Can this expression be simplified any further?

I am attempting to simplify the following expression with regards to obtaining a transfer function;

$$ I_2(s)(\dfrac{C(L_1+L_2)s^2+CR_2s+1}{CL_1s^2+1})(\dfrac{CL_1s^2+CR_1s+1}{Cs})-I_2(s)(\dfrac{CL_1s^2+1}{Cs})=E_i(s)\tag{1} $$

After obtaining a common denominator, expanding out the brackets and combining like terms, I get this:

$$I_2(s)\dfrac{(C^3L_1(L_1+L_2)s^5 +(C^3R_1L_1+C^3R_1L_2+C^3R_2L_1-C^2L_1^2)s^4+C^2(CR_1R_2+L_1+(L_!+L_2))s^3+(C^2R_1+C^2R2-2CL_1)s^2+Cs-1}{(CL_1s^2+1)(Cs)}=E_i(s)\tag{2}$$

My main problem is simplifying the $s^4$ and the $s^2$ terms further, i.e., eliminating their negative terms, as it currently stands. If, I substituted in values for these parameters, I am able to simplify it further. However, I am stuck, as it currently stands.

Any insight that anyone can provide will be greatly appreciated, as my Maths is a quite rusty and i'm just trying to get back into the swing of things.

I am ideally trying to reduce the numerator to only include positive terms!

  • 2
    $\begingroup$ Are you sure you made a common denominator correctly? At first glance I thought it was fine, but now I see the first term has two big factors and already has the full denominator, so doesn't have to get multiplied by $Cs$. I don't think you are supposed to get an $s^5$ term at all. Apologies if I'm wrong, but that's how it looks to me. $\endgroup$ Apr 9, 2021 at 1:53
  • $\begingroup$ The first term's denominator may be reduced to $C_s$. $$\require{cancel}\frac{C(L_1+L_2)s^2+CR_2s+1} {\cancel{CL_1S^2+1})} \frac{\cancel{(CL_1S^2+1}+CR_1s}{C_s}$$ $\endgroup$
    – Benp404
    Apr 9, 2021 at 2:57
  • $\begingroup$ @Ben, that seems ill-formed and inadvisable. $\endgroup$ Apr 9, 2021 at 3:26
  • 2
    $\begingroup$ You can't cancel the $A$ top and bottom in ${A+B\over A}$. $\endgroup$ Apr 9, 2021 at 3:35
  • $\begingroup$ It is not trivial. Please plot it. Then, we can work with e. g. local monotony (by calculating derivatives) and a specific value of the numerator, but I need a picture. $\endgroup$ Apr 9, 2021 at 14:16

1 Answer 1


The answer to my question is "YES" and these are the following steps to obtaining a simplified expression:

$$ I_2(s)(\dfrac{C(L_1+L_2)s^2+CR_2s+1}{CL_1s^2+1})(\dfrac{CL_1s^2+CR_1s+1}{Cs})-I_2(s)(\dfrac{CL_1s^2+1}{Cs})=E_i(s)\tag{1} $$

  1. Factor out the $I_2(s)$

$$ I_2(s)[(\dfrac{C(L_1+L_2)s^2+CR_2s+1}{CL_1s^2+1})(\dfrac{CL_1s^2+CR_1s+1}{Cs})-(\dfrac{CL_1s^2+1}{Cs})]=E_i(s)$$

  1. Cross-multiple the denominators of the two fractional terms being subtracted and mutilple the denominators

$$ I_2(s)[(\dfrac{(Cs)(C(L_1+L_2)s^2+CR_2s+1)(CL_1s^2+CR_1s+1)-(CL_1s^2+1)(Cs)(CL_1s^2+1)}{(CL_1s^2+1)(Cs)(Cs)})]=E_i(s) $$

  1. Expanding the brackets in the numerator, neglecting the $I_2(s)$ term

$$ C_s(C(L_1+L_2)s^2(CL_1s^2)+C(L_1+L_2)s^2(CR_1s)+C(L_1+L_2)s^2(1)+(CR_2s)(CL_1s^2)+(CR_2s)(CR_1s)+(CR_2s)(1)+(1)((CL_1s^2)+(1)(CR_1s)+(1)(1))-[(C^2L_1s^3)(CL_1s^2)+(C^2L_1s^3)(1)+(Cs)((CL_1s^2)+(Cs)(1)]=E_1(s) $$

  1. Multiplying out inner bracket

$$C_s(C^2L_1(L_1+L_2)s^4+C^2R_1(L_1+L_2)s^3+C(L_1+L_2)s^2+C^2R_2L_1s^3+C^2R_1R_2s^2+CR_2s+CL_1s^2+CR_1s+1)-[C^3L_1^2s^5+C^2L_1s^3+C^2L_1s^3+Cs]=E_1(s) $$

  1. Multiplying out outer bracket

$$C^3L_1(L_1+L_2)s^5 + C^3R_1(L_1+L_2)s^4 + C^2(L_1+L_2)s^3 + C^3R_2L_1s^4 + C^3R_1R_2s^3 + C^2R_2s^2 + C^2L_1s^3 + C^2R_1s^2+Cs) - C^3L_1^2s^5 - C^2L_1s^3 - C^2L_1s^3 - Cs = E_1(s) $$

  1. Multiplying out further

$$C^3L_1^2s^5 + C^3L_1L_2s^5 + C^3R_1(L_1+L_2)s^4 + C^2L_1s^3 + C^2L_2s^3 + C^3R_2L_1s^4 + C^3R_1R_2s^3 + C^2R_2s^2 + C^2L_1s^3 + C^2R_1s^2+Cs)- C^3L_1^2s^5 -C^2L_1s^3 - C^2L_1s^3 - Cs $$

  1. Cancelling out like terms

$$\cancel{C^3L_1^2s^5} + C^3L_1L_2s^5 + C^3R_1(L_1+L_2)s^4 + \cancel{C^2L_1s^3} + C^2L_2s^3 + C^3R_2L_1s^4 + C^3R_1R_2s^3 + C^2R_2s^2 + \cancel{C^2L_1s^3} + C^2R_1s^2 + \cancel{Cs}) - \cancel{C^3L_1^2s^5} - \cancel{C^2L_1s^3} - \cancel{C^2L_1s^3} - \cancel{Cs} $$

  1. Which leaves

$$ C^3L_1L_2s^5 + C^3R_1(L_1+L_2)s^4 + C^2L_2s^3 + C^3R_2L_1s^4 + C^3R_1R_2s^3 + C^2R_2s^2 + C^2R_1s^2 $$

  1. And therefore

$$ I_2(s)(\dfrac{(C^3L_1L_2s^5 + C^3R_1(L_1+L_2)s^4 + C^2L_2s^3 + C^3R_2L_1s^4 + C^3R_1R_2s^3 + C^2(R_1+R_2)s^2}{C^2L_1s^4+C^2s^2}) = E_i(s)$$

  1. Simplifying

$$ I_2(s)(\dfrac{(C^3L_1L_2s^5 + C^3(R_1(L_1+L_2) + R_2L_1)s^4 + C^2(CR_1R_2+L_2)s^3 + C^2(R_1+R_2)s^2}{C^2L_1s^4+C^2s^2}) = E_i(s)$$

  1. Finally, expressing in terms of $I_2(s)/E_i(s)$ I get the simplified transfer function:

$$ \dfrac{I_2(s)}{E_i(s)}=\dfrac{C^2L_1s^4+C^2s^2}{(C^3L_1L_2s^5 + C^3(R_1(L_1+L_2) + R_2L_1)s^4 + C^2(CR_1R_2+L_2)s^3 + C^2(R_1+R_2)s^2} \tag{2} $$


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