Data Communication · Lecture 5
Analog-to-Digital Conversion
Sampling, quantisation and encoding: how a voice becomes bits
Md Mahbub E Noor · Assistant Professor, Department of CSE, University of Barishal
Course: Data Communication · 2nd Year, 2nd Semester
Turning analog signals into bits: today's goals
- Describe the three steps of PCM: sampling, quantising, encoding
- Apply the Nyquist theorem and explain aliasing
- Quantise samples, assign codes and calculate the error
- Calculate SQNR, bit rate and required bandwidth
- Explain delta modulation and serial transmission modes
The conversion roadmap
| Lecture | Conversion | Example |
|---|---|---|
| 5 · today | Analog → digital | Voice → bits (PCM) |
| 6 | Digital → digital | Bits on Ethernet |
| 7 | Digital → analog | Bits on Wi-Fi |
| 8 | Analog → analog | Voice on FM radio |
Reading: Forouzan, Sections 4.2 and 4.3
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
2 / 38
Why and how we digitise: PCM
- Analog data must be digitised before it can travel as bits
- Pulse code modulation (PCM): the standard method, used in phones, CDs and recording
- Three steps: sample, quantise, encode
- The device that does it is a codec (coder-decoder)
The PCM encoder
Sample
measure at fixed times
Quantise
round to a level
Encode
level → n bits
Then line coding (Lecture 6) turns the bits into a digital signal.
| Why digitise? | Because… |
|---|---|
| Noise | repeaters regenerate clean bits; amplifiers cannot |
| Processing | compress, encrypt and error-check with software |
| One network | voice, video and data all become bits |
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Step 1: sampling
- Measure the signal's amplitude at equal intervals Ts
- Ideal sampling uses instant impulses: a mathematical model
- Natural sampling switches the signal on briefly
- Flat-top (sample and hold) reads a value and holds it while it is converted
Ideal: instant impulses
Natural: pulse tops follow
Flat-top: sample and hold
Sampling interval Ts (seconds) and sampling rate fs = 1 / Ts (samples per second). Flat-top, using a sample-and-hold circuit, is what real converters use.
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
4 / 38
How fast must we sample? The Nyquist theorem
- To rebuild a signal exactly, sample at at least twice its highest frequency
- Intuition: each cycle needs at least two samples, one up and one down
- Before sampling, a low-pass anti-aliasing filter removes frequencies above fs/2
- Sampling at the Nyquist rate or above loses no information
fs ≥ 2 × fmax
fmax = the highest frequency in the signal. 2 × fmax is the Nyquist rate.
| Signal | Highest frequency | Sampling rate used |
|---|---|---|
| Telephone voice | about 4 kHz (filtered) | 8,000 samples/s |
| CD audio | about 20 kHz (hearing limit) | 44,100 samples/s |
| Studio audio | about 20 kHz | 48,000 or 96,000 samples/s |
Engineers sample a bit above 2 fmax because real filters are not perfectly sharp.
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
5 / 38
Sampling too slowly: aliasing
- Below the Nyquist rate, the samples also fit a slower, false wave
- The receiver rebuilds that wrong wave: an alias
- Example: a 5 Hz wave sampled 6 times per second looks like 1 Hz
- Everyday sign: wagon wheels or fan blades that seem to spin backwards on video
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
6 / 38
Try it: break the Nyquist rule
Interactive
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
7 / 38
Worked examples: choosing the sampling rate
- (a) Telephone voice filtered below 4 kHz
- (b) A low-pass signal occupying 0 to 200 kHz
- (c) What does an undersampled 5 Hz wave look like?
- Always find the highest frequency first, then double it
(a) Voice, fmax = 4 kHz
fs ≥ 2 × 4,000
8,000 samples/s
(b) Low-pass signal, 0 to 200 kHz
fs ≥ 2 × 200 kHz
400,000 samples/s
(c) 5 Hz sampled at 6 samples/s
|5 − 6| = 1
alias at 1 Hz
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
8 / 38
Checkpoint 1 · 3 minutes
Check your understanding
Q1
Music contains frequencies up to 15 kHz. What is the minimum sampling rate?
Q2
A 7 Hz signal is sampled at 10 samples per second. Is there aliasing? What frequency appears?
Q3
Why must the anti-aliasing filter come before the sampler, not after?
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Step 2: quantisation
- Divide the range Vmin to Vmax into L equal zones of height Δ
- Each sample is replaced by the middle value of its zone
- The difference is the quantisation error, never more than Δ/2
- More levels → smaller Δ → smaller error, but more bits per sample
Signal vs quantised: 8 levels (dashed boundaries)
Δ = (Vmax − Vmin) / L
|error| ≤ Δ / 2
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
10 / 38
How good is it? Signal-to-quantisation-noise ratio
- Quantisation error acts like noise added to the signal
- SQNR compares signal power with quantisation-noise power
- Each extra bit halves Δ, so it quarters the noise power: +6 dB
- Use it to choose n for a required quality
SQNR(dB) = 6.02 n + 1.76
n = bits per sample; for a full-range sine-wave signal
| Bits per sample n | Levels L | SQNR | Typical use |
|---|---|---|---|
| 3 | 8 | 19.8 dB | Classroom examples |
| 8 | 256 | 49.9 dB | Telephone voice |
| 16 | 65,536 | 98.1 dB | CD audio |
| 24 | 16.7 million | 146 dB | Studio recording |
Rule of thumb: each extra bit adds about 6 dB. A telephone line needing SQNR ≥ 40 dB needs n ≥ (40 − 1.76)/6.02 = 6.35, so 7 bits.
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
11 / 38
Non-uniform quantisation: companding
- Speech is mostly quiet; uniform levels waste resolution on rare loud peaks
- Companding = compress before quantising, expand after decoding
- Small amplitudes get finer steps; large amplitudes coarser steps
- Telephone networks use 8-bit μ-law or A-law (ITU-T G.711)
μ-law curve (μ = 255) vs straight line: quiet sounds get more of the levels
Two standards
μ-law: North America and Japan
A-law: Europe and most other countries
Both: 8 bits per sample, about the same quality as 12-bit uniform quantisation for speech
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Try it: companding, small sounds get more levels
Interactive
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Step 3: encoding, and the resulting bit rate
- Each zone index is written as an n-bit binary code, with L = 2ⁿ
- Codes for successive samples are sent one after another: the PCM bit stream
- Bit rate = samples per second × bits per sample
- Minimum bandwidth of the digital signal ≈ n × the analog bandwidth
n = log₂ L bits per sample
bit rate = fs × n
Bmin ≈ n × Banalog
Last line: for a low-pass signal sampled at the Nyquist rate (B = fmax)
| Zone i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| Code (n = 3) | 000 | 001 | 010 | 011 | 100 | 101 | 110 | 111 |
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
14 / 38
Worked examples: PCM bit rate and bandwidth
- Telephone: voice filtered to 4 kHz, 8 bits per sample
- fs = 2 × 4 kHz = 8,000/s; bit rate = 8,000 × 8 = 64 kbps
- Minimum bandwidth: 8 × 4 kHz = 32 kHz, eight times the analog voice
- CD stereo: 44,100 × 16 × 2 = 1,411,200 bps
Telephone: 8,000 samples/s × 8 bits
fs × n
64 kbps
Bandwidth needed for that voice channel
n × B = 8 × 4 kHz
32 kHz (vs 4 kHz analog)
CD: 44,100 × 16 bits × 2 channels
fs × n × channels
1.411 Mbps
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
15 / 38
Try it: sample, quantise and encode
Interactive
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
16 / 38
Checkpoint 2 · 3 minutes
Check your understanding
Q1
Range −2 V to +2 V, 3 bits. What is Δ, and which code does a sample of 0.9 V get?
Q2
How many bits per sample give an SQNR of at least 60 dB?
Q3
A 10 kHz signal is sampled at the Nyquist rate with 10 bits. What is the bit rate?
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
17 / 38
Textbook example (Forouzan, Fig. 4.26): set up the quantiser
- Amplitudes lie between −20 V and +20 V; nine samples (ideal sampling)
- Choose the bits first: n = 3 gives L = 2³ = 8 levels
- Δ = (Vmax − Vmin)/L = 40/8 = 5 V
- Work in units of Δ: normalised value = amplitude / Δ, so the range is −4Δ to +4Δ
- Zone midpoints: ±0.5, ±1.5, ±2.5, ±3.5 (in units of Δ)
Samples (V): −6.1, 7.5, 16.2, 19.7, 11.0, −5.5, −11.3, −9.4, −6.0
7
6
5
4
3
2
1
0
+4Δ
+3Δ
+2Δ
+1Δ
+0Δ
-1Δ
-2Δ
-3Δ
-4Δ
Left: quantisation codes 0–7 and zone boundaries in units of Δ · red: samples · orange: quantised values
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
18 / 38
Fig. 4.26 solved: the five rows of the chart
| Sample | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Amplitude (V) | -6.1 | +7.5 | +16.2 | +19.7 | +11.0 | -5.5 | -11.3 | -9.4 | -6.0 |
| Normalised PAM (x/Δ) | -1.22 | +1.50 | +3.24 | +3.94 | +2.20 | -1.10 | -2.26 | -1.88 | -1.20 |
| Normalised quantised | -1.50 | +1.50 | +3.50 | +3.50 | +2.50 | -1.50 | -2.50 | -1.50 | -1.50 |
| Normalised error | -0.28 | +0.00 | +0.26 | -0.44 | +0.30 | -0.40 | -0.24 | +0.38 | -0.30 |
| Quantisation code | 2 | 5 | 7 | 7 | 6 | 2 | 1 | 2 | 2 |
| Encoded word | 010 | 101 | 111 | 111 | 110 | 010 | 001 | 010 | 010 |
x/Δ → zone z = ⌊x/Δ⌋ → quantised = z + 0.5 → error = quantised − x/Δ
code = z + L/2 = z + 4 → encoded word = code in 3-bit binary
Textbook typo: sample 1's error is printed as −0.38; it is −1.50 − (−1.22) = −0.28.
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
19 / 38
Choosing the bits: the same samples with n = 4
| Sample | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Normalised PAM (x/Δ) | -2.44 | +3.00 | +6.48 | +7.88 | +4.40 | -2.20 | -4.52 | -3.76 | -2.40 |
| Normalised quantised | -2.50 | +3.50 | +6.50 | +7.50 | +4.50 | -2.50 | -4.50 | -3.50 | -2.50 |
| Normalised error | -0.06 | +0.50 | +0.02 | -0.38 | +0.10 | -0.30 | +0.02 | +0.26 | -0.10 |
| Quantisation code | 5 | 11 | 14 | 15 | 12 | 5 | 3 | 4 | 5 |
| Encoded word | 0101 | 1011 | 1110 | 1111 | 1100 | 0101 | 0011 | 0100 | 0101 |
| n = 3 (book) | n = 4 | |
|---|---|---|
| Levels L = 2ⁿ | 8 | 16 |
| Δ = 40 / L | 5 V | 2.5 V |
| Largest possible error Δ/2 | 2.5 V | 1.25 V |
| SQNR = 6.02n + 1.76 | 19.82 dB | 25.84 dB |
| Bits for 9 samples | 27 | 36 |
Now Δ = 2.5 V, so x/Δ doubles, zones run −8 to +8, and code = ⌊x/Δ⌋ + 8.
One more bit: half the error, +6 dB, but 33% more bits to send.
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
20 / 38
Fig. 4.26, final step: the encoded words become a signal
0: 010
1: 101
2: 111
3: 111
4: 110
5: 010
6: 001
7: 010
8: 010
Unipolar NRZ: 1 = +V, 0 = 0 V for the whole bit interval (27 bits)
0
1
0
1
0
1
1
1
1
1
1
1
1
1
0
0
1
0
0
0
1
0
1
0
0
1
0
Polar NRZ: 1 = +V, 0 = −V
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
21 / 38
Solved example A: PCM of a sine wave
- Signal: s(t) = sin t, so period T = 2π s and f = 1/(2π) ≈ 0.159 Hz
- Samples every Ts = 2π/7 ≈ 0.898 s: 7 samples per cycle
- Check Nyquist: fs = 7/(2π) ≈ 1.11 Hz ≥ 2f ≈ 0.32 Hz ✓
- Quantise with n = 3 bits over −1 to +1: L = 8, Δ = 2/8 = 0.25
Samples (k = 0…7): 0.00, 0.78, 0.97, 0.43, −0.43, −0.97, −0.78, 0.00
Dashed blue: signal · red: samples · orange: quantised values · dashed lines: the 8 zones
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Example A solved: zones, codes and errors
| k | t | Sample | Zone i = ⌊(x − Vmin)/Δ⌋ | Quantised value | Code | Error |
|---|---|---|---|---|---|---|
| 0 | 0.000 | +0.00 | 4 | +0.125 | 100 | +0.125 |
| 1 | 0.898 | +0.78 | 7 | +0.875 | 111 | +0.095 |
| 2 | 1.795 | +0.97 | 7 | +0.875 | 111 | -0.095 |
| 3 | 2.693 | +0.43 | 5 | +0.375 | 101 | -0.055 |
| 4 | 3.590 | -0.43 | 2 | -0.375 | 010 | +0.055 |
| 5 | 4.488 | -0.97 | 0 | -0.875 | 000 | +0.095 |
| 6 | 5.386 | -0.78 | 0 | -0.875 | 000 | -0.095 |
| 7 | 6.283 | +0.00 | 4 | +0.125 | 100 | +0.125 |
PCM bit stream (24 bits)
100 111 111 101 010 000 000 100
Largest error 0.125 = Δ/2 · theory SQNR = 6.02 × 3 + 1.76 = 19.8 dB
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Solved example B: a custom waveform
- Samples at t = 0, 1, …, 7 (Ts = 1): 0, 2.2, 1.0, 0.5, −1.4, −0.8, 0.2, 1.5
- Choose the range from the signal, not just the samples: it reaches about −1.5 and +2.25
- Pick Vmin = −1.6, Vmax = +2.4 (a 4-unit span with a little margin)
- n = 3 bits: L = 8, Δ = 4/8 = 0.5
If we had used only the sample extremes (−1.4 to 2.2), the true peak of 2.25 would be clipped (overload).
The curve peaks near 2.25 and dips near −1.5, but the samples only reach 2.20 and −1.40
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
24 / 38
Example B solved: zones, codes and errors
| k | t | Sample | Zone i = ⌊(x − Vmin)/Δ⌋ | Quantised value | Code | Error |
|---|---|---|---|---|---|---|
| 0 | 0 | +0.00 | 3 | +0.150 | 011 | +0.150 |
| 1 | 1 | +2.20 | 7 | +2.150 | 111 | -0.050 |
| 2 | 2 | +1.00 | 5 | +1.150 | 101 | +0.150 |
| 3 | 3 | +0.50 | 4 | +0.650 | 100 | +0.150 |
| 4 | 4 | -1.40 | 0 | -1.350 | 000 | +0.050 |
| 5 | 5 | -0.80 | 1 | -0.850 | 001 | -0.050 |
| 6 | 6 | +0.20 | 3 | +0.150 | 011 | -0.050 |
| 7 | 7 | +1.50 | 6 | +1.650 | 110 | +0.150 |
PCM bit stream (24 bits)
011 111 101 100 000 001 011 110
All errors within ±Δ/2 = ±0.25 · bit rate = 1 sample/s × 3 bits = 3 bps
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Solved example C: a non-symmetric signal, start to finish
- A sensor voltage between about 0.4 V and 7 V: all positive, not centred on zero
- Fastest component: period 4.3 s, so fmax ≈ 0.23 Hz; sampling at fs = 1 sample/s (Ts = 1 s) beats 2 fmax ≈ 0.47 ✓
- Step 1–2: the PAM values are the sample heights (table below)
- Step 3: quantise over 0 to 8 V with n = 3: L = 8, Δ = 8/8 = 1 V
PAM values (k = 0…9): 4.15, 6.23, 6.91, 5.50, 4.07, 3.87, 3.45, 1.74, 0.42, 1.31 V
Dashed blue: signal · orange: PAM pulses (flat-top samples) · dashed lines: 1 V levels from 0 to 8 V
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Example C solved: from PAM values to codes
| k | t | Sample | Zone i = ⌊(x − Vmin)/Δ⌋ | Quantised value | Code | Error |
|---|---|---|---|---|---|---|
| 0 | 0 | +4.15 | 4 | +4.500 | 100 | +0.350 |
| 1 | 1 | +6.23 | 6 | +6.500 | 110 | +0.270 |
| 2 | 2 | +6.91 | 6 | +6.500 | 110 | -0.410 |
| 3 | 3 | +5.50 | 5 | +5.500 | 101 | +0.000 |
| 4 | 4 | +4.07 | 4 | +4.500 | 100 | +0.430 |
| 5 | 5 | +3.87 | 3 | +3.500 | 011 | -0.370 |
| 6 | 6 | +3.45 | 3 | +3.500 | 011 | +0.050 |
| 7 | 7 | +1.74 | 1 | +1.500 | 001 | -0.240 |
| 8 | 8 | +0.42 | 0 | +0.500 | 000 | +0.080 |
| 9 | 9 | +1.31 | 1 | +1.500 | 001 | +0.190 |
PCM bit stream (10 samples × 3 bits = 30 bits)
100 110 110 101 100 011 011 001 000 001
With Δ = 1 V: zone = ⌊x⌋, value = ⌊x⌋ + 0.5 · bit rate = 1 × 3 = 3 bps
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Example C, final step: the bits become a digital signal
k=0: 100
k=1: 110
k=2: 110
k=3: 101
k=4: 100
k=5: 011
k=6: 011
k=7: 001
k=8: 000
k=9: 001
Unipolar NRZ (Lecture 2's two-level rule): 1 = +V, 0 = 0 V for the whole bit interval
1
0
0
1
1
0
1
1
0
1
0
1
1
0
0
0
1
1
0
1
1
0
0
1
0
0
0
0
0
1
Polar NRZ: 1 = +V, 0 = −V (no zero level, so the average is closer to 0 V)
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
28 / 38
At the receiver: the PCM decoder
- Read n bits at a time and convert each code back to its level value
- Hold each value until the next sample: a staircase
- A low-pass filter smooths the staircase into an analog signal
- The result equals the original plus quantisation noise
Bits
011 111 101…
Levels
code → value
Hold
staircase
Filter
low-pass
Staircase from the hold circuit, smoothed by the low-pass filter
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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A simpler alternative: delta modulation
- PCM sends the value of each sample; DM sends only the change
- One bit per sample: is the signal above or below the staircase?
- Very simple hardware, but needs a much higher sampling rate than PCM
- Adaptive DM enlarges δ on steep slopes and shrinks it on flat parts
Each sample sends 1 bit: 1 = step up by δ, 0 = step down
| Problem | Cause | Fix |
|---|---|---|
| Slope overload | signal rises faster than δ per sample | larger δ or higher fs |
| Granular noise | steps jitter on flat parts | smaller δ |
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
30 / 38
Try it: delta modulation, step by step
Interactive
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
31 / 38
Sending the bits: parallel and serial transmission
- Parallel: n bits at once over n wires; fast but costly and short-range
- Serial: one bit at a time over one channel; used for almost all links
- Asynchronous serial frames each byte with start and stop bits
- Synchronous serial sends a continuous stream; isochronous guarantees a steady rate
Parallel: 8 bits at once, 8 wires
Serial: 1 bit at a time, 1 wire
Asynchronous byte: start bit (0) + 8 data bits + stop bit (1)
| Serial mode | Timing | Example |
|---|---|---|
| Asynchronous | start/stop bits per byte; gaps allowed | UART, old modems |
| Synchronous | continuous stream; frames carry timing | Ethernet, SONET/SDH |
| Isochronous | guaranteed fixed delivery rate | real-time audio and video |
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
32 / 38
Worked example: the cost of asynchronous framing
- Each 8-bit character gets 1 start bit and 1 stop bit
- So 10 bits travel for every 8 useful bits: 80% efficiency
- At 9,600 bps: 9,600 ÷ 10 = 960 characters per second
- Synchronous transmission avoids per-byte overhead, so it suits high speeds
Bits sent per character
1 start + 8 data + 1 stop
10 bits
Efficiency
8 useful ÷ 10 sent
80%
Characters per second at 9,600 bps
9,600 ÷ 10
960 chars/s
Md Mahbub E Noor
Lecture 5: Analog-to-Digital Conversion
University of Barishal
33 / 38
Try it yourself: PCM Lab
- 1 · Sampling: find where aliasing begins for a 5 Hz signal
- 2 · Quantise and encode: reproduce Fig. 4.26 and Examples A, B and C; change n and the range
- 3 · Delta modulation: provoke slope overload and granular noise
- 4 · Bit rate and 5 · Full pipeline: signal → PAM → codes → digital signal
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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See it in action: from voice to bits
- Watch a smooth signal being sampled at equal intervals
- See undersampling produce a false, slower wave
- Watch samples snap to levels and become a bit stream (Example A)
- When a “Pause & predict” card appears, answer before the video moves on
Click the video to play or pause · Press → for full screen
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Checkpoint 3 · 3 minutes
Check your understanding
Q1
Range 0 to 8 V, 4 bits. Quantise a sample of 5.3 V: zone, code, quantised value and error.
Q2
A voice signal is limited to 3.4 kHz and coded with 8 bits at the Nyquist rate. Bit rate?
Q3
Asynchronous transmission at 19,200 bps with 1 start and 1 stop bit: how many characters per second?
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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Key takeaways
- PCM = sample, quantise, encode; a codec does it
- Sample at fs ≥ 2 fmax, or aliasing creates false frequencies
- Δ = (Vmax − Vmin)/L; error ≤ Δ/2; SQNR ≈ 6.02n + 1.76 dB
- Bit rate = fs × n; voice = 64 kbps; companding improves quiet sounds
- Delta modulation sends 1 bit per sample; serial links are async, sync or isochronous
Next lecture
Lecture 6 · Digital-to-Digital Conversion
Line coding: how bits become voltage levels on a wire (NRZ, Manchester, AMI and more), plus block coding and scrambling.
Reading: Forouzan, Section 4.1 · Revisit Examples A, B and C in PCM Lab
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Lecture 5: Analog-to-Digital Conversion
University of Barishal
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