Figure 5-22

Digital to Analog Encoding

• 3 Characteristics of a sine wave – Amplitude– Frequency– Phase

• 3 Mechanisms for Modulating Digital Data into Analog Signal– ASK– FSK– PSK

Figure 5-23

Bit Rate and Baud Rate

• Bit rate – number of bits transmitted during 1 second

• Baud rate– number of signal units per second – determines the BW required to send the

signal

Bit rate = baud rate * # of bits represented by each signal unit

Bit Rate and Baud Rate

• An analog signal carries four bits in each signal element. If 1000 signal elements are sent per second, find the baud rate and the bit rate.

• The bit rate of a signal is 3000. If each signal element carries six bits, what is the baud rate?

ASK

Each shift represents a single bit

Figure 5-25

Bandwidth for ASK

Nbaud = baud ratefc = carrier frequency

BW of a signal is the total range of frequencies occupied by that signal

• Find the minimum BW for an ASK signal transmitting at 2000 bps. The transmission mode is half-duplex

• Solution: – In ASK bit rate and baud rate are the same. – An ASK signal requires a min BW equal to its

baud rate

• Given a BW of 10,000 Hz (1000 – 12,000 Hz), draw the full-duplex ASK diagram of the system. Find the carriers and the BWs in each direction. Assume there is no gap between the bands in 2 directions

• Solution:BW for each direction = 10,000/2 = 5000 Hz

The carrier frequencies can be chosen at the middle of the bands

fc(forward) = 1000 + 5000/2 = 3,500Hz

fc(backward) = 11,000 - 5000/2 = 8,500Hz

FSKFigure 5-27

Bandwidth for FSKFigure 5-28

• Find the minimum BW for an FSK signal transmitting at 2000 bps. The transmission mode is half-duplex and the carriers must be separated by 3,000 Hz

• Solution:BW = baud rate + (fc1 –fc0)The baud rate is the same as the bit rate

BW = 2000 + 3000 =5000 Hz

• Find the max bit rates for an FSK signal if the BW of the medium is 12,000 Hz and the difference between the carriers must be at least 2000 Hz. Transmission is in full-duplex mode.

• Solution:BW = baud rate + (fc1 –fc0)The BW for each direction is 6000 Hz Baud rate = 6000 –2000 = 4000 Baud rate = bit rateBit rate = 4000 bps

PSKFigure 5-29

PSKConstellation

Figure 5-30

4-PSKFigure 5-31

4-PSKCharacteristics

Figure 5-32

8-PSKCharacteristics

Figure 5-33

PSK BandwidthFigure 5-34

The minimum BW for PSK transmission is the same as that required for ASK transmission

Max baud rates of ASK and PSK are the same for a given BW, but the bit rates could be 2 or more times greater

(1) Find the BW for a 4-PSK signal transmitting at 2000 bps. Transmission is in half-duplex mode

(2) Given the BW of 5000Hz for an 8-PSK signal, what are the baud and bit rate?

(1) Solution:

Baud rate is half of the bit rate.

A PSK signal requires a BW equal to its baud rate

(2) Solution: Baud rate = 5000– Bit rate = 3 (5000) = 15, 000 bps

QAM

Quadrature Amplitude Modulation

• Combined ASK and PSK

• If there are x variations in phase and y variations in amplitude, it will give us x times y possible variations.

4-QAM and 8-QAM ConstellationsFigure 5-35

4 possible variations 8 possible variations

Time domain for 8-QAM Signal

Figure 5-36

16-QAM Constellation – Different configurations

16 out of 36/32 possible variations are utilized – to ensure readability

Greater ratio of phase shift to amplitude handles noise best1st figure – ITU-T recommendation2nd figure – OSI recommendation

16-QAM Constellation – Different configurations

Several QAM design link specific amplitudes with specific phases.

This means that even with the noise problems associated with amplitude shifting, the meaning of the shift can be recovered from phase information

Bit Rate and Baud Rate

Figure 5-38

Figure 5-38-continued

Bit Rate and Baud Rate

Bit and Baud Rate Comparison

Modulation Units Bits/Baud 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

EXAMPLES

(1) A constellation diagram consists of 8 equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate?

(2) Compute the bit rate for a 1000-baud 16-QAM signal.

(3) Compute the baud rate for a 72,000 bps 64-QAM signal.

(4) The data points of a constellation are at (4,0) and (6,0). Draw the constellation. Show the amplitude and phase for each point. Is the modulation ASK, PSK, or QAM? How many bits per baud can one send with this constellation?

(5) Repeat the exercise above if the data points are (4,5) and (8,10)

(1) Solution: – the constellation indicates 8-PSK with the

points 45 degrees apart– 3 bits are transmitted with each signal

element– 4800/3 = 1600 baud