Fourier series
Introducido| English | Significado |
|---|---|
| Fourier series/ˈfɔːrɪə ˈsɪəriːz/ | A way to write a periodic signal as a sum of sines and cosines at whole-number multiples of its frequency. |
| fundamental/ˌfʌndəˈmentl/ | The lowest-frequency component of a periodic signal. |
| harmonic/hɑːˈmɒnɪk/ | armónico |
| orthogonality/ˌɔːθəɡəˈnælɪti/ | Distinct sinusoids averaging to zero against each other, so each coefficient is independent. |
A distorted waveform, taken apart
- A distorted mains waveform heats a resistor exactly as a particular mix of pure sinusoids would. Which mix is not a guess; it can be computed.
- Any periodic signal is a sum of sinusoids. This lesson shows how to find the mix, and what it says about edges, filters and power.
The Fourier series
- For a signal of period $T$, the fundamental 基波 is $\omega_{0} = \dfrac{2\pi}{T}$ rad/s. The Fourier series 傅里叶级数 writes the signal as an average plus sinusoids at whole multiples of $\omega_{0}$.
- $a_{0} = \dfrac{1}{T}\displaystyle\int_{T} f\,dt$ is the average value. Each term at $n\omega_{0}$ is a harmonic 谐波.
Match each symbol to its meaning.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Every other quantity in the series is built from these four.
Mains at 50 Hz has a period of 20 ms. What is its fundamental ω₀, in rad/s? Give one decimal place.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\omega_{0} = \dfrac{2\pi}{0.02} \approx 314.2$ rad/s.
Orthogonality picks out each coefficient
- Two different harmonics multiplied together average to zero over a period. This orthogonality 正交性 means each $a_{n}$ and $b_{n}$ is found on its own, as the average of $f$ times the matching sinusoid.
- It works like a bank of tuned filters, each hearing only its own frequency.
The square wave
- A ±1 square wave with odd symmetry has all $a_{n} = 0$, and $b_{n} = \dfrac{4}{n\pi}$ for odd $n$.
- The amplitudes fall as $\dfrac{1}{n}$. Sharp edges need many harmonics, which is the whole story of filtering.
At $t = \dfrac{T}{4}$ every sine is ±1, and the sum is $\dfrac{4}{\pi}\left(1 - \dfrac{1}{3} + \dfrac{1}{5} - \dots\right)$. The bracket is Gregory's series for $\dfrac{\pi}{4}$, so the total is exactly 1, as the square wave demands.
What is the amplitude of the 5th harmonic of a ±1 square wave? Give three decimal places.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$b_{5} = \dfrac{4}{5\pi} \approx 0.255$, one fifth of the fundamental's $\dfrac{4}{\pi} \approx 1.273$.
At t = T/4, what does the square-wave series (4/π)(1 − 1/3 + 1/5 − …) add up to?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The alternating series of odd reciprocals sums to $\dfrac{\pi}{4}$; scaled by $\dfrac{4}{\pi}$ it gives exactly 1, the value the square wave demands there.
Truncation and the Gibbs overshoot
- Keep only a few harmonics and the corners round off: with $n \le 3$ the peak reaches $\dfrac{4}{\pi}\left(1 - \dfrac{1}{3}\right) \approx 0.85$, not 1.
- Add more and the sum hugs the square, except at each jump, where it overshoots by about 9% however many terms you take. This is the Gibbs effect.
A channel passes only the harmonics n ≤ 3 of a square wave. What comes out?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Sharp corners need many harmonics cancelling. At the peak the two terms give $\dfrac{4}{\pi}\left(1 - \dfrac{1}{3}\right) \approx 0.85$, not 1.
With enough harmonics, the overshoot at each jump of a square wave disappears.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The overshoot narrows towards the jump but stays at about 9% of the step, however many harmonics are added. That is the Gibbs effect.
Power in the harmonics
- A ±1 V square wave across 1 Ω delivers 1 W. Parseval's theorem says the harmonics' powers add up to the same total; each sinusoid of amplitude $A$ delivers $\dfrac{A^{2}}{2}$.
- The fundamental alone carries $\dfrac{1}{2}\left(\dfrac{4}{\pi}\right)^{2} = \dfrac{8}{\pi^{2}} \approx 81\%$; the third harmonic adds about 9%. A filter that keeps only the fundamental still delivers most of the power.
The first 20 odd harmonics give 0.990 W. The full sum is $\dfrac{8}{\pi^{2}}\left(1 + \dfrac{1}{9} + \dfrac{1}{25} + \dots\right)$, and the bracket is $\dfrac{\pi^{2}}{8}$, so the total is exactly 1 W.
A ±1 V square wave drives 1 Ω. What share of the power does the fundamental alone carry?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The fundamental's amplitude is $\dfrac{4}{\pi}$; its power is $\dfrac{1}{2}\left(\dfrac{4}{\pi}\right)^{2} = \dfrac{8}{\pi^{2}} \approx 0.81$ W of the 1 W total.
Three common slips
- Hertz for radians per second. The fundamental in the formula is $\omega_{0} = \dfrac{2\pi}{T}$, not $\dfrac{1}{T}$.
- Forgetting the ½ in harmonic power. A sinusoid's mean square is half its peak squared.
- Assuming the symmetry. Shift the time origin and the coefficients change; check the signal as placed.
Recap
- A periodic signal = average + harmonics at $n\omega_{0}$, with $\omega_{0} = \dfrac{2\pi}{T}$.
- Orthogonality finds each coefficient separately; square-wave amplitudes are $\dfrac{4}{n\pi}$ for odd $n$.
- Parseval: harmonic powers $\dfrac{A^{2}}{2}$ add to the total; the Gibbs overshoot never leaves the jumps.