1. Chapter 6. Digital Modulation Techniques
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2. Digital Communication NUML, Islamabad
6.1.1 Why Modulation
Modulation for Ease of Radiation: Requires antennas of EM wavelength.
Modulation for Multiplexing
To Overcome Equipment Limitations
For Frequency Assignment
To Reduce Noise and Interference
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3. Performance of Digital Modulation Scheme
lowest bit error rates,
performs well in multipath and fading conditions,
occupies a minimum bandwidth,
cost effective, and
minimum circuit complexity.
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4. Digital Communication NUML, Islamabad
Modulation Formats
Modulation format Application
MSK, GMSK GSM
BPSK Cable modems, satellite
QPSK Satellite, CDMA
8PSK Satellite
16QAM Microwave digital radio
32QAM Terrestrial microwave
64QAM Modems
256QAM Modems
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5. 6.2 Common Digital Modulation Formats
T: symbol duration
A: symbol energy
E = PT is energy of si(t) contained in symbol duration.
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6. Waveform Amplitude Coefficient
General form
Derivation:
s(t) = Acosωt
Or, s(t) = √2 Armscosωt = √(2 A2rms) cosωt
A2rms represents average power P (normalized to 1 ohm).
Therefore, s(t) = √2P cosωt
Replacing P watts by E joules/T seconds, we get
s(t) = √(2E/T) cosωt
where, parameter E: symbol energy; T: symbol duration time
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7. 6.3 Amplitude Shift Keying
Multiplication Operation ASK(t) = c(t) ・d(t)
Single-frequency Carrier: c = cosct
Unipolar periodic Data:
d(t)= 1/2 + (2/) X {cos0t –1/3 cos30t + 1/5 cos50t - ……}
ASK(t) = 1/2 cosct + 2/{cosct cos0t –1/3 cosct cos30t + 1/5 cosct cos50t - ..}
Using 2cosAcosB = cos(A – B) + cos(A + B)
ASK(t) = 1/2 cosct + 1/{ cos(c - 0)t + cos(c + 0)t
- 1/3 cos(c - 30)t – 1/3 cos(c + 30)t ……}
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8.  Digital Communication NUML, Islamabad c(t) fc + (fc + f0 ) 0 ~ f0
c(t) = cosct
(Carrier)
d(t)
(Unipolar binary input data)
c(t)
fc + (fc + f0 )
0 ~ f0
(Binary data output) 
LPF
PSTN
VASK(t)
BPF
(a)
ASK modulator
’d(t)
f0 = Fundamental frequency component = 1/2 bit rate, rb, (Hz)
ASK modulated signal m(t) cosct t fc
fc + f0
fc - f0
fc + 3f0
fc - 3f0
2f0
6f0
Signal power
Frequency
(c)
(b)
Carrier cosct
Modulating signal m(t) 1 Tb
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9. Signal Space Diagram for ASK Signals
Vector (or phasor) schematic: It presents only the information bearing phasor positions, relative to one another.)
Space signal diagrams for the ASK signals
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10. ASK Probability of Bit Error
Pe = Q( )
Q: complementary error function.
Eb: signal energy per binary symbol; and is equal to (A2/2)Tb.
N0: noise power spectral density
Eqn. can be written as
Pe = Q( )
where, A: bit amplitude
Tb: bit duration
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11. 6.4 Frequency Shift Keying
Frequency Shift: Difference between two frequencies.
Bit rate and baud rate are equal.
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12. 6.4.1 Generation of FSK Signal
Summing Two ASK Outputs:
[cos(2πf0t + θ)]
s0 =
s1 =
[cos(2πf1t + θ)]
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FSK Operation
VFSK(t) = cos1t・d(t) + cos2t・’d(t)
’d(t) is complement of original data signal, d(t). Mathematically, ’d(t) = 1 - d(t).
VFSK(t) = cos1t[1/2 + 2/(cos0t – 1/3cos30t )] + cos2t[1/2 - 2/(cos0t – 1/3cos30t )]
VFSK(t) = 1/2cos1t + 1/[cos(1 – 0)t + cos(1 + 0)t - 1/3cos(1 - 30)t – 1/3cos(1 + 30)t ] + 1/2cos2t + 1/[cos(2 – 0)t + cos(2 + 0)t – 1/3cos(2 - 30)t - 1/3cos(2 + 30)t ]
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FSK Bandwidth
Carson’s Rule:
BFSK= 2(B + Δf)
B: baseband bandwidth
Δf: frequency deviation
In case of rectangular pulses, baseband bandwidth B = R (R is data rate), equation becomes
BT = 2(R + Δf)
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15. Digital Communication NUML, Islamabad
300 bps FSK Modem Example
Figure illustrates frequency assignments that are used for two types of FSK modem to provide a full-duplex (two-way simultaneous) 300 bps link between two DTEs.
One set of frequency assignments is defined by EIA for Bell 103 modem and other by ITU-T for V.21 modem.
In such modems, fundamental frequency component associated with each carrier is 75 Hz. Hence, frequency shift of 200 Hz allows 50 Hz between the two primary sidebands
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Bandwidth Efficiency
1) Bandwidth Efficiency: For M-ary signals, bandwidth efficiency is
ηB = (2log2M)/M Hz
2) Spectral Efficiency
β = Rb/RT
Rb: bit transmission rate
RT: transmission bandwidth
β shows “how much data rate is achieved per unit Hz of transmission bandwidth”.
Example, spectral efficiency of binary ASK is 0.5 bps/Hz as its RT is double the bit transmission rate Rb.
3) Improving ASK Waveform Spectral Efficiency: Use higher values of M. Spectral efficiency of M-ary ASK waveform is
β = (log2M)/2 bps/Hz
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17. Spectral Efficiency Bandwidth
A V.92 modem for telephone network can transfer 56,000 bit/s downstream and 48,000 bit/s upstream over an analog telephone network. Due to filtering in telephone exchange, frequency range is limited to between 300 hertz and 3,400 hertz, corresponding to a bandwidth of 3400 − 300 = 3100 hertz.
1) Downstream spectral efficiency β = Rb/RT 56,000/3,100 = 18.1 (bit/s)/Hz downstream,
Upstream spectrum efficiency β = Rb/RT 48,000/3,100 = 15.5 (bit/s)/Hz upstream.
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18. FSK Bit Error Probability
FSK probability of a bit error Pe is given by
Pe = Q ( )
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19. Digital Communication NUML, Islamabad
Binary FSK Modulator
Oscillator
freq. = f1
freq. = f2
Electronic Switch 
Binary data input m(t)
Control line
FSK output
Demerit – Large Sidelobes/Large TX Bandwidth: Abrupt switching results in relatively large sidelobes outside the main spectral band of the signal and, consequently, this method requires a large transmission bandwidth.
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Continuous-phase FSK
binary 0 is transmitted as f0
binary 1 is transmitted as f1.
FSK modulator (VCO)
Binary input
Analog output
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6.4.3 PLL FSK Receiver
Voltage-controlled oscillator
Phase comp.
Amp
Binary data output
Analog FSK in
dc error voltage 
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