2

Im trying to do partial fraction but cant seem to get it right, and the examiner have not showed how he did the partial fraction, just the answer.

I want to partial fraction;

$$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$$

My solution so far:

$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})} = \frac{A}{s+\frac{1}{2}}+\frac{Bs+D}{(s+\frac{1}{2})^2+\frac{3}{4}} = \frac{A((s+\frac{1}{2})^2+\frac{3}{4})+(Bs+D)(s+\frac{1}{2})}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$

$\Rightarrow As^2+As+A+Bs^2+\frac{1}{2}Bs+Ds+\frac{1}{2}D=1$

Comparing coefficients gives me

$s^2:A+B=0$

$s^1:A+\frac{1}{2}B+\frac{1}{2}D=0$

$s^0:A+\frac{1}{2}D=1$

Which gives me $A=2, B=-2, C=-2$ which is wrong.

Can someone tell me what I'm doing wrong? Something tells me that its the $Bs+D$ that is wrong.

Thanks!

Sean Roberson
  • 9,110
uoiu
  • 593
  • 2
    Don't complete the square when you do partial fractions! Leave it as an irreducible quadratic. – Sean Roberson Oct 23 '22 at 18:08
  • So this problem is from an exercise in Laplacetransform, and the reason I'm writing $(s+\frac{1}{2})^2+\frac{3}{4}$ instead of $(s^2+s+1)$ is that we dont have any formulas to do Laplace inverse on the later expression. – uoiu Oct 23 '22 at 18:11
  • 3
    Why work in terms of $s$ at all? Make a substitution $t=s+\frac12$, do the (much easier) partial fraction decomposition there, then substitute back. – Steven Stadnicki Oct 23 '22 at 18:11
  • How will my $Bs+D$ term look like? I guess it wont be $Bt+D$ since we replaced $s$ – uoiu Oct 23 '22 at 18:18
  • 1
    Check the coefficients of $s^1$: it should be $D$ not $\frac12 D$ – David Quinn Oct 23 '22 at 18:20

1 Answers1

1

There are two approaches here - complete the square later, or as Steven Stadnicki suggests, do a change of variable to make your life easier. I'll show my intended solution of completing squares later.

Let $$F(s) = \frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}.$$

We can show that after re-expanding and clearing fractions, we can write $$F(s) = \frac{2}{(2s+1)(s^2 + s + 1)}$$ which makes our lives easier. Now do the usual expansion:

$$\frac{2}{(2s+1)(s^2 + s + 1)} = \frac{A}{2s+1} + \frac{Bs + C}{s^2 + s + 1}$$

and clearing fractions gives

$$2 = A(s^2 + s + 1) + (Bs + C)(2s+1).$$

One can equate coefficients here but I will opt to use the Heaviside cover-up method. With $s = -\frac{1}{2}$ we get $A = \frac{8}{3}.$ For $s = 0$ we obtain the equation $2 = \frac{8}{3} + C$ which gives $C = -\frac{2}{3}.$ Finally, as we have run out of "nice" $s$ values, we'll choose $s = 1$ which will give $B = -\frac{4}{3}$ (check this!). Thus our expansion is

$$F(s) = \frac{\frac{8}{3}}{2s+1} + \frac{-\frac{4}{3}s - \frac{2}{3}}{s^2 + s + 1}$$

and, after re-completing the square and normalizing,

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

The inverse Laplace transform should then be easy to compute (an exponential, a sine, and a cosine).

Sean Roberson
  • 9,110