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I have been trying this problem for a 1 week now but for some reason can not get my head around how to even approach this problem. $$ \frac{\omega K}{(sT+1)(s^{2}+ \omega^{2})} $$ I first tried using the most straight forward way of using: $$ \frac{\omega K}{(sT+1)(s^{2}+ \omega^{2})} = \frac{A}{(sT+1)} + \frac{B}{(s^{2}+ \omega^{2})} $$ And then I expanded this and tried to equate coefficients but could not get anywhere.

Could someone help me just start this problem?

Thank You in advance

1 Answers1

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$$\frac{a}{s T+1}+\frac{b+c s}{s^2+\omega ^2}\quad(*)$$ add the fractions $$\frac{s^2 (a+c T)+a \omega ^2+s (b T+c)+b}{(s T+1) \left(s^2+\omega ^2\right)}$$ numerator must be equal to the numerator of the given fraction $$\frac{\omega K}{(sT+1)(s^{2}+ \omega^{2})}$$ so we have

$ \left\{ \begin{array}{l} a \omega ^2+b=\omega K \\ c+b T=0 \\ a+c T=0 \\ \end{array} \right. $

which gives the solutions

$$a= T^2\,\color{red}{\frac{K \omega }{T^2 \omega ^2+1}},b= 1,\color{red}{\frac{K \omega }{T^2 \omega ^2+1}},c=-T\,\color{red}{\frac{K \omega }{T^2 \omega ^2+1}}$$

plug into $(*)$ $$\frac{T^2\color{red}{\left(\frac{K \omega }{T^2 \omega ^2+1}\right)}}{s T+1}+\frac{1\cdot\color{red}{\left(\frac{K \omega }{T^2 \omega ^2+1}\right)}- \color{red}{\left(\frac{K \omega }{T^2 \omega ^2+1}\right)} Ts}{s^2+\omega ^2}$$

Collect the common term $\dfrac{K \omega }{T^2 \omega ^2+1}$

and the given fraction as

$$\frac{\omega K}{(sT+1)(s^{2}+ \omega^{2})}=\frac{K \omega }{T^2 \omega ^2+1} \left(\frac{T^2}{s T+1}+\frac{1-s T}{s^2+\omega ^2}\right)$$

Hope it is clear

Raffaele
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  • Thank you for the solution, I understand the result up until you get the solution for a, b and c. Did you make each component the subject and then put it into the equation to get the results? –  Oct 15 '17 at 21:13
  • I collected the term in parenthesis which is common. I'll edit to make it clearer – Raffaele Oct 16 '17 at 07:44
  • I am still not sure where you got the common term T^{2}\omega^{2} + 1 –  Oct 21 '17 at 16:07
  • @MMS1986 I have rewritten part of the answer. Hope now is clear – Raffaele Oct 21 '17 at 16:26