Data Communication · Lecture 7
Digital-to-Analog Conversion
ASK, FSK, PSK and QAM: carrying bits on a carrier wave
Md Mahbub E Noor · Assistant Professor, Department of CSE, University of Barishal
Course: Data Communication · 2nd Year, 2nd Semester
From bits to a carrier: today's goals
- Relate bit rate, baud rate and the number of signal levels: N = S × log₂L
- Explain ASK, FSK and PSK and calculate their bandwidth
- Read and draw constellation diagrams
- Encode bit streams with 4-QAM and 8-QAM, step by step
- Explain why higher-order QAM needs a cleaner channel
Where we are
| Lecture | Conversion |
|---|---|
| 5 | Analog → digital |
| 6 | Digital → digital (baseband) |
| 7 · today | Digital data → analog signal |
| 8 | Analog → analog |
Reading: Forouzan, Section 5.1
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
2 / 29
Changing the carrier: three basic schemes, plus QAM
- A carrier is a sine wave at frequency fc, chosen to fit the channel's band
- The bits change one property of it: amplitude (ASK), frequency (FSK) or phase (PSK)
- QAM changes amplitude and phase together
- Needed whenever the channel is band-pass: radio, Wi-Fi, mobile, cable, DSL
ASK · amplitude
1
0
1
1
0
0
1
0
FSK · frequency
1
0
1
1
0
0
1
0
PSK · phase
1
0
1
1
0
0
1
0
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
3 / 29
Bit rate, baud rate and signal levels
- Each distinct signal element (a particular amplitude, frequency or phase) is one of L types
- Each element carries r = log₂L bits
- Bandwidth depends on the baud rate, so more levels mean more bits in the same bandwidth
- Textbook note: Example 5.2 is printed as 8000/10,000; it should read 8000/1000 = 8
S = N × (1 / r), r = log₂ L
N = bit rate · S = baud rate (signal elements per second) · L = number of different signal elements
Ex. 5.1: 4 bits per element, 1000 elements/s
N = S × r = 1000 × 4
4000 bps
Ex. 5.2: N = 8000 bps, S = 1000 baud
r = N / S = 8000 / 1000 = 8 → L = 2⁸
L = 256
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
4 / 29
Amplitude shift keying (ASK)
- The carrier's amplitude changes with the data; frequency and phase stay fixed
- Binary ASK uses two amplitudes; with one of them zero, it is on-off keying (OOK)
- Bandwidth B = (1 + d) S, centred on the carrier frequency
- Simple, but very sensitive to noise and fading; used in optical fibre (light on/off) and RFID
Binary ASK (OOK) · 1 = carrier, 0 = no carrier
1
0
1
1
0
0
1
0
B = (1 + d) × S
d (0 to 1) depends on the filtering; with d = 0, B equals the baud rate. The band is centred on fc.
Implementation: multiply a unipolar NRZ signal by the carrier. Weakness: noise changes amplitude easily.
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
5 / 29
Worked example: ASK in a 100 kHz band
- Ex. 5.3: bandwidth 100 kHz from 200 to 300 kHz, ASK with d = 1
- Put the carrier in the middle: fc = 250 kHz
- B = (1 + d) S = 2S = 100 kHz → S = 50 kbaud → N = 50 kbps
- Ex. 5.4: split the band for full duplex: 25 kbps each direction
Carrier: the middle of 200 to 300 kHz
(200 + 300) / 2
fc = 250 kHz
Bit rate: B = (1 + d) S = 2 S = 100 kHz
S = 50 kbaud, r = 1
N = 50 kbps
Full duplex (Ex. 5.4): two 50 kHz bands
each B = 50 kHz → S = 25 kbaud
25 kbps each way
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
6 / 29
Frequency shift keying (FSK)
- The carrier's frequency changes with the data; amplitude and phase stay fixed
- BFSK: two carrier frequencies, f₁ and f₀, 2Δf apart
- MFSK: L frequencies carry log₂L bits per element
- More robust to noise than ASK, but uses more bandwidth; early modems, Bluetooth (GFSK), pagers
Binary FSK · 1 = f₁ (higher), 0 = f₀ (lower)
1
0
1
1
0
0
1
0
B = (1 + d) S + 2Δf
2Δf = spacing between the two carrier frequencies; MFSK with L frequencies: B = L × S (d = 0)
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
7 / 29
Worked examples: BFSK and MFSK
- Ex. 5.5: same 200 to 300 kHz band, now with BFSK and 2Δf = 50 kHz
- B = 2S + 50 kHz = 100 kHz → S = 25 kbaud → N = 25 kbps (half of ASK's)
- Ex. 5.6: 3 bits at a time at 3 Mbps with a 10 MHz carrier
- L = 8 frequencies, S = 1 Mbaud, spaced 1 MHz apart → B = 8 MHz
Ex. 5.5: 100 kHz band, BFSK, d = 1, 2Δf = 50 kHz
2S + 50 = 100 → S = 25 kbaud
N = 25 kbps
Ex. 5.6: MFSK, 3 bits at a time, 3 Mbps
L = 2³ = 8 · S = 3 M / 3 = 1 Mbaud
8 frequencies
Ex. 5.6 bandwidth (d = 0, 1 MHz apart)
B = L × S = 8 × 1 MHz
B = 8 MHz
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
8 / 29
Checkpoint 1 · 3 minutes
Check your understanding
Q1
A signal uses 64 kinds of element at 5,000 baud. What is the bit rate?
Q2
Binary ASK must fit a 40 kHz band with d = 1. What is the maximum bit rate?
Q3
Why is FSK more resistant to noise than ASK?
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
9 / 29
Phase shift keying: BPSK
- The carrier's phase changes with the data; amplitude and frequency stay fixed
- BPSK: two phases, 0° and 180°
- Implemented by multiplying a polar NRZ signal (±1) by the carrier
- Noise changes amplitude more than phase, so PSK beats ASK; no second frequency, so it beats FSK on bandwidth
BPSK · 1 = phase 0°, 0 = phase 180°
1
0
1
1
0
0
1
0
I
Q
0
1
Constellation: two points, opposite each other
Same bandwidth as ASK: B = (1 + d) S
Very robust: the two points are as far apart as possible
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
10 / 29
Quadrature PSK (QPSK)
- QPSK sends 2 bits per signal element (a dibit) using four phases
- Built from two BPSK signals: one on an in-phase carrier (I) and one on a quadrature carrier (Q, shifted 90°)
- First bit → I (1 = +, 0 = −), second bit → Q: four combinations
- Ex. 5.7: 12 Mbps, d = 0 → S = 6 Mbaud → B = 6 MHz
I
Q
11
01
00
10
| Dibit | Phase |
|---|---|
| 11 | 45° |
| 01 | 135° |
| 00 | 225° |
| 10 | 315° |
Amplitude √2 for all points. Neighbours differ in one bit (Gray code).
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
11 / 29
Try it: ASK, FSK, PSK and QPSK side by side
Interactive
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
12 / 29
Reading a constellation diagram
- The horizontal axis is the in-phase carrier (I); the vertical axis is the quadrature carrier (Q)
- Each dot is one kind of signal element, labelled with the bits it carries
- Length of the line to the dot = amplitude; its angle = phase
- Ex. 5.8: OOK has points at 0 and 1; BPSK at −1 and +1; QPSK on a circle (previous slide)
I
Q
0
1
ASK (OOK)
I
Q
0
1
BPSK
Point = one signal element. Distance from the origin = peak amplitude. Angle from the I axis = phase. Projections on I and Q = the two carriers' amplitudes.
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
13 / 29
Checkpoint 2 · 3 minutes
Check your understanding
Q1
In QPSK (Forouzan labelling), which phase does dibit 10 use, and what are its I and Q?
Q2
QPSK at 20 Mbps with d = 1. What bandwidth is needed?
Q3
A constellation point is at I = −1, Q = −1. What are its amplitude and phase?
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
14 / 29
Quadrature amplitude modulation (QAM)
- QAM = ASK + PSK: change both amplitude and phase
- Two carriers (I and Q), each with several amplitude levels
- PSK alone is limited by how finely the receiver can tell phases apart; QAM spreads points over the plane
- Same bandwidth rule as ASK and PSK: B = (1 + d) S
I
Q
| Scheme | Points L | Bits/symbol |
|---|---|---|
| 4-QAM | 4 | 2 |
| 8-QAM | 8 | 3 |
| 16-QAM | 16 | 4 |
| 64-QAM | 64 | 6 |
| 256-QAM | 256 | 8 |
Left: 16-QAM, 4 amplitudes on each carrier
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
15 / 29
Solved example: 4-QAM, step by step
- Data: the first 12 bits of Example C (Lecture 5): 100110110101
- Step 1: split into dibits (2 bits per symbol): 10 01 10 11 01 01
- Step 2: map each dibit to a point using the rule (right)
- Step 3: read amplitude and phase from the point
- Step 4: send A cos(2πfc t + φ) for one symbol time
I
Q
11
01
00
10
Rule
1st bit → I: 1 = +1, 0 = −1
2nd bit → Q: 1 = +1, 0 = −1
A = √(I² + Q²), φ = angle of (I, Q)
Orange = points used in this example
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
16 / 29
4-QAM solved: dibits to signal elements
| Symbol k | Dibit | I | Q | Amplitude A | Phase φ | Signal element sent |
|---|---|---|---|---|---|---|
| 0 | 10 | +1 | -1 | √2 = 1.414 | 315° | 1.414 cos(2πfct + 315°) |
| 1 | 01 | -1 | +1 | √2 = 1.414 | 135° | 1.414 cos(2πfct + 135°) |
| 2 | 10 | +1 | -1 | √2 = 1.414 | 315° | 1.414 cos(2πfct + 315°) |
| 3 | 11 | +1 | +1 | √2 = 1.414 | 45° | 1.414 cos(2πfct + 45°) |
| 4 | 01 | -1 | +1 | √2 = 1.414 | 135° | 1.414 cos(2πfct + 135°) |
| 5 | 01 | -1 | +1 | √2 = 1.414 | 135° | 1.414 cos(2πfct + 135°) |
If N = 12 kbps
S = N / 2 = 6 kbaud
B = 6 kHz (d = 0)
Same data with BPSK
S = N = 12 kbaud
B = 12 kHz (d = 0)
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
17 / 29
4-QAM solved: the transmitted signal
10
01
10
11
01
01
315°
135°
315°
45°
135°
135°
Constant amplitude √2; the wave restarts at a new phase at each symbol boundary (dashed lines). Two carrier cycles are drawn per symbol.
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
18 / 29
Solved example: 8-QAM, step by step
- Data: all 30 bits of Example C's PCM codes: 100 110 110 101 100 011 011 001 000 001
- 8-QAM: L = 8, so 3 bits per symbol: each PCM code becomes exactly one symbol
- Constellation: 2 amplitudes × 4 phases (as in the course's earlier slides)
- Rule: first two bits choose the phase, the last bit chooses the amplitude
I
Q
000
001
010
011
100
101
110
111
| First 2 bits | Phase |
|---|---|
| 00 | 0° |
| 01 | 90° |
| 10 | 180° |
| 11 | 270° |
| Last bit | Amplitude |
|---|---|
| 0 | 1 |
| 1 | 2 |
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
19 / 29
8-QAM solved: tribits to signal elements
| k | Tribit | Phase bits | Amp. bit | I | Q | Signal element sent |
|---|---|---|---|---|---|---|
| 0 | 100 | 10 → 180° | 0 → 1 | -1 | 0 | 1 cos(2πfct + 180°) |
| 1 | 110 | 11 → 270° | 0 → 1 | 0 | -1 | 1 cos(2πfct + 270°) |
| 2 | 110 | 11 → 270° | 0 → 1 | 0 | -1 | 1 cos(2πfct + 270°) |
| 3 | 101 | 10 → 180° | 1 → 2 | -2 | 0 | 2 cos(2πfct + 180°) |
| 4 | 100 | 10 → 180° | 0 → 1 | -1 | 0 | 1 cos(2πfct + 180°) |
| 5 | 011 | 01 → 90° | 1 → 2 | 0 | +2 | 2 cos(2πfct + 90°) |
| 6 | 011 | 01 → 90° | 1 → 2 | 0 | +2 | 2 cos(2πfct + 90°) |
| 7 | 001 | 00 → 0° | 1 → 2 | +2 | 0 | 2 cos(2πfct + 0°) |
| 8 | 000 | 00 → 0° | 0 → 1 | +1 | 0 | 1 cos(2πfct + 0°) |
| 9 | 001 | 00 → 0° | 1 → 2 | +2 | 0 | 2 cos(2πfct + 0°) |
Check: I = A cos φ and Q = A sin φ. For 101: A = 2, φ = 180° → I = −2, Q = 0.
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
20 / 29
8-QAM solved: the transmitted signal
100
110
110
101
100
011
011
001
000
001
A1 180°
A1 270°
A1 270°
A2 180°
A1 180°
A2 90°
A2 90°
A2 0°
A1 0°
A2 0°
Example C at 1,000 samples/s
N = 3 kbps · S = N / 3
1,000 baud
Bandwidth with d = 0
B = S
1 kHz (BASK: 3 kHz)
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
21 / 29
Try it: QAM from bits to a signal
Interactive
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
22 / 29
More points, more bits, same bandwidth
- Bit rate = baud rate × log₂L, and the bandwidth depends only on the baud rate
- So moving from QPSK to 64-QAM triples the bit rate in the same channel
- Wi-Fi and LTE switch modulation many times a second as the signal changes
- The limit: Shannon (Lecture 3) and noise, next slide
| Modulation | Bits/symbol | Bit rate at 1 Mbaud | Typical use |
|---|---|---|---|
| BPSK | 1 | 1 Mbps | far from the Wi-Fi router, GPS |
| QPSK / 4-QAM | 2 | 2 Mbps | satellite TV, LTE edge |
| 16-QAM | 4 | 4 Mbps | LTE, Wi-Fi mid range |
| 64-QAM | 6 | 6 Mbps | Wi-Fi, cable TV |
| 256-QAM | 8 | 8 Mbps | Wi-Fi 5, cable modems |
| 1024-QAM | 10 | 10 Mbps | Wi-Fi 6, close to the router |
Same bandwidth for every row: only the number of points changes.
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
23 / 29
Why dense constellations need a clean channel
- Noise moves each received point away from where it was sent: a cloud around each point
- The receiver picks the nearest point (decision regions)
- More points → closer together → a smaller cloud is enough to cause errors
- So 16-QAM needs a higher SNR than QPSK, and 256-QAM higher still
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
24 / 29
Try it yourself: Modulation Lab
- 1 · ASK, FSK, PSK: type bits; see all three modulations
- 2 · QAM step by step: 4-, 8- or 16-QAM; table, constellation and signal; click a row to highlight it
- 3 · Noise: drag the SNR until the constellation breaks
- Buttons load Example C, the old-slide 8-QAM example and the 4-QAM example
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
25 / 29
See it in action: bits on a carrier
- Watch ASK, FSK and PSK carry the same bits
- See a constellation point's distance and angle become amplitude and phase
- Watch Example C's codes become 8-QAM symbols one by one
- When a “Pause & predict” card appears, answer before the video moves on
Click the video to play or pause · Press → for full screen
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
26 / 29
Checkpoint 3 · 3 minutes
Check your understanding
Q1
4-QAM (Forouzan labelling): encode 001110. Give each symbol's phase.
Q2
8-QAM (this lecture's mapping): what are the amplitude and phase of 111, and of 010?
Q3
Old-slide example: 8-QAM tribits sent at 24 bps. What is the baud rate, and how many symbols per second?
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
28 / 29
Key takeaways
- Digital modulation puts bits on a carrier: N = S × log₂L, and bandwidth follows S
- ASK changes amplitude, FSK frequency, PSK phase; B = (1 + d)S (FSK adds 2Δf)
- A constellation point's distance is amplitude, its angle is phase
- QAM combines amplitude and phase: 4-QAM (= QPSK) carries 2 bits, 8-QAM 3 bits, 16-QAM 4
- Denser constellations carry more bits but need a higher SNR
Next lecture
Lecture 8 · Analog-to-Analog Conversion
AM, FM and PM: putting an analog signal such as voice directly on a carrier, with spectra and bandwidth calculations.
Reading: Forouzan, Section 5.2 · Practise in Modulation Lab
Md Mahbub E Noor
Lecture 7: Digital-to-Analog Conversion
University of Barishal
29 / 29