Resonance & filters intro

YOU WILL BE ABLE TO

Locate series resonance and describe the circuit's behaviour there.
Compute Q and bandwidth, and interpret selectivity.
Analyse the RC low-pass and high-pass filters and their cutoff.
Read a frequency-response curve.

A series RLC's impedance is Z = R + j(ωL − 1/ωC): a fixed real part and an imaginary part that is a TUG-OF-WAR between the inductor (growing with ω) and the capacitor (shrinking). At exactly one frequency they cancel — and Lesson 1's ω₀ returns, wearing its true name:

\omega_0 = \frac{1}{\sqrt{LC}} \quad\Rightarrow\quad Z(\omega_0) = R

series resonance

At resonance the circuit is purely resistive: minimum impedance, maximum current, voltage and current in step. Off resonance, either the inductor or the capacitor dominates and the current falls away. How SHARPLY it falls is the quality factor Q — high-Q circuits respond enthusiastically at ω₀ and snub everything else, which is precisely how a radio picks one station out of the crowded air: the antenna hears everything; the tuned circuit answers to one frequency.

Q = \frac{\omega_0 L}{R}

\text{BW} = \frac{\omega_0}{Q}

quality factor · bandwidth

Normalised current vs frequency for two Q values: same ω₀, very different selectivity.

Lesson 3 ended with a frequency-dependent divider; here is its family. Take R and C in series. Output across the capacitor: at low ω the capacitor's impedance is huge, it takes nearly all the input, and the output follows — a LOW-PASS filter. Take the output across the resistor instead and the roles flip: a HIGH-PASS. Both change character around the same corner, the cutoff frequency, where |Z_C| = R and the output magnitude is 1/√2 of the input:

\omega_c = \frac{1}{RC}

f_c = \frac{1}{2\pi RC}

cutoff frequency

The RC filter pair: each passes what the other rejects, crossing at the 1/√2 cutoff.

WORKED EXAMPLE

Tuning a radio, with numbers

An AM tuner uses L = 100 µH and wants to receive 1 MHz (mid-band). Required C: ω₀² = 1/LC → C = 1/(ω₀²L) = 1/((2π × 10⁶)² × 10⁻⁴) ≈ 253 pF.
With series resistance R ≈ 12.6 Ω: Q = ω₀L/R = (2π × 10⁶ × 10⁻⁴)/12.6 ≈ 50.
Bandwidth: BW = f₀/Q = 10⁶/50 = 20 kHz — the tuner answers to a ~20 kHz slice around the station and ignores the rest of the band.
Turning the tuning knob varies C, sliding ω₀ along the dial. That knob is this equation.

C ≈ 253 pF, Q ≈ 50, BW = 20 kHz

WORKED EXAMPLE

Designing a cutoff

Design a low-pass to smooth audio noise above f_c ≈ 1.6 kHz, with R = 1 kΩ already fixed by the surrounding circuit.
C = 1/(2πf_cR) = 1/(2π × 1600 × 1000) ≈ 100 nF — a stock value, no accident: RC pairs are usually chosen so one lands on the shelf.
Sanity-check the corner: at 1.6 kHz the output is 0.707 of the input; at 16 kHz (10× above) it's ≈ 0.0995 — a tenth. The filter's "slope" is a factor of 10 per decade, a shape you'll formalise as −20 dB/decade in Microelectronics.

C ≈ 100 nF

PRACTICE

A series RLC uses L = 1 mH and C = 1 µF. Find ω₀ and f₀.

For practice problem 1 with R = 31.6 Ω, compute Q and the bandwidth in hertz.

Design an RC high-pass that blocks rumble below about 20 Hz, using C = 1 µF. What R sets f_c = 20 Hz, and what fraction of a 20 Hz signal still gets through?

Identify each as low-pass or high-pass, from reasoning alone: (a) output across C in a series RC; (b) output across R in a series RC; (c) output across L in a series RL.