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Determine $\alpha,\beta,\gamma,\delta$ in

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

I have come across the partial fraction and trying to decompose it. I have tried the methods used by multiplying out and making one variable to 0 to find the others etc but It does not work here. Any ideas how I could approach this?

String
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fred
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  • Can you show us what you have tried? If you really have substituted $s = ±5i, -3 ± 5i$ to make one of the terms zero, then there might be an arithmetic error. – Toby Mak Mar 05 '21 at 12:53
  • Is the question to be understood as: $$\frac{10(s+6)}{\left(s+3)^2+25\right}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$ Now determine $\alpha,\beta,\gamma,\delta$? – String Mar 05 '21 at 12:56
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    This already looks like a good approach to me. Combine the right two fractions into a single fraction, and you get some polynomial in $s$ with coefficients in $\alpha, \beta, \gamma, \delta$. Now you just need to find values for those variables so that each coefficient of the numerator has the correct value. This will just be a system of four linear equations in four unknowns. – Magma Mar 05 '21 at 12:57

3 Answers3

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Hint:Multiply both sides by the denominator of the l.h.s.: $$10s+60=(\alpha s+\beta)(s^2+25)+(\gamma s+\delta)\bigl((s+3)^2+25\bigr),\tag{1}$$ and set $s$ to be successively roots of each factor in the denominator (i.e. poles of the fraction), $5i$ and $-3+5i$, and identify the real and the imaginary parts of each side, to obtain two linear systems in $\alpha,\beta$ and in $\gamma,\delta$ respectively.

Bernard
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$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

$$10(s+6) =(\alpha s+\beta)(s^2+25)+ (\gamma s+\delta)((s+3)^2+25) $$

$$s=0:\;\; 60= 25\beta + 34 \delta \tag{1}$$ $$s=1:\;\; 70 =26(\alpha +\beta)+ 41(\gamma +\delta) \tag{2}$$ $$s=-1:\;\; 50 =26(-\alpha +\beta)+ 29(-\gamma +\delta) \tag{3}$$ $$s=2:\;\; 80 =29(2\alpha +\beta)+ 50(2\gamma +\delta) \tag{4}$$

I'm not going to write out the rest of the algebra, but it is simple from there to solve for $\alpha,\beta, \gamma , \delta$, if a bit tedious.

Blitzer
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Apply your method to proceed as follows

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)} =\frac{A}{s+5i} + \frac{B}{s-5i} +\frac{C}{s+3+i5}+ \frac{D}{s+3-i5}$$ Then \begin{align} &A = \lim_{s\to-5i} \frac{10(s+6)(s+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45+68i}{327}\\ &B = \lim_{s\to5i} \frac{10(s+6)(s-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45-68i}{327}\\ &C = \lim_{s\to-3-5i} \frac{10(s+6)(s+3+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45-41i}{327}\\ &D= \lim_{s\to-3+5i} \frac{10(s+6)(s+3-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45+41i}{327}\\ \end{align}

and \begin{align} & \lambda = A+B = -\frac{90}{327}\\ & \delta =(B -A)\>5i= \frac{680}{327}\\ & \alpha = C +D = \frac{90}{327}\\ & \beta =(C+D) \>3+(D -C ) \>5i= -\frac{140}{327}\\ \end{align}

Thus $$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)} =\frac{10}{327}\left( \frac{68-9s}{s^2+25} + \frac{9s-14}{(s+3)^3+25}\right)$$

Quanto
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