Worked examples: AM-radio tuner and a 50 Hz notch filter

AC Circuits · Resonance in RLC Circuits · Example

Example 1 — AM radio front-end for the 820 kHz broadcast channel

An AM radio is trying to pluck the 820 kHz channel (a local news station in the 540 – 1700 kHz band) out of the electromagnetic noise around the antenna. The front-end uses a series RLC with a fixed 250 µH antenna coil and a variable tuning capacitor. Size the capacitor to land on 820 kHz and compute the circuit's selectivity if its effective series resistance is 10 Ω.

  1. Resonance condition.
    f0=12πLCC=1(2πf0)2Lf_0 = \frac{1}{2\pi\sqrt{LC}} \Rightarrow C = \frac{1}{(2\pi f_0)^2 L}
    C=1(2π820,000)2250×106150pFC = \frac{1}{(2\pi \cdot 820{,}000)^2 \cdot 250 \times 10^{-6}} \approx 150\,\text{pF}
    A classic value — right inside the range of a typical variable "ganged" tuning capacitor.
  2. Quality factor.
    Q=2πf0LR=2π820,000250×10610129Q = \frac{2\pi f_0 L}{R} = \frac{2\pi \cdot 820{,}000 \cdot 250 \times 10^{-6}}{10} \approx 129
    High Q — which is what a radio needs to split 10 kHz- spaced channels cleanly.
  3. Bandwidth.
    BW=f0/Q=820,000/1296.4kHz\text{BW} = f_0 / Q = 820{,}000 / 129 \approx 6.4\,\text{kHz}
    Just narrow enough that the adjacent channel 10 kHz away is well into the −3 dB attenuation band. The selected station's audio (typically 5 kHz bandwidth on AM) fits inside the pass window.
  4. Current boost at resonance. Off resonance, X_L + X_C dominates and impedance is huge — virtually no signal current. At 820 kHz the reactances cancel, leaving just Z=10|Z| = 10 Ω. Even a microvolt-level signal at the antenna gets through to the mixer strong enough to detect.
Radio selectivity = high Q at the target frequency. Make Q bigger by reducing R (lower-loss coil), raising L, or reducing C. The design trade-off is that wider audio programmes (high-fidelity AM — yes it's a thing) need a broader bandwidth, which means a lower Q.

Example 2 — 50 Hz notch to kill mains hum in an audio cable

A sensitive audio preamp picks up 50 Hz hum from nearby mains wiring. A common fix is a series LC "trap" across the signal line to ground — low impedance at 50 Hz, high impedance elsewhere, so the hum shunts to ground without touching the music.

  1. Pick L × C for 50 Hz resonance. With a smallish inductor L=1L = 1 H (physically large, but this is audio not RF):
    C=1(2π50)2110μFC = \frac{1}{(2\pi \cdot 50)^2 \cdot 1} \approx 10\,\mu\text{F}
    Common audio-grade film capacitor value. Two 5 µF caps in parallel if you want to trim.
  2. Q and bandwidth. If the inductor's DC resistance is 20 Ω:
    Q=2π5012015.7,BW3.2HzQ = \frac{2\pi \cdot 50 \cdot 1}{20} \approx 15.7, \quad \text{BW} \approx 3.2\,\text{Hz}
    Very narrow — only mains and its immediate neighbourhood (say 48 – 52 Hz) get suppressed. Audio content at 20 Hz (sub-bass) and 100 Hz (second harmonic of mains, often a bigger problem!) passes unaffected.
  3. Stop-band attenuation. At exactly 50 Hz the trap is a near-short across the signal — the attenuation is set by the ratio of the source impedance (Thevenin-equivalent of the audio line) to the 20 Ω trap. With a typical low-impedance line driver the notch depth is 30 – 40 dB: audible hum drops to inaudibility.

For the 100 Hz second harmonic (where most mains-hum energy actually lives), you repeat the exercise: build a second trap tuned to 100 Hz. Stack as many as you need. Guitar amplifiers ship with pre-tuned traps in the input stage for exactly this reason.

Example 3 — The AM radio peak, drawn

Here's the Example-1 tuning circuit shown as a magnitude plot. At f0=820f_0 = 820 kHz the response peaks at 0 dB; the −3 dB points sit atf1816.8f_1 \approx 816.8 kHz and f2823.2f_2 \approx 823.2 kHz (bandwidth ≈ 6.4 kHz). At 830 kHz — just 10 kHz away — the adjacent channel is already down past −9 dB and well attenuated. That is the selectivity a high-Q resonator buys you.

0 dB-10 dB-20 dB-30 dB805810815820825830835frequency f (kHz)−3 dBf₀ = 820 kHzmagnitude [dB]
Magnitude response centred on the 820 kHz AM-broadcast channel with Q = 129. The −3 dB bandwidth is the frequency span between the two orange intercepts — about 6.4 kHz, narrow enough to split 10 kHz-spaced channels.

Sweep your own values in the Simulate stage or lock in the reflex on the Quiz.