modulation technique

1. 1
Digital modulation: ASK, PSK, BPSK,
QPSK, FSK, MSK

2. 2
Modulation of Digital Data
 Carrier Signal - High-frequency signal used for
digital-to-analog or analog-to-digital modulation
 May change any aspect of an analog signal to
represent digital data: amplitude, frequency, and
phase
 Modulation - process of changing one of the
characteristics of an analog signals based on
information in digital signal

3. 3
Digital-to-Analog Conversion

4. Types of methods
 Amplitude shift keying (ASK): some audio modem, optical
fiber, printer
 Frequency shift keying (FSK): Computer Modem, HF and
VHF radio
 Amplitude shift keying (PSK): Mobile, Satellite
(microwave) etc
Digital to Analog
Modulation
FSK
QAM
PSK
ASK
Bit rate is the number of bits transmitted per second. Baud rate is the
number of signal elements transmitted per second. Baud rate is less than or
equal to the bit rate

5. 5
Amplitude Shift Keying (ASK)
 Strength of carrier signal is varied to represent a
1 or 0
 Frequency and phase remain constant while
amplitude is changed
 Amplitude remains constant during bit duration
 Highly susceptible to noise interference since
noise usually affects amplitude
 Minimum bandwidth required is equal to baud
rate

6. 6
Amplitude Shift Keying (ASK)

7. 7
Relationship between baud rate and
bandwidth in ASK
BW = (1+d)Nbaud

8. 8
Example 3
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is half-
duplex.
Solution
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.

9. 9
Frequency Shift Keying (FSK)
 Frequency is varied to represent a 1 or 0
 Frequency during bit duration is constant
 Amplitude and phase remain constant
 Avoids most of noise problems of ASK; can
ignore voltage spikes
 Limited by physical capabilities of medium
 Bandwidth required is equal to baud rate plus
frequency shift

10. 10
Frequency Shift Keying (FSK)

11. 11
Relationship between baud rate and
bandwidth in FSK
BW = fc1-fc0+Nbaud

12. 12
Example 6
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.
Solution
For FSK
BW = baud rate + fc1 - fc0
BW = bit rate + fc1 - fc0 = 2000 + 3000 = 5000 Hz

13. 13
Phase Shift Keying (PSK)
 Phase of carrier is varied to represent 1 or 0
 Peak amplitude and frequency remain constant
 Phase remains constant during each bit duration
 Not as susceptible to noise degradation as ASK
or to bandwidth limitations of FSK

14. 14
Phase Shift Keying (PSK)
2-PSK or Binary PSK

15. 15
PSK constellation (Phase state diagram)
Constellation is a graphical representation of the phase
and amplitude of different bit combinations in digital-to-
analog modulation.

16. 16
4-PSK; 8-PSK
 4-PSK
 May utilize four variations of phase shift by 90 degrees
 Each phase shift represents 2 bits (dibit); technique is
referred to as 4-PSK
 Allows data transmission two times as fast as 2-PSK
 8-PSK
 Vary signal by shifts of 45 degrees; each shift may
then represent three bits (tribit) and send data three
times as fast

17. 17
The 4-PSK method (Q-PSK)

18. 18
The 4-PSK characteristics

19. 19
The 8-PSK characteristics

20. 20
Relationship between baud rate and
bandwidth in PSK
BW = (1+d)Nbaud

21. 21
Example 8
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.
Solution
For 4-PSK the baud rate is one –half of the bit rate (1 baud :
2 bit). The baud rate is therefore 1000. A PSK signal requires
a bandwidth equal to its baud rate . Therefore, the bandwidth
is 1000 Hz

22. 22
Example 9
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
Solution
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.

23. 23
Quadrature Amplitude Modulation (QAM)
 Combination of ASK and PSK so that a maximum
contrast between each signal unit (bit, dibit, tribit, and
so on) is achieved
 x number of variations in phase and y variations in
amplitude
 Number of phase shifts is always larger than number
of amplitude shifts due to amplitude susceptibility to
noise
 QAM is therefore less susceptible to noise than ASK
 Same bandwidth is required for ASK and PSK

24. 24
Credits
 All figures obtained from
Data Communications and Networking, 3rd edition by
Behrouz A. Forouzan. McGraw Hill Publishing, 2004

26. Intersymbol Interference (ISI)