0
$\begingroup$

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!

$\endgroup$
9
  • 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 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 at 2:57
  • $\begingroup$ @Ben, that seems ill-formed and inadvisable. $\endgroup$ Apr 9 at 3:26
  • 2
    $\begingroup$ You can't cancel the $A$ top and bottom in ${A+B\over A}$. $\endgroup$ Apr 9 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 at 14:16
0
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.