Why partial fraction decomposition of $\frac{1}{s^2(s+2)}$ is $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$?

Question

Can someone please explain why: $$\frac{1}{s^2(s+2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$$

And not:$$\frac{1}{s^2(s+2)}=\frac{A}{s^2}+\frac{B}{(s+2)}$$

I'm a bit confused where the extra s term comes from in the first equation.

Answer 1

The general result is the following.

Suppose that the degree of $p(s)$ is less than the degree of $q(s)$, and that $q(s)=q_1(s)q_2(s)$ where $q_1(s)$ and $q_2(s)$ have no common factor. Then there exist polynomials $r_1(s)$ and $r_2(s)$, with degrees less than $q_1(s)$ and $q_2(s)$ respectively, such that $$\frac{p(s)}{q(s)}=\frac{r_1(s)}{q_1(s)}+\frac{r_2(s)}{q_2(s)}\ .$$

In your case the denominator factorises as $s^2$ times $s+2$ so you have $$\frac1{s^2(s+2)}=\frac{As+B}{s^2}+\frac{C}{s+2}\ .$$ It is then usually more convenient (though not obligatory) to split up the first fraction, which gives your answer.

Note that you cannot, for the purposes of the above result, regard the denominator as $s$ times $s(s+2)$, because these polynomials do have a common factor.

Answer 2

That is because for $$\frac{as^2+bs+c}{s^2(s+2)}=\frac{A}{s^2}+\frac{B}{(s+2)},$$ the left hand side has three parameters $a,b,c$, but the right hand side only has two parameters $a,b$. And if you try to solve TWO values from THREE equations, it will usually lead to a contradiction. So a third term of the right is needed. Even though this is not obvious in your question, you should think 1 as a degree 2 polynomial.

Or more simply, consider the example $$ \frac{s+1}{s^2}=\frac{1}{s^2}+\frac{1}{s} $$

Answer 3

If your denominator has a factor of the form $(as+b)^n$ then to write partial fractions you should write all the powers up to $n$, i.e. $\frac{A}{as+b}+\frac{B}{(as+b)^2}+\cdots+\frac{Z}{(as+b)^n}$. In the case you showed, you have that $s^2$ is a factor of the denominator and that's why in partial fractions you should write the terms $\frac{A}{s}+\frac{B}{s^2}$.

Answer 4

One can immediately see why in this case the partial fraction expansion will lead to a nonzero coefficient for the $1/s$ term. The asymptotic behavior of the fraction for large $s$ is $\sim 1/s^3$. The singularity at $s = -2$ contributes a term proportional to $1/(s+2)$ to the partial fraction expansion, which for large $s$ behaves like $\sim 1/s$. This $\sim 1/s$ must be canceled out by the partial fraction expansion terms coming from the singularity at $s = 0$, this requires the presence of a contribution proportional to $1/s$.

By making this reasoning more precise we can get to the complete partial fraction expansion using only the contribution from the singularity at $s = -2$. The amplitude of the $1/(s+2)$ term in the partial fraction expansion is given by the factor that multiplies it in the fraction evaluated at $s = -2$, this is therefore equal to $1/4$. So the contribution to the partial fraction expansion coming from the singularity at $s = -2$ is:

$$\frac{1}{4(s+2)}$$

For large $s$ we can expand this in powers of $1/s$:

$$\frac{1}{4(s+2)} = \frac{1}{4 s}\frac{1}{1+\frac{2}{s}} = \frac{1}{4s} - \frac{1}{2 s^2} + \mathcal{O}\left(\frac{1}{s^3}\right)$$

The singularity at $s = 0$ will contribute terms to the partial fraction expansion whose large $s$ behavior will have to cancel out these first two terms, this means that this contribution to the partial fraction expansion is:

$$\frac{1}{2 s^2}-\frac{1}{4s} $$

The complete partial fraction expansion is thus given by:

$$\frac{1}{2 s^2}-\frac{1}{4s} + \frac{1}{4(s+2)} $$

Answer 5

You can do it step-by-step: $$\begin{align}\frac{1}{s^2(s+2)}&=\frac1s\cdot \frac{1}{s\cdot (s+2)}=\\ &=\frac1s\cdot \left(\frac As+\frac{B}{s+2}\right)=\\ &=\frac A{s^2}+\frac1s\cdot \frac B{s+2}=\\ &=\frac{A}{s^2}+\frac{C}{s}+\frac{D}{s+2}.\end{align}$$

Answer 6

Suppose we multiplied both sides of the second equation by $s^2(s-2)$, giving us the equivalent equation: $$1 = 0s^2 + 0s + 1= A(s+2) + Bs^2 = Bs^2 + As + 2A$$ Notice that on the RHS of this equation, that the constant term is not independent of the coefficient of the $s$ term. This dependency, in turn, causes a contradiction.

Now let's try the same thing with your first equation. This generates the equivalent equation, whose coefficients can be uniquely determined: $$\begin{align} 1 &= As(s+2) + B(s+2)+Cs^2\\ 0s^2+0s+1 &= (A+C)s^2+(2A+B)s+2B\\ \end{align}$$

The reason, of course, is because ${\{1,s,s^2 \}}$ represents an independent set that spans the set of all polynomials up to degree two, requiring three parameters to determine its coefficients uniquely.

I hope this helps.