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In the design of an electrical circuit, after solving some Laplace transforms (see stack exchange electrical engineering question here for full details), the following expression appears:

$$ ratio = \frac{ln(1+R_4/R_2)}{ln(1+R_3/R_2)+ln(1+R_4/R_2)} $$

Simplification when $R_3,R_4 << R_2$ is easy using Taylor series.

The question is, are there any simplifications applicable when $R_3+R_4=R_2$? Approximations with an error smaller than 5%, 10% are perfectly valid.

It could be also valid an approximation for any other fixed proportion between $R_3+R_4$ and $R_2$.

The objective is find a lineal approximation of the expression. By example, if $R_3+R_4 = R << R_2$ it is possible simplify as $ratio ~ = ~ R_4/R ~ - ~ R_4^2 / 2(R_3^2+R_4^2)+...$ being the first term the lineal control $R_4/R$ and second term the first distortion.

pasaba por aqui
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  • Can you show us how you derived that equation? – Jan Eerland May 12 '17 at 15:50
  • The origin of the expression is in the answer cited. In short, a capacitor charges in T_L time, discharges in T_H, charges again, ... . Both, T_L and T_H are log expressions in function of circuit resistors R_2, R_3 and R_4. The target is the relation between charging time and full period of charge-discharge cicle: T_L/(T_L+T_H). This one is the expression I like to simplify. Recall a capacitor is lineal with Laplace transfer function V(s)/I(s)=1/Cs, that in time space produces exponential expressions. – pasaba por aqui May 12 '17 at 15:56

2 Answers2

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Well, okay what do we have:

$$\frac{\ln\left(1+\frac{\text{R}_4}{\text{R}_2}\right)}{\ln\left(1+\frac{\text{R}_3}{\text{R}_2}\right)+\ln\left(1+\frac{\text{R}_4}{\text{R}_2}\right)}=\frac{\ln\left(1+\frac{\text{R}_4}{\text{R}_2}\right)}{\ln\left(\left(1+\frac{\text{R}_3}{\text{R}_2}\right)\left(1+\frac{\text{R}_4}{\text{R}_2}\right)\right)}=$$ $$\frac{\ln\left(\text{R}_2+\text{R}_4\right)-\ln\left(\text{R}_2\right)}{\ln\left(\text{R}_2+\text{R}_3\right)+\ln\left(\text{R}_2+\text{R}_4\right)-2\ln\left(\text{R}_2\right)}\tag1$$

Jan Eerland
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We can simplify by letting $r_3 = R_3/R_2, r_4 = R_4/R_2$. Further, suppose that $r_3 = x-y, r_4 = x+y$. The ratio is now $$\frac{\ln (1+r_4)}{\ln (1+(r_3+r_4)+r_3r_4)} = \frac{\ln ((1+x)+y))}{\ln ((1+x)^2-y^2))}.$$ For a fixed value of $x$ and with $y$ small, let $r=(y/(1+x))/\ln(1+x)$. Approximation is$$\frac{\ln(1+x) + y/(1+x)}{2\ln(1+x) - (y/(1+x))^2} = \frac{1+r}{2-r^2\ln(1+x)}.$$

Somos
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