TF EIM Chapt3

gain

Loop Theorem

$\Rightarrow V= I(R+X_{tot}) = I \left (R+ \frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} \right )$

or

$I= \frac{V_0 e^{i \omega t}}{\left (R+ \frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} \right )}$

Notice

When $\omega \approx \omega_{LC} = \sqrt{\frac{1}{LC}}$ then the AC signal is attenuated.

Looking at the Voltage divider aspect of the circuit

$V_{AB}=V_{out} = \frac{X_{tot} }{R + X_{tot}}V_{in}$

$\left |\frac{ V_{out}} {V_{in}}\right | = \sqrt{ \left [ \frac{X_{tot} }{R + X_{tot}} \right ] \left [ \frac{X_{tot} }{R + X_{tot}} \right ]^*}$

$= \sqrt{ \left [ \frac{\frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} }{\left (R+ \frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} \right )} \right ] \left [ \frac{\frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} }{\left (R+ \frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} \right )} \right ]^*}$

$= \sqrt{ \frac{ \omega^2 L^2 \omega_{LC}^4}{R^2(\omega_{LC}^2 - \omega^2) - \omega^2L^2 \omega_{LC}^4}}$

$= \sqrt{ \frac{ \omega^2 }{C^2R^2(\omega_{LC}^2 - \omega^2) - \omega^2}}$

Forest_Electronic_Instrumentation_and_Measurement