How Much Signal a Tuned Circuit Catches

A tuned circuit responds most strongly at its resonant frequency, and its power response is one over a quadratic: a Lorentzian. Completing the square turns the integral into an arctangent, which tells you exactly how much power gets through any band.

How it works

Integrals involving quadratics

In one sentence

To integrate $\frac{1}{ax^2 + bx + c}$ with no real roots, complete the square to $a\left((x - h)^2 + k^2\right)$ and use $\int\frac{du}{u^2 + k^2} = \frac{1}{k}\tan^{-1}\frac{u}{k} + C$ .

General formula

{\htmlData{sym=0}{P(\omega)}} = \frac{{\htmlData{sym=2}{\gamma}}^2}{(\omega - {\htmlData{sym=1}{\omega_0}})^2 + {\htmlData{sym=2}{\gamma}}^2}, \qquad \int_{\htmlData{sym=3}{a}}^{\htmlData{sym=4}{b}} P\,d\omega = {\htmlData{sym=2}{\gamma}}\left[\tan^{-1}\frac{\omega - {\htmlData{sym=1}{\omega_0}}}{\htmlData{sym=2}{{\gamma}}}\right]_{\htmlData{sym=3}{a}}^{\htmlData{sym=4}{b}}

Completing the square recenters the quadratic at its vertex; after a shift it's the standard arctangent integral.

What the symbols mean

$P(\omega)$
Power response
Fraction of the peak power passed at frequency ω.
$\omega_0$
Resonant frequency
Where the circuit responds most.
$\gamma$
Half-width
How far from resonance the response falls to half.
$a$
Band start
Lowest frequency of the band you're catching.
$b$
Band end
Highest frequency of the band.

Near resonance, the circuit's power response is $\frac{\gamma^2}{\omega^2 - 2\omega_0\omega + \omega_0^2 + \gamma^2}$ : one over a quadratic.
Complete the square: $\omega^2 - 2\omega_0\omega + \omega_0^2 + \gamma^2 = (\omega - \omega_0)^2 + \gamma^2$ .
Substitute $u = \omega - \omega_0$ and use $\int\frac{du}{u^2 + \gamma^2} = \frac{1}{\gamma}\tan^{-1}\frac{u}{\gamma}$ .
The total area is $\pi\gamma$ , and exactly half of it lies within $\pm\gamma$ of resonance: that band is the circuit's bandwidth.

Learn the math

Integrals Involving QuadraticsCalc II · Lesson 6, Integration Techniques Resonance and FiltersE&M · Lesson 25, AC Circuits