The Series RLC Circuit — Question 4

Question 4

The half-power bandwidth of a series RLC circuit has an exact simple form

A series RLC circuit is driven by a fixed rms voltage. Let $\omega_0=1/\sqrt{LC}$. At the half-power frequencies, the resistor receives half its resonant power.

1. Derive both positive half-power angular frequencies.

2. Find their difference and geometric mean.

Question 4 - Solution

The half-power bandwidth of a series RLC circuit has an exact simple form

(a) Method and calculation

$$P/P_{\max}=\frac{R^2}{R^2+(\omega L-1/(\omega C))^2}=1/2\Rightarrow|\omega L-1/(\omega C)|=R.$$$$\boxed{\omega_\pm=\frac{\sqrt{R^2+4L/C}\pm R}{2L}}.$$

(b) Interpretation and further calculation

$$\omega_+-\omega_-=R/L,\quad\omega_+\omega_-=1/(LC).$$$$\boxed{\Delta\omega=R/L,\quad\sqrt{\omega_+\omega_-}=\omega_0}.$$

Physical check

The geometric mean is exactly resonant; the arithmetic midpoint is only approximately resonant for narrow bandwidth.