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

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

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

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

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

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

$= \sqrt{ \frac{ \omega^2 }{\frac{1}{\omega_{RC}^2}(\omega_{LC}^2 - \omega^2)^2 - \omega^2}}$

$L = 33 \mu H$ and $C = 1 \mu F$ and R=200 $\Omega$

$\Rightarrow \omega = 174077.66$ rad/s or $\nu = \frac{\omega}{2 \pi} = 27705$ Hz

Q and Bandwidth

In the above circuit

$\nu = \omega/2 \pi = \frac{1}{2 \pi \sqrt{LC}}= 27705 Hz$

The inductors reactance at this resonance frequency is

$X_{L} = \omega L = 2 \pi \nu L = \frac{L}{\sqrt{LC}} = \sqrt{\frac{L}{C}} = \sqrt{\frac{33 \times 10^{-6} H}{1 \times 10^{-6}F}} = 5.7 \Omega$

Bandwith

The most common definition for the Bandwidth of this circuit is the frequency range over which the output decreases by 3 dB. This correspond to the frequency at which the circuits power is cut in half from the resonance frequency.

$P = I^2 R$

$P_{1/2} = (\sqrt{2}I)^2 R$

Forest_Electronic_Instrumentation_and_Measurement