UNIT IV
DIGITAL COMMUNICATION
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DIGITAL MODULATION TECHNIQUES
❖ The digital modulation techniques may be classified into two categories as under:
(i) Coherent techniques
(ii) Non-coherent techniques.
Dr. Bhuvana B P, ME., Ph.D.,
ASK
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Coherent techniques
❖ In the coherent digital modulation techniques, we have to use a phase synchronized carrier to be
generated at the receiver to recover the information signal.
❖ The frequency and phase of this carrier produced at the receiver should be synchronized with that at the
transmitter.
❖ Coherent techniques are complex but yield better performance.
Non-Coherent techniques
❖ In the non-coherent techniques, no phase synchronized local carrier is needed at the receiver.
❖ These techniques are less complex.
❖ However, the performance is inferior to that of coherent techniques.
Dr. Bhuvana B P, ME., Ph.D.,
ASK
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COHERENT BINARY FREQUENCY SHIFT KEYING (BFSK)
Generation of BFSK
Dr. Bhuvana B P, ME., Ph.D.,
FSK
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Dr. Bhuvana B P, ME., Ph.D.,
FSK
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Bandwidth of BFSK Signal
Dr. Bhuvana B P, ME., Ph.D.,
FSK
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BFSK Receiver: Coherent Detection of BFSK
Dr. Bhuvana B P, ME., Ph.D.,
FSK
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Geometrical Representation for BPSK Signals
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Bit Error Rate (BER) or Probability of Error
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Salient Features of BFSK
(i) BFSK is relatively easy to implement.
(ii) It has better noise immunity than ASK. Hence, the probability of error free reception of data is high.
Drawback of BFSK
(i) The major drawback is its high bandwidth requirement*. Therefore, FSK is extensively used in low
speed modems having bit rates below 1200 bits/sec.
Dr. Bhuvana B P, ME., Ph.D.,
FSK
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BINARY PHASE SHIFT KEYING (BPSK)
Generation of BPSK Signal
Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Reception of BPSK Signal : Coherent Detection
Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Bandwidth for BPSK Signal
Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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Salient Features of BPSK
(i) BPSK has a bandwidth which is lower than that of a BFSK signal.
(ii) BPSK has the best performance of all the three digital modulation techniques in presence of noise. It yields the
minimum value of probability of error.
(iii) Binary phase shift keying (BPSK) has a very good noise immunity.
Drawbacks of BPSK
(i) The recovered carrier is unchanged even if the input signal has changed its sign.
(ii) Therefore, it is not possible to determine whether the received signal is equal to b(t) or – b(t). Infact, this results in
ambiguity in the output signal.
Dr. Bhuvana B P, ME., Ph.D.,
BPSK
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DPSK
Generation of DPSK
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Generation of DPSK
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Detection of DPSK
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Evaluation of Bandwidth of DPSK Signal
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Salient Features
(i) DPSK does not need carrier at the receiver end. This means that the complicated circuitry for generation of local
carrier is not required.
(ii) The bandwidth requirement of DPSK is reduced as compared to that of BPSK.
Drawbacks
(i) The probability of error (i.e., bit error rate) of DPSK is higher than that of BPSK.
(ii) Because DPSK uses two successive bits for its reception, error in the first bit creates error in the second bit.
Therefore, error propagation in DPSK is more. On the other hand, in BPSK single bit can go in error since detection
of each bit is independent.
(iii) Noise interference in DPSK is more.
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Comparison of BPSK and DPSK
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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Example
1. Binary data stream 0 0 1 0 0 1 0 0 1 1 needs to be transmitted using DPSK technique. Prove that the
reconstruction of the DPSK signal by the technique discussed in the previous article is independent of the choice
of the extra bit.
Dr. Bhuvana B P, ME., Ph.D.,
DPSK
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QUADRATURE PHASE SHIFT KEYING (QPSK)
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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QUADRATURE PHASE SHIFT KEYING (QPSK)
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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QUADRATURE PHASE SHIFT KEYING (QPSK)
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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QUADRATURE PHASE SHIFT KEYING (QPSK)
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Mathematical representation of QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Offset QPSK (OQPSK) or Staggered QPSK Transmitter
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Dr. Bhuvana B P, ME., Ph.D.,
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Non-offset QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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The QPSK Receiver
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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The QPSK Receiver
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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The QPSK Receiver
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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The QPSK Receiver
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Bandwidth of QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Error Probability of QPSK System
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Advantages of QPSK
(i) Very good noise immunity.
(ii) Baud rate is half the bit rate therefore more effective utilization of the available bandwidth of the transmission channel.
(iii) Low error probability.
Due to these advantages, the QPSK is used for very high bit rate data transmission.
Drawback
The generation and detection of QPSK is quite complex.
QPSK is Better than PSK
The QPSK is better than PSK because of the following reasons:
(i) Due to multilevel modulation used in QPSK, it is possible to increase the bit rate to double the bit rate of PSK without
increasing the bandwidth.
(ii) The noise immunity of QPSK is same as that of PSK system.
(iii) Available channel bandwidth is utilized in a better way by the QPSK system than PSK system.
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Comparison of BPSK and QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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Difference between OQPSK and QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QPSK
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QUADRATURE AMPLITUDE SHIFT KEYING (QASK) OR QAM
Types of QAM
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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QUADRATURE AMPLITUDE SHIFT KEYING (QASK) OR QAM
4 QAM and 8 QAM Systems
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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QUADRATURE AMPLITUDE SHIFT KEYING (QASK) OR QAM
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Dr. Bhuvana B P, ME., Ph.D.,
QAM
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QUADRATURE AMPLITUDE SHIFT KEYING (QASK) OR QAM
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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QASK Transmitter
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Dr. Bhuvana B P, ME., Ph.D.,
QAM
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QASK Receiver
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Bandwidth of QASK System
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Dr. Bhuvana B P, ME., Ph.D.,
QAM
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COMPARISON OF QASK AND QPSK
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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COMPARISON OF ASK, FSK, PSK AND QAM
Sl.
No.
Parameter
ASK
FSK
PSK
QAM
Amplitude
Frequency
Phase
Amplitude and Phase
1.
Modulation Type
2.
Bits per symbol
One
One
One
N
3.
Number of possible symbols M= 2N
Two
Two
Two
M= 2N
4.
Detection method
Coherent
Non-Coherent
Coherent
Coherent
5.
Minimum Euclidean distance
√Eb
√2Eb
2√Eb
√0.4Es for M=16
6.
Minimum Bandwidth(BW)
2fb
4fb
2fb
2fb/N
7.
Symbol duration(Ts)
Tb
Tb
Tb
NTb
8.
Noise
High
Low
Low
High
9.
S/N Ratio
Low
High
High
Low
10.
Data Rate
Less
Less
High
High
Dr. Bhuvana B P, ME., Ph.D.,
QAM
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Problem: 1
For 16-PSK and a transmission system with a 10 kHz bandwidth, determine the
maximum bit rate.
Solution:
The bandwidth efficiency for 16-PSK is 4, which means that four bits can be
propagated through the system for each hertz of bandwidth. Therefore, the maximum bit
rate is simply the product of the bandwidth and the bandwidth efficiency, or
bit rate = 4 x 10,000 = 40,000 bps
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Problem: 2
For 32-PSK and a transmission system with a 5 kHz bandwidth, determine the
maximum bit rate.
Solution:
The bandwidth efficiency for 32-PSK is 5, which means that five bits can be
propagated through the system for each hertz of bandwidth. Therefore, the maximum bit
rate is simply the product of the bandwidth and the bandwidth efficiency, or
bit rate = 5 x 5,000 = 25,000 bps
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Problem: 3
For a Modulator, with an input data rate equal to 10 Mbps and a carrier frequency of 70
MHz, find the minimum double sided Nyquist bandwidth for 8-PSK and 16 QAM.
Solution:
For 8-PSK
The bit rate is equal to one-third of the input bit rate = 10 Mbps / 3 = 3.33 Mbps
The fastest rate of change and highest fundamental frequency presented to either
balanced modulator is fa = 3.33 Mbps / 2 = 1.667 Mbps
The output wave from the balance modulators is (sin 2πfat)(sin 2πfct)
0.5 cos 2π(fc – fa)t – 0.5 cos 2π(fc + fa)t
0.5 cos 2π[(70 – 1.667)MHz]t – 0.5 cos 2π[(70+ 1.667)MHz]t
0.5 cos 2π(68.333MHz)t - 0.5 cos 2π(71.667MHz)t
The minimum Nyquist bandwidth is B= (71.667 - 68.333) MHz = 3.333 MHz
The minimum bandwidth for the 8-PSK, B = 10 Mbps / 3 = 3.33 MHz
Again, the baud equals the bandwidth thus,
baud = 3.333 megabaud
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For 16-QAM
The bit rate is equal to one fourth of the input bit rate = fb / 4 = 10 Mbps / 4 = 2.5 Mbps
Therefore, the fastest rate of change and highest fundamental frequency presented to either
balanced modulator is fa = 2.5 Mbps / 2 = 1.25 MHz
The output wave from the balanced modulator is (sin 2πfat)(sin 2πfct)
0.5 cos 2π(fc– fa)t – 0.5 cos 2π(fc + fa)t
0.5 cos 2π[(70 – 1.25)MHz]t – 0.5 cos 2π[(70 +1.25)MHz]t
0.5 cos 2π(68.75MHz)t - 0.5 cos2π(71.25MHz)t
The minimum Nyquist bandwidth is B=(71.25 - 68.75) MHz = 2.5 MHz
The minimum bandwidth for the 16-QAM
B = 10 Mbps / 4 = 2.5 MHz
The symbol rate equals the bandwidth; thus,
symbol rate = 2.5 megabaud
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Problem: 4
For an 8-PSK(or 8-QAM) system, operating with an information bit rate of 24 kbps,
determine (a) baud, (b) minimum bandwidth, and (c) bandwidth efficiency.
Solution:
(a) baud
Baud= 24 kbps / 3 = 8000 baud (or) 8 kilobaud
(b) Bandwidth
BW= 24 kbps / 3 = 8000 Hz (or) 8KHz
(c) Bandwidth efficiency
Bη = 24, 000 / 8000= 3 bits per second per cycle of bandwidth
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Problem: 5
For an 4-PSK (4-QAM) system, operating with an information bit rate of 24 kbps,
determine (a) baud, (b) minimum bandwidth, and (c) bandwidth efficiency.
Solution:
(a) baud
Baud= 24 kbps / 2 = 12000 baud (or) 12 kilobaud
(b) Bandwidth
BW= 24 kbps / 2 = 12000 Hz (or) 12KHz
(c) Bandwidth efficiency
Bη = 24,000 / 12,000= 2 bits per second per cycle of bandwidth
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Problem: 6
What is the minimum bandwidth required for BPSK, QPSK, 8-PSK, and16- QAM systems
if the bit rate is 10 MBPS?
i) For BPSK(21=2), BW= 10Mbps / 1 = 10MHz
ii) For QPSK(22=4), BW= 10Mbps / 2 = 5MHz
iii) For 8-PSK(23=8), BW= 10Mbps / 3 = 3.33MHz
iv) For 16-QAM(24=16), BW= 10Mbps / 4 = 2.5MHz
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Dr. Bhuvana B P, ME., Ph.D.,
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