PCM and baseband transmission
도입됨| English | 의미 |
|---|---|
| quantisation/ˌkwɒntaɪˈzeɪʃn/ | 양자화 |
| raised-cosine | 상승 코사인 |
The microphone goes digital
- The link's microphone voltage now becomes bits, and the bits become pulses on a wire.
- How finely you measure sets the quality; the pulse shape sets how fast the wire can carry the bits.
Quantisation
- Quantisation 量化 rounds each sample to one of $2^{n}$ levels. The rounding error is spread evenly over one step, $\Delta$.
- Against a full-scale sine, the signal-to-quantisation-noise ratio is $SQNR \approx 6.02n + 1.76$ dB: about 6 dB per bit.
What SQNR does a 12-bit quantiser give for a full-scale sine, in dB? Give 1 decimal place.
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
$6.02 \times 12 + 1.76 = 74.0$ dB.
How many bits are needed for an SQNR of at least 60 dB?
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
$(60 - 1.76)/6.02 = 9.7$, so 10 bits (61.96 dB). 9 bits give only 55.9 dB.
Sampling and PCM
- Sample faster than twice the highest frequency, $f_{s} > 2W$, then code each sample in $n$ bits. That is PCM, the basis of the telephone network.
- The bit rate is $f_{s} \times n$.
Audio goes up to 20 kHz. Which sampling rate works?
난이도 1/5 · 100코인 + 오늘 첫 완료 시 작은 보너스 추가
The rate must be above 40 kHz. 44.1 kHz leaves room for the anti-alias filter; exactly 40 kHz leaves none.
Worked example: the telephone number
- 4 kHz voice sampled at 8 kHz with 8 bits: $8000 \times 8 = 64$ kbit/s.
- $SQNR = 6.02 \times 8 + 1.76 = 49.9$ dB from 256 levels. Each extra dB costs 1/6 of a bit per sample.
Wideband voice is sampled at 16 kHz with 16 bits per sample. What is the bit rate, in kbit/s?
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
$16\,000 \times 16 = 256\,000$ bit/s, four times the classic 64 kbit/s.
Pulse shaping
- Sharp pulses spread over a wide spectrum. A raised-cosine 升余弦 filter with roll-off $\alpha$ limits the bandwidth to $B = \dfrac{(1 + \alpha)R_{s}}{2}$.
- $\alpha = 0$ gives the minimum, $R_{s}/2$, but real filters cannot reach it. A larger $\alpha$ gives more room for timing errors.
A link sends 2 Msymbol/s through a raised-cosine filter with α = 0.25. What bandwidth does it need, in MHz?
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
$B = (1 + 0.25) \times 2/2 = 1.25$ MHz.
Worked example: the wire
- At $R_{s} = 1$ Msymbol/s with $\alpha = 0.35$: $B = 1.35 \times 0.5 = 0.675$ MHz.
- Manchester coding puts a transition in every bit for a strong clock, at the cost of about twice the bandwidth of plain NRZ.
Three common slips
- Sampling at exactly $2W$. It leaves no room for the filter; sample above it.
- SQNR for the whole chain's SNR. SQNR is only the rounding noise.
- $\alpha = 0$ as a design. It is a limit that real filters only approach.
An ideal α = 0 filter can be built, and real modems use it.
난이도 1/5 · 100코인 + 오늘 첫 완료 시 작은 보너스 추가
α = 0 needs a brick-wall filter and perfect timing. Real modems use about 0.2 to 0.35.
Match each choice to what it decides.
난이도 2/5 · 140코인 + 오늘 첫 완료 시 작은 보너스 추가
Each PCM design choice buys one thing and costs another.
Recap
- $SQNR = 6.02n + 1.76$ dB.
- Sample at $f_{s} > 2W$; the bit rate is $f_{s}n$.
- Raised cosine: $B = (1 + \alpha)R_{s}/2$.