Angle modulation (FM/PM)
Introduced| English | Meaning |
|---|---|
| Carson's rule | An approximate FM bandwidth estimate: twice the sum of peak frequency deviation and highest message frequency. |
| frequency modulation | frequency modulation |
| instantaneous frequency | The time derivative of signal phase divided by two pi, with phase measured in radians. |
| frequency deviation | The specified departure of instantaneous frequency from the unmodulated carrier, often quoted as its peak value. |
Clear through a thunderstorm
- AM crackles in a thunderstorm; FM stays clear. FM trades extra bandwidth for less noise.
- This lesson prices both sides of that trade: Carson's rule for the bandwidth, and the gain in signal-to-noise ratio.
The message moves the frequency
- Frequency modulation 频率调制 puts the message in the instantaneous frequency 瞬时频率: $f_{i} = f_{c} + k_{f}x(t)$.
- The largest swing is the frequency deviation 频偏 $\Delta f$. Against the message frequency it gives the index $\beta = \Delta f / f_{m}$.
An FM signal has Δf = 50 kHz and f_m = 10 kHz. What is β?
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$\beta = \Delta f / f_{m} = 50/10 = 5$.
Carson's rule
- Carson's rule 卡森公式 estimates the bandwidth: $B \approx 2(\Delta f + f_{m})$.
- It holds about 98% of the power. Narrowband FM ($\beta \ll 1$) shrinks to about $2f_{m}$, like AM.
A two-way radio uses Δf = 5 kHz with 3 kHz audio. What bandwidth does Carson's rule give, in kHz?
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$B \approx 2(5 + 3) = 16$ kHz.
For narrowband FM (β much less than 1), the bandwidth is about
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With $\Delta f \ll f_{m}$, Carson gives $2(\Delta f + f_{m}) \approx 2f_{m}$.
Where the lines come from
- The phase is $2\pi f_{c}t + \beta\sin(\omega_{m}t)$. Expanding it gives lines at $f_{c} \pm kf_{m}$ with Bessel-function weights $J_{k}(\beta)$.
- The significant lines stop near $k \approx \beta + 1$, which is exactly Carson's $2(\Delta f + f_{m})$.
With β = 5, about how many significant lines lie on each side of the carrier?
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They stop near $k = \beta + 1 = 6$, so 6 lines each side, spaced $f_{m}$ apart: $2 \times 6 \times 15 = 180$ kHz.
Worked example: FM broadcast
- $\Delta f = 75$ kHz and $f_{m} = 15$ kHz: $\beta = 5$ and $B = 2 \times 90 = 180$ kHz.
- Stations are spaced 200 kHz apart, so 20 kHz is left as a guard band.
Worked example: the price of quiet
- Above threshold, wideband FM improves the SNR by about $3\beta^{2}/2$: 37.5, or 15.7 dB, at $\beta = 5$.
- The price is 180 kHz, six times the 30 kHz AM needs for the same audio. Below an input SNR of about 10 dB the gain disappears.
Using 3β²/2, how much SNR gain, in dB, does β = 3 give? Give 1 decimal place.
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$3 \times 9/2 = 13.5$, and $10\log_{10}13.5 = 11.3$ dB.
Three common slips
- Carson as exact. It is a 98%-power estimate.
- $\Delta f$ for $B$. $\Delta f$ belongs to the modulator; $B$ belongs to the spectrum.
- Forgetting the threshold. Below it, FM's advantage is lost.
Raising β improves the output SNR even when the input SNR is very low.
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Below the FM threshold (about 10 dB input SNR) clicks and capture take over and the gain is lost.
Recap
- $\beta = \Delta f / f_{m}$.
- $B \approx 2(\Delta f + f_{m})$.
- More bandwidth buys more SNR, but only above threshold.
Match each quantity to its meaning.
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The modulator sets Δf; β and B follow from it and the message.