Transcript Dig-to-Analog_Encoding_ASK_PSK_FSK_QAM

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
BW of a signal is the total range of frequencies occupied by that signal
Nbaud = baud rate
fc = carrier frequency
• 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
Figure 5-27
FSK
Figure 5-28
Bandwidth for FSK
• 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 fullduplex mode.
• Solution:
BW = baud rate + (fc1 –fc0)
The BW for each direction is 6000 Hz
Baud rate = 6000 –2000 = 4000
Baud rate = bit rate
Bit rate = 4000 bps
Figure 5-29
PSK
Figure 5-30
PSK
Constellation
Figure 5-31
4-PSK
Figure 5-32
4-PSK
Characteristics
Figure 5-33
8-PSK
Characteristics
Figure 5-34
PSK Bandwidth
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.
Figure 5-35
4-QAM and 8-QAM Constellations
4 possible variations
8 possible variations
Figure 5-36
Time domain for 8-QAM Signal
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 best
1st figure – ITU-T recommendation
2nd 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
Figure 5-38
Bit Rate and Baud Rate
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,4QAM
Dibit
2
N
2N
8-PSK,8QAM
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