Voltage Transfer Function (Part 1)

Let the input voltage be $V_S = \sin(\omega t)$ . Since the source frequency is $\omega$ , inductive, resistive and capacitive reactances are:

$Z_L = jL\omega$ $Z_R = R$ $Z_{C1} = \frac{-j}{\omega C_1}$ $Z_{C2} = \frac{-j}{\omega C_2}$

In steady-state, the source voltage magnitude is defined, for convenience, as $V_S = 1 + j0$

Now, let the steady stare source current be $I$ , that through the inductive branch be $I_1$ and through the capacitive branch be $I_2$ . Then, according to Kirchoff's laws:

$-V_S + Z_{C1}I + Z_{C2}I_2 = 0$ $(R + Z_L)I_1 = Z_{C2}I_2$ $I=I_1 + I_2$

Solving for $I_1$ gives:

$I_1 = \frac{V_S}{Z_{C1} + \left( 1 + \frac{Z_{C1}}{Z_{C2}}\right)\left(R + Z_L\right)}$ $V_R = \frac{V_SR}{Z_{C1} + \left( 1 + \frac{Z_{C1}}{Z_{C2}}\right)\left(R + Z_L\right)}$

The above quantity $V_R$ is a complex number. According to the problem statement, this leads to the conclusion that:

$\alpha \mathrm{e}^{j \theta} = \frac{R}{Z_{C1} + \left( 1 + \frac{Z_{C1}}{Z_{C2}}\right)\left(R + Z_L\right)}$

Substituting all values and $\omega = 1$ , and computing $V_R$ leads to the following complex number:

$\alpha \mathrm{e}^{j \theta} =0.2308 -j0.1538$

This complex number has an argument:

$\theta = - \arctan\left(\frac{0.1538}{0.2308}\right)$ $\alpha = \sqrt{0.2308 ^2 + 0.1538^2}$

The answer turns out to be $\boxed{\alpha \theta \approx -0.1631}$