This document discusses various methods of modulating digital and analog data for transmission: 1. It describes digital-to-analog modulation techniques including amplitude shift keying (ASK), frequency shift keying (FSK), phase shift keying (PSK), and quadrature amplitude modulation (QAM). 2. It explains the relationships between bit rate, baud rate, and bandwidth for different modulation schemes. ASK, FSK, and PSK have baud rate equal to bit rate, while higher-order PSK and QAM can have higher bit rates through multiple bits per symbol. 3. Modems and standards like V.32, V.34, and V.90 are discussed in the context of mod

1. Modulation of Digital Data
1.
2.
3.
4.
5.
6.
7.
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
Phase Shift Keying (PSK)
Quadrature Amplitude Modulation (QAM)
Bit/Baud Comparison
Modems

2. Digital-to-analog modulation
Types of digital-to-analog modulation

3. Aspects to digital-to Analog conversion
Bit Rate / Baud Rate



Bit rate is the number of bits per second. Baud rate is the number
of signal units per second. Baud rate is less than or equal to the bit
rate.
Bit rate is important in computer efficiency
Baud rate is important in data transmission.
 Baud rate determines the bandwidth required to send signal


Baud rate = bit rate / # bits per signal unit
An analog signal carries 4 bits in each signal unit. If 1000 signal units are
sent per second, find the baud rate and the bit rate
 Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000
bps

The bit rate of a signal is 3000. If each signal unit carries 6 bits, what is
the baud rate?
 Baud rate = 3000/6 =500 bauds/sec

4. Amplitude Shift Keying (ASK)
 A cos(2Π f c t ) binary1
s (t ) = 
0 binary 0

On/Off keying
The strength of the carrier signal is varied to represent binary 1
and 0.
Frequency and phase remains the same.
Highly susceptible to noise interference.
Used up to 1200 bps on voice grade lines, and on optical fiber.

5. Relationship between baud rate and bandwidth in ASK
BW = (1 + d) * Nd
Find the minimum bandwidth for an ASK signal transmitting at 2000
bps. The transmission mode is half-duplex.

In ASK the baud rate and bit rate are the same. The baud rate is therefore
2000. An ASK signal requires a minimum bandwidth equal to its baud rate.
Therefore, the minimum bandwidth is 2000 Hz.

6. Full duplex ASK
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz), if draw the full-duplex ASK
diagram of the system. We can find the carriers and the bandwidths in each direction.
Assume there is no gap between the bands in the two directions.

For full-duplex ASK, the bandwidth for each direction is
 BW = 10000 / 2 = 5000 Hz
 The carrier frequencies can be chosen at the middle of each band
 fc (forward) = 1000 + 5000/2 = 3500 Hz
 fc (backward) = 11000 – 5000/2 = 8500 Hz

7. Frequency Shift Keying
Frequency of the carrier is varied to represent digital data (binary 0/1)
Peak amplitude and phase remain constant.
Avoid noise interference by looking at frequencies (change of a signal)
and ignoring amplitudes.
Limitations of FSK is the physical capabilities of the carrier.
f1 and f2 equally offset by equal opposite amounts to the carrier freq.
In MFSK more than 2 freq are used, each signal element represents
more than one bit
 A cos(2Π f 1t ) binary1
s (t ) = 
 A cos(2Π f 2 t ) binary 0

8. Relationship between baud rate and bandwidth in
FSK
FSK shifts between two carrier frequencies
FSK spectrum = combination of two ASK spectra centered on f c1 and fc0.
BW = fc1-fc0 + Nbaud

9. FSK Examples (cont.)
•
What is the Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex mode, and the
carriers are separated by 3000 Hz.
• Because For FSK
BW = baud rate + fc1 − fc0
BW = bit rate + fc1 − fc0 = 2000 + 3000 = 5000 Hz
•
What is the maximum bit rates for an FSK signal if the bandwidth of
the medium is 12,000 Hz and the difference between the two carriers
is 2000 Hz. Transmission is in full-duplex mode.
– Because the transmission is full duplex, only 6000 Hz is allocated for each
direction.
– BW = baud rate + fc1 − fc0
– Baud rate = BW − (fc1 − fc0 ) = 6000 − 2000 = 4000
– But because the baud rate is the same as the bit rate, the bit rate is 4000
bps.

10. Phase Shift Keying
Phase of the carrier is varied to
represent digital data (binary 0 or
1)
Amplitude and frequency
remains constant.
If phase 0 deg to represent 0,
180 deg to represent 1. (2-PSK)
PSK is not susceptible to noise
degradation that affects ASK or
bandwidth limitations of FSK

11. 4-PSK (QPSK) method

12. 8-PSK
We can extend, by varying the the signal by shifts of 45 deg
(instead of 90 deg in 4-PSK)
With 8 = 23 different phases, each phase can represents 3 bits
(tribit).

13. Relationship between baud rate and bandwidth in PSK
Bandwith similar to ASK, but data rate can 2 or more times greater.
What is the bandwidth for a 4-PSK signal transmitting at 2000 bps. Transmission is in
half-duplex mode.

For PSK the baud rate is the same as the bandwidth, which means the baud rate is
5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.
Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate?

For PSK the baud rate is the same as the bandwidth, which means the baud rate is
5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

14. Quadrature Amplitude Modulation
PSK is limited by the ability of the equipment to distinguish
between small differences in phases.

Limits the potential data rate.
Quadrature amplitude modulation is a combination of ASK and
PSK so that a maximum contrast between each signal unit (bit,
dibit, tribit, and so on) is achieved.


We can have x variations in phase and y variations of amplitude
x • y possible variation (greater data rates)
Numerous variations. (4-QAM, 8-QAM)
# of phase shifts > # of amplitude shifts

15. 8-QAM and 16-QAM
Second example, recommendation of OSI.
not all possibilities are used, to increase
readability of signal, measurable differences
between shifts are increased
First example handles noise best
Because of ratio of phases to amplitudes
ITU-T recommendation.

16. Bit Baud comparison
Assuming a FSK signal over
voice-grade phone line can send
1200 bps, it requires 1200 signal
units to send 1200 bits (each
frequency shift represents one
bit, baud rate 1200)
Assuming 8-QAM, baud rate is
only 400 to achieve same data
rate.
Modulation
Units
Bits/Ba
ud
Baud
rate
Bit
Rate
ASK, FSK, 2-PSK
Bit
1
N
N
4-PSK, 4-QAM
Dibit
2
N
2N
8-PSK, 8-QAM
Tribit
3
N
3N
16-QAM
Quadbit
4
N
4N
32-QAM
Pentabit
5
N
5N
64-QAM
Hexabit
6
N
6N
128-QAM
Septabit
7
N
7N
256-QAM
Octabit
8
N
8N

17. Bit Baud comparison (examples)
A constellation diagram consists of eight equally spaced points on a circle. If
the bit rate is 4800 bps, what is the baud rate?

The constellation indicates 8-PSK with the points 45 degrees apart. Since 2 3
= 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is
4800 / 3 = 1600 baud
What is the bit rate for a 1000-baud 16-QAM signal.

A 16-QAM signal has 4 bits per signal unit since log216 = 4. Thus, (1000)(4)
= 4000 bps
Compute the baud rate for a 72,000-bps 64-QAM signal.

A 64-QAM signal has 6 bits per signal unit since log2 64 = 6.

Therefore, 72000 / 6 = 12,000 baud

18. Phone modems

19. V-series (ITU-T Standards)
V-32.





Uses a combined modulation and encoding technique: Trellis coded
Modulation.
Trellis = QAM + a redundant bit
5 bit (pentabit) = 4 data + 1 calculated from data.
A signal distorted by noise can arrive closer to an adjacent point than the
intended point (extra bit is therefore used to adjust)
Less likely to be misread than a QAM signal.
2400 baud • 4 bit = 9600 bps

20. V-series (cont.)
V-32.bis


First ITU-T standard to support 14,400 bps transmission.
Uses 128-QAM
 (7 bits/baud with 1 bit for error control)
 at a rate 2400 baud 2400 * 6 = 14,400 bps

Adjustment of the speed upward or downward depending on the quality of the line or
signal
V-34 bis

Bit rate of 28,800 bps with 960-point constellation to 1664-point constellation for a bit
rate of 33,600 bps.

21. V-series (cont.)
V-90


Traditionally modems have a
limitation on data rate (max. 33.6
Kbps)
V-90 modems can be used (up to
56Kbps) if using digital signaling.
 For example, Through an
Internet Service Provider (ISP)

V-90 are asymmetric
 Downloading rate (from ISP to
PC) has a 56 Kbps limitation.
 Uploading rate (from PC to IST)
has a 33.6 Kbps limitation.

22. V-90/92 (cont.)
V-90


In uploading, signal still to be
sampled at the switching station.
Limit due to noise sampling.
Phone company samples at 8000
times/sec with 8 bits (including bit
error)
 Data rate = 8000 * 7 = 56Kbps

In download, signal is not affected
by sampling.
V-92



Speed adjustment
Upload data at 48Kbps.
Call waiting service

23. Analog-to-analog modulation
Why analog to analog conversion?

24. Amplitude modulation
The total bandwidth required for AM can be determined from the bandwidth of
the audio signal: BWt = 2 x BWm.

25. AM band allocation
Bandwidth of an audio signal is 5KHz. An AM radio station needs at least a
minimum bandwidth of 10 KHz.
AM stations are allowed carrier freq between 530 and 1700 KHz.
If One station uses 1100 KHz the next one uses 1110 KHz

26. Frequency modulation
The bandwidth of a stereo audio signal is
usually 15 KHz. Therefore, an FM
station needs at least a bandwidth of 150
KHz. The minimum bandwidth is at least
200 KHz (0.2 MHz).

27. FM band allocation