University Physics — Year 1 · Bachelor Year 1

9Filters and Transfer Functions

Inside a loudspeaker cabinet a coil and a capacitor decide, without a single transistor, that the bass goes to the big cone and the treble to the small dome. An oscilloscope’s “AC” button removes a slowly drifting offset and keeps the wiggle on top of it. A power supply turns a pulsating rectified mains into a steady voltage with one capacitor. All three are filters: linear circuits that treat each frequency differently. Because any periodic signal is a sum of sinusoids, knowing what a circuit does to every sinusoid — its transfer function — is knowing what it does to everything. This chapter defines transfer functions, draws their Bode diagrams for the first- and second-order circuits of the previous chapters, and uses them to predict how a square wave comes out of a filter.

9.1 Transfer function and Bode diagram

Definition 9.1 (Transfer function)

A linear circuit with an input voltage $\underline U_e$ and an output voltage $\underline U_s$ (taken with nothing connected at the output) has the transfer function

\underline H(j\omega) = \frac{\underline U_s}{\underline U_e}, \qquad G(\omega) = \abs{\underline H}, \qquad \varphi(\omega) = \arg\underline H,

the gain $G$ (ratio of amplitudes) and the phase shift $\varphi$ of the output relative to the input. A filter is such a circuit used for its frequency dependence: a sinusoid of amplitude $E$ at $\omega$ comes out as $G(\omega)E\cos(\omega t + \varphi(\omega))$ .

Definition 9.2 (Decibels; Bode diagram)

The gain in decibels is $G_{\mathrm{dB}} = 20\log_{10}G$ : a factor $10$ is $20\,\mathrm{dB}$ , a factor $2$ is $6\,\mathrm{dB}$ , $1/\sqrt2$ is $-3\,\mathrm{dB}$ . The Bode diagram plots $G_{\mathrm{dB}}$ and $\varphi$ against $\log_{10}\omega$ (or $f$ ). The cutoff frequency is where $G$ falls to $G_{\max}/\sqrt2$ ( $-3\,\mathrm{dB}$ from the maximum); the passband is the range of frequencies within $3\,\mathrm{dB}$ of the maximum. A decade is a factor $10$ in frequency, an octave a factor $2$ .

Method 9.3 (Nature of a filter at a glance)

Before computing, look at the limits: at very low frequency a capacitor is an open circuit and a coil a wire; at very high frequency the reverse. Replace each and read whether the output is the input (passes) or zero (blocked). Passes low, blocks high: low-pass; the reverse: high-pass; blocks both: band-pass; passes both: band-stop. Then compute $\underline H$ by a voltage divider.

9.2 First-order filters

Proposition 9.4 (First-order low-pass)

The RC divider with output across $C$ (or the RL divider with output across $R$ ) has

\underline H = \frac{1}{1 + j\omega/\omega_c}, \qquad \omega_c = \frac{1}{RC} \ \Big(\text{or } \frac{R}{L}\Big), \qquad G = \frac{1}{\sqrt{1 + (\omega/\omega_c)^2}}, \qquad \varphi = -\arctan\frac{\omega}{\omega_c} .

Bode asymptotes: $G_{\mathrm{dB}} = 0$ for $\omega \ll \omega_c$ , a straight line falling $20\,\mathrm{dB}$ per decade for $\omega \gg \omega_c$ , meeting at $\omega_c$ where the true curve is $-3\,\mathrm{dB}$ and $\varphi = -45^\circ$ . For $\omega \gg \omega_c$ , $\underline H \approx \omega_c/(j\omega)$ : the output is proportional to the integral of the input — the filter is an integrator there.

Proof. Voltage divider:

\underline H = \frac{1/jC\omega}{R + 1/jC\omega} = \frac{1}{1 + jRC\omega} .

Far above cutoff, $1 + j\omega/\omega_c \approx j\omega/\omega_c$ , and dividing a complex amplitude by $j\omega$ is integrating (Proposition 8.3): $u_s \approx \omega_c\int u_e\,\dd t$ . $20\log(\omega_c/\omega)$ loses $20\,\mathrm{dB}$ per tenfold $\omega$ . ∎

Proposition 9.5 (First-order high-pass)

The CR divider with output across $R$ (or LR with output across $L$ ) has

\underline H = \frac{j\omega/\omega_c}{1 + j\omega/\omega_c}, \qquad G = \frac{\omega/\omega_c}{\sqrt{1 + (\omega/\omega_c)^2}}, \qquad \varphi = \frac{\pi}{2} - \arctan\frac{\omega}{\omega_c},

rising $20\,\mathrm{dB}$ per decade below $\omega_c$ , flat above, $-3\,\mathrm{dB}$ and $+45^\circ$ at $\omega_c$ . For $\omega \ll \omega_c$ , $\underline H \approx j\omega/\omega_c$ : the output is proportional to the derivative of the input.

Proof. $\underline H = R/(R + 1/jC\omega) = jRC\omega/(1 + jRC\omega)$ ; multiplying by $j\omega$ is differentiating. ∎

Bode diagrams of the first-order low-pass (blue) and high-pass (red): gain in decibels with the asymptotes (dashed) meeting at the cutoff, where the true gain is -3\, dB, and phase, which crosses 45 at the cutoff. — Bode diagrams of the first-order low-pass (blue) and high-pass (red): gain in decibels with the asymptotes (dashed) meeting at the cutoff, where the true gain is $-3\,\mathrm{dB}$ , and phase, which crosses $\mp45^\circ$ at the cutoff.

Example 9.6 (AC coupling)

An oscilloscope’s AC input is a high-pass with $f_c \approx 10\,\mathrm{Hz}$ : a $1\,\mathrm{kHz}$ signal riding on a $5\,\mathrm{V}$ offset comes through at $-0.0002\,\mathrm{dB}$ with its offset removed. But a $50\,\mathrm{Hz}$ square wave is visibly distorted — its flat tops sag — because at $50\,\mathrm{Hz}$ the filter is only one decade above its cutoff and the signal’s low harmonics are shifted in phase (Exercise 9.12).

9.3 Second-order filters

Proposition 9.7 (The series RLC as three filters)

With $x = \omega/\omega_0$ , $\omega_0 = 1/\sqrt{LC}$ and $Q = \sqrt{L/C}/R$ , the series RLC fed by $\underline U_e$ gives, according to where the output is taken:

\begin{align*} \text{across } C:&\quad \underline H = \frac{1}{1 - x^2 + jx/Q}, \\ \text{across } R:&\quad \underline H = \frac{jx/Q}{1 - x^2 + jx/Q} = \frac{1}{1 + jQ(x - 1/x)}, \\ \text{across } L:&\quad \underline H = \frac{-x^2}{1 - x^2 + jx/Q}: \end{align*}

a low-pass, a band-pass and a high-pass of second order. Far from $\omega_0$ the low-pass and high-pass fall by $40\,\mathrm{dB}$ per decade, the band-pass by $20\,\mathrm{dB}$ per decade on each side; the band-pass has its maximum $1$ at $\omega_0$ and the bandwidth $\Delta\omega = \omega_0/Q$ of Theorem 8.9.

Proof. Voltage dividers with the total impedance $R + j(L\omega - 1/C\omega)$ ; multiply numerator and denominator by $jC\omega$ and use $LC\omega_0^2 = 1$ , $RC\omega_0 = 1/Q$ . The band-pass is the current resonance of the previous chapter read as a voltage across $R$ . ∎

Proposition 9.8 (Second-order low-pass: the role of $Q$ )

For the low-pass, $G = 1/\sqrt{(1 - x^2)^2 + x^2/Q^2}$ : for $Q > 1/\sqrt2$ the gain rises above $1$ near $\omega_0$ (resonance, peak $\approx Q$ for large $Q$ ); for $Q = 1/\sqrt2$ it is “maximally flat”, $G = 1/\sqrt{1 + x^4}$ , exactly $-3\,\mathrm{dB}$ at $\omega_0$ ; for smaller $Q$ it droops early. The phase runs from $0$ to $-\pi$ , through $-\pi/2$ at $\omega_0$ .

Proof. $(1 - x^2)^2 + x^2/Q^2 = 1 + x^4 + x^2(1/Q^2 - 2)$ : the middle term vanishes for $Q^2 = \tfrac12$ , is negative (peak) above, positive (droop) below; the behavior near $x = 1$ was analyzed in Proposition 8.10. ∎

Second-order filters from the series RLC. Left, the low-pass (output across C) for three Q: a resonance peak, the maximally flat response, or an early droop, all joining the -40\, dB/ decade asymptote (dashed). Right, the band-pass (output across R): a peak of width _0/Q falling 20\, dB/ decade on each side. — Second-order filters from the series RLC. Left, the low-pass (output across $C$ ) for three $Q$ : a resonance peak, the maximally flat response, or an early droop, all joining the $-40\,\mathrm{dB}/\mathrm{decade}$ asymptote (dashed). Right, the band-pass (output across $R$ ): a peak of width $\omega_0/Q$ falling $20\,\mathrm{dB}/\mathrm{decade}$ on each side.

Example 9.9 (A crossover)

A loudspeaker’s woofer and tweeter, each $8\,\Omega$ , share the amplifier through a low-pass (a series coil) and a high-pass (a series capacitor) cut at $2\,\mathrm{kHz}$ : $L = R/\omega_c = 0.64\,\mathrm{mH}$ , $C = 1/R\omega_c = 10\,\text{µ}\mathrm{F}$ . At $2\,\mathrm{kHz}$ each driver gets $-3\,\mathrm{dB}$ , i.e. half the power, and $G_L^2 + G_H^2 = 1$ at every frequency: the power is shared, not lost. The weekend problem designs a steeper version.

9.4 Filtering a periodic signal

Theorem 9.10 (Fourier decomposition)

A periodic signal of period $T = 2\pi/\omega$ (reasonably regular) is the sum of a constant and of sinusoids at the multiples of $\omega$ :

s(t) = c_0 + \sum_{n = 1}^{\infty} c_n\cos(n\omega t + \varphi_n),

where $c_0 = \langle s\rangle$ is its average value; the term $n = 1$ is the fundamental, the others the harmonics, and the amplitudes $c_n$ form the spectrum. A square wave of amplitude $\pm E$ has $c_n = 4E/(n\pi)$ for odd $n$ and $0$ for even $n$ ; a triangle wave of amplitude $\pm E$ has $c_n = 8E/(n\pi)^2$ for odd $n$ .

Proof. Admitted at this level. ∎

Remark 9.11 (What to take from it)

The theorem and the computation of the $c_n$ belong to the Year 2 mathematics course. Here it is a tool: sharp corners need many harmonics (the square wave’s fall off only as $1/n$ ), smooth signals need few (the triangle’s as $1/n^2$ ). A signal’s harmonic content is what an instrument’s timbre is made of, and what a filter acts on.

Proposition 9.12 (Linear filtering of a periodic signal)

Through a filter of transfer function $\underline H$ , the signal of the theorem becomes

s_{\text{out}}(t) = G(0)\,c_0 + \sum_{n \geq 1} G(n\omega)\,c_n\cos\!\big(n\omega t + \varphi_n + \varphi(n\omega)\big):

each harmonic is scaled by the gain and shifted by the phase at its own frequency, then the pieces are added. Hence:

a low-pass with $\omega_c \ll \omega$ keeps essentially $c_0$ : it averages (smoothing, ripple removal), and turns a square wave into a small triangle (integration);
a high-pass with $\omega_c \ll \omega$ removes $c_0$ and passes the rest (AC coupling); with $\omega_c \gg \omega$ it differentiates (spikes at the edges of a square wave);
a band-pass tuned to $n\omega$ with $Q \gg 1$ extracts the $n$ -th harmonic alone: a sinusoid out of a square wave.

Proof. Superposition for a linear circuit (Theorem 6.18) applied to the sum of sinusoidal inputs, each treated by Definition 9.1. ∎

A square wave through a first-order low-pass. Left: its spectrum (odd harmonics in 1/n) and the spectrum after a filter cut at the fundamental. Right: the time signal for _c = (red, exponential arcs) and _c = 10 (orange, rounded corners only); with _c the output would be a small triangle around the average. — A square wave through a first-order low-pass. Left: its spectrum (odd harmonics in $1/n$ ) and the spectrum after a filter cut at the fundamental. Right: the time signal for $\omega_c = \omega$ (red, exponential arcs) and $\omega_c = 10\omega$ (orange, rounded corners only); with $\omega_c \ll \omega$ the output would be a small triangle around the average.

Example 9.13 (Numbers)

A $1\,\mathrm{kHz}$ square wave of amplitude $5\,\mathrm{V}$ : harmonics of $6.37\,\mathrm{V}$ at $1\,\mathrm{kHz}$ , $2.12\,\mathrm{V}$ at $3\,\mathrm{kHz}$ , $1.27\,\mathrm{V}$ at $5\,\mathrm{kHz}$ . Through an RC low-pass cut at $10\,\mathrm{Hz}$ ( $\tau = 16\,\mathrm{ms}$ ) the output is a triangle of amplitude $ET/4\tau \approx 80\,\mathrm{mV}$ around zero; through a band-pass tuned to $3\,\mathrm{kHz}$ with $Q = 20$ , a $2.1\,\mathrm{V}$ sinusoid at $3\,\mathrm{kHz}$ with only a few percent of the neighbors (Exercise 9.8).

9.5 Cascading filters

Proposition 9.14 (Cascade and loading)

Two filters in cascade have $\underline H = \underline H_1\underline H_2$ — gains multiply, decibels and phases add — provided the second does not load the first, i.e. draws negligible current from its output. Otherwise the divider relations must be rewritten for the combined network. An operational-amplifier follower (Chapter 10) between stages makes the product rule exact.

Proof. $\underline H_1$ was defined with an open output; if the next stage’s input impedance is much larger than the first stage’s output impedance (its Thévenin resistance), the output voltage is unchanged and the second stage sees exactly $\underline U_{s1}$ as its input. ∎

Example 9.15 (Two RC stages)

Two identical RC low-passes back to back do not give $1/(1 + j\omega/\omega_c)^2$ : the second stage’s $R$ loads the first. The exact result is $\underline H = 1/[1 - (\omega/\omega_c)^2 + 3j\omega/\omega_c]$ , $-9.5\,\mathrm{dB}$ at $\omega_c$ instead of the $-6\,\mathrm{dB}$ the product rule promises, though both reach the same $-40\,\mathrm{dB}/\mathrm{decade}$ slope eventually.

9.6 Exercises

Exercise 9.1 ★

Convert to decibels: gains $0.5$ , $10$ , $1/\sqrt2$ , $100$ . Convert to ratios: $-40\,\mathrm{dB}$ , $+6\,\mathrm{dB}$ . Two filters in cascade (no loading) give $-6\,\mathrm{dB}$ and $-20\,\mathrm{dB}$ at some frequency: overall gain?

Solution

Solution of Exercise 9.1.

$0.5 \to -6.0\,\mathrm{dB}$ ; $10 \to 20\,\mathrm{dB}$ ; $1/\sqrt2 \to -3.0\,\mathrm{dB}$ ; $100 \to 40\,\mathrm{dB}$ . $-40\,\mathrm{dB} \to 0.01$ ; $+6\,\mathrm{dB} \to 2.0$ . Cascade: $-6 - 20 = -26\,\mathrm{dB}$ , a ratio $0.05$ .

Exercise 9.2 ★

RC low-pass, $R = 10\,\mathrm{k}\Omega$ , $C = 16\,\mathrm{nF}$ : cutoff frequency, and the gain in decibels at $100\,\mathrm{Hz}$ , $1\,\mathrm{kHz}$ , $10\,\mathrm{kHz}$ , $100\,\mathrm{kHz}$ .

Solution

Solution of Exercise 9.2.

$f_c = 1/(2\pi RC) = 1/(2\pi \times 10^4 \times 1.6\times10^{-8}) = 1.0\,\mathrm{kHz}$ . Gains $-10\log(1 + (f/f_c)^2)$ : $100\,\mathrm{Hz}$ : $-0.04\,\mathrm{dB}$ ; $1\,\mathrm{kHz}$ : $-3\,\mathrm{dB}$ ; $10\,\mathrm{kHz}$ : $-20\,\mathrm{dB}$ ; $100\,\mathrm{kHz}$ : $-40\,\mathrm{dB}$ .

Exercise 9.3 ★

Without computing, give the nature (low-, high-, band-pass) of: (a) $R$ in series, $C$ across the output; (b) $C$ in series, $R$ across the output; (c) $L$ in series, $C$ across the output; (d) $L$ and $C$ in series, $R$ across the output.

Solution

Solution of Exercise 9.3.

(a) low-pass ( $C$ shorts the output at high $f$ ); (b) high-pass ( $C$ opens at low $f$ ); (c) second-order low-pass ( $L$ opens at high $f$ , $C$ shorts); (d) band-pass ( $C$ blocks low $f$ , $L$ blocks high $f$ ).

Exercise 9.4 ★

A $1.0\,\mathrm{kHz}$ square wave of amplitude $5.0\,\mathrm{V}$ : give the frequencies and amplitudes of its first three non-zero harmonics. Why does it have no even harmonics?

Solution

Solution of Exercise 9.4.

$4E/n\pi$ : $6.37\,\mathrm{V}$ at $1\,\mathrm{kHz}$ , $2.12\,\mathrm{V}$ at $3\,\mathrm{kHz}$ , $1.27\,\mathrm{V}$ at $5\,\mathrm{kHz}$ . The square wave is antisymmetric about a half-period shift ( $s(t + T/2) = -s(t)$ ), which even harmonics are not.

Exercise 9.5 ★★

A CR high-pass cut at $10\,\mathrm{kHz}$ receives a triangle wave of amplitude $1.0\,\mathrm{V}$ at $100\,\mathrm{Hz}$ . Explain why it behaves as a differentiator and give the shape and amplitude of the output.

Solution

Solution of Exercise 9.5.

At $100\,\mathrm{Hz}$ and its useful harmonics, $\omega \ll \omega_c$ : $\underline H \approx j\omega/\omega_c$ , a derivative divided by $\omega_c$ . The triangle’s slope is $\pm 4Ef = \pm400\,\mathrm{V}/\mathrm{s}$ , so the output is a square wave of amplitude $400/(2\pi \times 10^4) = 6.4\,\mathrm{mV}$ .

Exercise 9.6 ★★

An RC low-pass cut at $10\,\mathrm{Hz}$ receives a $1.0\,\mathrm{kHz}$ square wave of amplitude $\pm1.0\,\mathrm{V}$ . Show that the output is a triangle and compute its amplitude.

Solution

Solution of Exercise 9.6.

$\omega \gg \omega_c$ : $u_s \approx (1/\tau)\int u_e\,\dd t$ , $\tau = 1/(2\pi \times 10) = 16\,\mathrm{ms}$ . The integral of $\pm E$ is a triangle rising by $ET/2$ per half period: swing $ET/2\tau$ , amplitude $ET/4\tau = 1.0 \times 10^{-3}/(4 \times 0.016) = 16\,\mathrm{mV}$ around zero.

Exercise 9.7 ★★

A series RLC low-pass with $L = 10\,\mathrm{mH}$ , $C = 100\,\mathrm{nF}$ : give $f_0$ ; find $R$ for the maximally flat response; give the gain in decibels at $f_0$ and at $10f_0$ .

Solution

Solution of Exercise 9.7.

$f_0 = 5.03\,\mathrm{kHz}$ . $Q = \sqrt{L/C}/R = 1/\sqrt2$ : $R = \sqrt2 \times 316 = 447\,\Omega$ . At $f_0$ : $-3\,\mathrm{dB}$ ; at $10f_0$ : $1/\sqrt{1 + 10^4} \to -40\,\mathrm{dB}$ .

Exercise 9.8 ★★

A band-pass ( $Q = 20$ ) tuned to $3.0\,\mathrm{kHz}$ receives the $1\,\mathrm{kHz}$ square wave of Exercise 9.4. Compute the gain at $1$ , $3$ and $5\,\mathrm{kHz}$ , and the output amplitudes of these harmonics; describe the output.

Solution

Solution of Exercise 9.8.

$G = 1/\sqrt{1 + Q^2(x - 1/x)^2}$ with $x = f/3$ : $1\,\mathrm{kHz}$ : $x = 1/3$ , $Q(x - 1/x) = -53$ , $G = 0.019$ ; $3\,\mathrm{kHz}$ : $G = 1$ ; $5\,\mathrm{kHz}$ : $x = 5/3$ , $21$ , $G = 0.047$ . Output: $2.1\,\mathrm{V}$ at $3\,\mathrm{kHz}$ , with $0.12\,\mathrm{V}$ at $1\,\mathrm{kHz}$ and $0.06\,\mathrm{V}$ at $5\,\mathrm{kHz}$ — a nearly pure $3\,\mathrm{kHz}$ sinusoid.

Exercise 9.9 ★★

Two identical RC low-passes ( $R = 1.0\,\mathrm{k}\Omega$ , $C = 159\,\mathrm{nF}$ ) in cascade. Compute the exact transfer function (the second loads the first) and the gain at $\omega_c = 1/RC$ ; compare with the product of the two separate gains.

Solution

Solution of Exercise 9.9.

Nodes: with $a = jRC\omega$ , the first capacitor sees $R$ upstream and $R + 1/jC\omega$ downstream: $\underline H = 1/(1 + 3a + a^2)$ , i.e. $1/[1 - (\omega/\omega_c)^2 + 3j\omega/\omega_c]$ . At $\omega_c$ : $1/3j$ , $G = 1/3$ , $-9.5\,\mathrm{dB}$ ; the product of two separate gains would be $1/2$ , $-6\,\mathrm{dB}$ . $f_c = 1.0\,\mathrm{kHz}$ .

Exercise 9.10 ★★★

Sketch the Bode asymptotes of $\underline H = (1 + j\omega/\omega_1)/ (1 + j\omega/\omega_2)$ with $\omega_1 = 100\,\mathrm{rad}/\mathrm{s}$ , $\omega_2 = 1 \times 10^{4}\,\mathrm{rad}/\mathrm{s}$ : slopes, corners, the high-frequency plateau in decibels, and the phase at $\sqrt{\omega_1\omega_2}$ .

Solution

Solution of Exercise 9.10.

Below $\omega_1$ : $0\,\mathrm{dB}$ ; between $\omega_1$ and $\omega_2$ : rising $20\,\mathrm{dB}/\mathrm{decade}$ ; above $\omega_2$ : flat at $20\log(\omega_2/\omega_1) = 40\,\mathrm{dB}$ . Phase $\arctan(\omega/\omega_1) - \arctan(\omega/\omega_2)$ : at $\omega = 10^3$ , $\arctan 10 - \arctan 0.1 = 84^\circ - 6^\circ = 78^\circ$ .

Exercise 9.11 ★★★

A full-wave rectified sine ( $50\,\mathrm{Hz}$ mains) has average $2E/\pi$ and a first harmonic at $100\,\mathrm{Hz}$ of amplitude $4E/3\pi$ . An RC low-pass must reduce the $100\,\mathrm{Hz}$ ripple below $1\%$ of the average: find the required cutoff frequency and time constant. Why is a capacitor alone (with the load resistance) the usual choice?

Solution

Solution of Exercise 9.11.

Far above cutoff, $G \approx f_c/f$ ; the ripple ratio after filtering is

\frac{(4E/3\pi)(f_c/100)}{2E/\pi} = \frac{2}{3}\,\frac{f_c}{100} < 0.01,

so $f_c < 1.5\,\mathrm{Hz}$ and $\tau = 1/2\pi f_c > 0.1\,\mathrm{s}$ . The load itself is the $R$ : a capacitor across it costs nothing in voltage drop and power, whereas a series resistor would waste both.

Exercise 9.12 ★★★

The AC coupling of an oscilloscope is a high-pass with $C = 0.10\,\text{µ}\mathrm{F}$ and $R = 160\,\mathrm{k}\Omega$ . Compute $f_c$ . A square wave is applied: show that each flat top becomes a decaying exponential and compute the sag (fraction lost over a half period) at $50\,\mathrm{Hz}$ and at $1\,\mathrm{kHz}$ .

Solution

Solution of Exercise 9.12.

$f_c = 1/(2\pi RC) = 1/(2\pi \times 1.6\times10^5 \times 10^{-7}) = 10\,\mathrm{Hz}$ , $\tau = RC = 16\,\mathrm{ms}$ . During a flat top the capacitor charges through $R$ : the output decays as $\eu^{-t/\tau}$ from each edge. Sag over $T/2$ : $1 - \eu^{-T/2\tau}$ : $50\,\mathrm{Hz}$ ( $T/2 = 10\,\mathrm{ms}$ ): $1 - \eu^{-0.63} = 47\%$ ; $1\,\mathrm{kHz}$ ( $T/2 = 0.5\,\mathrm{ms}$ ): $3\%$ .

Inside a two-way loudspeaker: the crossover’s coils and capacitors are the low-pass and high-pass filters that send bass to the woofer and treble to the tweeter.

9.7 Problem: The loudspeaker crossover

Problem 9.1

Weekend problem — a woofer, a tweeter, a coil and a capacitor: how a passive crossover splits the music, why the simplest one leaks bass into the tweeter, and what a real voice coil does to the design

The woofer and the tweeter are first modeled as pure resistances $R = 8.0\,\Omega$ . The amplifier is an ideal voltage source $\underline U_e$ . The crossover frequency is $f_c = 2.0\,\mathrm{kHz}$ .

Part I — First-order crossover. The woofer is fed through a series coil $L$ , the tweeter through a series capacitor $C$ .

Define the transfer function of each branch (voltage across the driver over $\underline U_e$ ) and the gain in decibels.
Show that the woofer branch is a first-order low-pass and give its cutoff $\omega_c$ in terms of $R$ and $L$ .
Compute $L$ for $f_c = 2.0\,\mathrm{kHz}$ .
Show that the tweeter branch is a first-order high-pass; give $\omega_c$ and compute $C$ .
Give each driver’s gain in decibels at $f_c$ , at $200\,\mathrm{Hz}$ and at $20\,\mathrm{kHz}$ .
Show that $G_L^2 + G_H^2 = 1$ at every frequency. What does this mean for the power drawn from the amplifier?
Give the phase of each branch at $f_c$ ; by how much do the two drivers’ voltages differ in phase there?
What are the slopes of the two responses, in dB per octave, far from $f_c$ ?

Part II — Second-order crossover. The woofer is now fed through a series coil $L$ with a capacitor $C$ in parallel with the driver.

Show that $\underline H = 1/(1 - LC\omega^2 + jL\omega/R)$ .
Identify $\omega_0$ and $Q$ in terms of $L$ , $C$ , $R$ .
For $Q = 1/\sqrt2$ show that $G = 1/\sqrt{1 + (\omega/\omega_0)^4}$ and that the gain is $-3\,\mathrm{dB}$ at $\omega_0$ .
Compute $L$ and $C$ for $f_0 = 2.0\,\mathrm{kHz}$ and $Q = 1/\sqrt2$ .
Compute the gain in decibels at $4\,\mathrm{kHz}$ and $8\,\mathrm{kHz}$ ; deduce the slope in dB per octave.
The tweeter branch swaps the roles of $L$ and $C$ (series $C$ , shunt $L$ ): write its transfer function without new calculation.
Compute the phase of each branch at $f_0$ . What must be done to the tweeter’s wiring, and why?

Part III — Music through the crossover.

A cello note at $220\,\mathrm{Hz}$ carries harmonics up to the 20th. Which harmonics go mostly to the tweeter, with the first-order crossover?
What happens to the 9th harmonic?
A $1.0\,\mathrm{kHz}$ square wave is used as a test signal. With the first-order crossover, give the gain of each branch for its fundamental, third and fifth harmonics.
The amplifier develops a $1\,\mathrm{V}$ DC offset by accident. Which driver is protected, and by what?
With the first-order crossover, what fraction of the voltage and of the power of a $500\,\mathrm{Hz}$ tone reaches the tweeter? Same with the second-order. Why does this matter for the tweeter’s survival?

Part IV — A real voice coil. The woofer is actually $R = 8.0\,\Omega$ in series with a voice-coil inductance $L_{vc} = 0.50\,\mathrm{mH}$ .

Compute the woofer’s impedance (modulus) at $2\,\mathrm{kHz}$ and at $8\,\mathrm{kHz}$ .
Recompute the first-order woofer gain at $2\,\mathrm{kHz}$ and $8\,\mathrm{kHz}$ with this impedance, and compare with the ideal $-3\,\mathrm{dB}$ and $-12.3\,\mathrm{dB}$ . What happened to the roll-off?
A “Zobel network” — $R_z = R$ in series with $C_z = L_{vc}/R^2$ — is connected in parallel with the woofer. Show that the combination behaves as a pure resistance $R$ at all frequencies, and compute $C_z$ .
The amplifier delivers $50\,\mathrm{W}$ into $8\,\Omega$ : what rms voltage is that, and what power does each driver receive at $f_c$ (first-order crossover)? At $500\,\mathrm{Hz}$ , how much reaches the tweeter?
Summarize the trade-offs: slope, phase, component count, protection of the tweeter, and the need for a Zobel network.

Solution

Solution of Problem 9.1.

1. $\underline H = \underline U_{\text{driver}}/\underline U_e$ ; $G_{\mathrm{dB}} = 20\log\abs{\underline H}$ .

2. $\underline H_L = R/(R + jL\omega) = 1/(1 + j\omega/\omega_c)$ , $\omega_c = R/L$ .

3. $L = R/\omega_c = 8.0/(2\pi \times 2000) = 0.64\,\mathrm{mH}$ .

4. $\underline H_H = R/(R + 1/jC\omega) = jRC\omega/(1 + jRC\omega)$ , $\omega_c = 1/RC$ ; $C = 1/(R\omega_c) = 9.9\,\text{µ}\mathrm{F} \approx 10\,\text{µ}\mathrm{F}$ .

5. At $f_c$ : $-3\,\mathrm{dB}$ each. $200\,\mathrm{Hz}$ ( $x = 0.1$ ): woofer $-0.04\,\mathrm{dB}$ , tweeter $-20\,\mathrm{dB}$ . $20\,\mathrm{kHz}$ ( $x = 10$ ): woofer $-20\,\mathrm{dB}$ , tweeter $-0.04\,\mathrm{dB}$ .

6. $G_L^2 + G_H^2 = (1 + x^2)/(1 + x^2) = 1$ : with equal loads the two powers add up to the power a single $8\,\Omega$ driver would take — shared, not lost.

7. $-45^\circ$ (woofer) and $+45^\circ$ (tweeter): $90^\circ$ apart; at $f_c$ the two cones do not move together.

8. $\pm 6\,\mathrm{dB}$ per octave ( $20\,\mathrm{dB}$ per decade).

9. Divider between $jL\omega$ and $R \parallel 1/jC\omega = R/(1 + jRC\omega)$ : $\underline H = R/[R + jL\omega(1 + jRC\omega)] = 1/(1 - LC\omega^2 + jL\omega/R)$ .

10. $\omega_0 = 1/\sqrt{LC}$ ; $L\omega/R = x/Q$ gives $Q = R/L\omega_0 = R\sqrt{C/L}$ .

11. $\abs{\underline H}^{-2} = (1 - x^2)^2 + x^2/Q^2 = 1 + x^4 + x^2(1/Q^2 - 2) = 1 + x^4$ for $Q^2 = \tfrac12$ ; at $x = 1$ , $G = 1/\sqrt2$ .

12. $L = R/(Q\omega_0) = 8\sqrt2/(1.257\times10^4) = 0.90\,\mathrm{mH}$ ; $C = 1/(L\omega_0^2) = 7.0\,\text{µ}\mathrm{F}$ .

13. $x = 2$ : $1/\sqrt{17}$ , $-12.3\,\mathrm{dB}$ ; $x = 4$ : $1/\sqrt{257}$ , $-24.1\,\mathrm{dB}$ : $12\,\mathrm{dB}$ per octave.

14. Exchanging $L$ and $C$ exchanges $x \leftrightarrow 1/x$ (and $Q$ stays): $\underline H_H = -x^2/(1 - x^2 + jx/Q)$ .

15. At $x = 1$ : $\underline H_L = Q/j = -jQ$ ( $-90^\circ$ ), $\underline H_H = jQ$ ( $+90^\circ$ ): the drivers are in phase opposition at the crossover and would cancel acoustically; the tweeter is wired with its polarity reversed.

16. Harmonics above $2\,\mathrm{kHz}$ : $n \geq 10$ ( $2.2\,\mathrm{kHz}$ and up) go mostly to the tweeter.

17. $9 \times 220 = 1980\,\mathrm{Hz} \approx f_c$ : split equally, $-3\,\mathrm{dB}$ to each.

18. Woofer ( $G_L$ ): $x = 0.5$ : $0.89$ ; $x = 1.5$ : $0.55$ ; $x = 2.5$ : $0.37$ . Tweeter ( $G_H$ ): $0.45$ , $0.83$ , $0.93$ .

19. The tweeter: its series capacitor blocks DC. The woofer takes the offset ( $1\,\mathrm{V}$ across $8\,\Omega$ : $125\,\mathrm{mA}$ , $0.125\,\mathrm{W}$ of heat and a displaced cone).

20. First order, $x = 0.25$ : $G_H = 0.25/\sqrt{1.0625} = 0.24$ ( $-12\,\mathrm{dB}$ ), power fraction $6\%$ . Second order: $0.0625/\sqrt{1 + 0.0039} = 0.06$ ( $-24\,\mathrm{dB}$ ), $0.4\%$ . A tweeter rated a few watts receives $6\%$ of a loud bass note with the first-order design — it can burn.

21. $\underline Z = 8 + jL_{vc}\omega$ : $2\,\mathrm{kHz}$ : $8 + 6.3j$ , $\abs{\underline Z} = 10.2\,\Omega$ ; $8\,\mathrm{kHz}$ : $8 + 25.1j$ , $26.3\,\Omega$ .

22. $\underline H = \underline Z/(\underline Z + jL\omega)$ : $2\,\mathrm{kHz}$ : $10.2/\abs{8 + 14.3j} = 10.2/16.4 = 0.62$ ( $-4.2\,\mathrm{dB}$ ); $8\,\mathrm{kHz}$ : $26.3/\abs{8 + 57.3j} = 0.45$ ( $-6.9\,\mathrm{dB}$ ) instead of $-12.3\,\mathrm{dB}$ : the rising coil impedance fights the series coil, and the roll-off flattens to a few dB per octave.

23. Admittances: $1/(R + jL_{vc}\omega) + jC_z\omega/(1 + jRC_z\omega)$ ; with $C_z = L_{vc}/R^2$ the second term is $jL_{vc}\omega/[R(R + jL_{vc}\omega)]$ , and the sum is $(R + jL_{vc}\omega)/[R(R + jL_{vc}\omega)] = 1/R$ . $C_z = 0.50\times10^{-3}/64 = 7.8\,\text{µ}\mathrm{F}$ .

24. $U = \sqrt{PR} = \sqrt{400} = 20\,\mathrm{V}$ rms. At $f_c$ each driver gets half: $25\,\mathrm{W}$ . At $500\,\mathrm{Hz}$ the tweeter gets $6\%$ : $3\,\mathrm{W}$ .

25. First order: two parts, power-complementary, gentle $6\,\mathrm{dB}/\mathrm{oct}$ slopes, $90^\circ$ between drivers, bass leaks into the tweeter. Second order: four parts, $12\,\mathrm{dB}/\mathrm{oct}$ , $180^\circ$ at crossover (invert the tweeter), far better tweeter protection. Either way the woofer’s inductance must be tamed by a Zobel network for the design to hold.

Terms defined in this chapter

See all 393 terms in the glossary