---
title: "Filters and Transfer Functions"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 9
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/9-filters-and-transfer-functions
---

# Chapter 9 — Filters 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) with one capacitor. All three are *[filters](#def-b1-filters-transfer-functions-H)*: [linear circuits](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) that treat each frequency differently. Because any periodic [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) is a sum of sinusoids, knowing what a circuit does to every sinusoid — its [transfer function](#def-b1-filters-transfer-functions-H) — is knowing what it does to everything. This chapter defines [transfer functions](#def-b1-filters-transfer-functions-H), draws their [Bode diagrams](#def-b1-filters-transfer-functions-bode) 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](#def-b1-filters-transfer-functions-H).

## 9.1 Transfer function and Bode diagram

**Definition 9.1 (Transfer function).**

A [linear circuit](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) with an input [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $\underline U_e$ and an output [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal)) 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $E$ at $\omega$ comes out as $G(\omega)E\cos(\omega t + \varphi(\omega))$.

**Definition 9.2 (Decibels; Bode diagram).**

The [gain](#def-b1-filters-transfer-functions-H) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#prop-b1-dc-circuits-dividers).

## 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](#def-b1-filters-transfer-functions-H) is an integrator there.

**Proof.** [Voltage divider](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#prop-b1-dc-circuits-dividers):

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

Far above cutoff, $1 + j\omega/\omega_c \approx j\omega/\omega_c$, and dividing a [complex amplitude](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-complex) by $j\omega$ is integrating ([Proposition 8.3](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#prop-b1-sinusoidal-impedance-rules)): $u_s \approx \omega_c\int u_e\,\dd t$. $20\log(\omega_c/\omega)$ 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-f08f56c3b0a8.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-551e635144f9.svg)

*[Bode diagrams](#def-b1-filters-transfer-functions-bode) of the first-order low-pass (blue) and high-pass (red): [gain](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) with the asymptotes (dashed) meeting at the cutoff, where the true [gain](#def-b1-filters-transfer-functions-H) 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-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](#def-b1-filters-transfer-functions-H) is only one decade above its cutoff and the [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal)’s low harmonics are shifted in phase ([Exercise 9.12](#exo-b1-filters-transfer-functions-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](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#thm-b1-sinusoidal-impedance-resonance).

**Proof.** [Voltage dividers](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#prop-b1-dc-circuits-dividers) with the total [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) across $R$. ∎

**Proposition 9.8 (Second-order low-pass: the role of QQQ).**

For the low-pass, $G = 1/\sqrt{(1 - x^2)^2 + x^2/Q^2}$: for $Q > 1/\sqrt2$ the [gain](#def-b1-filters-transfer-functions-H) 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](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#prop-b1-sinusoidal-impedance-overvoltage). ∎

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-ca3bcba9b685.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-083103bb08c8.svg)

*[Second-order filters](#prop-b1-filters-transfer-functions-rlc) 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $c_n$ form the *spectrum*. A square wave of [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\pm E$ has $c_n = 4E/(n\pi)$ for odd $n$ and $0$ for even $n$; a triangle wave of [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\pm E$ 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) need few (the triangle’s as $1/n^2$). A [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal)’s harmonic content is what an instrument’s timbre is made of, and what a [filter](#def-b1-filters-transfer-functions-H) acts on.

**Proposition 9.12 (Linear filtering of a periodic signal).**

Through a [filter](#def-b1-filters-transfer-functions-H) of [transfer function](#def-b1-filters-transfer-functions-H) $\underline H$, the [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-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](#def-b1-filters-transfer-functions-H) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-linear) ([Theorem 6.18](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-superposition)) applied to the sum of sinusoidal inputs, each treated by [Definition 9.1](#def-b1-filters-transfer-functions-H). ∎

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-a910f44bcccb.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/fig-28451d640fd6.svg)

*A square wave through a first-order low-pass. Left: its spectrum (odd harmonics in $1/n$) and the spectrum after a [filter](#def-b1-filters-transfer-functions-H) cut at the fundamental. Right: the time [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $5\,\mathrm{V}$: 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $ET/4\tau \approx 80\,\mathrm{mV}$ 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](#exo-b1-filters-transfer-functions-8)).

## 9.5 Cascading filters

**Proposition 9.14 (Cascade and loading).**

Two [filters](#def-b1-filters-transfer-functions-H) in cascade have $\underline H = \underline H_1\underline H_2$ — [gains](#def-b1-filters-transfer-functions-H) multiply, [decibels](#def-b1-filters-transfer-functions-bode) 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](https://one-course.com/books/physics/3/en/chapter/10-the-operational-amplifier#ch-b1-operational-amplifier)) between stages makes the product rule exact.

**Proof.** $\underline H_1$ was defined with an open output; if the next stage’s input [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) is much larger than the first stage’s output [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) (its Thévenin resistance), the output [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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](#def-b1-filters-transfer-functions-bode): [gains](#def-b1-filters-transfer-functions-H) $0.5$, $10$, $1/\sqrt2$, $100$. Convert to ratios: $-40\,\mathrm{dB}$, $+6\,\mathrm{dB}$. Two [filters](#def-b1-filters-transfer-functions-H) in cascade (no loading) give $-6\,\mathrm{dB}$ and $-20\,\mathrm{dB}$ at some frequency: overall [gain](#def-b1-filters-transfer-functions-H)?

**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](#def-b1-filters-transfer-functions-bode), and the [gain](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) at $100\,\mathrm{Hz}$, $1\,\mathrm{kHz}$, $10\,\mathrm{kHz}$, $100\,\mathrm{kHz}$.

**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](#def-b1-filters-transfer-functions-H) $-10\log(1 + (f/f_c)^2)$: $100\,\mathrm{Hz}$: $-0.04\,\mathrm{dB}$; $1\,\mathrm{kHz}$: $-3\,\mathrm{dB}$; $10\,\mathrm{kHz}$: $-20\,\mathrm{dB}$; $100\,\mathrm{kHz}$: $-40\,\mathrm{dB}$.

**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 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $5.0\,\mathrm{V}$: give the frequencies and [amplitudes](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of its first three non-zero harmonics. Why does it have no even harmonics?

**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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $1.0\,\mathrm{V}$ at $100\,\mathrm{Hz}$. Explain why it behaves as a differentiator and give the shape and [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of the output.

**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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $400/(2\pi \times 10^4) =
6.4\,\mathrm{mV}$.

**Exercise 9.6 ★★.**

An RC low-pass cut at $10\,\mathrm{Hz}$ receives a $1.0\,\mathrm{kHz}$ square wave of [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\pm1.0\,\mathrm{V}$. Show that the output is a triangle and compute its [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal).

**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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $ET/4\tau = 1.0 \times 10^{-3}/(4 \times 0.016) = 16\,\mathrm{mV}$ 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](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) at $f_0$ and at $10f_0$.

**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](#exo-b1-filters-transfer-functions-4). Compute the [gain](#def-b1-filters-transfer-functions-H) at $1$, $3$ and $5\,\mathrm{kHz}$, and the output [amplitudes](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of these harmonics; describe the output.

**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](#def-b1-filters-transfer-functions-H) (the second loads the first) and the [gain](#def-b1-filters-transfer-functions-H) at $\omega_c = 1/RC$; compare with the product of the two separate [gains](#def-b1-filters-transfer-functions-H).

**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](#def-b1-filters-transfer-functions-H) 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](#def-b1-filters-transfer-functions-bode), and the phase at $\sqrt{\omega_1\omega_2}$.

**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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $4E/3\pi$. An RC low-pass must reduce the $100\,\mathrm{Hz}$ ripple below $1\%$ of the average: find the required [cutoff frequency](#def-b1-filters-transfer-functions-bode) and [time constant](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder). Why is a capacitor alone (with the load resistance) the usual choice?

**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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) drop and power, whereas a series [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-filters-transfer-functions/img-4accfe2f09c0.jpg)

*Inside a two-way loudspeaker: the crossover’s coils and capacitors are the low-pass and [high-pass filters](#prop-b1-filters-transfer-functions-highpass1) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-sources) $\underline U_e$. 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 $C$.

1. Define the [transfer function](#def-b1-filters-transfer-functions-H) of each branch ( [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) across the driver over $\underline U_e$ ) and the [gain](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) .
2. Show that the woofer branch is a first-order low-pass and give its cutoff $\omega_c$ in terms of $R$ and $L$ .
3. Compute $L$ for $f_c = 2.0\,\mathrm{kHz}$ .
4. Show that the tweeter branch is a first-order high-pass; give $\omega_c$ and compute $C$ .
5. Give each driver’s [gain](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) at $f_c$ , at $200\,\mathrm{Hz}$ and at $20\,\mathrm{kHz}$ .
6. Show that $G_L^2 + G_H^2 = 1$ at every frequency. What does this mean for the power drawn from the amplifier?
7. Give the phase of each branch at $f_c$ ; by how much do the two drivers’ [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) differ in phase there?
8. 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.

9. Show that $\underline H = 1/(1 - LC\omega^2 + jL\omega/R)$ .
10. Identify $\omega_0$ and $Q$ in terms of $L$ , $C$ , $R$ .
11. For $Q = 1/\sqrt2$ show that $G = 1/\sqrt{1 + (\omega/\omega_0)^4}$ and that the [gain](#def-b1-filters-transfer-functions-H) is $-3\,\mathrm{dB}$ at $\omega_0$ .
12. Compute $L$ and $C$ for $f_0 = 2.0\,\mathrm{kHz}$ and $Q = 1/\sqrt2$ .
13. Compute the [gain](#def-b1-filters-transfer-functions-H) in [decibels](#def-b1-filters-transfer-functions-bode) at $4\,\mathrm{kHz}$ and $8\,\mathrm{kHz}$ ; deduce the slope in dB per octave.
14. The tweeter branch swaps the roles of $L$ and $C$ (series $C$ , shunt $L$ ): write its [transfer function](#def-b1-filters-transfer-functions-H) without new calculation.
15. 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.**

16. 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?
17. What happens to the 9th harmonic?
18. A $1.0\,\mathrm{kHz}$ square wave is used as a test [signal](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) . With the first-order crossover, give the [gain](#def-b1-filters-transfer-functions-H) of each branch for its fundamental, third and fifth harmonics.
19. The amplifier develops a $1\,\mathrm{V}$ DC offset by accident. Which driver is protected, and by what?
20. With the first-order crossover, what fraction of the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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 $L_{vc} = 0.50\,\mathrm{mH}$.

21. Compute the woofer’s [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) (modulus) at $2\,\mathrm{kHz}$ and at $8\,\mathrm{kHz}$ .
22. Recompute the first-order woofer [gain](#def-b1-filters-transfer-functions-H) at $2\,\mathrm{kHz}$ and $8\,\mathrm{kHz}$ with this [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) , and compare with the ideal $-3\,\mathrm{dB}$ and $-12.3\,\mathrm{dB}$ . What happened to the roll-off?
23. 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 $C_z$ .
24. The amplifier delivers $50\,\mathrm{W}$ into $8\,\Omega$ : what rms [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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?
25. Summarize the trade-offs: slope, phase, component count, protection of the tweeter, and the need for a Zobel network.

**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](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) fights the series coil, and the roll-off flattens to a few dB per octave.

**23.** [Admittances](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance): $1/(R + jL_{vc}\omega) + jC_z\omega/(1 + jRC_z\omega)$; 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.