Parad0x88 said:
That's an interesting formula you got there! I didn't see that in my book
These come from the standard form for the response to a second order differential equation.
If you find the voltage across the capacitor in a series RLC circuit you will find the transfer function looks like:
##\frac{\frac{1}{LC}}{s^2+s\frac{R}{L}+\frac{1}{LC}}##
Here you can see if R=0, the response has poles on the imaginary axis so there will be oscillation of frequency ##\frac{1}{\sqrt LC}##The standard normalized (gain=1 at dc) second order response is:
##\frac{wn^2}{s^2+s(\frac{wn}{Q})+wn^2}##
Comparing the factor multiplying s, you can find Q in terms of R, L and wn.You have an equation for Q in terms of energy but if you did not know that you could look at this the long way. The natural response in the time domain (impulse response, inverse laplace) is a damped sinusoid meaning e
-atsin(wt) where a and w are functions of Q and wn (look it up if needed).
Since this response is the voltage across the capacitor in a
series circuit, or so I have assumed, the current through the circuit is the current through the capacitor i=Cdv/dt. When the voltage waveform is at a peak, i=0. This means the total energy stored in the circuit at these peak times is just 0.5CV
2 since 0.5Li
2=0. So energy at some time t0 will be 0.5CV
2 and then one period later, 10% less. The e
-at term will tell you how much the voltage on the capacitor has decreased one period later so that you will have another equation for L,C. This is where that energy equation for Q comes from.
