Resonancia
Introducido| English | Significado |
|---|---|
| resonance/ˈrezənəns/ | resonancia |
| bandwidth/ˈbændwɪdθ/ | ancho de banda |
| half-power points | half-power points |
One station out of many
- A radio aerial picks up every station at once. Turn the tuning knob and one comes through clearly while the rest fade.
- The tuner makes one frequency special by cancelling two reactances there. This lesson shows how sharp that choice is and what it costs.
The reactances cancel
- A series RLC has $Z = R + j\left(\omega L - \dfrac{1}{\omega C}\right)$. At resonance 谐振 the two reactances are equal and cancel: $\omega_{0} = \dfrac{1}{\sqrt{LC}}$.
- There $Z = R$: purely resistive and at its minimum, so the current from a fixed voltage source is at its maximum and in phase with the voltage.
L = 1 mH and C = 1 nF. What is the resonant frequency f₀, in kHz?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\omega_{0} = \dfrac{1}{\sqrt{10^{-3} \times 10^{-9}}} = 10^{6}$ rad/s, and $f_{0} = \dfrac{\omega_{0}}{2\pi} = 159.2$ kHz.
At series resonance the current is in phase with the source voltage.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The reactances cancel, so $Z = R$ is purely resistive and the angle between V and I is zero.
Voltage magnification
- The inductor and capacitor voltages are each $Q$ times the source, where $Q = \dfrac{\omega_{0}L}{R}$. They are 180° apart, so they cancel in the loop.
- So a 10 V source can put hundreds of volts across a capacitor. Check the rating of both parts.
$R = 5\,\Omega$, $L = 0.1\,\text{H}$, $C = 10\,\mu\text{F}$: $\omega_{0} = 1000\,\dfrac{\text{rad}}{\text{s}}$ and $Q = \dfrac{1000 \times 0.1}{5} = 20$. From 10 V: $I = 2\,\text{A}$, $V_{L} = 100 \times 2 = 200\,\text{V}$, and $V_{C} = \dfrac{2}{0.01} = 200\,\text{V}$ too.
A series RLC with Q = 20 at resonance on a 10 V source has V_C equal to —
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$I = \dfrac{10}{R}$ flows; $V_{C} = \dfrac{I}{\omega_{0}C} = Q \times 10 = 200$ V. $V_{L}$ is also 200 V, equal and opposite: they cancel, while each alone is twenty times the source.
At resonance V_L = 200 V and V_C = 200 V. What does a voltmeter across the L and C together read?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The phasors point in opposite directions, so their sum is zero. All of the source voltage appears across R.
Bandwidth and half-power points
- Away from $\omega_{0}$ the current falls. At the half-power points 半功率点 $|Z| = \sqrt{2}R$, so the current is $\dfrac{1}{\sqrt{2}}$ of its peak and the power is half.
- The bandwidth 带宽 between them is $BW = \dfrac{R}{L} = \dfrac{\omega_{0}}{Q}$. For the example, $BW = 50\,\dfrac{\text{rad}}{\text{s}}$: half-power at 975.3 and 1025.3 rad/s.
A series RLC has ω₀ = 1000 rad/s and Q = 20. The half-power frequencies are
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$BW = \dfrac{\omega_{0}}{Q} = 50$ rad/s. The exact points sit slightly unevenly at 975.3 and 1025.3, 50 apart, with $\omega_{1}\omega_{2} = \omega_{0}^{2}$.
Worked example: rejecting a neighbour
- A neighbouring signal at 900 rad/s, two bandwidths below. There $\omega L = 90\,\Omega$ and $\dfrac{1}{\omega C} = 111.1\,\Omega$.
- $|Z| = |5 - j21.1| = 21.7\,\Omega$, so $I = \dfrac{10}{21.7} = 0.46\,\text{A}$: under a quarter of the 2 A at resonance. Higher Q makes this rejection stronger.
R = 10 Ω and L = 50 mH. What is the bandwidth, in rad/s?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$BW = \dfrac{R}{L} = \dfrac{10}{0.05} = 200$ rad/s. Halving R halves the bandwidth and sharpens the tuning.
The parallel tank is the dual
- L and C in parallel resonate at the same $\omega_{0}$, but now the branch currents cancel. The impedance is at its maximum, not its minimum.
- Each branch carries about Q times the line current, circulating between L and C. Oscillators use the parallel tank to set their frequency.
At resonance, match each circuit to its behaviour.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
The parallel tank is the dual of the series circuit, with voltages and currents swapping roles.
Three common slips
- $f_{0}$ for $\omega_{0}$. $\dfrac{1}{\sqrt{LC}}$ is in rad/s; divide by $2\pi$ for hertz.
- Checking one element. $V_{L}$ and $V_{C}$ must both come out as $Q$ times the source.
- "Resonance maximises power." It minimises $|Z|$; the current peaks only for a stiff voltage source.
Recap
- Series resonance: $\omega_{0} = \dfrac{1}{\sqrt{LC}}$, $Z = R$, current at its peak.
- $V_{L} = V_{C} = Q \times$ source, with $Q = \dfrac{\omega_{0}L}{R}$; $BW = \dfrac{\omega_{0}}{Q}$ between the half-power points.
- The parallel tank is the dual: maximum impedance, circulating current.