the term on the right side and develop it:

$$\frac{k_1}{s+2} + \frac{k_2}{s+11} + \frac{k_3}{s+8} = \frac{k_1(s+11)(s+8) + k_2(s+2)(s+8) + k_3(s+2)(s+11)}{(s+2(s+11)(s+8)}$$

Now you have to develop and arrange the numerator by collecting the terms in $s^2$, $s$ and the constant terms, as functions of $k_i$:

The numerator becomes:

$$k_1(s^2 + 19s + 88) + k_2(s^2 + 10s + 16) + k_3(s^2 + 13s + 22)$$

That is

$$s^2(k_1 + k_2 + k_3) + s(19k_1 + 10k_2 + 13k_3) + 88k_1 + 16k_2 + 22k_3$$

Now it's about to check TERM BY TERM with your initial function: you see there are NO terms in $s^2$ hence it must be

$$k_1 + k_2 + k_3 = 0$$

You see the term in $s$ must have coefficient $1$ so you need

$$19k_1 + 10k_2 + 13k_3 = 1$$

And eventually the constant is $13$ so you need

$$88k_1 + 16k_2 + 22k_3 = 13$$

Note that this gives you a system of three equations in three unknown. Can you go on from here? This will give you the $k_i$ terms.

Answer 2

$\frac{k_1}{s+2}+\frac{k_2}{s+11}+\frac{k_3}{s+8} =\frac{k_1(s+11)(s+8)+k_2(s+2)(s+8)+k_3(s+2)(s+11)}{(s+2)(s+11)(s+8)}$

so multiply out the numerator on the right hand side and find the co-coefficients of $s^2$, $s^1$ and $s^0$. Compare these with the coefficients in $s+13 = 0s^2+1s^1+13s^0$

That gives you three simultaneous equations in three unknowns, which you can solve.

For example for the coefficient of $s^2$ you get $k_1s^2+k_2s^2+k_3s^2$ so $k_1+k_2+k_3=0$. I will leave the others to you.

Answer 3

A simple trick is to write:

$\begin{align*} \frac{s+13}{(s+2)(s+11)(s+8)} &=\frac{k_1}{s+2}+\frac{k_2}{s+11}+\frac{k_3}{s+8} \end{align*}$

Multiply e.g. by $s + 8$ and compute the limit as $s \to - 8$:

$\begin{align*} \lim_{s \to -8} \frac{s+13}{(s+2)(s+11)} &= \lim_{s \to -8} \frac{k_1 (s + 8)}{s+2} + \lim_{s \to -8} \frac{k_2 (s + 8)}{s+11} + \lim_{s \to -8} k_3 \\ \frac{-8 + 13}{(-8 + 2) (- 8 + 11)} &= k_3 \\ k_3 &= - \frac{5}{18} \end{align*}$

Answer 4

$$ \frac{s+13}{(s+2)(s+11)(s+8)} =\frac{k_1}{s+2}+\frac{k_2}{s+11}+\frac{k_3}{s+8} $$

Another trick is to multiply through by $(s+2)(s+11)(s+8)$

$$ s+13 = k_1(s+11)(s+8)+k_2(s+2)(s+8)+k_3(s+2)(s+11) $$

Note, for example, that (s+8) is a factor of two out of three of the terms on the right. So, if we let $s=-8$ and simplify, we get

\begin{align} 5 &= -18 k_3 \\ k_3 &= -\dfrac{5}{18} \end{align}

The other two coefficients can be found in the same way.