Difference between revisions of "TF EIM Chapt3"

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|<math>\Rightarrow \omega = 174077.66</math> rad/s  or <math>\nu = \frac{\omega}{2 \pi} = 27705</math> Hz
 
|<math>\Rightarrow \omega = 174077.66</math> rad/s  or <math>\nu = \frac{\omega}{2 \pi} = 27705</math> Hz
 
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|}
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==== Q and Bandwidth====
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In the above circuit
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: <math>\nu = \omega/2 \pi = \frac{1}{2 \pi LC}</math>
  
 
[[Forest_Electronic_Instrumentation_and_Measurement]]
 
[[Forest_Electronic_Instrumentation_and_Measurement]]

Revision as of 05:43, 2 February 2011

gain

Loop Theorem

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

or

[math] I= \frac{V_0 e^{i \omega t}}{\left (R+ \frac{i \omega L}{1- \frac{\omega^2}{\omega_{LC}^2}} \right )}[/math]
Notice
When [math]\omega \approx \omega_{LC} = \sqrt{\frac{1}{LC}}[/math] then the AC signal is attenuated.

Looking at the Voltage divider aspect of the circuit

[math]V_{AB}=V_{out} = \frac{X_{tot} }{R + X_{tot}}V_{in}[/math]
[math]\left |\frac{ V_{out}} {V_{in}}\right | = \sqrt{ \left [ \frac{X_{tot} }{R + X_{tot}} \right ] \left [ \frac{X_{tot} }{R + X_{tot}} \right ]^*}[/math]
[math] = \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 ]^*}[/math]
[math] = \sqrt{ \frac{ \omega^2 L^2 \omega_{LC}^4}{R^2(\omega_{LC}^2 - \omega^2)^2 - \omega^2L^2 \omega_{LC}^4}}[/math]
[math] = \sqrt{ \frac{ \omega^2 }{C^2R^2(\omega_{LC}^2 - \omega^2)^2 - \omega^2}}[/math]
[math] = \sqrt{ \frac{ \omega^2 }{\frac{1}{\omega_{RC}^2}(\omega_{LC}^2 - \omega^2)^2 - \omega^2}}[/math]



[math]L = 33 \mu H[/math] and [math]C = 1 \mu F [/math] and R=200 [math]\Omega[/math]
TF EIM LC paral Gain.png
[math]\Rightarrow \omega = 174077.66[/math] rad/s or [math]\nu = \frac{\omega}{2 \pi} = 27705[/math] Hz

Q and Bandwidth

In the above circuit

[math]\nu = \omega/2 \pi = \frac{1}{2 \pi LC}[/math]

Forest_Electronic_Instrumentation_and_Measurement