# Finding a transfer function by voltage division

If I have the following circuit:

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? $$T(s)=\frac{-Z_2(s)}{Z_1(s)+Z_2(s)}=\frac{-R_1}{R_1+R_2+L \cdot s}$$

• 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 $T(s)=-T_0\frac{1}{1+\frac{s}{\omega_p}}$ – Verbal Kint Nov 2 '19 at 18:38
• @VerbalKint But why don't we put a negative sign in front of R2 and L then? – user164324 Nov 2 '19 at 18:48
• 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. – Verbal Kint Nov 2 '19 at 20:24
• How does voltage division relate to this question? $T\small (s)$ is (output current)/(input current). – Chu Nov 2 '19 at 20:58

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\$$.