While solving Laplace transform using Partial fraction expansion. I have confusion in solving partial fraction for complex roots.

I have this equation $$ \frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$

Please anyone help to tell me to understand the steps for solving partial fraction for complex roots


It is the same story as with real roots.

Consider $$A=\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}$$ the first partial fraction decomposition gives $$A=\frac{s+1}{s^2+2 s+10}+\frac{1}{s+2}$$ So, now, just focus on $$B=\frac{s+1}{s^2+2 s+10}=\frac{s+1}{(s+1+3i)(s+1-3i)}=\frac \alpha {(s+1+3i)}+\frac \beta {(s+1-3i)} $$ Reducing to same denominator $$s+1=\alpha(s+1+3i)+\beta (s+1-3i)=(\alpha+\beta)(s+1)+3(\alpha-\beta)i$$ Idenitfyng the real and imaginary parts than gives $$\alpha+\beta=1 \, \, \, \quad \alpha-\beta=0$$ that is to say $\alpha=\beta=\frac 12$.


To clarify the first partial fraction decomposition $$\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}=\frac a{s+2}+\frac {b+cs}{s^2+2s+10}$$ Reducing to same denominator and removing the common denominator $$2s^2+5s+12=a(s^2+2s+10)+(b+cs)(s+2)$$ Expanding the rhs and grouping terms $$2s^2+5s+12=(a+c)s^2+ (2 a+b+2 c) s+(10 a+2 b)$$ Comparing the coefficients $$a+c=2 \quad, 2a+b+2c=5 \quad, 10a+2b=12$$ Solving the three equations for the three unknowns leads to $a=b=c=1$.

  • $\begingroup$ Here I want to ask 2 things, 1) how you separated numerator after the first partial fraction decomposition gives: 2) how can I make pairs of $s^2+2s+10$ in the form of $i$? $\endgroup$ – Shinning Eyes Nov 26 '15 at 7:39
  • $\begingroup$ The first point is the standard decomposition. For the second, just solve the quadratic equation as usual. $\endgroup$ – Claude Leibovici Nov 26 '15 at 7:59
  • $\begingroup$ I am asking about $2s^2 + 5s +12$ how you decomposed in to $s+1$? $\endgroup$ – Shinning Eyes Nov 26 '15 at 9:48
  • $\begingroup$ $$\frac {2s^2+5s+12} {(s^2+2s+10)(s+2)}=\frac a{s+2}+\frac {b+cx}{s^2+2s+10}$$ Reduce to same denominator and identify the coefficients for the same power of $x$. This will give $a,b,c$. $\endgroup$ – Claude Leibovici Nov 26 '15 at 9:57
  • $\begingroup$ how can we find the values of $a,b, and c$? $\endgroup$ – Shinning Eyes Nov 26 '15 at 10:39

You would write it as

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

and use the formula $$L(e^{at}\cos(bt))={s-a\over (s-a)^2+b^2}$$


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