TF EIM Chapt3: Difference between revisions
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:<math>\tan(\phi) = \frac{X_{out}}{R_L} = \frac{\omega L - \frac{1}{\omega C}}{R}= \frac{\omega^2 - \omega_o^2}{\omega_0^2R}</math> | :<math>\tan(\phi) = \frac{X_{out}}{R_L} = \frac{\omega L - \frac{1}{\omega C}}{R}= \frac{\omega^2 - \omega_o^2}{\omega_0^2R}</math> | ||
If <math>\omega < \omega_0</math> then the phase shift is negative | |||
If <math>\omega > \omega_0</math> then the phase shift is positive | |||
[[Forest_Electronic_Instrumentation_and_Measurement]] | [[Forest_Electronic_Instrumentation_and_Measurement]] | ||
Revision as of 04:25, 2 February 2011
Phase shift
[math]\displaystyle{ X_C = \frac{-i}{\omega C} \Rightarrow }[/math] Voltage lags current for a capacitor
[math]\displaystyle{ X_L = i \omega L \Rightarrow }[/math] Voltage leads current for Inductor
- [math]\displaystyle{ \tan(\phi) = \frac{X_{out}}{R_L} = \frac{\omega L - \frac{1}{\omega C}}{R}= \frac{\omega^2 - \omega_o^2}{\omega_0^2R} }[/math]
If [math]\displaystyle{ \omega < \omega_0 }[/math] then the phase shift is negative
If [math]\displaystyle{ \omega > \omega_0 }[/math] then the phase shift is positive