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If I have the following circuit:

enter image description here

I know that I can find the transfer function by doing voltage division. However, in this example, where does the negative sign in front of R1 come from? $$T(s)=\frac{-Z_2(s)}{Z_1(s)+Z_2(s)}=\frac{-R_1}{R_1+R_2+L \cdot s}$$

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  • \$\begingroup\$ I would say because of \$i_2\$ direction compared to that of \$i_1\$? To stick to a low-entropy format, you can factor \$\frac{R_1}{R_1+R_2}\$ which is your dc gain (\$s=0\$) while the denominator will be in the form of \$1+\frac{s}{\omega_p}\$. Your transfer function correctly expressed is thus \$T(s)=-T_0\frac{1}{1+\frac{s}{\omega_p}}\$ \$\endgroup\$ – Verbal Kint Nov 2 '19 at 18:38
  • \$\begingroup\$ @VerbalKint But why don't we put a negative sign in front of R2 and L then? \$\endgroup\$ – user164324 Nov 2 '19 at 18:48
  • \$\begingroup\$ If you simulate this circuit, \$i_2\$ circulates in the same direction as \$i_1\$, it's leaving \$L\$. So its value is \$R_1.i_1\$ divided by the sum of \$R_1\$ plus the impedance made of \$R_2\$ and \$sL\$ in series. If you decide to consider \$i_2\$ in the opposite direction, you add a minus sign in front of its definition. \$\endgroup\$ – Verbal Kint Nov 2 '19 at 20:24
  • \$\begingroup\$ How does voltage division relate to this question? \$T\small (s)\$ is (output current)/(input current). \$\endgroup\$ – Chu Nov 2 '19 at 20:58
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If you apply the current divider formula, and observe the direction of \$i_2\$, the result is:

\$T(s) = -\frac {Z_T}{Z_2}\$ where

\$Z_T =\frac{R_1(R_2+sL)}{R_1+R_2+sL}\$ and

\$Z_2 = R_2 + sL\$

so

\$T(s) = -\frac {R_1}{R_1+R_2+sL}\$

The negative sign only arises because \$i_2\$ is defined in the opposite direction to \$i_1\$.

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