Signals, spectra and bandwidth
Introduit| English | Sens |
|---|---|
| bandwidth/ˈbændwɪdθ/ | bande passante |
| sideband | bande latérale |
Spectrum is sold by the kilohertz
- A radio regulator sells spectrum by the kilohertz. Whether a signal fits its licence depends on its spectrum.
- The spectrum depends on the waveform's shape. This lesson links the two.
Harmonics
- A square wave of amplitude 1 is a sum of odd harmonics with amplitudes $\dfrac{4}{\pi k}$: the 3rd at 1/3 of the fundamental, the 5th at 1/5.
- Each harmonic's power goes as its amplitude squared, so the tail falls as $1/k^{2}$ in power. Bandwidth 带宽 is a choice of how much tail to keep.
A square wave's fundamental has amplitude 1.273. What is the amplitude of its 7th harmonic? Give 3 decimal places.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
Odd harmonics fall as $1/k$: $1.273/7 = 0.182$.
Match each harmonic of a square wave to its amplitude, relative to the fundamental.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
A square wave has only odd harmonics, each at 1/k of the fundamental.
Power kept
- By Parseval's theorem, the fundamental alone carries $\dfrac{1.273^{2}/2}{1} = 81\%$ of the square wave's power.
- Adding the 3rd harmonic brings it to 90%. Every extra harmonic costs bandwidth and adds less.
What is the power of the 3rd harmonic as a fraction of the fundamental's power? Give 3 decimal places.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
Power goes as amplitude squared: $(1/3)^{2} = 1/9 = 0.111$.
Worked example: through a 5 kHz low-pass
- A 1 kHz square wave passes a 5 kHz low-pass filter: the 1st (1.273) and 3rd (0.424) harmonics pass; the 5th (0.255) and above are blocked.
- 90% of the power remains. The edges are rounder, but the signal is still clearly there.
A 1 kHz square wave passes a 4 kHz low-pass filter. How many harmonics get through?
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The 1st (1 kHz) and 3rd (3 kHz) pass; the 5th (5 kHz) does not. Only odd harmonics exist.
Modulation doubles the width
- Mixing with a carrier shifts the spectrum; it adds no new shape. Each side of the carrier gets a copy: a sideband 边带.
- So a message of width $W$ on a carrier occupies $2W$ of radio spectrum.
A 3 kHz voice signal is mixed onto a 1 MHz carrier. How much spectrum does it occupy, in kHz?
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Two sidebands of 3 kHz each: 997 to 1003 kHz, so $2W = 6$ kHz.
Worked example: the radio licence
- A 4 kHz signal on a 100 kHz carrier spans 96–104 kHz: 8 kHz of spectrum. An 8 kHz licence is exactly filled.
- Grant only 4 kHz and you must send one sideband (SSB). No filter can squeeze a message below its own width $W$.
A licence grants only 3 kHz for that 3 kHz voice signal. What can you do?
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
One sideband carries the whole message in W. Filtering the voice itself below 3 kHz would delete part of it.
Three common slips
- "Bandwidth" without a definition. −3 dB, 98% power or first null: say which.
- Filtering below the message width. That deletes part of the message.
- Amplitude for power. Harmonics fall as $1/k$ in amplitude but $1/k^{2}$ in power.
The harmonics of a square wave fall as 1/k in power.
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
They fall as 1/k in amplitude, so 1/k² in power. The tail is weaker than the amplitudes suggest.
Recap
- Shape decides the spectrum; bandwidth is how much tail you keep.
- The fundamental holds 81% of a square wave's power.
- Modulation shifts and doubles: $2W$.