EC8395 COMMUNICATION ENGINEERING UNIT III

Digital Modulation :
High-frequency analog carriers are modulated by relatively low frequency
digital information signals.
Types :
ASK : If the information signal is digital and the amplitude (V) of
the carrier is varied proportional to the information signal.
FSK : If the frequency (f) is varied proportional to the information signal
PSK : If the phase of the carrier (θ) is varied proportional to the information
signal
QAM : If both the amplitude and the phase are varied proportional to the
information signal

Applications:
(1) Relatively low-speed voice-band data communications modems,
such as those found in most personal computers
(2) high-speed data transmission systems, such as broadband digital
subscriber lines (DSL)
(3) digital microwave and satellite communications systems
(4) cellular telephone Personal Communications Systems

Bit rate is simply the number of bits transmitted during one second and is
expressed in bits per second. Hartley’s law is
I ∝ B * T
where I information capacity (bits per second)
B bandwidth (hertz)
T transmission time (seconds)
Baud rate is the rate of change of signal on transmission medium after
encoding and modulation have occurred.
Bandwidth efficiency : Bandwidth efficiency is the ratio of the transmission
bit rate to the minimum bandwidth required for a particular modulation

AMPLITUDE SHIFT KEYING
Process of changing amplitude of the carrier signal with respect to the
binary information or digital signal

ASK– On-Off keying
Bandwidth = Bit rate = baud rate

AMPLITUDE SHIFT KEYING
• The amplitude shit keying is also called on-off keying(OOK). This is the
simplest digital modulation technique.
• The binary input data is converted to uni-polar NRZ signal.
• A product modulator takes this NRZ signal and carrier signal.
• The binary data sequence ‘d’ is given to the NRZ level encoder.
• This NRZ level encoder converts the input binary sequence to the signal
suitable for product modulator.
• The product modulator also accepts a sinusoidal carrier of frequency fc.
• The output o the product modulator is passed through a bandpass filter for
bandwidth limiting.
• The output of the bandpass filter is the ASK signal.

FREQUENCY-SHIFT KEYING - Changing frequency of
the carrier signal with respect to the binary information or digital signal.
Frequency Deviation:
𝛥𝑓 =
𝑓𝑚 −𝑓𝑠
2
Bandwidth
B =2(Δf+ fb)
Bit rate = Baud rate = 𝑓𝑏

FSK modulator
Level Shifter
Level Shifter
Inverter
Input
High Frequency
Low Frequency
BFSK

FSK Demodulator (Non – Coherent detector)
FSK Demodulator (Coherent detector)
Envelope
Detector
Envelope
Detector
Band pass Filter
FL = fs
Band pass Filter
FH = fm
comparator
Message
bits
FSK
Local Oscillator
FL = fs
Local Oscillator
FH = fm
FSK Decision
Device
S1
S2
1 If S1=1
0 If S2 = 1

Frequency Shift Keying - PLL

Phase Shift Keying
Phase of the analog carrier is varied with respect to the input binary digital
signal.
The number of output phases is defined by M = 2N
Binary Phase-Shift Keying (N=1, M = 2)
Two phase changes are possible which are separated by 1800
One phase represents a logic 1, and the other phase represents a logic 0.

Phase Shift Keying
Bandwidth = 2fa or fb
Where fa is the maximum fundamental frequency of binary input
fb = bit rate = baud rate = 2fa
LSB = fc-fa or fc-fb/2 and USB = fc+fa or fc+fb/2
BPSK Receiver

For a BPSK input signal of sin ωct (logic 1),
Output = (sin ωct)(sin ωct) = sin2 ωct
=(1-cos 2ωct)/2
=1/2 - cos 2ωct/2 cos ωct is filtered by LPF
= ½
For a BPSK input signal of -sin ωct (logic 0),
Output = (-sin ωct)(sin ωct) = -sin2 ωct
=-(1-cos 2ωct)/2
=-1/2 + cos 2ωct/2 cos ωct is filtered by LPF
=- ½

Quaternary Phase-Shift Keying or Quadrature PSK
• QPSK is an M-ary encoding scheme N= 2 and M=4.
• With QPSK, four output phases are possible for a single carrier frequency.
• Binary input data are combined into groups of two bits, called dibits
• 00, 01, 10, and 11 (+45°, +135°, -45° and -135°)
• +sin ωct+cos ωct, +sinω ωct-cos ωct, -sin ωct+cos ωct, and -sin ωct -cos ωct.

QPSK - Phasor diagram

Bandwidth = fb /2
fb = bit rate = 4fa Baud = 2 Bit rate
LSB = fc - (fb)/4 and USB = fc+ (fb)/4
QPSK Receiver

• The carrier recovery circuit reproduces the original transmit carrier
oscillator signal

QUADRATURE-AMPLITUDE MODULATION
• Digital information is contained in both the amplitude and the phase of the
transmitted carrier.
• Positions of the signaling elements on the constellation diagrams are
optimized to achieve the greatest distance between elements
8-QAM transmitter(M = 8 & N= 3)

I & Q – Polarity
C – Magnitude (1.307, 0.541)
Let the input be 001 Q = 0 I =0 C =1
I channel
2 to 4 O/P = -1.307
Product Modulator = -1.307sin ωct
Q channel
2 to 4 O/P = -1.307
Product Modulator = -1.307cos ωct
Summer O/P = -1.307sin ωct -1.307cos ωct
Bandwidth = fb/3

8 QAM Receiver

Let the input be -1.307sin ωct -1.307cos ωct
I channel
Product detector = (-1.307sin ωct -1.307cos ωct) sin ωct
Product Modulator = -1.307sin2 ωct – 1.307cos ωct sin ωct
= -1.307(1-cos 2ωct)/2 – 1.307[sin0-sin2 ωct]/2
=-1.307/2
Q channel
Product detector = (-1.307sin ωct -1.307cos ωct) cosωct
Product Modulator = -1.307cos2 ωct – 1.307cos ωct sin ωct
= -1.307(1+cos 2ωct)/2 – 1.307[sin0-sin2 ωct]/2
=-1.307/2
O/P = 001

16 QAM
N= 4, M = 16
I&Q - Polarity Q’ & I’ – Magnitude (logic 1=0.821 V logic 0=0.22 V)
Bandwidth = fb/4

1. Determine the baud and minimum bandwidth necessary to
pass a 10 kbps binary signal using amplitude shift keying
Baud rate = Bandwidth = fb
= 10,000

Determine (a) the peak frequency deviation, (b) minimum bandwidth, and (c)
baud for a binary FSK signal with a mark frequency of 49 kHz, a space
frequency of 51 kHz, and an input bit rate of 2 kbps

For a BPSK modulator with a carrier frequency of 70 MHz and an input bit
rate of 10 Mbps, determine the maximum and minimum upper and lower
side frequencies, determine the minimum Nyquist bandwidth, and calculate
the baud.
.

For a QPSK modulator with an input data rate (fb) equal to 10 Mbps and a
carrier frequency of 70 MHz, determine the minimum double-sided Nyquist
bandwidth (fN) and the baud.

For a quad bit input of I = 0, I’ = 0, Q = 0, and Q’ = 0 (0000), determine the
output amplitude and phase for the 16-QAM modulator
.

For a 16-QAM modulator with an input data rate (fb) equal to 10 Mbps and a
carrier frequency of 70 MHz, determine the minimum double-sided Nyquist
frequency (fN) and the baud
Bandwidth = fb/4 = baud rate
= 2.5 Mhz and 2.5 baud

EC8395 COMMUNICATION ENGINEERING UNIT III

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