Amplitude modulation
도입됨| English | 의미 |
|---|---|
| envelope detector | 포락 검波器 |
| amplitude modulation | 진폭 변조 |
| modulation index | 변조 지수 |
The simplest carrier system
- AM radio is the simplest way to put sound on a carrier. Its power budget closes neatly on paper.
- That budget explains why broadcasters moved on, and why the cheap envelope detector is still in every radio.
The envelope is the message
- In amplitude modulation 幅度调制, $s(t) = A[1 + m\,x(t)]\cos(\omega_{c}t)$. The envelope follows the message $x(t)$.
- The modulation index 调制指数 $m$ sets the depth. It must stay at or below 1.
What happens when the modulation index goes above 1?
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With m > 1 the term 1 + m·x(t) goes negative, so the envelope no longer follows the message.
Spectrum and power
- A tone at $f_{m}$ gives the carrier plus two sidebands at $f_{c} \pm f_{m}$, each of amplitude $mA/2$.
- So the sidebands together carry $\dfrac{m^{2}}{2}P_{c}$, and the total is $P_{c}\left(1 + \dfrac{m^{2}}{2}\right)$.
A 1 MHz carrier is modulated by a 2 kHz tone. Where are the sidebands?
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The sidebands sit at $f_{c} \pm f_{m}$.
Worked example: the power budget
- A 100 W carrier at $m = 0.5$: sidebands $\dfrac{0.25}{2} \times 100 = 12.5$ W, total 112.5 W. Only 11.1% of the power carries the message.
- Even at $m = 1$: 50 W in sidebands out of 150 W, 33.3%. Two thirds of the power goes into a carrier with no information.
A 200 W carrier is modulated at m = 0.8. What is the total transmitted power, in W?
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Sidebands: $\dfrac{0.64}{2} \times 200 = 64$ W. Total $200 + 64 = 264$ W.
For that transmitter, what percentage of the total power is in the sidebands? Give 1 decimal place.
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$64/264 = 24.2\%$. Even at the limit $m = 1$ it is only 33.3%.
The envelope detector
- The envelope detector 包络检波器 is a diode and an RC circuit. It charges on each carrier peak and discharges slowly between them.
- Its time constant must satisfy $\dfrac{1}{\omega_{c}} \ll \tau \le \dfrac{\sqrt{1 - m^{2}}}{m\,\omega_{m}}$: slow for the carrier, fast for the message.
A receiver's carrier is 455 kHz. What is 1/ω_c, the lower limit for the detector's τ, in µs? Give 2 decimal places.
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$1/(2\pi \times 455 \times 10^{3}) = 0.35$ µs. τ must be well above this to bridge the carrier peaks.
Worked example: choosing τ
- $f_{c} = 100$ kHz, $f_{m} = 5$ kHz: $1/\omega_{c} = 1.6$ µs and $1/\omega_{m} = 31.8$ µs. $\tau = 10$ µs works.
- At $m = 0.9$ the upper bound falls to $\dfrac{\sqrt{0.19}}{0.9} \times 31.8 = 15.4$ µs: less than half the room.
With f_m = 5 kHz and m = 0.5, what is the detector's upper bound on τ, in µs? Give the nearest whole number.
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$\dfrac{\sqrt{0.75}}{0.5} \times 31.8 = 55$ µs. Shallow modulation leaves more room; deep modulation leaves less.
Three common slips
- $m > 1$. The envelope folds over and the detector produces nonsense.
- Unclear efficiency. Say whether it is against the total or the carrier power.
- τ from the carrier alone. The message sets the upper bound.
In AM, most of the transmitted power is in the carrier, which carries no information.
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At least two thirds of the power stays in the carrier, even at m = 1. That waste is why AM broadcasting gave way to other methods.
Recap
- The envelope is the message; $m \le 1$.
- Sidebands carry $\dfrac{m^{2}}{2}P_{c}$; efficiency is at most 33%.
- The detector's τ sits between the carrier period and the message.