Fourier series of switched waveforms

Intent

• Split a periodic waveform into its dc component and its harmonics.
• Compute the harmonic amplitudes and the rms value of a rectangular pulse train, the waveform at a converter’s switch node.
• Explain why a low-pass filter recovers the dc component of the switch-node voltage.

Theory

Periodic signals as sums of sinusoids

A periodic signal is the sum of its average value, the dc component, and sinusoids at integer multiples of its fundamental frequency, the harmonics [Alexander & Sadiku 2017, Ch. 17]. The amplitudes of the harmonics, plotted against frequency, form the signal’s amplitude spectrum [Alexander & Sadiku 2017, Ch. 17].

The rectangular pulse train

The switch node of a buck converter carries a pulse train: the input voltage for the fraction $D$ of every period and zero for the rest [Erickson & Maksimović 2020, Ch. 2]. Its dc component is its average (average switch-node voltage), and its harmonics have the amplitudes

$$a_h = \frac{2 V_\mathrm{pk} \left|{\sin{\left(\pi h D \right)}}\right|}{\pi h}$$

$a_h$

amplitude of harmonic h

$D$

duty ratio

$V_\mathrm{pk}$

pulse amplitude

$h$

harmonic order

Amplitude of the h-th harmonic of a rectangular pulse train✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal rectangular pulse train between 0 and V_pk with duty ratio D
• periodic steady state

Notes

the dc component is $D V_\mathrm{pk}$; the harmonic amplitudes fall as $1/h$ and vanish where $h D$ is an integer

Source

Alexander & Sadiku 2017, Ch. 17 (The Fourier Series)

Derivation

step-by-step sympy derivation

Equation id

fourier.pulse.harm · equations.yaml entry

The amplitudes fall in inverse proportion to the harmonic order, and a harmonic vanishes whenever $h D$ is an integer: at $D = 0.5$ all the even harmonics are missing.

The rms value

The rms value is the square root of the mean of the square over one period [Alexander & Sadiku 2017, Ch. 11]. For the pulse train only the pulse contributes to that mean:

$$V_\mathrm{rms} = V_\mathrm{pk} \sqrt{D}$$

$V_\mathrm{rms}$

rms voltage

$D$

duty ratio

$V_\mathrm{pk}$

pulse amplitude

RMS value of a rectangular pulse train✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal rectangular pulse train between 0 and V_pk with duty ratio D

Source

Alexander & Sadiku 2017, Ch. 11 (AC Power Analysis)

Derivation

step-by-step sympy derivation

Equation id

fourier.pulse.rms · equations.yaml entry

The loss in a resistance depends on the rms value of its current, not on the average (conduction loss), and for a narrow pulse the rms value is much larger than the average.

Why the output filter works

A buck converter’s L-C filter passes the dc component of the switch-node voltage and attenuates its harmonics, the lowest of which lies at the switching frequency. What remains at the output is the average, $D V_g$, plus a small ripple [Erickson & Maksimović 2020, Ch. 2].

Worked example

An example pulse train, as found at the switch node of a buck converter:

Rectangular pulse-train example

Symbol	Value	How it is obtained
$V_\mathrm{pk}$ pulse amplitude	12 V	given
$D$ duty ratio	0.25	given
$h$ harmonic order	1	given
$\langle v_s \rangle$ average switch-node voltage	3 V	DC component of the switch-node voltage (buck)
$a_h$ amplitude of harmonic h	5.4 V	Amplitude of the h-th harmonic of a rectangular pulse train
$V_\mathrm{rms}$ rms voltage	6 V	RMS value of a rectangular pulse train

The pulse train’s dc component is 3 V, and its fundamental has an amplitude of 5.4 V, larger than the dc component itself at this duty ratio. Its rms value is 6 V.

Try it

▶ Open in the equation explorer: Amplitude of the h-th harmonic of a rectangular pulse train
Preset from the example pulse-train

Sweep $D$: the fundamental is largest at $D = 0.5$. Then set $h$ to 2 and sweep again: the second harmonic vanishes at $D = 0.5$ and is largest at 0.25 and 0.75.

Bench exercise

On a buck evaluation board, measure the switch-node voltage with a differential probe, or a probe with a short ground connection, and use the oscilloscope’s FFT function. Read the dc value and the amplitudes of the first few harmonics, and compare them with the equations at the measured duty ratio and input voltage. Check whether the oscilloscope displays each harmonic’s peak amplitude or its rms value, which is smaller by a factor of $\sqrt{2}$.

Gotchas

• The harmonics do not stop at the switching frequency. Those of an ideal pulse train fall only as $1/h$, so they extend far above the switching frequency; this matters for filters, interference and sampling.
• Average and rms are different. Two currents with the same average can have very different rms values, and resistive losses follow the rms value [Alexander & Sadiku 2017, Ch. 11].
• The rms value of a sinusoid is its amplitude divided by $\sqrt{2}$; that of a pulse train is not. Use the rms value of the actual waveform [Alexander & Sadiku 2017, Ch. 11].

Go deeper

• Lecture notes 6.003 Lecture Notes — D. Freeman, MIT OpenCourseWare · retrieved 2026-09-23
  Lecture slides: the Laplace transform (lecture 6), frequency response and Bode plots (lectures 9 and 11), Fourier series (lecture 15).
• Course 6.003 Signals and Systems (Fall 2011) — D. Freeman, MIT OpenCourseWare · retrieved 2026-09-23
  Signals and systems with video lectures: Laplace transform, frequency response, Bode plots and Fourier series.

Quiz

1. A pulse train switches between $0$ and $10\,\mathrm{V}$ with $D = 0.3$. What is its dc component?

   1. $0.3\,\mathrm{V}$
   2. $3\,\mathrm{V}$
   3. $5\,\mathrm{V}$
   4. $7\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The dc component is the average: the pulse height times the fraction of the period it lasts.

2. What is the rms value of a pulse train between $0$ and $10\,\mathrm{V}$ at $D = 0.25$?

   1. $2.5\,\mathrm{V}$
   2. $5\,\mathrm{V}$
   3. $7.07\,\mathrm{V}$
   4. $10\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The mean square is $0.25 \times 100\,\mathrm{V^2}$, and its square root is $5\,\mathrm{V}$, twice the average.

3. At $D = 0.5$, which harmonics of a rectangular pulse train are missing?

   1. None
   2. The odd harmonics
   3. The even harmonics
   4. All harmonics above the third
   Answer and explanation

   Correct answer: C A harmonic vanishes where $h D$ is an integer; at $D = 0.5$ that is every even $h$.

4. Apart from the zeros, how do the harmonic amplitudes of a rectangular pulse train fall with the harmonic order $h$?

   1. As $1/h$
   2. As $1/h^2$
   3. Exponentially
   4. They do not fall
   Answer and explanation

   Correct answer: A The amplitude is $2 V_\mathrm{pk} |\sin(\pi h D)|/(\pi h)$: the envelope falls as $1/h$.

5. What is the amplitude of the fundamental of a pulse train between $0$ and $12\,\mathrm{V}$ at $D = 0.5$?

   1. $6\,\mathrm{V}$
   2. $7.64\,\mathrm{V}$
   3. $12\,\mathrm{V}$
   4. $3.82\,\mathrm{V}$
   Answer and explanation

   Correct answer: B With $h = 1$ and $D = 0.5$ the sine term is 1, so the amplitude is $2 \times 12/\pi$.

6. Why does a buck converter's output equal the average of its switch-node voltage?

   1. The L-C filter passes the dc component and strongly attenuates the harmonics, the lowest of which is at the switching frequency
   2. The switch-node voltage has no harmonics
   3. The diode removes the harmonics
   4. The load resistor averages the voltage
   Answer and explanation

   Correct answer: A The output keeps the dc component and a small residue of the harmonics: the ripple.