Modulation of Digital Data

1. Digital-to-Analog Conversion2. Amplitude Shift Keying (ASK)3. Frequency Shift Keying (FSK)4. Phase Shift Keying (PSK)5. Quadrature Amplitude Modulation (QAM)6. Bit/Baud Comparison7. Modems

Digital-to-analog modulation

Types of digital-to-analog modulation

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) Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 Bit rate = 1000 x 4 = 4000 bps= 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/secBaud rate = 3000/6 =500 bauds/sec

Amplitude Shift Keying (ASK)

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.

00

1)2cos()(

binary

binarytfAts c

On/Off keying

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.

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

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

0)2cos(

1)2cos()(

2

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binarytfA

binarytfAts

Relationship between baud rate and bandwidth in FSK

FSK shifts between two carrier frequencies

FSK spectrum = combination of two ASK spectra centered on fc1 and fc0.

BW = fc1-fc0 + Nbaud

FSK Examples (cont.)

• 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 + fcBW = baud rate + fc11 fc fc00

– Baud rate = BW Baud rate = BW (fc (fc11 fc fc00 ) = 6000 ) = 6000 2000 = 4000 2000 = 4000

– But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

• 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 + fBW = baud rate + fc1c1 f fc0c0

BW = bit rate + fc1 BW = bit rate + fc1 fc0 = 2000 + 3000 = 5000 Hz fc0 = 2000 + 3000 = 5000 Hz

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

4-PSK (QPSK) method

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).

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.

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

8-QAM and 16-QAM

First example handles noise bestBecause of ratio of phases to amplitudesITU-T recommendation.

Second example, recommendation of OSI.not all possibilities are used, to increase readability of signal, measurable differencesbetween shifts are increased

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.

ModulationModulation UnitsUnitsBits/Bits/BaudBaud

Baud Baud raterate

Bit Rate

ASK, FSK, 2-PSKASK, FSK, 2-PSK Bit 1 N N

4-PSK, 4-QAM4-PSK, 4-QAM Dibit 2 N 2N

8-PSK, 8-QAM8-PSK, 8-QAM Tribit 3 N 3N

16-QAM16-QAM Quadbit 4 N 4N

32-QAM32-QAM Pentabit 5 N 5N

64-QAM64-QAM Hexabit 6 N 6N

128-QAM128-QAM Septabit 7 N 7N

256-QAM256-QAM Octabit 8 N 8N

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 23 = 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

Bit Baud comparison (examples)

Phone modems

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

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.

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.

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

Analog-to-analog modulation

Why analog to analog conversion?

Amplitude modulation

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

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

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).

FM band allocation