2018-10-26

作者: carpediemmlf | 来源:发表于2018-12-10 06:09 被阅读0次

$\omega_{acc} = \omega_0 ( {1 - \frac{1}{2Q^2}} ) ^ {-\frac{1}{2}}$

for completeness, listing the resonance frequency for the three quantities in question

$\omega_{A} = \omega_0 ( {1 - \frac{1}{2Q^2}} ) ^ {\frac{1}{2}}$

$\omega_{velo} = \omega_0$

$\omega_{acc} = \omega_0 ( {1 - \frac{1}{2Q^2}} ) ^ {-\frac{1}{2}}$

notice $\omega_a \cdot \omega_{acc} = {\omega_0}^2$ by comparing with the resonance frequency for displacement as discussed last time

$\mathcal{Re}\{\bf A\} \mathcal{Re}\{\bf B\} = \frac{1}{2} \mathcal{Re}\{ {\bf A} {\bf B^{\dagger}} \}$ is true when $\bf A$ and $\bf B$ have the same time dependence (e.g., both propagating forward/backward in time) in their complex power representation

Power resonance and bandwidth

Consider the half power points, whose difference gives the bandwidth
$\Delta \omega = \omega_{+} - \omega_{-} = \gamma$

$Q = \frac{\omega_0}{\gamma} = \frac{\omega_0}{\Delta \omega}$

$\omega_0 = \frac{1}{\sqrt{LC}}$

$\gamma = \frac{R}{L}$

$Q = \frac{\omega_0}{\gamma} = \frac{1}{R}\sqrt \frac{L}{C}$

$\ddot q + \gamma \dot q + {\omega_0}^2 q = \frac {V} {L}$

$Z = \frac{F}{v}$

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